Government U of Tam Hnadu n t 0 ..uC h 3 Inhuman- M F...t V. Crime Department of School Education A P u ._D 1cl0 a LL:10 n U n d e .1 Government Distribution of Tamilnadu of Free Textbook (NOT FOR SALE) Programme © Government of Tami! First Edition Nadu 2011 Reprint 2012 CHAIRPERSON Dr. R. Murthy Associate Professor Department of Mathematics PresidencyCollege (Autonomous) Chennai 600 005. REVI EWIE RS Dr. G. P.Youvciresi S. Ranganathan AssociateProfessor Ramanuian Institutefor Advanced Study in Mathematics Universityof Madras, Chennai-600 005. Headmaster (Retd) Sri Gopal Naidu Hr. Sec. School Peelamedu Coimbatore 641 004. AUTHORS V. Munian S. Panneer Headmaster Chennai High School Jafferkhanpet,Chennai - 600 083. Govt. Hr. Sec. School M.G.R. Nagar, Chennai - 600 078. V. S. Sankara Narayanan S. Balm B.T. Assistant B.T. Assistant St. Joseph's Anglo Indian Boys Hr. Sec. School Vepery, Chennai 600 007. Govt. Girls Hr. Sec. School Vil|upuram605602. K. Selvamian Shcmtha Suresh Vice Principal Gurunanak Mat. B.T.Assistant Hr. Sec. School CSl Jessie Moses Mat. Velachery,Chennai - 600 042. V. James Hr. Sec. School Anna Nagar, Chennai - 600 040. LaserTypeset, Layout and Wrapper Design: Abraham TextbookPrinting Tamilnadu Textbook Corporation, College Road, Chennai 600 006. Price Se.-ivam P. G. Assistant : Rs. This book has been printedon 80 G.S.M. Maplitho Paper Printed by Offset at : (ii) :1 It is gratifying to note that education as a whole and school education in particular witness marked changesin the state of Tamil Nadu resulting in the implementationof uniform curriculum for all streams in the school education system. This is a golden opportunity given by the Government of Tamil Nadu which must be utilized for the over all improvement of education in Tamil Nadu. Mathematics, the queen of all sciences,remains and will remain as a subject with great charm having an intrinsic value and beauty of its own. It plays an indispensable role in sciences,engineering and other subjects as well. So, mathematical knowledge is essential for the growth of science and technology, and for any individual to shine well in the field of ones choice. In addition, a rigorous mathematical training gives one not only the knowledge of mathematics but also a disciplined thought process, an ability to analyze complicated problems. Thiruvalluvar, the prophetic Tamil poet, had as far back as at least two thousand years ago, underlined the importance and the value of mathematical education by saying, -@gmwwm ~ Kural (392) We need the power and prowess of mathematics to face and solve the ever increasingcomplexproblemsthat we encounterin our life. Furthermore,mathematicsis a supremely creative force and not just a problem solving tool. The learners will realize this fact to their immense satisfaction and advantage as they learn more and more of mathematics. Besides, a good mathematical training is very much essential to create a good work force for posterity. The rudiments of mathematics attained at the school level form the basis of higher studies in the eld of mathematics and other sciences.Besideslearning the basicsof mathematics,it is also important to learn how to apply them in solving problems. (iii) Deeper understanding of basic principles and problem solving are the two important components of learning mathematics. This book is a step in this direction. It is intendedto help the studentsto graspthe fundamentalsof mathematicsand apply them in problem solving. It also fosters an informed awarenessof how mathematics develops and works in different situations. With this end in view, the chapters are arranged in their natural and logical order with a good number of worked out examples. Each section of a chapter is designed in such a way as to provide the students the much needed practice which promotes a thorough understanding of the concepts. We suggest that before going into the problems, the teachers and the students get themselves acquainted with the underlying mathematical ideas and their connections which precede the set of problems given in the exercises. However, be it remembered that mathematics is more than the science of numbers. The teacher in the classroom is the most important person whose help and guidance are indispensable in learning mathematics. During the stage of transition from basic mathematics to higher mathematics, the teachers have a more signicant role to play. In this context we hope that this book serves the purpose and acts as a catalyst. To reap the maximum benefit out of this, the teacher should necessarily strive for a two- way communication.This endeavourwill undoubtedly pavethe way for learner-centered activities in the class rooms. Moreover, this text book is aimed at giving the students a space to explore mathematics and develop skills in all directions. As we have mentioned already, there are two parts in learning mathematics. One is learning the basics and the other is applying the basics in problem solving. Going through the examples in the text does help in understanding the methods; but learning basics, solving exercise problems on onesown and then trying to create new related problems alone will help consolidate onesmathematical knowledge. We learn Mathematics by doing Mathematics. We would be grateful for suggestions and comments from experts, teachers and students for the improvement of this book. Murthy R Chairperson. (iv) Functions and Sets I. of Serie and Sequences II. SYLLABUS Introduction Properties of operations on sets De Morgans lawsveri cation using examples, Venn diagrams. ' . Formulafor n(AUBUC) Functions Expected Learning Outcomes Transactional To revise the basic con- cepts on Set operations To understand the properties of operations of sets commutative, associative, and distributive restricted to three sets. To understand the laws of complementation of sets. To understand De Mor- Teaching Strategy gans laws and demonstrating them by Venn diagram Use Venn as well. diagrams for all To determine the remain- der and the quotient of the given polynomial using Synthetic Division Compare with operations on fractions Method. To determine the factors of the given polynomial using Synthetic Division Method. Able to understand the difference between GCD and Compare with the LCM, of rational expres- square root operation on numerals. sion. Algebra III. Able to simplify rational expressions (Simple Problems), To understand square roots. To understand the standard Help students visualize the nature of roots algebraically and graphically. form of a quadratic equation . To solve quadratic equations (only real root) by factorization, by completing the square and by using quadratic formula. Able to solve word problems based on quadratic equations. Able to correlate relation- ship between discriminant and nature of roots. Able to Form quadratic equation when the roots are given. i. Introduction Using of rectangular array of and formation iii. Addition Able to recognize the types numbers. and subtraction iv. Multiplication V. Matrix equation Matrices IV. Able to identify the order ii. Types of matrices of matrices of matrices Able to add and subtract Using real life the given matrices. To multiply a matrix by a scalar, and the transpose of situations. a matrix. To multiply the given matrices (2x2; 2x3; 3x2 Matrices). Using matrix method solve the equations of two variables. (Vi) Arithmetic operations to be used i. ii. Introduction Revision :Distance be- tween two points iii. Section formula, Mid point formula, Centroid formula iv. Area of a triangle and quadrilateral V. Straight line To recall the distance Simple geometri- between two points, and locate the mid point of two cal result related given points. To determine the point of division using section to triangle and quadrilaterals to be veried as applications. formula. To calculate the area of a the form triangle. To determine the slope of a line when two points are given, equation is given. To nd an equation of line with the given information. Able to nd equation of a line in: slopeintercept form, point slope form, two point form, intercept y = mx + c to be Geometry Coordinate V. taken as the start- ing point form. To nd the equation of a straight line passing through a point which is (i) parallel (ii) perpendicular to a given straight line. i. Basic proportionality theorem (with proof) ii. Converse of Basic proportionality theorem (with proof) iii. Angle bisector theorem (with proof internal case To understand the theo- rems and apply them to solve simple problems. adopted. only) iv. Converse of Angle bisector theorem (with proof internal case only) v. Similar triangles (theorems without proof) vi. Pythagoras theorem and TangentChord theorem (without proof) i. ii. Introduction Identities iii. Heights and distances transformation techniques tobe Geometry VI. Paper folding symmetry and Formal proof to be given Drawing of gures Step by step logical proof with diagrams to be explained and Able to identify the Trigonometric identities and apply them in simple problems. To understand trigonometric ratios and applies them to calculate heights and discussed Trigonometry VII. distances. (not more than two right triangles) By using Alge- braic formulae Using trigonometric identities. The approximate nature of values (vii) to be explained Introduction To determine volume and '1. Surface AreaandVolume of Cylinder, Cone, Sphere, Hemisphere, Frustum surface area of cylinder, cone, sphere, hemisphere, ustum Volume and surface area Mensuration Surface area and volume of combined gures Use 3D models to create combined shapes Use models and of combined gures (only two). Some problems restricted pictures ad teaching aids. to constant Volume. Choose examples Geome Practica VIII. . Invariant volume from real life situations. Introduction Able to construct tangents Construction of tangents to circles. to circles Construction of Triangles . Construction of cyclic quadrilateral To introduce algebraic vericaAble to construct triangles, tion of length of given its base, vertical tangent segments. angle at the opposite vertex and Recall related (a) median (b) altitude Able to construct a cyclic quadrilateral properties of angles in a circle IX. before construction. Recall relevant theorems in theo- retical geometry . Introduction Quadratic graphs Some special graphs Able to solve quadratic equations through graphs Able to apply graphs to solve word problems Interpreting skills also to be taken care of graphs of quadratics to precede algebraic treatment. Real life situations to be introduced. Recall Measures of central tendency Measures of dispersion Statistics Coefficient of variation XI. To recall Mean for grouped and ungrouped data situation to be avoided). To understand the concept of Dispersion and able to nd Range, Standard Use real life situa- tions like performance in exami- nation, sports, etc. Deviation and Variance. Able to calculate the coefcient of variation. Introduction To understand Random Diagrams and investigations and Events Mutually on coin tossing, die throwing and Exclusive, Complementary, certain and impossible picking up the 3.Probabilitytheoretical ap- experiments, Sample space proach Addition Theorem on Probability Probability XII. events. To understand addition Theorem on probability and apply it in solving some simple problems. (viii) cards from a deck of cards are to be used. CONTENTS ETS AND FUNCTBQNS lntroduction Sets iiiiw Operations on Sets Properties of Set Operations De Morgan's Laws Cardinality of Sets 12 16 Relations 19 Functions 20 I<°°I\I§>~ SEQUENCES AND 2.1. lntroduction SERIES OF REAL NUMBERS 34-67 34 2.2. 2.3. 2.4. Sequences Arithmetic Sequence Geometric Sequence 35 38 43 2.5. Series 49 ALGEBRA 3.1 lntroduction 3.2 3.3 3.4 3.5 3.6 3.7 3.8 System of Linear Equations in Two Unknowns Quadratic Polynomials Synthetic Division Greatest Common Divisor and Least Common Multiple Rational Expressions Square Root Quadratic Equations MATRICES 68-3 17 68 69 80 82 86 93 97 101 11 8-139 4.1 4.2 lntroduction Formation of Matrices 1 18 1 19 4.3 4.4 4.5 4.6 4.7 Types of Matrices Operation on Matrices Properties of Matrix Addition Multiplication of Matrices Properties of Matrix Multiplication 121 125 128 130 132 GEOMETRY I 71 -195 6.1 Introduction 171 6.2 6.3 6.4 Basic Proportionality and Angle Bisector Theorems Similar Triangles Circles and Tangents 172 182 189 TRIGONOMETRY 196-218 7.1 Introduction 196 7.2 7.3 Trigonometric Identities Heights and Distances 196 205 MENSURATION 8.1 8.2 8.3 8.4 219-248 Introduction Surface Area Volume Combination PRACTICAL of Solids GEOMETRY 219 219 230 240 249-266 9.1 Introduction 249 9.2 9.3 9.4 Construction of Tangents to a Circle Construction of Triangles Construction of Cyclic Quadrilaterals 250 254 259 GRAPHS 267-278 10.1 Introduction 267 10.2 10.3 Quadratic Graphs Some special Graphs 267 275 STATISTICS 279-293 1 1.1 Introduction 279 1 1.2 Measures of Dispersion 280 PROBABILITY 299-316 12.1 Introduction 299 12.2 12.3 Classical Definition of Probability Addition theorem on Probability 302 309 Answers 31 7-327 Miscellaneous problems 328-329 Bibliography 330-331 Question 331 -334 Paper Design SETS AND FUNCTIONS A set2":M4191that allowsitself t0 hethoughtof at a One GeorgCezhtor 1.1 Introduction The concept of set is one of the fundamental concepts in mathematics. The notation and terminology of set theory is useful in every part of mathematics. So, we may say that set theory is the language of mathematics. This subject, which originated from the works of (1815-1864) and * ii (1845-1918) in the later part of 19th century, Eonénesoom _ has had a profound inuence on the development of all branches of mathematics in the 20th century. It has helped in unifying many disconnected ideas and thus facilitated the advancement In ttveehipjgyzhtholx [ of mathematics. class IX, we have learnt the concept of set, some .z}zterzzetio7z.: iczhidc, operations like union, intersection and difference of two sets. Here, we shall learn some more concepts relating to sets and W amt another important concept in mathematics namely, function. First let us recall basic denitions with some examples. We denote all positive integers(natural numbers)by N and all real numbersby R. 1.2 Sets A set is a collection of well-dened in a set are called elements Here, welldened objects. The objects or members means that of that set. the criteria for deciding if an object belongs to the set or not, should be dened without confusion. For example, the collection of all tall people in Chennai does not form a set,becausehere, the deciding criteria tall people is not clearly dened. Hence this collection does not dene a set. Notation We generally use capital letters like A, B, X, etc. to denote a set. We shall use small letters like x, y, etc. to denote elements of a set. We write x E Y to mean x is an element of the set Y. We write t 65 Y to mean tis not an element of the set Y. Examples (i) The set of all high school students in Tamil Nadu. (ii) The set of all students either in high school or in college in Tamil Nadu. (iii) The set of all positive even integers. (iv) The set of all integers whose square is negative. (v) The set of all people who landed on the moon. Let A, B, C, D and E denote the sets dened in (i), (ii), (iii), (iv), and (v) respectively. Note that square of any integer is an integer that is either zero or positive and so there is no integer whose square is negative. Thus, the set D does not contain any element. Any such set is called an empty set. We denotethe empty setby qb. (i) A set is said to be a nite set if it contains only a nite number of elements in it. (ii) A set which is not nite is called an innite set. Observe that the set A given above is a nite set, whereas the set C is an innite set. Note that empty set contains no elements in it. That is, the number of elements in an empty set is zero. Thus, empty set is also a nite set. (i) If a set Xis finite, then we define the cardinality of X to be the number of elements in X . Cardinality of a set X is denoted by n(X).t (ii) If asetX isinnite,thenwedenote thecardinality of X byasymbol oo. Now looking at the sets A, B in the above examples, we see that every element of A is also an element of B. In such caseswe say A is a subset of B. Let us recall Subset some of the denitions that we have learnt in class IX. Let X and Y be two sets. We say X is a subset of Y if every element of X is also an element of Y. That is, X is a subset of Y if z E X implies z E Y. It is clear that every set is a subset of itself. If X is a subset of Y, then we denote this by X E Y. 701/9Std. Mfbéidliff Set Equality Two sets X and Y are said to be equal if both contain exactly same elements. In such a case, we write X = Y. That is, X = Y if and only if X E Y and Y9 X. Equivalent Sets Two nite sets X and Y are said to be equivalent if n(X) = n(Y). Forexample, let P = {x|x2-x - 6 = 0} andQ = {3,-- 2}. It is easyto seethat both P, Q contain sameelementsand so P = Q. If F = {3, 2}, then F, Q are equivalent sets but Q yéF. Using the concept of function, one can dene the equivalent of two innite sets Power Set Given a set A, let P(A) denote the collection of all subsets of A. The set P(A) is called the power set of A. If n(A) = m, then the numberof elementsin P(A) is given by n(P(A)) = 2'. For example,if A = {a,b,c}, then P(A) = {q5,{a},{b},{c}.{a,b},{a,c},{b,c},{a,b,c}} and hence n(P(A)) = 8. One possibility is to put all the elements together from both sets and create a new set. Another possibility is to create a set containing only common elements from both sets.Also, we may create a set having elements from one set that are not in the other set. Following denitions give a precise way of formalizing these ideas. We include Venn diagram next to each denition to illustrate 1.3 it. Operations Y on sets Let X and Y be two sets. We dene the following new sets: (i) Union XUY={z|zEX (read as X Note that X U Y contains or zEY} union Y) if all the elements XUY A of X and all the Fig. 1.1 elements of Y and the Fig. 1.1 illustrates this. Itis (ii) clearthat XE intersection X XUYandalsoYQXUY. Y 2:{z I Z 6 X and z E Y} (read as X intersection Y) Note that X F)Y contains only those elements which belong to both X and Y and the Fig. 1.2 illustrates this. It is trivial (iii) that X Y Q X and also X Fig.1.2 Y E Y. Set difference X \ Y = {z I z E X but z EEY} (read as X difference Y) Note that X \ Y contains only elements of X that are not in Y and the Fig. 1.3 illustrates this. Also, some authors use A B for A\B. We shall use the notation A\B which is widely used in mathematics for set difference. (iv) Symmetric Difference (read as X X A Y = ( X \ Y ) U ( Y \ X) symmetric difference Y). X A Y contains all elements X Note that in X U Y that are not in X F) Y. Sets and F;/mtiom (V) Complement U \X If X E U, whereU is a universalset,then is called the complement of X with respect to U. If underlying universal set is xed, then we denote U \ X by X ' andis calledcomplement of X. "fbec.l%:7e.2*er:::e;;> setA\B camalzaoV be vicwastl (vi) the :;:a}n2*g>l<:;n<::3t MS with i°CS§T)C:i315 to /3. Disjoint sets Two sets X and Y are said to be disjoint if they do not have any common element. That is, X and Y are disjoint if X Y = (,5. It is clear that n(A UB) = n(A) + n(B) ifA andB are disjoint nite sets. Now, we shall see some examples. Let A = {x | x is a positive integerlessthan 12}, C = {- 2, - 1,0, 1,3, 5, 7}. Now let us nd the following: (i) B = {1, 2, 4, 6, 7, 8, 12,15} and AUB={x|xEA or x¬B} = {x|x is a positive integerlessthan 12,or x = 12,or 15} = {1,2, 3, 4,5,6, 7, 8, 9,10,11,12,l5}. (ii) CB={y|yECandyEB}={1,7}. (iii) A\C={x|xEA but x¬EC}={2,4,6,8,9,10,11}. (iv)AAC=(A\C)U(C\A) ={2,4,6,8,9,10,l1}U{-2,1,0} ={-2,-1,0,2,4,6,8,9,10,1 (v) Let U = {x | x is an integer} be the universal set. Note that 0 is neither positive nor negative. Therefore, 0 GEA. Now, A = U\A = {x :x is an integer but it should not be in A} = {x | x is either zero or a negative integer or positive integer greater than or equal to 12} = {---,- 4,- 3, - 2, - l,O}U{12, 13,14,15,---} = {---,-- 4,~- 3,- 2,-- l,0,12,13,14,15,---}. Let us list out some useful results. Let U be an universal set and A, B are subsets of U . Then the following hold: (1) A\B=AnB (iii) A\B=A<==>AnB=q5 (V) (A\B)nB=g/5 01/9Std. Mai/?8I7Zdli¬X (ii) B\A=BA (iv) (A\B)UB=AUB (vi) (A\B)U(B\A)=(AUB)\(AB) Let us state some properties of set operations. 1.4 Properties of set operations For any three sets A, B and C, the following hold. (i) Commutative property (a) A U B = B U A (b) A O B = B F)A (ii) (iii) (set union is commutative) (set intersection is commutative) Associative property (a) A U (B U C) = (A U B) U C (b) A O(B 0 C) = (A 0 B) n C (set union is associative) (set intersectionis associative) Distributive property (a) A 0 (B U C) = (A DB) U (A D C) (b) A U (B F)C) = (A U B) 0 (A U C) (intersectiondistributesover union) (union distributesover intersection) Mostly we shall verify these properties with the given sets. Instead of verifying the above properties with examples, it is always better to give a mathematical proof. But this is beyond the scope of this book. However, to understand and appreciate a rigorous mathematical proof, let us take one property and give the proof. (i) Commutative property of union In this part we want to prove that for any two sets A and B, the sets A U B and B U A are equal. Our denition of equality of sets says that two sets are equal only if they contain same elements. First we shall show that every element of A U B, is also an element of B U A. Let z E A U B be an arbitrary element. Then by the denition we have z EA or z EB. forevery Since 5 zEAUB of union of A and B That is, ==> ZEA or ZEB =:> ZE B 01 Z G A => z EBUA by the denition of BUA. is true for every z E A U B, the above work shows that every element of A UB is also is an element of B U A. Hence, by the denition of a subset,we have (A U B) Q (B U A). Next, we consider an arbitrary y E B U A and show that this y is also an element of A U B. Now, forevery yEBUA==> ...> yeB or yeEA y E A or y E B => yEAUB bythedenitionof AUB. . .. Since IE is true for every y E B U A, the above work shows that every element of B U A is also an element of A U B. Hence, by the denition of a subset, we have (B U A) Q (A U B). So, we have shown that (A U B) Q (B UA) and (B UA) Q (A U B). This can happen only when (A U B): (B U A). One could follow above steps to prove other properties listed above by exactly the same method. Sets and Fmmz'om[ About proofs in Mathematics In mathematics, a statement is called a true statement if it is always true. If a statement is not true even in one instance, then the statement is said to be a faise statement. For example, let us consider a few statements: (i) Any positive odd integeris a prime number (ii) Sum of all anglesin a triangle is 180° (iii) Every prime numberis an odd integer (iv) For any two setsA and B, A \ B = B \ A Now, the statement (i) is false, though very many odd positive integers are prime, because integers like 9,15, 21, 45 etc. are positive and odd but not prime. The statement (ii) is a true statement because no matter which triangle you consider, the sum of its angles equals 180°. The statement (iii) is false, because 2 is a prime number but it is an even integer. In fact, the statement (iii) is true for every prime number except for 2. So, if we want to prove a statement we have to prove that it is true for all instances. If we want to disprove a statement it is enough to give an example of one instance, where it is false. The statement (iv) is false. Let us analyze this statement. Basically, when we form A\B we are removing all elements of B from A. Similarly, for B\A. So it is highly possible that the above statement is false. Indeed, let us consider a case where A = {2,5,8}and B ={5,7,~ 1}. In this case, A\B ={2,8} and B\A ={7,- 1} and we have A \ B 74B \ A. Hence the statement given in (iv) is false. Example 1.1 For the given sets A = {- 10,0,1, 9, 2,4, 5} and B = {- 1, 2, 5, 6, 2,3,4}, verify that (i) set union is commutative. Also verify it by using Venn diagram. (ii) set intersection is commutative. Also verify it by using Venn diagram. Sarrimzm (i) Now,AUB= {~10,0,1,9,2,4,5}U{~ 1,-2,5,6, 2,3,4} ={10,Also, BUA 5,6, 9} = {- 1,- 2,5,6,2,3,4}U{-10,0,1,9,2,4,5} ={10,- Thus, from 2,- 1,0,1,2,3,4, 2,-1,0,1,2,3,4,5,6,9} and ji we have veried that A U B = B U A. ByVenn diagram, we have A. UB = B UA Fig. 1.7 (ii) Let us verify that intersection is commutative. Now, AB ={-~10,0,1,9,2,4,5}{ 1, 2,5,6, 2,3,4} = {2,4, 5}. Also, BOA ={1,-- 2,5,6,2,3,4}{~10,0,1,9,2,4,5} = {2,4, 5}. From 3 and 1;, we have A D B = B DA for the given sets A and B. By Venn diagram, we have Hence, it is veried. Example 1.2 Given, A = {1, 2, 3,4, 5}, B = {3,4, 5, 6} andC = {5,6, 7, 8}, show that (i) A U (B U C) = (A UB) U C. (ii) Verify (i) using Venndiagram. .§0£mfic3n (i) Now, BUC= {3, 4,5,6}U{5, 6,7,8}={3, 4,5,6,7,8} AU(BUC) Now, From (ii) = {1, 2, 3, 4, 5}U{ 3, 4, 5, 6, 7, 8}: {1, 2, 3, 4, 5, 6, 7, 8} A UB = {1, 2, 3, 4, 5} U {3, 4, 5, 6} = {1,2,3,4,5,6} (AUB)UC = {1,2,3,4,5,6}U{5,6,7,8} = {1, 2, 3, 4, 5, 6, 7, 8} and 5:, we have AU(BUC) = (AUB)U C. Using Venn diagram, we have Fig. 1.9 Thus, from and 1,we have veried that the set union is associative. Sam and Functions Example 1.3 Let A = {a,b,c,d}, B = {a,c,e} andC = {a,e}. (i) Showthat A D (B D C) = (A HB) D C. (ii) Verify (i) using Venn diagram. Sssisstiazy (i) We are given A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. We needto show A 0 (B D C) = (A DB) 0 C . So, we rst considerA D (B 0 C). Now, B 0 C = {a,c,e} {a,e} A (B D C) = {a,b,c,d} Next, we shall nd A B (A B) Now (ii) and C = {a,b,c,d.} = {a, e} ; thus, {a,e} = {a}. {a,c,e} = {a,c}. Hence = {a,c} {a,e} = {a} give the desired result. Using Venn diagram, we have Fig.1.1O Thus, from aLi and , it is veried that A H(B (1C) = (A n B) D C Example 1.4 Given A = {a, b,c,d, e}, B = {a, e, i,0,u} and C = {c,d,e,u}. (i) Showthat A\ (B \ C) 75(A \ B) \C . 70!/9 Std. Mat/mmzlicx (ii) Verify (i) using Venn diagram. Saimizm (i) First let us ndA\(B\C).To do so, consider (B\C) = {a,e,i,0,u}\{c,d,e,u} = {a,i,0}. Thus, A\ (B \ C) = {a,b,c,d,e}\{a,i,0} = {b,c,d,e}. Next, we nd (A \ B) \C. A \ B = {a,b,c,d,e}\{a,e,i,0,u} Hence, (A \ B) \C = {b,c,d}\{c,d,e,u} = {b,c,d}. = {I9}. From at and ;§ 1» we seethat A\ (B \ C) %(A \ B) \C. Thus, the set difference (ii) is not associative. Using Venn diagram, we have Fig. 1.11 From 2 and i, it is Veried that A \ (B\C) ¢ (A \ B) \ C. Sn; and Fmwtiom Example 1.5 Let A = {0,1,2,3,4},B = {1, 2,3,4,5,6} andC = {2,4,6,7}. (i) Showthat A U (B 0 C) = (A UB) 0 (A U C). (ii) Verify using Venndiagram. Saitsiékrrz (i) First, we ndAU(B0C). Consider B O C = {1, - 2, 3,4, 5, 6} D {2, 4, 6, 7} AU(BC) Next, consider = {0,1,2,3,4}U{4,6}= A UB = {0, 1,2,3,4} = {AUC = {4, 6}; {0,1,2,3,4,6}. U {1, ~2, 3,4,5,6} 2,0,1,2,3,4,5,6}, = {0,1,2,3,4}U{2,4,6,7} = {0, 1,2,3,4, 6,7}. Thus,(A UB) n (A U C)= {- 2, 0, 1,2, 3, 4, 5, 6} H {0, 1,2, 3, 4, 6, 7} = {0,1,2,3,4,6}. From (ii) and ,we get AU(B C) = (A UB)(AU C). Using Venn diagram, we have From2:and it isveriedthat A u (Bn C)= (Au B)n (Au C) 7Oil: Std. Mat/mmzlicr H92 EXample1.6 For A:{x|-3Sx<4,xER},B= {x|x< 5,xEN} and C: {- 5, -3, - 1,0,1,3}, ShowthatA(BUC)= (AnB)u(AnC). $}:£z;airmFirst note that the set A contains all the real numbers (not just integers) that are greater than or equal to 3 and less than 4. On the other hand the set B contains all the positive integers that are less than 5. So, A={x| 3Sx < 4,xER} ; thatis,A consists of allreal g-"--M-=-"-'"--numbers from 3 upto 4 but 4 is not included. Also, B={x|x<5,xEN}={1, BUC ={1,2,3,4}U{-- ={1,2,3,4,-5,~3 An(BuC) =An{1,2,3,4,-5,-3,-1,0} ={-3,-l,0,l,2,3}. Next, to nd (AB)U(AC),we consider AB and ={x| -3Sx<4,xElR}{l,2,3,4} ={l,2,3}; ADC={x|--3Sx<4,xElR}{-5 6. Verify the commutative property of set intersection for A = {l,m,n,0, 2,3,4,7} 7. andB = {2,5,3, -- 2, m,n,0,p}. For A = {x|xisaprimefactorof42}, B = {x|5 i<,B. Let us look at an example. Suppose that a cell phone store sells three different types of cell phones and we call themC1,C2,C3.Letusalsosuppose thatthepriceof C1is i 1200,priceof C2is ? 2500and priceof C3is ?2500. WetakeA = { C1,C2,C3} andB ={1200, 2500}. In thiscase,A XB={(C1,1200), (C1,2500),(C2,1200), (C2,2500),(C3,1200), (C3,25O0)} but B >B be afunction given by f(x) = 2x + l. Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow diagram and (iv) a graph. Satgiims A= {0, 1, 2, 3 }, B = { 1, 3, 5, 7, 9 }, f(x) = 2x +1 f(0)=2(0)+1=1,f(1)=2(1)+1=3, Oz/9Std. Mat/mmlics f(2)=2(2)+1=5,f(3)=2(3)+1=7 The given function f can be represented as a set of ordered pairs as f = { (0, 1)»(1, 3), (2, 5)» (3, 7)} Let us represent f using a table as shown below. Let us represent f by an arrow diagram. We draw two closed curves to represent the sets A and B. Here each element of A and its unique image element in B are related with an arrow. We are given that f= {(x»f(x))|x EA} = {(0,1),(1,3),(2,5),(3, 7)}. Now, the points (0, 1), (1, 3), (2, 5) and (3, 7) are plotted on the plane as shown below. The totality of all points represent the graph of the function. 1.8.3 . Fig. 1.26 Types of functions Based on some properties of a function, we divide functions into certain types. Let f :A ~+B be a function. The function one-one function if it takes different elements f is called an of A into different elements of B. That is, we say f is one-one if u 7Ev in A always imply f(u) ;£f(v). In other words f is one-one if no element in B is associated with more than one element in A. A one-one function is also called an injective function. The above gure represents a one-one function. Sets and F;/mtiom A function f : A » B is said to be an onto function if every element in B has a pre-image in A. That is, a function f is onto if for each b E B, there is atleastone elementa E A, suchthat f(a) = b. This is same as saying that B is the range of f. An onto function is also called a surjective function. In the above gure, f is an onto function. A function f : A ~+B is called a one-one and onto or a bijective function if f is both a one-one and an onto function. Thus f : A » B is one-one and onto if f maps distinct elements of A into distinct images in B and every element in B is an image of some element in A. ARA R: isonto andonly i:f,rangeiiioff A. tft=A~+ Bisoeroné andem»ifandonlyiff =fimP1iesai= 02inA and;everyte,lemCntttin.B hasexactlyonepre-image . L . ' r If :A; Bisaibijieetive function andif/1andBarenitesets,then thecardinalitiesp A iyiof andiB aresame. IniFigip_l1,2p9i, thefunction one;Ion§§.andi;onto. x AAA If f :.Ai_;13 igaeibijieictive function, then Aand Bareiiequivalentitsetis Aif lone.-one and ontoiifunctiopn also called one-onecorrespondencttegt RAif A function f : A » B is said to be a constant function if every element of A has the same image in B. Range of a constant function is a singleton set. LetA={x,y,u,v,l},B={3,5,7,8, 10, 15}. Fig. 1.30 The function f : A -» B dened by f(x) = 5 for every x E A is a constant function. The given gure represents a constant function. Let A be a non-empty set.A function f : A + A is called an identity function of A if f(a) = a for all a E A. That is, an identity function maps each element of A into itself. For example,let A = R. The function f : R > R be dened by f(x) = x for all x E R is the identity function on R. Fig.l.3l represents thegraph oftheidentity function onR. 701/9Std. Mat/yeirzalics Fig1.31 Example 1.21 Let A = { 1, 2, 3, 4, 5 }, B = N andf:A+B be denedby f(x) =x2. Find the range of f. Sz:{:t¬§03? Now, Identify the type of function. A = { 1, 2, 3, 4, 5 }; B = { 1, 2, 3, 4, } Given f : A > B andf(x) = x2 f(1) = 12:1; f(2) =4 ;f(3) =9; f(4) = 16; f(5) =25- Range of f = { 1, 4, 9,16, 25} Since distinct elements are mapped into distinct images, it is a one-one function. However, the function is not onto, since 3 E B but there is no x E A such that f(x) = x2= 3. Example 1.22 A function f : [1, 6) -» R is dened as follows f(x)= 1+x, 1Sx<2 2x-1, 2Sx<4 3962-~10, 4Sx<6 Find the value of (1) f(5) (Here, [1,6)={xe1R:1Sx< (ii) f(3) (iV) f(2) -f(4) 6}) (iii) f(1) (V) 2f(5) - 3f(1) S(;!1:2£m: (i) Let us ndf(5). Since 5lies between 4and 6,we have touse f(x) =3x2 - 10. Thus, f(5) = 3(52)- 10 = 65. (ii) To nd f(3), note that 3 lies between 2 and 4. So, we use f(x) = 2x - 1 to calculate f(3). Thus, (iii) f(3) = 2(3) - 1 = 5. Let us nd f(1). Now, 1 is in the interval 1S x < 2 Thus, we have to use f(x) = 1 + x to obtain f(1) = 1 + 1 = 2. Set: and F;/mtiom (iv) f(2) ~'f(4) Now, 2 is in the interval 2 S x < 4 and so, we use f(x) = 2x - 1. Thus, f(2) =2(2) - 1 = 3. Also,4 is in theinterval4 S x < 6. Thus,weusef(x) = 3x2- 10. Therefore, f(4) =3(42)- 10 = 3(16)~10 = 48 ~10 = 38. Hence, f(2) f(4) (V) =3 38 = 35. To calculate 2 f(5) 3 f(1), we shall makeuse of the valuesthat we have already calculatedin (i) and (iii). Thus, 2 f(5) ~ 3 f(1)= 2(65) - 3(2) = 130 --6 = 124. State whether each of the following arrow diagrams dene a function or not. Justify yO1l1' answer. Q x "Z For the given function F= { (1, 3), (2, 5), (4, 7), (5, 9), (3, 1) }, write the domain and range. Let A ={10,11,12,13,14};B = { O, 1, 2, 3, 5 } and :A>B , i = 1,2,3. State the type of function for the following (give reason): (1) )§= { (10,1),(11,2),(12,3),(13,5),(14,3) } (11) f2 ={(10,1),(11,1),(12,1),(13,1),(14,1)} (111)f3 = { (10,0),(11,1),(12,2),(13,3),(14,5) } If X= { 1, 2, 3, 4, 5 }, Y= { 1, 3, 5, 7, 9 } determine which ofthe following relations from X to Y are functions? Give reason for your answer. If it is a function, state its type. (i) R1={(x,y)|y=x+2, x¬X, y¬Y} (ii) R2= {(1, 1),(2, 1),(3,3),(4,3),(515)} (iii) R3= { (1, 1),(1,3),(3,5),(3,7),(5,7) } (iV)R4= { (1,3):(2,5),(4,7),(5,9),(3, 1)} If R: {(a, 2),( 5,b), (8, c), (d,values of a, b, c and d. 701/9Std. Mat/mmlics 1)} represents the identity function, nd the A={-2,-1,1,2} and f: {Bisdenedbyf(x)=x--2,then the range of f is (A){l,4,5} 18. (B){l,2,3,4,5} 20. (D){3,4,5} If f(x) = x2 + 5, then f(-- 4) = (A) 26 19. (C) {2,3,4} (B) 21 (C) 20 (D) -20 If the range of a function is a singleton set, then it is (A) a constant function (B) an identity function (C) a bijective function (D) an one-one function If f : A > B is a bijective function and if n(A) = 5 , then n(B) is equal to (A) 10 (B) 4 (C) 5 (D) 25 Points. to :Remem'ber ) SETS E]Aset isacollection ofwell defined objects. Setunion is commutativeand associative. A Set intersection Set difference is commutative and associative. is not commutative. Set difference is associative only when the sets are mutually disjoint. Distributive Laws S Au(BnC)= (AuB)n(AuC) An(BuC)=(AnB)u(AnC) De Morgans Laws for set difference A\(BuC) = (A\B)n(A\C) A\(BnC)= (A\B)u(A\C) De Morgans Laws for complementation. (AUB)=A'r)B'(AnB)':A'LJB' Formulae for the cardinality of union of sets n(AUB) = n(A) + n(B) - n(AB) n(AUBUC) FUNCTIONS CI The cartesian product of Awith B is dened as A>R denedby g(k) = ak,V k E N. Thesymbol V meansfor all. If the generalterm akof a sequence {ak}f°is given, we can construct the whole sequence.Thus, a sequence is a function whose domain is the set{ 1, 2, 3, , }of natural numbers, or some subset of the natural numbers and whose range is a subset of real numbers. Eample . Write the rst three terms in a sequencewhose 11term is given by C = n(n + l)(2n 6 +1) ,V n CN S§¬9§££§§¬;*£§ Here, C="(n+1)6(2n +1),VnCN For n=1, C1 = =. Mr ":1 Cf:m2+2@+u__x?GLj_ Mmwn=a %=3G2m(?UVw4 Hence, the rst three terms of the sequenceare 1, 5, and 14. In the above example, we were given a formula for the general term and were able to nd any particular term directly. In the following example, we shall see another way of generating a sequence. Example 2.2 Write the rst ve terms of each of the following sequences. . a,,_ (1) a1:-1, (m 3=g=1am 701/9Std. Mat/yeimzlicx a=n+12, n>land Vn¬N q=q4+Q_ 29 n=1n« Szziasfivgz (i) Given a1 =-1and an =:_{:12, 11 >1 2 2 +12 a_ 3 a_ 4 a_ 5 4 a2 = 3 + 2 3 = 4 + 2 4 5 + 2 _. L 4= 5 1 20 _. L 20 1 6 = 120 _. ___1__ 120 = 7 '. Therequired terms ofthesequence are 1, 1 4 (ii) 1 840 1 20 1 nd 120a 1 840' GiventhatF1 =F2 = 1 andF =Fn_1+Fn_2,forn = 3,4,5,---. Now,F1=1,F2=1 F3=F2+F1=1+1=2 F4=F3+F2=2+1=3 F5=F4+F3=3+2=5 The rst ve terms of the sequenceare 1, 1, 2, 3, 5. 3. Find the 18and 25*terms of the sequencedened by n(n + 3), if n E N and n is even a 4. n +1 Find the 13and 16*terms of the sequencedened by I9: Vl 5. 22" ,if n¬N andn isodd. 2 ll {n, ifnEN and niseven 2 . . n(n+2), if nEN and n isodd. Find the rst ve terms of the sequencegiven by a1=2,a2=3+a1andan=2an1+5 6. forn>2. Find the rst six terms of the sequencegiven by a=a=a=1anda=a 1 2.3 2 Arithmetic 3 n nl +a sequence or Arithmetic n2 forn>3. Progression (A.P.) In this section we shall see some special types of sequences. A sequencea1a2 as" --,a_n, is called an arithmetic sequence if an 2:an+d, EN where disaconstant. Here aliscalled thefirstterm and the constantd is calledthecommon difference. Anarithmetic sequence is also called an Arithmetic Progression (A.P.). A Exampies (i)2,5,8,11, 14,isan A.P. because a1 =2and the common difference d=3. (ii) -4, -4, -4, -4, (iii) is anA.P. becausea1= -4 and d = 0. 2, 1.5, 1, 0.5, O, - 0.5, - 1.0, - 1.5,--- is anA.P. becausea1=2andd= -0.5. The general form of an A.P. Let us understand the general form of an A.P. Suppose that a is the rst term and d isthecommon difference of anarithmetic sequence {ak}Z:1. Then,wehave a1=aanda Forn= 1,2,3 n+1 =a+d,Vn¬N. n we get, a2=a1+d=a+d=a+(2 a3=a2+d=(a+d)+d= Thus,wehavean= a + (n --1)d for everyn E N. So, a typical arithmetic sequenceor A.P. looks like a, a+cl. 51+ 2d, a+3d, , a+(n l)d, a+na.~-- Also, the formula for the general term of an Arithmetic sequenceis of the form inz a + (Fl l)cZ foreveryn E N. A(i) Remember asequencemay also beanitesequence.So, if anA.P. has only nterms) AL0' thenthelasttermI isgivenbyea (!éL',-ifl )d LL L l . helps ustondthe (ii) La -2 canalso berewrittenas Fl: (1 ~l~ Lnumber ofterms when thefirst,thelastterm andthecommonidifference aregiven. .2(iii) Threeconsecutive terms ofanA.P. maybetaken. as -ed,m,m+ d A5 (iv)LFour consecutive terms ofanA.P. may betaken asms-3d,Lm4-d.m+ d. m+ 3d 3. withcommondifference 2d. 3 A L LL A LL AL L .(v) AnA.P.remains an A.P. ifgeach of terms: isaddedt orsubtracted byaL same 5 constant.A L ALL A LA L (vi)AnA.P. remains anA.P. ifeach ofitsterms ismultippliedor ydividedbypa nonzero.[ V 7constant-_- Example 2.3 Which of the following sequencesare in an A.P.? (1). (ii)3m-l,3m-3,3m- iiizfsesi (i) Let tn,nENbe the n"term ofthe given sequence. 1 S0 2; 3 ,2 "-4 5 ,3 -3 7 _ 4 tt_6 3 Since (ii) 2 _ 2 I 5 3 12 - 15 10 4_30-28_ 2 7 5 35 _ 2 15 2 35 I2- tl aét3- t2, thegivensequence is not anA.P. Given 3m ~1,3m - 3,3m 5,---. Here Also, t1=3ml,t2=3m3,t3=3m5,---. t2 t1 (3m 3) (3m 1) = 2 t3 t2 (3m 5) (3m 3) = 2 Hence, the given sequenceis an A.P. with rst term 3ml and the common dierence -2. Sequences andreex of realm/when Example 2.4 Find the rst - term and common __ ._ -- difference of the A.P. Lilim. (1) 5929 19 49 - (11)29696929 96 :§v{}ig'{g§§{}i3 (i) First term a=5,and the common difference d=2-5= -3. (ii) a=%and thecommon difference d =%~%= 56%? =%. Example 2.5 Findthe smallestpositiveintegern suchthat tnof thearithmeticsequence 20,19%, 18%,isnegative? ,siizsm6si Here wehave a=20,d=19% ~20=--%. Wewantto nd therst positiveintegern suchthat tn < 0. This is sameas solving a + (n - 1)d < 0 for smallest n E N. That issolving20+(n~1)(- <0 forsmallest nEN. Now, (12 --1)(231-) <--20 _-=> (rz --1)Xi}> 20 (The inequality isreversed onmultiplying both sides by-1) £_&= n .._1>20><3_3 2 263. This implies n>26% +1.That is,n>27-31 =27.66 Thus, the smallestpositive integer n E N satisfying the inequality is n = 28. Hence, the 28 term ,t 28is the rst negativeterm of the A.P. Example 2.6 In a ower garden, there are 23 rose plants in the rst row, 21 in the second row, 19 in the third row and so on. There are 5 rose plants in the last row. How many rows are there in the ower garden? EogisfémsLet n be the number of rows in the ower garden . The numberofrose plants in the 15,2, 3 ,--- , 11rows are 23, 21, 19, respectively. Now, tk--tk_1=-~2 for k=2,---,n. Thus, the sequence 23, 21, 19, Oz/9Std. Mat/mmzlics , 5 is in an A.P. ,5 Wehave a=23, n _l-a d=-2,andl=5. d +1 _5-23 ___2+1 _ 10. So, there are 10 rows in the ower garden. Example 2.7 If a person joins his work in 2010 with an annual salary of ?30,000 and receives an annual increment of ?6OOevery year, inwhich year,will his annualsalary be ?39,000? §w§z££§:mSuppose that the persons annual salary reaches ?39,000 Annual salary of the person in 2010, 2011, 2012, ?30,000, ?30,600, ?31,200, in the 11 year. , [2010 +(n ~ 1)] will be , ?39000 respectively. First note that the sequenceof salaries form an A.P. To nd the required number of terms, let us divide each term of the sequenceby a xed constant 100. Now, we get the new sequence 300, 306, 312, , 390. Here a = 300, d = 6, l= 390. So,n=l;a+1 _390--300 6 +1 _9O 6+1 _ 16 Thus, 16annualsalaryof the personwill be 1'3 9,000. '. His annual salary will reach i39,000 in the year 2025. Example 2.8 Three numbers are in the ratio 2 : 5 : 7. If the rst number, the resulting number on the substraction of 7 from the second number and the third number form an arithmetic sequence, then nd the numbers. .§mm§:2:2Let the numbers be 2x, 5x and 7x for some unknown x,(x 750) By the given information, we have that 2x, 5x ~ 7, 7x are in A.P. (5x--7)-~2x=7x-~(5x-7) ==>3x-~7 = 2x+7 and so x= 14. Thus, the required numbers are 28, 70, 98. 1. The rst term of an A.P. is 6 and the common difference is 5. Find the A.P. and its general term. 2. Find the common difference and 15 term ofthe A.P. 125, 120, 115, 110, . 3. Which term ofthe arithmetic sequence 24,23-411: 22%, 21%,.is3? Sequences and reex ofreal m/when? Findthe12htermoftheA.P. @3555, . Find the 17 term ofthe A.P. 4, 9, 14, How many terms are there in the following Arithmetic Progressions? ___1, __:35 6 T... 2 3 j 3 '(ii) 7, 13, 19, 205. If 9* term of an A.P. is zero, prove that its 29 term is double (twice) the 19 term. The 10and 18terms of an A.P. are 41 and 73 respectively.Find the 27 term. 10. Find n so that the 11terms of the following two A.P.s are the same. 1, 7,13,19,--- and 100, 95, 90, 11. How many two digit numbers are divisible by 13? A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the tenth year. Assuming that the production increases uniformly by a xed number every year,nd the numberof TVs producedin the rst year and in the 15* year. 12. A man has saved ?640 during the rst month, ?720 in the second month and ?800 in the third month. If he continueshis savings in this sequence,what will be his savingsin the 25* month? 13. The sum of three consecutive terms in an A.P. is 6 and their product is -120. Find the three numbers. 14. Find the three consecutive terms in an A. P. whose sum is 18 and the sum of their squares is 140. 15. If m times the mthterm of an A.P. is equal to n times its 11term, then show that the (m+n)hterm of the A.P. is zero. 16. A person has deposited ?25,000 in an investment which yields 14% simple interest annually. Do these amounts (principal + interest) form an A.P.? If so, determine the amount of investment after 20 years. 17. If a,b,c areinA.P.thenprovethat(a- c)2 = 4(b2- ac). 18. If a,b,c areinA.P.thenprove thatLbc ca L, ab L arealsoinA.P. 19. If a2,b2,c2arein A.P.thenshowthat 20. Ifax=by=cZ,x ¢0,y¢0,z¢0and b2=ac,then show that L,%are inA.P. 1 , 1 b+C c+a , 1 l 'A.P. Cl+bareaS0m Ly 9 2.4 Geometric Sequence or Geometric Progression (G.P.) Asequencea1,a2, a3,---,an, is calleda an = anrd, nE N, where r isanon-zero constant. Here,allisthersttermand theconstant r is calledthe A geometric sequenceis alsocalleda Let us consider some examples of geometric sequences. (i) 3,6, 12,24,---. a A sequence {an}1°° is ageometric sequence if "+1 = r 75O, rt63N. 61 Now, %=%=% =27E 0.Sothe given sequence isageometric sequence. -- L <1) 9 _. L L __ L .. 27 81 243 1 1 1 Here,wehave 2: = -E: 9 §= 27 $750. 81 Thus, the given sequenceis a geometric sequence. The general form of a G.P. Let us derive the general form of a G.P. Suppose that a is the rst term and r is the common ratioof ageometric sequence {ak}:°_ 1. Then,wehave a a1=aand Cl+1=r forn¬N. 11 Thus, For an+1 = r an for n E N. n=l,2,3 we get, _ _ __ a2 a1rarar 2-1 a3 = azr = (ar)r = ar2 =ar 3-1 2 a4 =a3r= (ar )r=ar 3 =ar 4-1 Following the pattern, we have a=a rt Thus, n2) r=ar IL-1- n 1 r=(ar a = ar for every n E N, gives nthterm of the G.P. So, a typical geometric sequenceor G.P. looks like 2 5:, ar, ar, 3 ar, /1~l ar n ,ar Thus , the formula for the general term of a geometric sequenceis zFl ==ar1,n=l,2,3,---. Sequence: andreexof realnumbers Suppose we are given the rst few terms of a sequence,how can we determine if the given sequence is a geometric sequenceor not? If I mr,vrsEN,where r isanon-zero constant, then { tn isinG.P. H L If theratio ofanytermother than therstterm toitspreceding term ofasequence is a non-zero: constant, thenitis a geometric sequence. , ' doA geometricsequence remains ageometric sequence if eachtermis multipliedor dividedby a nonzeroconstant.0 LL L L L L LL L LL LThree consecutive terms inaGPmay betaken as%,cl,arwithcommon ratio r.) ~Four consecutive terms inaG.P may betaken as.%, 0%, ar,ar. e 0(here, thecommon ratioisr 0notras above) Le Example 2.9 Which of the following sequencesare geometric sequences (1)5,10,15,20,-~. (ii) 0.15,0.015,0.0015,---. (iii) , ¢21,3,3¢21,---. t§g§e§§i:m (i) Considering the ratios oftheconsecutive terms, wesee that %aé Thus, there is no common ratio. Hence it is not a geometric sequence. -0.015 _ 0.0015 _ (11) Weseethat 0.15 0.015 =L 10. Since thecommon ratio is%,the given sequence isageometric sequence. (iii)Now, /2_7_1 = = = =/.3.Thus, the common ratio is«/3. Therefore, the given sequence is a geometric sequence. Example 2.10 Find the common ratio and the general term of the following geometric sequences. (i) %, . (ii) 0.02, 0.006, 0.0018, . .3T10§s£:fig2 it (i) Given sequence is a geometric sequence. l l The commonratio is given by r = -2- = -1 = 6 Thus, r =~:2;5= 5 7N0:/; Std.Mat/yeiizalicr t1 2 The rsttermofthesequence is So, the general term ofthesequence is -1 tn=ar" ,n= 1,2,3,---. => tn=§(§~)"'1, n== 1,2,3,-~ (ii) The common ratio of the given geometric sequence is r: 0.006 = 03 = 3 0.02 10' The rst term of the geometric sequenceis 0.02 So, the sequencecan be represented by .3. "-1 .. tn__ (o.o2)(10) , n_..1,2,3,Example 2.11 The 4termofageometric sequence is%and theseventh term is Find the geometric sequence. V * I égeezeswss Given thatt4__g2 andt7__81. Using theformula tn= ar"_1, n = 1,2,3, .forthegeneral term wehave, 16 t4_ar3=§2 andt7=ar6=81. Note that in order to nd the geometric sequence,we need to nd a and r. By dividing t7 by £4we obtain, Thus, r3=287 = which implies r= Now, t4=%> ar3 = ==> a(287) = a= Hence, therequired geometric sequence is (1,ar,arz,ar3, a/-1,ar",-~ e»%(%>» ea» Example 2.12 The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria presentin the culture initially, how many bacteriawill be presentat the end of 14hour? ..§:;»§zs§i(:e2 Note that the number of bacteria present in the culture doubles at the end of successive hours. Sequences andrerierof realnumbers Number of bacteria present initially in the culture = 30 Number of bacteria present at the end of rst hour = 2 (30) Number of bacteriapresentat the end of secondhour = 2 (2(30)) = 30(22) Continuing in this way, we see that the number of bacteria present at the end of every hour forms a G.P. with the common ratio r = 2. Thus,if in denotesthenumberof bacteriaafterit hours, tn= 30(2")isthegeneral termof theG.P. Hence, thenumber ofbacteria attheendof14 hourisgiven by r14= 30(214). Example 2.13 An amount ?500 is deposited in a bank which pays annual interest at the rate of 10% compoundedannually. What will be the value of this depositat the end of 10year? S{}§£§3f0i§ The principal is?500. So, the interest forthis principal forone year is5O0(11()%) =50. Thus, the principal for the 2nd year = Principal for 1st year + Interest = 500 +500(-115%) =50o(1+ 116%) Now, theinterest forthesecond year =<500n_ 1. The amountat the end of (nl)h year = Principal for the nth year. Thus, the amountin the accountat the end of 10*year =?500(1+ -I16Q0-)1°= ?500(%)°. 0122Sid. Mat/mrzalicr Example 2.14 The sum ofrstthree terms ofageometric sequence is% and their product is 1. Find the common ratio and the terms. §£}§§£!§f§§§ Wemay take therstthreeterms ofthegeometric sequence as-3-, a,ar. Then, %+a+ar=% _1_ _13 r2+r+1>_13 "(r'"1> 12 ""< r 12 Also, Q =... (r )(a)(arz1 => a Substituting a = - 1 in = -1 a = - 1 we obtain, r +rr+1)2Q (__ 1)< 12 2 :2. 12r2+ 12r+ 12 =-~13r 12r2+ 25r +12 = 0 (3r + 4)(4r + 3) = 0 Thus, r =--%or-% When r =-g and a= 1,theterms are 33-, When r = ~-2anda= 1,weget5%, -1,72-, which isinthe reverse order. Example 2.15 If a, b, c, d are in geometric sequence,then prove that (b~c)2+(ca)2+(db)2 =(a-d)2 S§}f£l§f§}!?Given a, b, C,d are in a geometric sequence. Let r be the common ratio of the given sequence. Here, the rst term is a. Thus,b=ar, c=ar, 2 d=ar 3 Now,(b - (3)2+ (c a)2+ (d - b)2 = (ar - ar2)2+ (arz a)2 + (ar3- ar)2 = a2[(r-- r2)2+ (r2- 1)2+ (r3- r)2i Find out which of the following sequencesare geometric sequences.For those geometric sequences,nd the common ratio. (1) 0.12, 0.24, 0.48,---. 9129 '9/5! 2/59 Luna 5 m 4. (ii)0.004, 0.02, 01,.--. (iii) ~§, .2f*7, 1 1 2, (vi) 4,-2, 1,L (iv) 12,1 Findthe10termandcommonratioofthegeometric sequence Z, E,1,- 2,. If the 4 and 7 terms of a G.P.are 54 and 1458respectively,nd the G.P. Inageometric sequence, thersttermisL3 andthesixthtermisL729 ndtheG.P. Which term of the geometric sequence, - .4. - 123 (1)5 2 5 25 "S 15625? (ii) 1, 2, 4, 8,---, is 1024 .7 If the geometric sequences 162, 54, 18,---. and 2 2 2 §T, 27, -9-,--have their11* term equal, nd the value of n. The fth term of a G.P. is 1875. If the rst term is 3, nd the common ratio. 10. Thesumofthree terms ofageometric sequence isand theirproductis 1.Findthe 10 common ratio and the terms. If the product of three consecutive terms in G.P. is 216 and sum of their products in 11. pairs is 156, nd them. Find 12. the rst three consecutive terms in G.P. whose sum is 7 and the sum of their reciprocals is1 4 The sum of the rst three terms of a G.P. is 13 and sum of their squaresis 91. Determine 13. the G.P. If 1'1 000 is deposited in a bank which pays annual interest at the rate of 5% compounded annually, nd the maturity amount at the end of 12 years . 14. A company purchases an ofce copier machine for ?50,000. It is estimated that the copier depreciates in its value at a rate of 15% per year. What will be the value of the 15. copier after 15 years? If a, b, c, d are in a geometric sequence,then show that (a-~b+c)(b+c+d)=ab+bc+cd. If a, b, c, d are in a G.P., then prove that a + b, b + c, c + d, are also in G.P. oi}Zi§;7c274W;;;7Zz§Z} WWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW ii 2.5 Series Let us consider the following problem: A person joined a job on January 1, 1990 at an annual salary of ?25,000 and received an annual increment of ?500 each year. What is the total salary he has received upto January 1, 2010? First of all note that his annual salary forms an arithmetic sequence 25000, 25500, 26000, 26500,---, (25000 + l9(500)). To answer the above question, we need to add all of his twenty years salary. That is, 25000 + 25500 + 26000 + 26500 + + (25000 + 19(500)). So, we need to develop an idea of summing terms of a sequence. An expression of additionof termsofa sequence is calleda series. 5If a seriesconsistsonlya nite numberof terms,it is calleda finite series. If a seriesconsistsofinnite numberofterms of a sequence, it iscalledaninnite series.i Consider a sequenceS = {an2:1 of realnumbers. Foreach n E N we denethe partialsumsby S = a1+a2+,--- + an,n = 1,2,3,---.Then{Sn}°° is thesequence of n=l partialsumsof the givensequence {an ;°_1. ppjp .The;orderedppairip({a;+}f=I iscalledinniteseries ofterms so:the irigeqqrieeei if inniteseries istdenoted bysal +a2i~:1~ a3 ,orsimply zany thesymbol Zj lstandspforsurnmation andis pronounced asisigrna. .iiii":1 Well, we can easily understand nite series (adding nite number of terms). It is impossible to add all the terms of an innite sequenceby the ordinary addition, since one could never complete the task. How can we understand (or assign a meaning to) adding innitely many terms of a sequence? We will learn about this in higher classes in mathematics. For now we shall focus mostly on nite series. In this section , we shall study Arithmetic series and Geometric series. 2.5.1 Arithmetic series An arithmetic series is a series whose terms form an arithmetic sequence. Sum of rst n terms of an arithmetic sequence Consider an arithmetic sequencewith rst term a and common difference d givenby a, a + d, a + 2d,..., a + (n 1)d,---. Let S be the sumof rst n termsof the arithmeticsequence. Sequences andreexof realm/when Thus, S":-a+(a+d)+(a+2d)+-~(a+(n-1)d) ==> Sn=na+(d+2d+3d+---+(n-1)d) =na+d(1+2+3+--+(n-1)) So, we can simplify this formula if we can nd the sum 1+2+--- +(n- 1). This is nothing but the sum of the arithmetic sequence 1, 2, 3, ---, (n - 1). So, rst we nd the sum 1+ 2 + + (n - 1) below. Now, let us nd the sum of the rst n positive integers. Let Sn=1+2+3+--+(n--2)+(n-1)+n. Weshallusea smalltrick to nd theabovesum.Notethatwe canwrite Snalsoas Sn=n+(nl)+(n-2)+---+3+2+l. Adding E3and we obtain, 2Sn=(n+1)+(n+1)+---+(n+l)+(n+1). Now, how many (n + l) are there on the right hand side of is? There are n terms in eachof 1 and 171We merely addedcorrespondingterms from V:and Thus, there must be exactly n such (n + 1) s. Therefore, EV: simpliesto 2S" = n (n + 1). Hence, the sum of the rst 11positive integers is given by =n(n+1)_ S'1 2 So, l+2+3+---+11 =n(n+l)_ 2 This is a useful formula in nding the sums. Now, let us go back to summing rst n terms of a general arithmetic sequence. We have already seen that S,,=na+[d+2d+3d+--=na+d[1+2+3+--- +(n--1)d] +(n~1)] = na+d-n(n2 1)using (4) =gm +(n-1)d] 01/7Std. Mat/yeiizalicx Hence, we have S = %[a+ (a+(n- 1)d)]=%(rstterm+lastterm) =%(a+l). ThesumS of therst n termsof anarithmeticsequence with rst term a is given by (i) S:g~[2a +(H 1)d]if thecommon difference aisgiven. (ii) Sn= gm+1),ifthe1ast term 1isgiven. Example 2.16 Find the sum of the arithmetic series 5 + 11 + 17 + .S70i:;t§'m2 Given that the series 5 + 11 + 17 + Notethat Now, a =5, d=11~5 = 6, + 95. + 95 is an arithmetic series. l=95. n = Hence, thesumSfl = +a] S16 =%[95 +5]=8(100) =800. Example 2.17 Find the sum of the rst 211terms of the following series. 12.22+32-42+.-. .§v!zmanWe want to nd 12- 22 + 32- 42 + =1--4+9--16+25---~ to 2n terms to2nterms = (1 ~ 4) + (9 - 16) + (25 - 36) + = -3 + (--7) + (-- 11) + to :1terms. (after grouping) n terms Now, the above series is in an A.P. with rst term a = - 3 and common difference d = -- 4 Therefore, therequired su =%[2a +(n- 1)d] E[2<-3) + (n -1><-4)] 2 =2. 2[ 6__ 4n+4]:n_ 2[ 4n_ 2] = "'2" (Zn+ 1)= n(2n+1). Sequences andreexof realnumbers Example 2.18 In an arithmetic series, the sum of rst 14 terms is - 203 and the sum of the next 11 terms is -572. Find the arithmetic Ss;§2m7w: Given that = series. S14= -203 %[2a +13d] = 203 ==> ==> 7[2a +13d] = -203 2a + 13d = -29. Also, the sum of the next 11 terms = - 572. Now, S25= S14+ (~--572) Thatis, S25= -203 -- 572 = -775. =..=,%[2a+24d]= -775 ==> 2a+24d=-31><2 ==> Solving a + 12d and = -31 we get, a = 5 and d = - 3. Thus, the required arithmetic seriesis 5 + (5 -- 3) + (5 + 2(~3)) + Thatis,theseriesis 5+214-7----. Example 2.19 How many terms of the arithmetic series 24 + 21 + 18 + 15 + continuously so that their sum is 351. .§§a§mz':.m In the given arithmetic series, a = 24, Let us nd n such that S ll = d = -~ 3. 351 Now, S = -21[2a+(n~ 1)d]=-351 That is,%[2(24)+(n-1)(-3)] = -351 => l21[48-3n+3] = -351 =-.> =2 n(51-~ 3n) = -702 n2-17n-234=0 (rz- 26)(n + 9) = 0 n = 26 or n = -- 9 Here n, being the number of terms needed, cannot be negative. Thus, 26 terms are needed to get the sum - 351. Oz/9Std. Mat/mmlics , be taken Example 2.20 Find the sum of all 3 digit natural numbers, which are divisible by 8. §§z;£5l£a§a,e§rz The three digit natural numbers divisible by 8 are 104, 112, 120, , 992. Let Sndenote their sum.Thatis, S = 104+ 112+ 120+ 128+, Now, the sequence 104, 112, 120, Here, a = 104, d = 8 n _ l-a d +1 _ , 992 forms an A.P. and 1: 992-.~ + 992. 8 104 992. +1 =-§§+1=112. Thus, Sm= gm+1]=172[1o4 +992] =56(1096) =61376. Hence, the sum of all three digit numbers, which are divisible by 8 is equal to 61376. Example 2.21 The measures of the interior angles taken in order of a polygon form an arithmetic sequence.The least measurement in the sequence is 85°. The greatest measurement is 215°. Find the number of sides in the given polygon. £¢;§zg;f:;;:Let n denote the number of sides of the polygon. Now, the measures of interior angles form an arithmetic sequence. Let the sum of the interior angles of the polygon be S =a+(a+d)+(a+2d)+ ---+l ,wherea=85andZ=215. Wehave, S: %[l +a] Weknowthatthe sumof the interioranglesof a polygonis (n ~2)><180°. Thus, S = (n ~2)><180 From3:, wehaveg[l +a]=(n2)><180 => %[2l5+85]=(n--2)> n = 12.. Hence, the number of sides of the polygon is 12. Exercise 2.4 1. Find the sum of the rst (i) 75 positive integers (ii) 125 natural numbers. Find the sum of the rst 30 terms of an A.P. whose nthterm is 3 + 2n. 3. Find the sum of each arithmetic (i) 38+35+32+---+2. series (ii)6+51~+41+--25terms. 4 2 Sequences and series ofreal m/when? 10. 11. Findthe S for the followingarithmeticseriesdescribed. (1) a=5, n=3o, 1:121 (ii) a=50, n=25, d=--4 Find the sum ofthe rst 40 terms ofthe series 12 - 22 + 32- 42 + 12. In an arithmetic series, the sum of rst 55. Find the arithmetic . 11 terms is 44 and that of the next 11 terms is series. In the arithmetic sequence 60, 56, 52, 48,--- , starting from the rst term, how many terms are needed so that their sum is 368? Find the sum of all 3 digit natural numbers, which are divisible by 9. Find the sum of rst 20 terms of the arithmetic is 2 more than three times 13. series in which 3rdterm is 7 and 7 term its 3 term. Find the sum of all natural numbers between 300 and 500 which are divisible by 11. Solve: 1+ 6+ 11 + 16+ +x --148. 20. A gardener plans to construct a trapezoidal shaped structure in his garden. The longer side of trapezoid needs to start with a row of 97 bricks. Each row must be decreased by 2 bricks on each end and the construction should stop at 25 row. How many bricks does he need to buy? 2.5.2 Geometric series A series is a geometric series if the terms of the series form a geometric sequence. 2 l . . Let a, ar, ar ,---, ar" ,ar",--- be a geometricsequence wherer aé0 is the common ratio. We want to nd the sum of the rst rz terms of this sequence. LetSn=a+ar+ar2+---+ar" If r = 1, thenfrom 3 it followsthat S = na. For r 751, using we have IS = r(a+ar+ar2+ +ar"_1) = ar+ar2+ar3+ +ar". Now subtracting from *, we get SnrSn = (a+ ar+arz+ + ar"_1)-(ar+ arz+ + ar") ==> Sn(l-~r) = a(l - r") Hence, we have Sna(11:rrn) ,since r7&1 The sum of the rst :1terms of a geometric series is given by 5: if rséi na if where a is the rst r 2 l. term and r is the common ratio. Example 2.22 Find the sum of the rst 25 terms of the geometric series 16-48+144432+---. .§7a£zzzi:m Here, a=16,r =-% =~3 ¢1.Now, S=a(11:;n), r761. So, we have S25 ~16(11~ ((__)_16(1;|1*3 ) =4(l+325) __ __ 25 25 Seq;/ewes andreexof realnumbers Example 2.23 Find S"for eachof the geometricseriesdescribedbelow: (i)a=2, t6=486, n=6 (ii)a=2400,r=3, n=5 f5?a£;ze'3§423@ (i) Here a=2,t6 =486, n=6 Now:6.-.2(r)5=486 ..=> r = 243 Now, S'1= i Thus, (ii) if r¢1 1 6 __ S6 _ 32(36~1) - "1 Here a = 2400, Thus, r-- r=3. r =- 3, n= 5 a(r5 -1) . S5 _ -7-:*i If 7751 : 2400[(-3)5-- 1] (.3)...1 Hence,S5= %+00(1+35) = 600(1 +243)= 146400. Example 2.24 In the geometric series 2 + 4 + 8 + ---, starting from the rst term how many consecutive terms are needed to yield the sum 1022? 5es§zm*742§§ Given the geometric series is 2 + 4 + 8 + . Let n be the number of terms required to get the sum. Here a = 2, r = 2, S = 1022. To nd 11,let us consider Sn [:"__11]ifr¢1 (2)[énj11]2(2 1) But Sn=1022 andhence2(2~ 1): 1022 => 21=511 2:. 2=512=29. Thus, n=9. Example 2.25 The rst term of a geometric series is 375 and the fourth term is 192. Find the common ratio and the sum of the rst Oz/9Std. Mat/mmlics 14 terms. §§e:!z:2.:fe;§e Let a be the rst term and r be the common ratio of the given G.P. Giventhat a = 375, t4 = 192. Now, tn= ar"_1 :4= 375r3 __.-> 375r3= 192 r3 :2: r3= » r3 3;: ==>r =g,which isthe required common ratio. Now, sn=a:_'_11]if 375 ii) 1 = =<~1>xsxsvs[<%> 11 5 =(37s)(5)[1 =1875[1 - "=/Ia ]if.*s%.1:iI1St¢ad{0f5,, [.7r,.71511,.]ifr{ Example 2.26 A geometric series consists of four terms and has a positive common ratio. The sum of the rst two terms is 8 and the sum of the last two terms is 72. Find the series. .§m:e£ie2:z Letthesumof thefourtermsof thegeometric series bea + ar + arz+ ar3and r > 0 Given that a + ar = 8 and ar2 + ar3 = 72 Now, arz+ ar3= r2(a + ar) = 72 .=> r2(8) =72 Since r > O, Now, we have r = :3 r = 3. a+ar=8 Thus, the geometric series is --> a=2 2 + 6 + 18 + 54. Example 2.27 Find the sum to n terms of the series 6 + 66 + 666 +--- ,§es§e:=zfm9 Note that the given series is not a geometric series. Weneedtond Sn=6+66+666+--tonterms Sn=6(1+11+111+--- tonterms) =g(9+99+999 + tonterms) (Multiply and divide by9) =%[(1o 1)+(100-1)+(1ooo1)+ tonterms] =%[(1o +102 +103 + rzterms)n] Example 2.28 An organisation plans to plant saplings in 25 streets in a town in such a way that one sapling for the rst street, two for the second, four for the third, eight for the fourth street and so on. How many saplings are needed to complete the work? ..§g.sEm£{m The number of saplings to be planted for each of the 25 streets in the town formsa G.P.Let S bethetotal numberof saplingsneeded. Then, Here, S" =1+2+4+8+l6+--a=l, r=2, to 25terms. rz.-=25 S": (1r"-1 rwli _ 525(D [2--1] 2~1 2 25__1 Thus,thenumberof saplings to beneeded is 225- 1. 1. 5 Findthesumoftherst20terms ofthegeometric series i5 + K + K5 + . 2. Find thesum ofthe rst27terms ofthe geometric series %+217 +811 + . 3. Find S for eachof the geometricseriesdescribedbelow. (i)a=3,t8=384,n=8. 4. Find the sum of the following nite series (i) 1 + 0.1+ 0.01+ 0.001+ 5. (ii)a=5,r=3,n=12. +(O.l)9 (ii) 1 + 11+ 111+ to 20 terms. How many consecutive terms starting from the rst term of the series (i) 3 + 9 + 27 + would sum to 1092? (ii) 2 + 6 +18 + would sum to 728 ? 6. The second term ofageometric series is3and thecommon ratio is Find thesum of rst 23 consecutive terms in the given geometric series. 7. A geometric series consists of four terms and has a positive common ratio. The sum of the rst 8. two terms is 9 and sum of the last two terms is 36. Find the series. Find the sum of rst (i)7+77+777+---. 9. n terms of the series (ii) 0.4+0.94+0.994+---. Supposethat ve people are ill during the rst week of an epidemic and each sick person spreadsthe contagious diseaseto four other people by the end of the second week and so on. By the end of 15week, how many people will be affectedby the epidemic? 7"oi;'If_"'s}7;2i*2{4";};};Zz'£-2} wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww ii 10. A gardener wanted to reward a boy for his good deeds by giving some mangoes. He gave the boy two choices. He could either have 1000 mangoes at once or he could get 1 mango on the rst day, 2 on the second day, 4 on the third day, 8 mangoes on the fourth day and so on for ten days. Which option should the boy choose to get the maximum number of mangoes? ll. A geometric series consists of even number of terms. The sum of all terms is 3 times the sum of odd terms. Find the common 12. ratio. If S1,S2 andS3are thesumof rst :1,2n and3ntermsof a geometricseriesrespectively, thenprovethat S1(S3 - S2)= (S2- S1)2. 2.5.3 Special series kn} :13 andif .1<=l /(=1 We have alreadyusedthe symbol 2 for summation. Let us list out some examples of nite series represented by sigma notation. We have derived that 1+ 2 + 3 + + n = n(n+ 1) . 2 This can also be obtained usingA.P. witha=1,d=l andZ=nasS = %(a+l) = %(l+n). Seq;/enter andrmlexof realm/when ll n(n Hence, usingsigma notation wewriteit as Z k = T2 + 1) . k=1 Let us derive the formulae for (1)§":(2k-1),(ii) :18 and (iii) :19. k=1 k=1 /(=1 1r(;¬1i°: (1) Letusnd:":(2k-1): 1+3+5+ +(2n-1). k=1 This is anA.P. consistingofn termswith a = 1, d = 2, l = (2n - 1). Sn=%(1+2n-1)=n2 (Sn=%(a+l)) Thus, Z":(2k-1)=n2 k=1 (ii) We know that =3 a3-b3 =(a-b)(a2+ab+b2). k3-(k-1)3=k2+k(k1)+(k-1)2 (takea=k andb=k1) k3--(k-1)3=3k2--3k+1 Whenk=1, 13-o3=3(1)2-3(1)+1 Whenk = 2, Whenk = 3, 23-13 = 3(2)2- 3(2)+1 33 23= 3(3)2- 3(3) + 1. Continuing this,wehave whenk=n, n3~(n~1)3=3(n)2-3(n)+1. Adding the above equations corresponding to k = 1,2,---,n column-wise, we obtain n3=3[12+22+ +n21-3[1+2+ Thus, 3[12+22+ +n2]=n3+3[1+2+ " 2 3n(n +1) 3L;1k 2 =fl+ 3 +n]~n - Fl Znlkz : n(n+1)(2n +1)' 6 k=1 A 10;/9 Std Mat/ymzaticr +n]+n (111) ii? = 13+23+---n3 lietl usobserve the following pattern. 13= 1 = (1)3 13+23= 9 =(1+2)2 13+23+33= 36 =(1+2+3)2 13+23+33+43= 100=(1+2+3+4)3. Extending this pattern to n terms, we get 13+23+33+---+113 =[1+2+3+---+n]3 Thus, (i) LThesum ofthe natural numbers,k = (ii) Thesum oftherst11 oddnatural numbers, 3 1 1)_ n2;3-V2 * Thesumof rstn oddnatural numbers (whenthelastterm]isgiven)is * .1 :z+12 3 Thelsumeogflsquaresofrst naturalsinumbers; 1 of thefirst inM naturalglnumbersa. VVis )3i3The13sumof:;ubes33 333;; L3 3 1 im=nm @f 2 % Example 2.29 Find the sum of the following series (i) 26 + 27 + 28 + + 60 (ii) 1 + 3 + 5 + to25 terms (iii) 31 + 33 + + 53. .§¢s13s£z§m§ (i)Wehave 26+27+28+---+60: (1+2+3+---+60)-~(1+2 60 25 =:n-Z1}; (ii) Here,n= 25 1+3+5+--to25terms=252 =625. (Zn:(2k-1)=n2) 2:1 (iii)31+33+---+53 =(1+3+5+---+53)-(1+3+5+---+2 =<53;1>2<;*>2 = 272 152 = 504. EXample2.30 Find the sum ofthe following series 2 (1)12+ 22+ 32+ + 252 (111) 12+ 32+ 52+ (ii) 122+132+142+ + 35 + 512. S§§§£§§£E?£ 25 (1) Now, 12 +22+ 32 + +252 =Zn2 1 225(25 +é)(50+ 1) (k:::1k2 I n(n+1)6(2n +1) ) _ (25)(26)(51) 6 12+ 22+ 32+ + 252 =5525. (11) Now, 122+132+142+ = (12+ 22+ 32+ 35 ll 1 1 + 352 + 352)(12+ 22+ 32+ : znz __Zn2 = 35(35+1)(70 +1) 11(12)(23) 6 = (35)(36)(71) 6 (11)(12)(23) 6 = 14910 ~ 506= 6 14404. + 112) = 2112 - 4Zn2 1 1 _ 51(51+1)(102+1)4 X 25(25+1)(50+1) 6 6 _ (51)(52)(103)4 25(26)(51) 6 = 45526 X - 22100 = 6 23426. Example 2.31 Find the sum of the series. (1)13+23+33+---+203 (11)113+123+133+---+283 Sammia 20 (1) 13 +23 +33 +---203 =2113 1 2 n : <2912_g__~+;.12) using 2,; : k=1 =(20>2<21)2 =(210)2 =44100. (11) Nextweconsider113+ 123+ n(n2+ 1)]? + 283 =(13+23+33+---+283) 28 -(131-23+---+103) 10 :Zn3_Zn3 1 1 _[28(28+1)]2 [10(10+1)]2 2 2 = 4063- 553 = (406+ 55)(4o6- 55) = (461)(351) = 161811. Example 2.32 Findthevalueofk, if 13+ 23+ 33+ + k3: 4356 Sezgmfwi Note that k is a positive integer. Given that 13+ 23+ 33+ + k3 = 4356 2 ...>-43566><6><11><11 Taking square root, we get k(k+1) 7-66 _ 2. k3+k~-132=o =5 (k+12)(k-~11)=O Thus, k = 11, since kis positive. ExaInple2.33 (1)lf1+2+3+---+n =120,11nd13+23+33+---n (11) 1113+231-33-+ +n3 = 36100,thennd1+2+3+ +n. Seaisrtfeie (1) GiVen1+2+3+---+n=12O i.e. rz(n+ 2 1) 2 120 2 13+23+---+n3==1202 =14400 (11) Given13+ 23+ 33+ + n3= 36100 2 =3 Thus, =36100=19><19><10><10 1+2+3+---+n=190. Example 2.34 Find the total area of 14 squares whose sides are 11cm, 12cm, , 24cm, respectively. Sm.'mi1*m Theareas of thesquares formtheseries 112+ 122+ Totalareaof 14squares=112+122+132+ = (12+ 22+ 32+ 24 + 242 + 242 + 242)~(12+ 22+ 32+ 10 ZZHZWZHZ 1 1 _ 24(24 +1)(48 + 1) 10(10 +1)(20 + 1) 6 _(24)(25)(49) 6 (10)(11)(21) 6 = 4900 6 385 = 4515 sq. cm. Exercise 2.6 1. Find the sum of the following series. (1) 1+2+3+...+45 (11) 162+172+182+---+252 (iii) 2+4+6+--. (iv) +100 7+14+21--- +490 +102) Findthe value of kif 13-1-23+33+---+k3== (1) 13+23+33+---+k3=6084 (ii) If1+2+3+---+p=17l, thennd 13+23+33+---+p If13+23+33+---+k3=8281,thennd1+2+3+---+k. .°.":>." 3 Find the total area of 12 squareswhose sides are 12cm, 13cm, ---, 23 cm respectively. Find the total volume of 15 cubes whose edges are 16cm, 17cm, 18cm, , 30cm respectively. sez. Choose 1. the correct answer. Which one of the following is not true? (A) A sequenceis a real valued function dened on N. (B) Every function represents a sequence. (C) A sequencemay have innitely many terms. (D) A sequencemay have a nite number of terms. The 8 term ofthe sequence1, 1, 2, 3, 5, 8, (A) 25 (B) 24 is (C) 23 (D) 21 The next term of21O inthesequence -in%,, 516, is (A) L24 (B) L22 (C) L30 (D) L18 If a, b, c, I, m are in AP, then the value of a -~ 4b + 6c - 41+ m is (A) 1 (B) 2 (C) 3 (D) 0 Ifa,b,care inA.P. then g: [C7 isequal to (A) 1b (B) AC (C) QC (D) 1 If the nterm of a sequenceis 100n+10,then the sequenceis (A) an A.P. (B) a G.P. (C) a constant sequence (D) neither A.P. nor G.P. Ifal,a2, a3,--are inA.P. such that %=%,then the13term ofthe A.P. is (A)g (B)0 (C)12a; (D)14a: 7 If thesequence al,a2,a3, is in A.P., thenthesequence a5,aw,als, is (A) a G.P. (B) an A.P. (C) neither A.P nor G.P. If k+2, 4k6, 3k2 are the three consecutive terms of an AP, (A) 2 (B) 3 (C) 4 (D) a constant sequence then the value of k is (D) 5 Ifa, b, c, I, m. n are in A.P., then 3a+7, 3b+7, 3c+7, 3l+7, 3m+7, 3n+7 form (A) a G.P. (B) an A.P. (C) a constant sequence (D) neither A.P. nor G.P Sequences and reex ofreal numbers] 11. If the third term of a G.P is 2, then the product of rst 5 terms is (A) 52 (B) 25 (c) 10 12. lfa,b,care inGP,then 2 (A) b 13. 2 (B) a L (C) C +10,15x (D) b +15 (B) a G.P. (C) a constant sequence The sequence-3, -3, 3,--(A) an A.P. only 15. isequal to If x, 2x + 2, 3x + 3 are in GP, then 5x,10x (A) an A.P. 14. A (D)15 form (D) neither A.P. nor a G.P. is (B) a G.P.only (C) neither A.P. nor G.P (D) both A.P. and G.P. If the product of the rst four consecutive terms of a G.P is 256 and if the common ratio is 4 and the rst term is positive, then its 3rd term is (A)8 (B)T15 (C)-31; ((3)16 16. InaG.P, t2=4}andt3=-g.Then the common ratio is (A)-1- (B)»g (c)1 (D)5 17. lfxaé0, then1 + secx+ sec2x + sec3x + sec4x + secsxis equalto (A) (1 + secx)(seczx + sec3x + sec4x)(B) (1 + secx)(1+ seczx+ sec4x) (C) (1 secx)(secx + sec3x + secsx)(D) (1 + secx)(1+ sec3x + sec4x) 18. If thenthtermofanA.P.isin= 3- 5n,then thesum oftherstntermsis (A)%[15n] (B)n(1Sn) 19. (C)%(1+5n) (D)%(1+n) Thecommonratio ofthe G.P. am_", am,am is (A) a" (B) a" (C) a" (D) a" 20. Ifl +2+3+...+n=kthen13+23+--- +n3isequalto (A)If (B)/<3 (C)k(k+2 1) (D) (k+ 1)3 o"ZZ;§7zzi4wZ;/9"";ZZzZ} WWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW M D A sequenceof real numbers is an arrangementor a list of real numbers in a specic order. CIThesequence givenby F1= F2= 1 and F = Fn_1+ Fn_2,n = 3,4, is calledthe Fibonacci sequencewhich is nothing but 1, 1, 2, 3, 5, 8, 13, 21, 34, --,arl , is calledanarithmeticsequence if 0 +1 = C1,, + d, n E N whered is a constant. Hereal is calledtherst termandtheconstantd is calledthe D A sequencea17a 2 common a 37 difference. The formula for the generalterm of anA.P. is tn = a + (n -- 1)d Vn E N. A sequence a19a2,a3,. --, an,--- is calleda geometricsequence if an = anr, wherer aé0, n E N wherer is a constant. Here,al is therst termandtheconstant r is calledthe common ratio.Theformula forthegeneral termof aG.P.is tn= ar"_1,n = 1,2,3,. An expression of addition of terms of a sequenceis called a series,If the sum consists only nite number of terms, then it is called a nite series. If the sum consists of innite number of terms of a sequence,then it is called an innite series. ThesumS of therst :1termsof an arithmeticsequence with rst terma andcommon difference disgiven byS,,=121-[2a + (n- 1)d]=-3-(a + l),where Zisthelastterm. The sum of the rst :1terms of a geometric series is given by S={£L(Z'::'__1..)_ =_._(_L:_"_n_)_, ifr751 n where a is the rst r-1 na 1-~r if term and r is the common r = 1. ratio. It Thesum oftherst nnatural numbers,2 k = n(n+1)_ 2 k= 1 The sum ofthe rst11odd natural numbers, :n:(2k -~1)=n2 k=1 The sum of rst 12odd natural numbers (when the last term I is given) is 1+3+5+--+l=(%). fl Thesumof squares ofrst n natural numbers,Zk k= 1 Fl n(n +1)(2n +1) 2 -----6 Thesum ofcubes ofthefirstnnatural numbers, Zk3= k=1 Do you know? . = n(n+1)r_ 2 V A Mersennenumber,namedafter Marin Mersenne,is a positiveintegerof the form M =21--1, wherep isa positiveinteger.If M isa prime,thenit is calleda Mersenne primelnterestingly,if 2 - 1 is prime,thenp is prime.Thelargestknownprimenumber 243m6°9 - 1 is a Mersenne prime. Sequencer andmm of realnumbers ALGEBRA T/ye/amaanMimi /aasneverinvented a /aI90arsaw'ag mac/9z'ae equalta algebra Aaz/907" aaéaowa 3.1 Introduction Algebra is an important and a very old branch of mathematics which deals with solving algebraic equations. In third century, the Greek mathematician Diophantus wrote a book fa , ,3: which contained a large number of practical problems. In the sixth and seventh centuries, Indian mathematicians like Aryabhatta and Brahmagupta have worked on linear equations and quadratic equations and developed general methods of solving them. The next major development in algebra took place in ninth century by Arab mathematicians. In particular, A1-Khwarizmis book entitled Compendium on calculation by completion and balancing was an important milestone. There he used the word aljabra - which was latinized into algebra - translates as competition or restoration. In the 13th century, Leonardo Fibonaccis books on algebra was important and inuential. Other highly inuential works on algebra were those of the Italian mathematician Luca Pacioli (1445-1517), and of the English mathematician Robert Recorde (1510-1558). In later centuries Algebra blossomed into more abstract and in 19th century British mathematicians took the lead in this effort. Peacock (Britain, 1791-1858) was the founder of axiomatic thinking in arithmetic and algebra. For this reason he is sometimes called the Euclid of Algebra. Del\/{organ (Britain, 1806-1871) extended Peacocks work to consider operations dened on abstract symbols. In this chapter, we shall focus on learning techniques of solving linear system of equations quadratic equations. I and 3.2 System of linear equations in two unknowns In class IX, we have studied the linear equation ax + b = 0, a 75O, in one unknown x. Let us consider a linear equation ax + by = c, where at least one of a and b is non-zero, in two unknownsx andy. An orderedpair (x0,yo) is called a solution to the linear equationif the Valuesx = x0, y = yo satisfy the equation. Geometrically, the graph of the linear equation ax + by = c is a straight line in a plane. So each point (x, y) on this line correspondsto a solution of the equation ax + by = c. Conversely, every solution (x, y) of the equation is a point on this straight line. Thus, the equation ax + by = c has innitely many solutions. A set of nite number of linear equations in two unknowns x and y that are to be treated together, is called a system of linear equations in x and y. Such a system of equations is also called simultaneous equations. Anordered pair(x0, yo)iscalled asolution toalinear system intwoVariables if the values x = x0, y = yo satisfy all the equationsin the system. A system of linear equations alx + bl y = cl azx + b2y = C2 in two variables is said to be (i) consistent if at least one pair of values of x and y satises both equations and (ii) inconsistent if there are no Valuesof x and y that satisfy both equations. In this section, we shall discuss only a pair of linear equations in two variables. Let us consider alinear system alx +bly =C1 azx + b2y = C2 in two variables x andy, where any of the constantsa1,b1,a2 and b2 can be zero with the exception that each equation must have at least one variable in it or simply, a12+b127é0 , a22+b227é0. Geometrically the following situations occur. The two straight lines representedby (i) may intersect at exactly one point (ii) may not intersect at any point (iii) may coincide. and Aégebm If (i) happens, then the intersecting point gives the unique solution of the system. If (ii) happens, then the system does not have a solution. If (iii) happens, then every point on the line corresponds to a solution to the system. Thus, the system will have innitely many solutions in this case. Now, we will solve a system of linear equations in two unknowns using the following algebraic methods (i) the method of elimination 3.2.1 Elimination (ii) the method of cross multiplication. method In this method, we may combine equations of a system in such a manner as to get rid of one of the unknowns. The elimination of one unknown can be achieved in the following ways. (i) Multiply ordivide the members ofthe equations bysuch numbers as tomake the coefcients of the unknown to be eliminated numerically equal. (ii) Then, eliminate by addition if the resulting coefcients have unlike signs and by subtraction if they have like signs. Example 3.1 Solve 3x- 5y =l6 , 2x+ 5y = 31 5;§:.sin'z§i2:z The given equations are 3x - 5y = -16 2x + 5y = 31 Note that the coefcients of y in both equations are numerically equal. So, we can eliminate y easily. Adding 1 and I ?,we obtain an equation 5x = 15 That is, x = 3. Now, we substitute x = 3 in Substituting x = 3 in For J5: to solve for y. we obtain, => 3(3) 5y = -16 y = 5. Now, (3, 5) a is solution to the given system because and 2]? are true when x = 3 andy = 5 as from and (Livwe get, 3(3) 5(5) = -16 and 2(3) +5(5) = 31. btaining equationin onlyionevariable isanimportant step in indingthe ,3 solution. BWeobtainedequationsf onevariable eliminatingthe variable y._ Sothis(method ofsolving asystem by:eliminating oneofthevariablesirst, iscalled ?method ofelimination. . 701/9Std. Mal/éidliff Example 3.2 The cost of 11pencils and 3 erasersis ? 50 and the cost of 8 pencils and 3 erasersis ? 38. Find the cost of eachpencil and eacheraser. S:;fm:§m: Let x denotethe cost of a pencil in rupeesand y denotethe cost of an eraserin rupees. Then according to the given information we have 11x + 3y = 50 8x + 3y = 38 Subtracting :3; from we get, 3x = 12 which gives x = 4. Now substitute x= 4 in isto nd the value of y. We get, 11(4)+3y =50 i.e., y=2. Therefore, x = 4 and y = 2 is the solution of the given pair of equations. Thus, the cost of a pencil is 3 4 and that of an eraseris ? 2. T Itisalways bettercheck that-the obtainedsyvaluesp satisfy the jbothlequiationsj ~ Example 3.3 Solve by elimination method 3x + 4y = -25, 2x - 3y = 6 S{}f£f!§s}£¬ The given system is 3x +4y=-25 2x - 3y = 6 To eliminate the variable x, let us multiply 28i><2 => 6x+8y X -3 -_-> Now, adding -6x+ and 3x+ 4(That is, .aby -3 to obtain =-50 9y =-18 we get, 17y = Next, substitute y = - 4 in z by 2 and 68 which gives y = - 4 to obtain 4) =-25 x =- 3 Hence, the solution is (-3, -4 ). Aégebm Example 3.4 Using elimination method, solve 101x + 99y = 499, 99x + 101)?= 501 .§7r;§;§§£:s;z The given system of equations is 101x + 99y = 499 99x +101y = 501 0 Here, of course we could multiply equations by appropriate numbers to eliminate one ofthe variables. However, note that the coefcient ofxinone equation isequal tothe coefcient of y in the other equation. In such a case, we add and subtract the two equations to get a new system of very simple equations having the same solution. Adding and we get 200x + 20032= 1000. Dividing by 200 we get, Subtracting ij x+y =5 from 13,we get 2x - 2y = -2 which is same as x Solving and E,we get y = x = 2, y = 3. Thus, the required solution is ( 2, 3 ). Example 3.5 Solve 3(2x + y) = 7902; 3(x + 3y) = llxy using elimination method .%;«;;m:§m: The given system of equations is 3(2x + y) = 7907 3(x + 3y) =11xy Observe that the given system is not linear because of the occurrence of xy term. Also, note that if x = 0, then y = 0 and vice versa. So, (0, 0) is a solution for the system and any other solution would have both x 7E0 and y 750. Thus, we consider the casewhere x % 0, y ¢ 0. Dividing both sides of each equation by xy, we get 2...: =7 y x and 3+i=u xy 2+2: x y To eliminate b, we have Subtracting ><2 ==> 18a + 6b = 22 from is we get, ~ 15a = -15. That is, a = 1. Substituting a=l in51? ;weget, I)= Thus, a=l and b= When a=l, wehave %=1. Thus,x=l. .1._ .2. WhenI): .2. 3,wehaVey3. Thus,y= .3. 2. Thus, the system has two solutions ( 1,»; )and (0,0). Aliter The given system of equations can also be solved in the following way. Now, 3(2x + y) = 7902 l3(x + 3y) = llxy Now, L31 :1X 2 ==> => 15y = l5xy l5y(lx) = 0. Thus, x =1 andy = 0 Whenx= 1,wehave y=§andwheny=O, wehavex=0 Hence, the two solutions are ( 1,-3)and (O,O). Solve each of the following system of equations by elimination method. 1. x+2y=7, x2y=l 2. 3x+y=8, 3. x+%=4,§+2y=5 5x+y=l0 4. llx7y=xy,9x~4y=6xy 5. x3+§=2° 2+5=15,x¢0,y¢o 6.8x-3y--Sxy, 6x-5y=-2xy xyx y xy 7. l3x+lly=70, llx+l3y=-74 8. 65x~33y=97, 33x-65y=l 9. -1-5-+-2~=l7,~1+~1~=~Z-§~,x7EO,y7E0 lO.~2-+-2-==-1-,-3~+-2-=0,x7é0,y7é0 x y x y 5 x 3y 6 x y Cardinality of the set of solutions of the system of linear equations Let us consider the system of two equations a1x+b1y+c1=O azx + bzy + C2: 0 where thecoefcients arerealnumbers suchthat 6112 + b12 750 , 6122 + b22 7% O. Aégebm Let us apply the elimination method for equating the coefcients of y. Now, multiply equation E by b2 and equation1;?by bl, we get, bzalx + bzbly + b2cl= 0 blalx + blbly + blcz = 0 Subtracting equation from 5%we get (bzal - bl a2)x bl C2 bzcl .s x Substituting the value of x in either provided albz azbl 750 or and solving for y, we get y -51% __ azbl , provided albzallbl75 0. Thus, we have x albz Mazblandy albz __ azbl , albzazbl 75 0. Here, we have to consider two cases. , ,, . b (T5552: {3} albl a2bl7E 0.Thatis, i ¢ 1. a2 b2 In this case, the pair of linear equations has a unique solution. {};z:§c iéi} albz azbl=0.That is,%=% ifa2 75 0and b2 75 0. 2 2 Inthiscase, let ai , bl = /lbl a2 "- 21b2 = /1.Thenal = Razz Now, substitutingthe valuesof al and b1in equation J we get, A(a2x+ by) + cl= 0 It is easily observed that both the equations and can be satised only if cl = RC2-==> If cl = /lC2,any solution of equation LIT:will also satisfy the equation . b . . . 1 and vice versa. . So, if -31=:-51=E-1=/l;then there are innitely many solutions tothepair of 2 2 2 linear equations given by 3and If cl 75RC2, thenany solutionof equation will not satisfyequation Ci , then the pair of linear Hence, if- a(12 ... 5b2aéC2 has no solution. w7m0w;/; Std.Matbmzalics andvice versa. equations given by ; j and L3:; Now,weisummarisieithe aboveidiscusisionyjbiz iiForthe7syste[im V Tf ° soifif y as ..a,"x[lpazyyi-ji~lsc2:p#;O,,wl1¢re .e[a,2,.,+,bf¢_o .a,lp?.+*b,y2ieo7¢y 0..y TT T--jlinlaiz 7 swans. anif if 1 lhelthe smmofequationshasinmeayma i,i(iii)iIif .4-; }§_1,;¢l%,,7the,ni.thee syste::i ofieq11atj¢:;s'has,n¢ :.{oru:ig:;.%i i ,_yq_2,'_ _ 2, , I , 2 , 3.2.2 Cross multiplication , , p\ , I \\ , I \\ , I \ , I method While solving a pair of linear equations in two unknowns x and y using elimination method, we utilised the coefcients effectively to get the solution. There is another method called the cross multiplication method, which simplies the procedure. Now, let us describe this method and see how it works. Let us consider the system a1x+b1y+c1= 0 We have already established that the system has the solution x_ b1C2'b2C1 _ C1a2"C2a1 Thus, we can write x _ 1 Y _ 1 Let us write the above in the following form x _ 3 _ 1 Thearrows between thetwonumbers indicate thattheyaremultiplied, thesecond product (upward arrow)istobesubtracted fromtherst product (downward arrow). Aégebm Method of solving a linear system of equations by the above form is called the x _ Y _ 1 9 Note that in the representation blcz - bzcl or claz - czal may be equalto 0 but albz - 61219175 0. Hence,forthe,systemof equations 1a1xi+ b1yc+C1= 0 L L Hi i a2x+ib2y+c2=0_l (iy , if blcz-,b,c1p::eo,*andi a1ib,2i-a,b1;é 0,thenix=itOii (ii) aif claz~czali=0 and 511192 e a2I9p1,c% (),theny =,O~ Hereafter, we shall mostly restrict ourselves to the system of linear equations having unique solution and nd the solution by the method of cross multiplication. Example 3.6 Solve 2x +7y5=0 3x + 8y = -11 S{9f££3.f£}1$ The given system of equations is 2x + 7y 5 = 0 3x + 8y +11 = 0 For the cross multiplication method, we write the coefcients as x _ Y _ 1 H"""Wegal <7)<11> - <8)<5) <-5><-3)-~<2><11> <2>(8> ~<-~ 3>(7>' Thatis, if7 =~_-317=§%. i.e.,x=1:;1-7-, y=--3-7:]-. - - Q Hence, thesolution <3 , .. L37 Example 3.7 Using cross multiplication method, solve 3x + 5y = 25 7x + 6y = 30 Saiszrfrzzr The given system of equations is 3x + 5y -25 = 0 7x + 6y3O = 0 Now, writing the coefcients for cross multiplication, we get W0:/7 Std.A4.;;?.¬I7Zdl'Z.6.S' W WW W WW IWW W WI W WW W WW W WWwmwlm X _> .150+150 y 175+90 i¬., i0 = -85 y = -17 18-351 1' Thus, we have x = O, y = 5. Hence, the solution is (0, 5). iHéfe; ,-~ ?= a--PM isfoim¢a.x%7-94=lTOi§ Thué islioiiantit.ati.on.e7ai.1[d,ifiisi 7 Hnotpdiivision byZero; eisialwaysitruethatidivision ZerQ7iS notdenedf >A Example 3.8 In a two digit number, the digit in the unit place is twice of the digit in the tenth place. If the digits are reversed, the new number is 27 more than the given number. Find the number. 5S%;§m5':;;z Let x denote the digit in the tenth place and y denote the digit in unit place.. So, the number may be written as 10x + y in the expanded form. (just like 35= 10(3) +5) When the digits are reversed, x becomes the digit in unit place and y becomes the digit in the tenth place. The changed number, in the expanded form is l0y + x. According to the rst condition, we have y = 2x which is written as 2x - y = 0 Also, by second condition, we have (10): + x) - (10x + y) = 27 Thatis, -9x+9y Adding equations and =27 ==> -x+y=3 2, we get x= 3. Substituting x = 3 in the equation j, we get y = 6. Thus,the given number is (3><10) + 6 = 36. Example 3.9 A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3,wegetg11 butif thenumerator is increased by 8 andthedenominator is 2 doubled, wegetv5. Findthefraction. .i'E§{:§z;£§:;m Letthefraction be According tothegiven conditions, wehave -~- => 11x=6y-18 and and 5x+40=4y Aégebm So, we have 11x-6y+ 18 =0 5x-4y+40=0 On comparing the coefcients of { wehave and a1=l1, b1=6, Thus, with alx + bl y + C1= 0, azx + by + c2 = 0, c1= 18; a1b2 6Z2b1_ (11)( 4) a2=5, (5)( 6) b2=4, c2=4O. 14% 0. Hence, the system has a unique solution. Now, writing the coefcients for the cross multiplication, we have x y 1 18., ____> x _ y .-240+72 " x ~168 _ 90-«440 _ y _ -350 -14 1 44+30 1 Thus,x=@ =12; y=% =25.Hence, the fraction is Example 3.10 Eight men and twelve boys can nish a piece of work in 10 days while six men and eight boys can nish the same work in 14 days. Find the number of days taken by one man alone to complete the work and also one boy alone to complete the work. Si;s!m:}:2:s Let x denote the number of days needed for one man to nish the work and y denote the number of days needed for one boy to nish the work. Clearly, x aéOand y 750. 1 So,onemancancomplete 36partoftheworkinonedayandoneboycancomplete ;partofthe work inone day. The amount ofwork done by8men and 12boys inone day is Thus,we haveg + -12-= -L x y 10 The amount ofwork done by6men and 8boys inone day is Thus,we haveQ + § = L x y 14 4a+6b____1__ 20 = O. 6a+8b= T11=>3a+4b-218 =0. o}if_i§"2Zz."54";};9i",?Z;"§-Z} WWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW W Writing the coefcients of (3) and (4) for the cross multiplication, we have a _ b _ 1 - a 2 Thus,wehave -"3+1_":L+L 16__18.1.e., M1fl 14 - 5 20 Z Thatis, 1 a 140,b 7 Z 70 b 2 L "2 140 1 280 Thus, wehavex=%=140,y=%=280. Hence, one man can nish the work individually in 140 days and one boy can nish the work individually in 280 days. ercise3. 1. Solve the following systems of equations using cross multiplication method. (1) 3x + 4y = 24, 20x 11y = 47 3x 5y_ (1102 32. x y_13 2,3+2-6 (ii) 0.5x + 0.832= 0.44, 0.8x + 0.632= 0.5 - i__£__.. (1v)x y 2 §_ 2,x+y13 Formulate the following problems as a pair of equations, and hence nd their solutions: (i) One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, nd the numbers. (ii) The ratio of income of two personsis 9 : 7 and the ratio of their expenditure is 4 : 3. If eachof them managesto save?' 2000 per month, nd their monthly income. (iii) A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number. (iv) Threechairsandtwo tablescost? 700 andve chairsandthreetablescost 3 1100. What is the total cost of 2 chairs and 3 tables? (v) In a rectangle, if the length is increased and the breadth is reduced each by 2 cm thentheareais reduced by 28cm2.If thelengthis reduced by 1 cmandthe breadth increased by2cm, thentheareaincreases by33cm2.Findtheareaof the rectangle. (vi) A train travelled a certain distance at a uniform speed. If the train had been 6 km/hr faster, it would have taken 4 hours less than the scheduled time. If the train were slower by 6 km/hr, then it would have taken 6 hours more than the scheduled time. Find the distance covered by the train. we-»-I Aégebm 3.3 Quadratic polynomials Apolynomial ofdegree ninthevariable xisaoxn + a1x"_1 + a2x"_2 + + an_1x + an where a075O and a1,a2,a3,...,a,,are real constants. Apolynomial of degree two is called a a i L and is normally written as p(x) = axz+ bx+ c,wherea %O, bandcarerealconstants. Realconstants arepolynomials of degree zero. Forexample, x2+ x + 1,3x2- 1,--3x2 + 2x- -7arequadratic polynomials. 2 3 Thevalueof a quadratic polynomialp(x) = axz+ bx+ C at x = k is obtainedby replacing x by k in p(x). Thus,thevalueof p(x) at x = k is p(k) = akz+ bk+ c. 3.3.} Zeros of a polynomial Consider a polynomial p(x). If k is a real number such that p(k) = 0, then k is called a of the polynomial p(x). For example, thezerosof thepolynomial q(x)= x2-- 5x+ 6 are2 and3because q(2)= Oandq(3)= 0. 3.3.2 Relationship between zeros and coefcients of a quadratic polynomial In general, if erandB arethezerosof thequadratic polynomial p(x) = axz+ bx+ c, a yéO, then by factor theoremwe get, x 01and x B are the factors of p(x). Therefore, axz+ bx+ c = k(x a)(x B), where k is anonzeroconstant. = k[x2~(a + [>)x+ a3] i. may Std.Matbemaiicx Comparing thecoefcients of x2,x andtheconstant termonbothsides, weobtain a=k, b =-k(a+[>) and c=kaB 3Thebasic relationships between thezeros andthecoefcients of p(x)= dx2+ bx+ care f . : a+ Sumo Z6108 , _ b ; _ coefficient of x . a t coefficient ofx2 product ofzeros : ab C i wnstanttem a T coefficient of x2' Example 3.11 Find the zerosof the quadraticpolynomialx2+ 9x+ 20, andverify the basic relationships between the zeros and the coefcients. §¬}f£Sf§.:}!? Let p(x) = x2+ 9x+ 20= (x + 4)(x+ 5) So,p(x)=O ==>(x+4)(x+5)=O Thus, = (4+4)(4+5) p(4) x=-4orx=~5 = O and p(5) = (5+4)(5+5) = Hence, the zeros of p(x) are 4 and -5 Thus, sum of zeros = -9 and the product of zeros = 20 From the basic relationships, we get thesumof thezeros Coefficient of x coefficient 9 = 9 of x2 1 term _ 20 = 20 product of the zeroS_ constant coefficient of x2 1 Thus, the basic relationships are veried. t* M quadraticpolynomial p(x)'= axz+bxi+cmay have atmost twofizeros. Now,forany a O,ia(x2e(C! + +013) isya, polynomial Withizeros aand,Since T wecanchoose anynonzeroa,thereareinnitelymanyquadratic polynomialstiwith , zeros c?and B. ;F Example 3.12 Find a quadratic polynomial if the sum and product of zeros of it are 4 and 3 respectively. Szs£:a:§.wzs Let a! and B be the zerosof a quadraticpolynomial. Giventhat a +3 = -4 and ab = 3. Oneof thesuchpolynomials is p(x) = x2- (a + B)x+ ab =x2--(~4)x+3=x2+4x+3 Aégebm Example 3.13 Find aquadratic polynomial with zeros atx=211and x=-1. §::§5:£5}'m Let C? and Bbe the zeros ofp(x) A.%%te:* The required polynomial isobtained Using the relationshipbetweenzerosand directly as follows: coefcients, wehave p(x) =(x...L)(x+1) 2 4 p(x) =x -(a+b>x+a[> _ 2 1 1 = x2+ ix - L. x ~(»4~~1)x+(~4-)(--1) 4 4 2 x2+ix ~L Anyother polynomial withthedesired property 4 4 is obtained by multiplying p(x) by any non- Itisapolynomial with zeros 21f and -1. 21 F 1. -~lp1iiis aplisoeapolynomial with izeros Zn and Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefcients. (1)x2-2x~8 (V)x2- 15 2. zero real number. (ii) 4x2-4x+1 (iii) 6x2-3-7x (vi) 3x2-- 5x + 2 (vii) 2x2- 2/Ex +1 (iv) 4x2+8x (viii) x2 + 2x 143 Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively. (i) 3,1 (V)1 3.4 (ii)2,4 (vi) 4 (iii)0,4 (vii) -% (iv) /2 (viii)f3,2 Synthetic division We know that when 29 is divided by 7 we get, 4 as the quotient and l as the remainder. Thus, 29 = 4(7) + 1. Similarly one can divide a polynomial p(x) by another polynomial q(x) which results in getting the quotient and remainder such that p(x) = (quotient)q(x) + remainder That is, 1906)= s(x)q(x) + r(x), where deg r(x) < degq(x). This is called the E If q(x)= x + a, then deg r(x) = 0. Thus, r(x) is a constant. Hence, p(x)= + a) + r, where r is a constant. Now ifwe put x = a in the above,we have p(a) = s(-a)(~a + a) + r ==>r = p( (1). Thus, if q(x) = x + a, then the remainder can be calculatedby simply evaluating p(x) at x = a. Oz/9Std. Mat/yeiimlzks .. If p(x) is the dividend and q(x) is the divisor, then by division algorithm we write, p(x) = s(x)q(x) + ow,wehavethefollowingresults. (i) If q(x) is linear , then r(x)= r is a constant. (ii) If deg q(x) l (i.e.,q(x)islinear), thendeg p(x)= l + deg s(x) (iii) If p(x) is dividedby x + a, thentheremainderis p(-- a). (iv)If r =0,wesayq(x)divides _ equivalemnptlqy isafactor of Remarks Let us explain the method of synthetic division with an example. Let p(x) = x3+ 2x2x - 4 bethedividendandq(x) = x + 2 bethedivisor. We shall nd the quotient s(x) and the remainder r, by proceeding as follows. Step i Arrange the dividend and the divisor according to the X3 + X2 x descending powers ofxand then write the coefcients ofK \\ dividend in the rst row (see gure). Insert 0 for missing l 2 4 ~l ~4 terms. 352;;2 Find out the zero of the divisor. Step3 Put 0 for the rst entry in the 2 row. Complete the entries of the 2 row and 3 row as shown below. 1 2 -1 o><(v2) 0 -4 1><(,2) 4+2 =-2 Remainder. Whenp(x) isdivided by x -~3,thequotient is x2+ 4x+ Sandtheremainder is 12. Example 3.15 Ifthe quotientondividing2x4+ x3- 14x2~-19x + 6 by 2x+ 1 is x3+ axz~bx--6. Find the values of a and I), also the remainder. §£§§§£f§§}f§ Let p(x) = 2x4+ x3- 14x2 19x + 6. Given that thedivisor is 2x+1.Write2x+1=0.Then x=% The zero ofthe divisor is--5-. > Remainder So,2x4 +x3-14x2 -19x+6=(x+~%){2x3 -14x12}+12 =(2x+1)%(2x3 -14x~12)+12 Thus, thequotient is£(2x314x12)x3 7x 6andtheremainder is12. But,givenquotientis x3+ axz- bx- 6. Comparing thiswith the quotientobtained we get, a = O and b = 7. Thus, a = 0, b = 7 and the remainder is 12. ercise3.4 1. Find the quotient and remainder using synthetic division. (1) (x3+x2 . 3x+ 5) + (x- 1) (ii) (3x3- 2x2+ 7x- 5) + (x+ 3) (iii) (3x3+ 4x2-10x+ 6 )+( 3x- 2) (iv) (3x3- 4x2- 5) + (3x+1) (v) (8x - 2x2 + 6x 5)+(4x + 1) 2. (vi) (2x- 7x3 13x2+ 63x~48)+(2x - 1) lfthequotient ondividingx4+ 10x3 + 35x2+ 50x+ 29byx + 4 isx3- axz+ bx+ 6, then nd a, b and also the remainder. 3. If the quotienton dividing,8x42x2+ 6x-. 7 by 2x+ 1 is 4x3+ pxz- qx + 3, then nd p , q and also the remainder. Oz/9Std. Mat/mmlics 3.4.1 Factorization using synthetic division We have already learnt in class IX, how to factorize quadratic polynomials. In this section, let us learn, how to factorize the cubic polynomial using synthetic division. If we identify one linear factor of cubic polynomial p(x) , then using synthetic division we get the quadratic factor of p(x). Further if possible one can factorize the quadratic factor into two linear factors. Hence the method of synthetic division helps us to factorize a cubic polynomial into linear factors if it can be factorized. Forany x ais i2(ii)q.pxp, arisiaifactor for f O.eC(l?actor3theio1'ern e (iii) xiéeilli isa7factorofp(x) andonlyifthe ofcoefcients) ofp(x)is 1 3) (iv)at+11 isafactorof if and if sum ofthecoefcients ofeven powers ofx, including constant ispiequalto sum ofthecoefcients of povvers: ofx.11 Example 3.16 (i) Provethatx 1 is a factorof x3 6x2+ 11x 6. (ii) Provethatx + 1 is a factorof x3+ 6x2+ 11x+ 6. 5§£3i£if§{}F§ (1) Let p(x) = x3-. 6x2+ 11x- 6. p(1) = l 6 + 11 6 = 0. (note that sum ofthe coefcients is 0) Thus, (x - 1) is a factor of p(x). (11) Let q(x)=x3+6x2+l1x+6. q(-1)=1+6ll+6=O. Hence, x+lisafactorofq(x) Example 3.17 Factorize 2x3-~3x2-- 3x + 2 into linear factors. .5:e:.mmLet p(x) = 2x3-. 3x2-- 3x+ 2 Now, p(l) = ~ 2 % 0 (notethatsumof thecoefcientsis not zero) .. (x ~ 1) is not a factor of p(x). However,p( 1) 2( 1)3 3( 1)2 3( 1) + 2 = 0. So, x + 1 is a factor of p(x). We shall use synthetic division to nd the other factors > Remainder Thus, p(x)= (x + l)(2x2 5x + 2) Now, 2x2-5x+2 =2x2-4x-x+2 =(x~2)(2x-- 1). Hence,2x3- 3x2- 3x+ 2 = (x + l)(x- 2)(2x-1). Aégebm Example 3.18 Factorizex3~-3x2- 10x + 24 .£m:ii¬:mLetp(x) = x3- 3x2 10x + 24. Sincep(1) 75O and p(-- 1) 7%0, neitherx +1 nor x - 1 is a factor ofp(x). Therefore, we have to search for different values of x by trial and error method. When x = 2, p(2) = 0. Thus, x - 2 is a factor ofp(x). To nd the other factors, let us use the synthetic division. > Remainder. The other factor is x2 -- x -~12. Now, x2-x 12 = x2-4x+3x- 12=(x--4)(x+3) Hence,x3- 3x2- 10x + 24= (x ~2)(x + 3)(x-- 4) Exercise 1. Factorize each of the following polynomials. (i) x3 2x2- 5x + 6 (ii) 4x3- 7x + 3 (iv) 4x3- 5x2+ 7x- 6 (vii) 2x3~9x2 + 7x+ 6 (x) 2x3+11x2-7x--6 35 3.5 (v) x3~7x + 6 (viii) x3-- 5x+ 4 (xi) x3+x2+x-14 (iii) x3- 23x2+ 142x 120 (vi) x3+ 13x2 + 32x+ 20 (ix) x3~10x2 - x + 10 (xii) x3-5x2-2x+24 Greatest Common Divisor (GCD) and Least Common Multiple (LCM) 3.5.1 Greatest Common Divisor (GCD) The Highest Common Factor (HCF) or Greatest Common Divisor (GCD) of two or more algebraic expressions is the expression of highest degree which divides each of them without remainder. Consider the simple expressions (i) a4,a3,a5,a6 (ii) a3b4, absC2,6121976 In (i), notethata,a2,a3arethedivisors of all these expressions. Outof them,a3is thedivisorwithhighest power. Therefore a3istheGCDof theexpressions a4,a3,a5,a6. In (ii), similarly,onecaneasilyseethat61124 is theGCDof 613 b4,absC2,a2b7c. If the expressions have numerical coefcients, nd their greatest common divisor, and prex it as a coefcient to the greatest common divisor of the algebraic expressions. Let us consider a few more examplesto understandthe greatestcommon divisor. Oz/9Std. Mat/mrmlicx Examples 3.19 FindtheGCDofthefollowing:(1)90,150,225 (ii) 15x4y3z5, 12x2y7z2 (iii) 6(2x2- 3x - 2), 8(4x2+ 4x+1),12(2x2+ 7.x+ 3) f§¬}§££fiE3§? (i) Let us write the numbers90, 150 and 225 in the product of their prime factors as 90=2><3><3><5 ,150=2><3><5><5 and225=3><3><5><5 From the above 3 and 5 are common prime factors of all the given numbers. Hence the GCD = 3 X 5 =15 We shall use similar technique to nd the GCD of algebraic expressions. 272 Nowletustakethegivenexpressions 15x4y3 Z5and12xyz. 3 Herethecommon divisors ofthegivenexpressions are3,x2,y (iii) andz. 2 Therefore, GCD= 3 ><105+84 105=1 x 34 21, 84:4X21__O 8 (or) 1 4 105 924 84 105 21 840 84 84 21 84 84 0 the polynomials f(x) ofand 21 is the GCD 924g(x) and are 105not easily factorable,then it will be a difcult problem. Similar technique works with polynomials when they have GCD. Let f(x) andg(x) betwo nonconstantpolynomialswith deg(f(x)) Z deg(g(x)). Wewant to nd GCD of f(x) andg(x). If f(x) and g(x) canbe factoredinto linear irreduciblequadratic polynomials, then we can easily nd the GCD by the method which we have learnt above. If However, the following method gives a systematicway of nding GCD. Step 1 First, divide f(x) by g(x) to obtain f(x) = g(x)q(x)+ r(x) where q(x) is the quotient and r(x) is remainder,so deg(g(x)) > deg(r(x)) If the remainder r(x) is 0, then g(x) is the GCD of x) and g(x). Step 2 If the remainderr(x) is non-zero,divide g(x) by r(x) to obtain g(x) = r(x)q(x) + I;(x) where r1 is the remainder.So deg r(x) > deg I; If the remainder r1(x) is 0, then r(x) is the required GCD. Step 3 If r1(x) is non-zero,then continuethe processuntil we get zero asremainder. The remainderin the last but one stepis the GCD of f(x) and We write GCD(f(x) , g(x)) to denotethe GCD of the polynomials f(x) and g(x) Example 3.20 FindtheGCDof thepolynomials x4+ 3x3-- x - 3 andx3+ x2~~ 5x + 3. S}}§z§££°1m Letf(x) = x4+ 3x3-~x -3 and g(x) = x3+ x2- 5x + 3 Heredegree of f(x) > degree of Divisoris x3+ x2- 5x+ 3 x+2 4 x+3x+0x 3 2 x-3 xl 2 x+2x-3 3 x+x5x+3 2 x4+x3-5x2+3x 2x3+5x2-4x-3 2x3+2x2-1Ox+ 1311111111111 Letf(x) = 3x4+ 6x3 12x2- 24x = 3x(x3+ 2x24x -. 8). Let g(x)= 4x4+ 14x3 + 8x2- 8x = 2x(2x3 + 7x2+ 4x- 4) Letusnd theGCDforthepolynomials x3 + 2x2-4x -- 8 and2x3 + 7x2+ 4x- 4 We choosethe divisor to be x3 + 2x2~ 4x -~8. 2 x3+2x24x8 2x3+7x2+4x--4 x 2 x2+4x+4 2x3+4x2-8x-16 x3+4x2+4x 3x2+l2x+l2 -2x2-8x-8 (x2+4x+4) --2x2-8x-8 |> remainder(7é 0) Common x3+2x2-4x-8 0 > remainder factor of 3x and 2x is x. Thus, GCD(f(x), g(x))= x(x2+ 4x+ 4). 1. 2. Find the greatest common divisor of (1) 7x2yz4 , 21x2 y5z3 (11) x2y, x3y,x2y2 (111) 25bc4d3,35b2c5,45c3d (iv)35x5y3z4, 49x2yz3, l4xy2z2 Find the GCD of the following (i) C2~d2, c(c d) (ii)x4~27a3x,(x ~3a)2 (111) m2- 3m (iv)x2+ 14x+ 33, x3+ 10x2llx 18, m2+ 5m+ 6 (v) x2+ 3xy+ 2312, x2+ 5xy+ 6y2 (Vi)2x2-«x- 1, 4x2 + 8x+ 3 (Vll)x2-x~ 2, x2+x-6, 3x2 l3x+ 14 (Vlll)x3x2+x- 1, x 1 (ix) 24(6x4-x3 - 2x2),20(2x6+ 3x5+ x4) (x) (a - l)5 (a + 3)2, (a ~ 2)2(a - l)3 (a + 3)4 3. Find the GCD of the following pairs of polynomials using division algorithm. (1) x3- 9x2 + 23x- 15, 4x2- 16x+12 (11)3x3+18x2+ 33x+18, 3x2+ 13x+10 (111) 2x3+ 2x2+ 2x+ 2, 6x3+ 12x2+ 6x+12 (iv) x3~3x2+4x-12, x+x3+4x2+4.x 3.5.3 Least Common Multiple (LCM) The least common multiple of two or more algebraic expressions is the expression of lowest degree which is divisible by each of them without remainder. For example, consider the . . 4 3 6 simple expressions a ,a ,a . Aégebm 6 7 8 . 3 Now, a , a , a ,--- are common multiples of a ,a 4 6 and a . Of all thecommon multiples, the leastcommon multipleis a6 Hence LCMof a4,a3,a6is a6. Similarly, a3b7istheLCMof a3b4, abs,a2b7. We shall consider some more examples of nding LCM. Example 3.22 Find the LCM of the following. (1) 90,150,225 (ii) 35a2c3b, 42a3cb2, 30ac2b3 (iii) (a 1)5(a + 3)2,(a - 2)2(a-1)3(a + 3)4 (iv) x3+ 323, x3- 323, x4+ xzyz+ y4 .i§'{s§mf{m (1)Now,90=2><3><3><5 =21><32><51 150 = 2><3><5><5=21><31><52 225 = 3><3><5><5=32><52 Theproduct 21X 32X 52= 450istherequired LCM. (ii) Now, LCM of35, 42 and 30 is 5 X 7 X 6 = 210 Hence, therequired LCM= 210Xa3XC3Xb3= 210a3 C3 193. (iii) Now,LCM of(a -1)5(a + 3)2,(a ~ 2)2(a -1)3(a + 3)4is (a -1)5(a + 3)4(a 2)2. (iv) Let us rst nd the factors for each of the given expressions. x3+ y3= (x + y)(x2-~xy+ yz) x3~y3 = (x --y)(x2+xy +322) x4+ xzyz+ y4 : (x2+ y2)2__xzyz : (x2+ xy + y2)(x2__xy + yz) Thus, LCM= (x + y)(x2-xy + y2)(x-y)(x2+ xy+ yz) : (x3+ y3)(x3__Y3): x6_ y6_ Find the LCM of the following. 1. x3y2,xyz 2. 3x2yz,4x3y3 3. azbc, bzca,czab 4. 66a4b2c3, 44a3b4c2, 24a2b3c4 5. am, am, am 7. 3(a~1), 2(a1)2, (a2- 1) 9. (x+4)2(x- 3)3, (x-1)(x+4)(x- 6. x2y+xy2, x2+xy 8. 2x2-18322, 5x2y+15xy2, x3+27y3 3)2 10. 10(9x2 + 6902 + yz), 12(3x2 - Sxy- 2372), 14(6x4 + 2x3). Oz/7Std. Mat/mizalics 3.5.4 Relation between LCM and GCD We know that the product of two positive integers is equal to the product of their LCM and GCD. For example,21 X 35 = 105 X 7, where LCM (21,35) =l05 and GCD (2l,35) = 7. In the sameway, we have the following result: The product of any two polynomials is equal to the product of their LCM and GCD. That is, f(x) >c)) >< GCD (x) , g(x)). Let us justify this result with an example. Letf(x) = 12(x4- x3)andg(x)= 8(x4~3x3 + 2x2)betwopolynomials. Now,f(x)= 12(x4-x3)=22><3><(x 1) Also, g(x)= 8(x4- 3x3+ 2x2)= 23><(x -1)><(x - 2) From } and we get, LCM(x) ,g(x)) = 23>< 31><(x -1)><(x . 2) = 24x3(x-~1)(x.. 2) GCD> = 4x2< 4x2(x -1) = 96x5(x-1)2(x-2) Also, f(x)><8x2(x--1)(x~2) = 96x5(x1)2(x 2) From and we obtain,LCM >< Thus, the product of LCM and GCD of two polynomials is equal to the product of the two polynomials. Further, if f(x), g(x) and one of LCM and GCD are given, then the other can be found without ambiguity becauseLCM and GCD are unique, except for a factor of -1. Example 3.23 The GCD of x4 + 3x3+ 5x2+ 26x + 56 and x4 + 2x3-~4x2-~x+28isx2+5x+7. Find their LCM. .§}:¬r;t§:m Let f(x) = x4+ 3x3+ 5x2+ 26x+ 56 andg(x)= x4+ 2x3- 4x2 x + 28 Given thatGCD= x2+ 5x+ 7.Also,wehaveGCD>< Thus, f(x)> LCM = (x2- 2x+ 8)> _ (x6x3+1 _ (x3 +1)(x3 1)(x+ 1)_(x31)(x+1) Hence, g(x) =(x3 --1)(x +1). 1. x3+1 Find the LCM of each pair of the following polynomials. (i) x2~5x+6, x2+4x- 12whoseGCDisx2. (ii) x4+ 3963+ 6962+ 5x+ 3, x4+ 2x2+x+ 2 whose GCDis x2+x+ 1. (iii) 2x3+15x2+ 2x-- 35, x3+ 8x2+ 4x- 21 whoseGCDis x + 7. (iv) 2x3- 3x2- 9x+ 5, 2x4x3 ~-1Ox2- 11x+ 8 whoseGCDis 2x 1. 2. Find the other polynomial q(x) of eachof the following, given that LCM and GCD and one polynomial p(x) respectively. (i) (x +1)2(x + 2)2, (x +1)(x + 2), (x + 1)2(x + 2). (ii) (4x + 5)3(3x- 7)3, (4x + 5)(3x- 7)2, (4x + 5)3(3x - 7)2. (iii) (x4__y4)(x4+ xzyz+ y4) x2__yzax4 __y4_ (iv) (x3~4x)(5x+ 1), (5x2+ x), (5x3- 9x2-- 2x). (v) (x-1)(x-«2)(x23x+3), (x-1), (x3-4Lx2+6x3). (vi) 2(x+1)(x2--4), (x+ 1), (x+1)(x-2). 701/9Std. Mfbéidliff 3.6 Rational expressions A rationalnumberis denedasa quotient, of two integersm andn 750. Similarly p " arational expression isaquotient q oftwopolynomials p(x)and q(x),where q(x)isa non zero polynomial. p(x) Every polynomial p(x) is a rational expression,since p(x) can be written as where l isthe constant polynomial. 1 x However, a rational expression need not be a polynomial, for example x2+l is a rational expression but not a polynomial. Some examples of rational expressions are 3 2x+/9x+x+l 23x+2x2+«/§x+5_ x+x-x/§ 3.6.1 Rational expressions in lowest form If the two polynomials p(x) and q(x) havethe integer coefcients suchthat GCD of p(x) and q(x) is 1, then we saythat S isarational expression initslowest terms. If a rational expression is not in its lowest terms, then it can be reduced to its lowest termsby dividing both numeratorp(x) and denominatorq(x) by the GCD of p(x) and Let us consider some examples. Example 3.25 Simplify the rational expressions into lowest forms. - 5x+20 -- x3-5x2 (1)79+28 (11) 3x3 +2x4 2 __ 9x +12x-5 2 __ (x~l)(x-2x-3) .§0f£¬¬f(3I§ - (1) N° 5x+2o_5(x+4)_5 7x +28 7(x+4) 7 NOW X3 *' 5x2_ x2(xN _ X""5 3x3+ 2x4 x3(2x+ 3) x(2x4' 3) (iii) Let p(x)=6x2-- 5x+ 1: (2x- l)(3x- 1)and q(x) = 9x2+ 12x5: (3x + 5)(3x- 1) Therefore, (iv) Let P06) _ (ZX-1)(3x*-'1)_ 2x-- 1 q(x) (3x + 5)(3x -~ 1) 3x + 5 f(x) = (x - 3)(x2- 5x+ 4)= (x 3)(x l)(x 4) and ?§¥%§¥ Simplify the following into their lowest forrns. (iv)xx::297 (V) (Hint: x4 +x2 +1=(x2 +1)2 --x2) 4 (ix) (x-3)(x2 - 5x+4) (X) (x 8)(x2 +5x-~50)(Xi)4x2 +9x+5 (x-4)(x2~ 2x- 3) (x+10)(x2--13x+4O) 8x2+6x- 5 (xii)(x1)(x-2)(x29x+14) (x 7)(x2- 3x + 2) 3.6.2 Multiplication and division of rational expressions If ;q(x) 75 Oand ; 75 Oare two rational expressions, then >.<>-<> Z3*i&> QhagM) Example3.26 32 5 3 3 2 2 3 2 ulpyl 9Z4yx4y2 ()a2+2ab+b2 y "19 x2-4 x2+2x+4 Salzazirm 32 5 32 5 (0 NOW xy4X2;7z2_(xy4)(247z2)_3z_ 9z xy (9z)(xy) 9 .. 3 3 2.. 2 +b( 2» b+b2) +19 a-~b (11) 20+?2><___; (( )l))(b) ><((?_(b) )a2ab+b2. a+2ab+b I + H a 3 2 3 3 NOW x2-~8 Xx+2x+4 x2+6x+8_ X2-"22X(x:4)(x+2) x-4 x-2 x+2x+4 _ (x--2)(x2+2x+4)X(x+4)(x+2) _x+4 (x+2)(x"'2) x2+2x+4 f§(;iz:t:}).¥z 4.76-4+.7C*~l_ X(x+1)_ 4 x2__,1 x+l (x+l)(x--1) (x-1) x»-1' -- 3 2 2 x-l;x+x+1_(x-1)(x 0) x+3 +x+1) x+3 3x+9 3(x+3) _ X x2_,_x_,_1 301) .9C2*"1+.9C2~"4.9C"'5 x2-25 x2+4x-5 (x+5)(x-~5) (x-5)(x+1) _(x"'1)(x"1) _ x2-2x+l_ (x-5)(x'~5)x2-lOx+25 1. Multiply the following and write your answer in lowest terms. - 2 x-2x 3x+6 (1) H2 Xx-2 2 x -2x+4 (111) x2...x...20 X x3+g 2 2 x~8l x+6x+8 (H) x2-4Xx2--5x--36 2 x -3x--10 2 -- - x2--16 x2--4 x2-4x+l6 (IV)x2-~3x+2Xx3+64Xx2--2x-8 2 4 (V) 3.9C2+2X"~1X2.9C2"'3.9C-'2 22.96-'1 X2x+5x-3 2x-8x X x-2x x--x~2 3x+5x-~2 x+2x+4 2. Divide the following and write your answer in lowest terms. - x ; x 2 -- (1) x+1x2-1 x~-25 (H) x2-49x+7 x+7x+l0 x-4x-77 (V) 2x2+l3x+l5+2x2-x-6 x2+3x-10 2 x-36;x+6 9x2-16 x2-4x+4 (Vii) 2x:+5x3+2x:+xl 2 x-2x-~15 _ 3x2-2xl 2 3x ".9C'*°'4. 4x--4 2x+9x+9 2x+x~3 3.6.3 Addition and subtraction of rational expressions If ( ) r( ) are any two rational expressionswith q(x) yéO and s(x) 750, then and we dene the sum and the difference (subtraction) as 1706) + r(x) _ p(x)-s(x)iq(x)r(x) c1(x)" s(x) Example 3.28 c1(x)-s(x) 533353353333 (D x+2+x~-1 _ (x+2)(x--2)+(x--1)(x+3)_ 2x2+2x--7 x+3 x--2 (x+3)(x--2) x2+x__6 .9C+1+ 1 _ (x+1)2+(x1)2_ (x1)2 x+1 (x-~1)2(x+1) _ 22 2x2+2 (x-1)2(x+1) 2 _x3--:2i:x+1 (111) x2x-6 x2+2x-24_(x3)(x+2) + x2-9 x2-~x-12 (x+6)(x-4) + (x+3)(x"'3)(x+3)(x"4) _x+2+x+6 x+3 _x+2+x+6 x+3 Example 3.29 3 x+3 What rational expression should beadded to x2" 1 toget x +2 3952355321: Let p(x) be the requiredrational expression. 3 3 _2x+8 x+3 3 2 x2 + 2 ? 2 Given thatx2"'1 +p(x)= --------295 "x + 3 x+2 x2+2 3 2 3 p(x)_2x x2-X-r3 _x2-1 +2 x +2 _ 2x3-x2+3-~x3+l x2+2 _ x3~~x2-+4 x2+2 Example 3.30 Simplify <2x " 1 x+1>+x+2as aquotient oftwo polynomials inthe simplest form. x~1 2x+l x+l S:'2£mi;m Now, 2x'" 1 x+ 1 + x+ 2 x -~ 1 2x + 1 x + 1 (2x- 1)(2x+1)~-(x+1)(x- (x-~1)(2x+l) 1) _(4x2-1)-*(x2-1)+x+2_ (x1)(2x+l) x+l 354.2 +x+l 3x2 (x-1)(2x+l) +x+2 x+1 _ 3x2(x+1)+(x+2)(x-~1)(2x+ 1) _ 5x3+6x2-3x~2 (x2-1)(2x+1) 2x3+x2-2x-1 1. Simplify the following as a quotient of two polynomials in the simplest form. - x 3 8 (1) X-2+2-"x 2 2 x-2-x6+x2+2x24 x-9 x-x-12 01/9Std. Mfbeidlitf -- x+2 x3 (H) x2+3x+2+x2-2x-3 2x-2 x~7x+1O +x-2x--15 2x+3 2 2 2 (V)2x2»-5x+32x-~7x-4 x-3x+2 (Vi) x-4 2x2-3x-2 -- 2x+5 x2+l x2+6x+8 3x2 1 2 x--1lx+3O x2-x20 1 2 (V111 x+1'1x2_.1 (x-1) (V1111 x1+3x+2+x1+5x+6 x2+4x+3' 3 3 2. Which rational expression should beadded to x2'" 1 toget? 3. Which rational expression should be subtracted from x 2x-1 +2 2 x +2 get2x25x+17 ' 4. IfP=x x+yQ=yhd1--~~-2Q. x+yten H P-Q p2___Q2 3.7 Square root Let a E R be a non negativereal number.A squareroot of a, is a real numberI) such thatb2= a, Thepositivesquare rootof ais denoted by 2/; or «/3. Eventhoughboth ( 3)2 = 9and(+ 3)2= 9aretrue,theradicalsign f isusedto indicate thepositive square rootofthenumberunderit.Hencex/5 = 3.tSimilarly,weihavex/l2}l = 11, = 100. In the same way, the ;. of any .:ei::=.:-. is an expression whose square is equal to the given expression. In the case of polynomials, we take 1/ (p(x))2 = [,where |p(x)I ={__pp(::)1ff pp(Z:)Z§mi;2n (i) (ii) (iii) 3/4x2+ 20907 + 25322 = ,/ (2x+ 5y)2 = |(2x+ 5y)| /x6+-1;--2= /(9c3-L3>2=<9c3-L3) x x x First, let us factorize the polynomials 6x2-~x-~2=(2x+1)(3x--2);3x2 2x2-x-1=(x-1)(2x+1) Now, x/(6x2-x -- 2)(3x2 - 5x+ 2)(2x2~-x~-1) = ¢(2x +1)(3x - 2)><(3x-- 2)(x 1)><(x-1)(2x +1) = /(2x + l)2(3x. 2)2(x_-1)2 = |(2x+1)(3x - 2)(x -1)} Find the square root of the following (1)196a6b8c10 (ii) 289(a- b)4(b~c)6 (iv) (x ~y)2 + 4902(v) 121x8y6 + 81x4y8 Find the square root of the following: (1) 16x224x + 9 (iii) (x + 11)2 . 44x 64(a + b)4(x - y)8(b-c)6 (Vi)25(x +y)4 (a-b)6(b +c)10 Tond (1) ¢66564 2 5 8 (ii) \/ 9x4+ 12x3 + 10x2 + 4x+ 1 Letp(x)=9x4+12x3+1Ox2+4x+1 266564 3x2+2x+1 45 508 Therefore, x/66564= 258and x/9x4+12x3+1Ox2+ 4x+1 = 13x2 + 2x+1] emarks ' ,, . , , (i[)fwhnefwritingTtheipee1ynemia16 in { v.fo1?.6misSin1g tertnsgj6 p6owjcrs6:of*6x; Vinsertjzerose M: (ii)Theabove method eeanjbe» eemeaeeewaeeehe 6fe1e1owg:ng( preeiedye; ;;§ [1 .ee(ea:+ bL+ec)2Q[ 2L7 LI 6Th6eref(5fe6,6.it isétrriattér 6(iffnd:ingL lsuitabb[1é:e6ea,f6b :6 ° c)2}=.e6Ta2wI+ b26+66c:2ie+f2db +_2Lci L ' *6 ~ 1;ga2'+eb?+:2ab_+f2ac +' 2bcj+ge2 1:: 66 j 6 6 6V6e%e»(~3x¢)23%e(6x2Le+[A2x)e(2;¢e)ee+ff(§x? 4;xi6..ge[1e)6(1*)iefV" ;*1 fl;66 i: +1 13x912x '6 e1jI6,{wh:e:e 393 ;;,6:V 2xendfe 5:61 nee,9x4+12x3+1ox2 n LVeA1;eejrJ;" Tojf1Lnd6 the%sequae¢}?root;rs:,wm;e£ e9x46+12x3 10x2 +66439/+ei if ; 6 6V ;+*nx z_)?7=; m2x4+2mnx3+ +(e362ce)](~3x);+ tzcsxm2+e212 ee 6 :L~V%fV$a:2:L§+<*(:%¬: b) 1 Vew;'hefcV6a be ;:36x, L 6AVL *:Lr:rA_ : 9 6 Aégebm Example 3.33 Findthesquare rootof x4- 10x3+ 37x2~ 60x+ 36. S£)§isf§i3§l Given polynomial is already in descending powers of x. x2 - 5x + 6 x2 x4 - 10x3+ 37x2- 60x + 36 4 .96 2x2-- 5x - 10x3 + 37x2 -10x3 + 25x2 2x21Ox+6 12x2~6Ox+36 12x2- 60x + 36 0 Thus,x/x4-10x3+ 37x2~60x + 36 = |(x2~ 5x+ 6)! Example 3.34 Findthesquare rootof x4 6x3+ 19x2- 30x+ 25 §};>§:mm: Let us write the polynomial in ascending powers of x and nd the square root. 5-3x+x2 5 25 --30x+ l9x2-~6x3+x4 25 -30x + 19x2 -30x +9x2 lO3x 10-6x+x2 10x2 -6x3 +x4 10x2- 6x3+ x4 Hence, thesquare rootofthegivenpolynomial is ixz~-3x + 5I Example 3.35 If m- nx + 28x2+ 12x3 + 9x4isaperfect square, then nd the Values of m and n. _37rn*7§z{§m: Arrange the polynomial in descending power of x. 9x4+12x3 + 28x2- nx + m. Oz/9Std. Mat/mmlics Now, 3x2+ 2x + 4 9x4+ 12x3+ 28x2- nx + m 9x 4 6x2+ 2x 6x2+4x+424x2nx+m 24x2+16x+l6 Since the given polynomial is a perfect square, we must have n = -16 and m = 16. 1. Find thesquare rootofthe 2. (1) x--4x3+ l0x2-- 12x+9 (ii) (iii) 9x4- 6x3+ 7x2 2x+ 1 (iv) 4 + 25x2- 12x- 24x3 + 16x4 4x4+8x3+8x2+4x+1 Find the values of a and b if the following polynomials are perfect squares. (1) 4x4- l2x3+37x2+ax+b (iii) ax + bx3+ lO9x2- 60x+ 36 3.8 bydivision method. Quadratic (ii) x4x3+ l0x2-a.x+b (iv) ax - bx3 + 40x2+ 24x+ 36 equations Greek mathematician Euclid developed a geometrical approach for nding out lengths which in our present day terminology, are solutions of quadratic equations. Solving quadratic equations in general form is often credited to ancient Indian Mathematicians. In fact, Brahma Gupta (A.D 598 - 665) gave an explicit formula to solve a quadratic equation of the formaxz+ bx= c. LaterSridhar /lcharya (1025 A.D)derived a formula, nowknownasthe quadratic formula, (as quoted by Bhaskara II) for solving a quadratic equation by the method of completing the square. In this section, we will learn solving quadratic equations, by various methods. We shall also see some applications of quadratic equations. A quadratic equation inthevariable xis anequation of theformax2+ bx+ c =0 , where a, 19, C are real numbers and a 760. In fact, any equationof the form p(x) = 0, where p(x) is a polynomial of degree2, is a quadratic equation, whosestandard formis axz+ bx+ c = 0, a 7EO. Forexample, 2x2- 3x + 4 = O,1- x + x2= 0 aresomequadratic equations. Aégebm 3.8.} Solution of a quadratic equation by factorization method Factorization method can be used when the quadratic equation can be factorized into linear factors. Given a product, if any factor is zero, then the whole product is zero. Conversely, if a product is equal to zero, then some factor of that product must be zero, and any factor which contains an unknown may be equal to zero. Thus, in solving a quadratic equation, we nd the values of x which make each of the factors zero. That is, we may equate each factor to zero and solve for the unknown. Example 3.36 Solve 6x2 5x - 25 Szgirrsrirszz Given =0 6x2~ 5x - 25 = 0. First, let us nd a and [3 such that oz +3 = -5 and ab = 6 ><(25) = -150, where-5 is the coefcient of x. Thus, we get oz = -15 and B = 10. Next, 6x2- 5x- 25 = 6x2- 15x+ 10x- 25= 3x(2x- 5) + 5(2x- 5) = (2x - 5)(3x + 5). Therefore, the solution set is obtained from 2x - 5 = O and 3x + 5 = 0 =1 Thus, x 2 , x=%. Hence, solution set is{--%, Example 3.37 Solve 6 l 7x"21 +l_O x26x+9 x29 _S£§i1aff{P§3 Given equation appearsto be a non-quadratic equation. But when we simplify the equation, it will reduce to a quadratic equation. 6 1 1 N°W7(x-3)2+(x+3)(x3) O __> 6(x2-9)7(x+3)+7(x-3)_0 7(x-3)2(x+3) :5 6x2 54 42-0 ...;. x2 16-0 Theequation x2= 16isquadratic andhence wehavetwovalues x = 4 andx= 4. Solution set is {~~ 4, 4} Example 3.38 Solve =3~4x,3~4x>0 .§%;Z£;z*ia:zGiven x/ 24 - 10x = 3 - 4x Squaringon bothsides,we get, 24 -- 10x= (3 - 4x)2 :5 Oz/9Std. Mfbeidlitf 16x2-l4x-15=O =5 16x224x+l0x~l5=0 _=>(8x+5)(2x 3)=0whichgives x=%or% Whenx = 1, 3~4x = 3--4 12 < 0 andhence, x= Q isnotasolution oftheequation. 2 2 When x=~§, 3--4x> 0and hence, the solution set is Solve the following quadratic equations by factorization method. (i) (2x+3)2-8l=0 (ii) 3x2-~5x--l2=O (iii) /§x2+2x~.3/3:0 (iv)3(x2~6)=x(x+7)-~3 (v)3x-%= (vi)x+%=%6 (vii)xifl +xjgl =% (viii)a2b2x2-(a2+b2)x+ 1=o (ix) 2(x +1)2 ~5(x + 1) = 12 (x) 3(x ~4)2- 5(x --4) = 12 3.8.2 Solution of a quadratic equation by completing square From (x+%)2 =x2+bx+(3)2, note that the last term isthe square ofhalf thecoefficient ofx.Hence, thex2+bxlacks only the term (2)2 ofbeing thesquare 2 ofx+%.Thus, ifthesquare ofhalf thecoefcient ofxbeadded toanexpression oftheform x2+ bx , the resultis the squareof a binomial.Suchan additionis usuallyknownas completing the square. In this section, we shall nd the solution of a quadratic equation by the method of completing the square through the following steps. iep 1 If thecoefcientof x2 is 1,goto step2. If not,dividebothsidesof theequation by thecoefcientof x2.Getall thetermswithvariableononesideof equation. $593.2 2 Find half the coefcient of x and square it. Add this number to both sides of the equation. To solve the equation, use the square root property: x2 = t=> x = «/7 or x = - «/7 wheretis non-negative. Aégebm Example 3.39 Solve thequadratic equation 5x2- 6x- 2 = 0bycompleting thesquare. Se?§£§f§i}§? Givenquadratic equation is 5x2~6x _=> 2 =O 6 x2 - 5x --3- 2 = 0 => x2- (Divide onbothsides by5) =g ( -:51isthe half ofthe coefcient ofx) 3 9 _ 9 2 1 2_ 9 ==> x2 2(5)x+25 25+5 (add(5)25onboths1des) 3 2 _ 19 =2 (x" 5) r 27 ==> x-§ = i- / %-g-(take square root onboth sides) ,_,19:34_r/i§_ Thus, wehave x=-§~ 5 Hence, the solution set is 5 {3+«1F9 3-43} 5 5 ' Example 3.40 Solvetheequation azxz-- 3abx+ 2b2= Obycompleting thesquare Seézstérss There is nothing to prove if a = 0. For a 75O, we have azxz~ 3abx + 2192= 0 => x2__i ax+2_b2 dz: 0 319 _ --2192 _.>x2 2(2a)x dz 3b 9192_ 9192 2192 => x2 2< 261 )x+4612 402 a2 :> (x 319 >2 _ 9192 2-8192__> (x 2319 >2 _4612 I92 2 4a2 ==> x~%=i2]97 Therefore, thesolution set is{%, ==> x= 3b2Eb 3.8.3 Solution of quadratic equation by formula method In this section, we shall derive the quadratic formula, which is useful for nding therootsof a quadratic equation.Consider a quadratic equationaxz+ bx+ c = 0, a 750. We rewrite the given equation as x2+-%x+-2=0 == ::>x2+2(2]9a>x:% 2 Adding (2]9a)2 : fizz both sides we get, x2+2(%)x +(Zia? =L ..% 0;/7Std. Mat/mrmlics b 2_ 192-4 Thatis, (.X'+"""> if 2 :_____> x+b2a _i/b-:lac_i«/I92--4ac 4a 2a -b4_r«/b2-4ac So, we have x= 2a bx//72-4616} The solution set is{I9+\/b24ac 2a 2a ' The formula given in equation r is known as quadratic formula. Now, let us solve some quadratic equations using quadratic formula. Example 3.41 Solvetheequation xi1+x_+2_2 x:4,wherex+l76O,x+27$0and x + 4 750 using quadratic formula. §z25££!§@§? Note that the given equation is not in the standard form of a quadratic equation. - 1 2 _ Consider x+1+x+2 - 1 _ 4 x+4 2 1 _ 2x+4-x-~4 Thatls x+l 2x+4 x+2i 2(x+4)(x+2) x1Il 2i(x+2)x(x+4)l x2+6x+8 = 2x2+2x Thus,wehave x2- 4x~8 = 0,whichis aquadratic equation. (The above equation can also be obtained by taking LCM ) Using the quadratic formula we get, _ 4ix/164(1)("'8) _4:/E " Thus, 2(1) 2 x=2+2/§or22/3 Hence,the solutionsetis {2 - 2«/§,2 + 2/3} l Solve the following quadratic equations by completing the square . (1) x2+6x-7=0 (ii) x2+3x+l=O (iii) 2x2+ 5x- 3 = 0 (iv) 4x2+ 4bx- (a2- 192)= 0 (v)x2~(/§+1)x+f=0 (vi)5;_j"17=3x+2 .A@Mm 2. Solve the following quadratic equations using quadratic formula. (i) x2-7x+12=O (ii) 15x2-11x+2=O (iii) x+31C=2-15 (iv) 3a2x2 -.abx 2192 =0 (V) a(x2+ 1)= x(a2+ 1) (vi) 36x2-~12ax+ (a2- 192) =0 (vii) '1 + x""3 = x + 1 -- 4 (viii) a2x2+(a2b2)x-b2=0 3 3.8.4 Solution of problems involving quadratic equations In this section, we will solve some simple problems expressed in words and some problems describing day-to-day life situations involving quadratic equation. First we shall form an equation translating the given statement and then solve it. Finally, we choose the solution that is relevant to the given problem. Example 3.42 Thesumofanumber anditsreciprocal is5L. Findthenumber. 5 .§7:;im*§:m Letxdenote the required number. Then itsreciprocal is316- Bythe given condition, x+916 5; > 962;" 1 256 So, 5x2-~26x+5 = 0 :.-.> 5x2-~25x--x+5 = 0 Thatis, (5x-l)(x--5)=0=> x=5or% Thus, the required numbers are5,%. Example 3.43 The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq. cm, then nd its base and altitude. f§¬3§§i§i(}§$ Let the altitude of the triangle be x cm. By the given condition, the base of the triangle is (x + 4) cm. Now, the area ofthe triangle = £(base) xheight Bythegiven condition+ = 48 =», ==> x2+4x--96=0 x = - 12 or =3, (x+l2)(x-8)=0 8 But x = - 12is not possible (since the length should be positive) Therefore, x = 8 and hence, x + 4 = 12. Thus, the altitude of the triangle is 8 cm and the base of the triangle is 12 cm. Oz/9Std. Mat/mmlics Example 3.44 A car left 30 minutes later than the scheduled time. In order to reach its destination l50km away in time, it has to increase its speed by 25 krn/hr from its usual speed. Find its usual speed. ryfsgréaw Let the usual speed of the car be x km/hr. Thus, the increasedspeedof the car is (x + 25) km/hr Total distance =150 km; Time taken =. Speed Let T1 and T2 be the time taken in hours by the car to cover the given distance in scheduled time and decreasedtime (as the speed is increased) respectively. Bythegiven information 71 --T2=% " 150 x 150 1 x + 25 x Z L x+25 2 "4150 x(x+25) 2 =..>x2+25x-7500 Thus, x = 75 or - 100, _ (30minutes =%hr) but = 0 ==> (x+ 100)(x~75)= 0 x = -~100 is not an admissible value. Therefore, the usual speed of the car is 75 km/hr. l. The sum ofanumber and itsreciprocal is Find the number. 2. The difference of the squaresof two positive numbers is 45. The square of the smaller number is four times the larger number. Find the numbers. 3. A farmer wishes to start a 100 sq.m rectangular vegetable garden. Since he has only 30m barbed wire, he fences the sides of the rectangular garden letting his house compound wall act as the fourth side fence. Find the dimension of the garden. 4. A rectangular eld is 20m long and 14m wide. There is a path of equal width all around it having an area of 111 sq. metres. Find the width of the path on the outside. 5. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 krn/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train. 6. The speed of a boat in still water is 15 km/hr. It goes 30 km upstream and return downstream to the original point in 4 hrs 30 minutes. Find the speed of the stream. 7. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his sons age. Find their present ages. 8. A chess boardcontains 64equalsquares andtheareaof eachsquare is 6.25cmz. A border around the board is 2 cm wide. Find the length of the side of the chessboard. gm-«»-I Aégebm 9. A takes 6 days less than the time taken by B to nish a piece of work. If both A and B together can nish it in 4 days, nd the time that B would take to nish this work by himself. 10. Two trains leave a railway station at the same time. The rst train travels due west and the second train due north. The rst train travels 5km/hr faster than the second train. If after two hours, they are 50km apart, nd the average speed of each train. 3.8.5 Nature of roots of a quadratic equation The roots oftheequation axz +bx+c= Oare given byx = ~bi-x/b2~4ac_ 2a If b2- 4ac>O, wegettwodistinct realroots X:-b+«/b2~4ac 2a and x= -b-~«/I92-~4ac 2a ' If [924ac =0,then the equation has two equal roots x=3-5. If 192 -- 4ac< 0, thenV b2- 4ac is notarealnumber. Therefore thereis noreal root for the given quadratic equation. So,evidently thenature of rootsdepends onthevalues of b2- 4ac.Thevalueof the expression [92-~4acdiscriminates thenature of therootsof axz+ bx+ c = Oandsoit is called the discriminant of the quadratic equation and denoted by the symbol A. A 1 I C. Example 3.45 Determine the nature of roots of the following quadratic equations (i) x2llxlO=0 (ii)4x2~28x+49=0 (iii)2x2+5x+5=0 .§'r)lz§tfmz Foraxz+ bx+ C= 0, thediscriminant, A = b2- 4ac. (i) Here,a=l;b=ll andc=lO. Now,thediscriminant is A = b2-- 4ac ( 11)2 4(l)( 10) 121 + 410:161 Thus, A > 0 .Therefore, the roots are real and unequal. (ii) Here, a = 4, b = -28 and c = 49. Now,thediscriminant is A = b2 4ac = ( 28)2~4(4)(49) =0 Since A = 0, the roots of the given equation are real and equal. (iii) Here, a = 2, b = 5 and c = 5. Now,thediscriminant A = b2- 4ac = (5)2- 4(2)(5) = 25 40 = -15 Since A< O, the equation has no real roots. Example 3.46 Provethattherootsof theequation (a - b + c)x2+ 2(a-~b)x+ (a- b --c) = 0 are rational numbers for all real numbers a and b and for all rational c. Saiizeiiresz Letthegivenequation beof theformAx2+ Bx+ C = 0. Then, A=a--b+c,B=2(a-b) andC=a-b--c. Now, thediscriminant of Ax2+ Bx+ C = Ois B2 4AC [2(a b)]2 4(a b + c)(a b c) = 4(a - b)2-4[(a - b) + c][(a - b) - C] 4(a b)2 4[(a b)2 c2] A = 4(a~-b)2~4(a b)2 + 402= 402,aperfect square. Therefore, A> O and it is a perfect square. Hence, the roots of the given equation are rational numbers. Example 3.47 Findthevalues of k sothattheequation x2- 2x(1+ 3k)+ 7(3+ 2k)= 0 hasreal and equal roots. sieiseieiiee Thegivenequation is x2~ 2x(1+ 3k)+ 7(3+ 2k)= 0. Lettheequation bein theform axz+ bx+ c = 0 Here, a = 1, b =- 2(3k+1), c := 7(3 + 2k). Now,thediscriminant is A = b2~4ac = (-2(3k +1))2 - 4(1)(7)(3 + 2k) = 4(9k2+ 6k+ 1)-- 28(3+ 2k)= 4(9k2-. 8k-. 20) Given that the equation has equal roots. Thus, =», 9k2-8k20 A =0 =0 =-> (k-- 2)(9k+ 10) = 0 Thus,k=2,-1§Q. Aégebm 1. 2. Determine the nature of the roots of the equation. (1) x2-~8x+l2=0 (ii) 2x2-~3x+4=O (iii) 9x2+12x+4=O (iv) 3x2--2«/gx+2=O (V) %x2 %x +1=0 (vi)(x2a)(x 219) =4ab Find the values of k for which the roots are real and equal in each of the following equations. (i)2x2--10x+k=0 (ii)12x2+4kx+3=0 3. (iii) x2+2k(x-~2)+5=O (iv) (k+1)x2-2(k1)x+1=0 Showthattherootsof theequation x2+ 2(a+ b)x+ 2(a2+ 192) = Oareunreal. 4. Showthattherootsof theequation 3p2x2 - Zpqx+ q2= Oarenotreal. 5. Iftherootsoftheequation (a2+ b2)x2 ~2(ac + bd)x+ C2+ d2= 0, where a,b,cand d ¢ 0,areequal, prove that-21= 6. Show that the roots of the equation (x a)(x b) + (x - b)(x - C) + (x --c)(x - a) = 0 are always real and they cannot be equal unless a = b = c. 7. If theequation (1+ m2)x2 + Zmcx+ C2- a2= O hasequalroots,thenprovethat c2 = a2(1+ m2). 3.8.6 Relations between roots and coefcients of a quadratic equation Consider a quadratic equationaxz+ bx+ c = 0, wherea , b , c arerealnumbers and a 750. The roots of the given equationare C!and B , where 2a 2a Then, the sum ofthe roots,ct+B "b+ 2% " 4616 +" b" zba ""4616 / _ b _ 2 coefficient a andthe product of roots, a / of x coefficient of x2 ~b+x/b2~~4ac X -b-«/b2-4ac 2a 2a _ b2- (192 --~4ac) _ 4a(: 4a2 constant 4a2 term coefficientof x2 Oz/9Std. Mat/mrmlicx 2 Therefore, if a, B aretherootsof axz+ bx+ c = 0, then (i) thesumoftheroots,ct+ B 2 -ba (ii) theproductof roots, a 2 ~93 Formation ofquadratic equation Wheii roots are given Let a and B be the roots of a quadraticequation. Then (x a/) and (x B) are factors. >=o => x2(0z+B)x+az[5=O Thatis, x2-~(sumof roots)x+ product of roots= 0 3 cg sVThiefgri¢ T are ijgnnjt¢1iyi roots}: f Example 3.48 If oneof therootsof theequation3x2 10x + k = O is thennd theotherroot _1_ 39 and also the value of k. SafisrianThegivenequation is 3x2~10x + k = 0. Let the two roots be 01and B . _1_(_)_ " +5 Z -(--10): 3 3 Substituting ct=%in weget3= Also, ab= ==> k=3 Thus, the other root 3 = 3 and the value of k = 3. Example 3.4.9 If thesumandproduct of therootsof thequadratic equation axz 5x+ c = 0 are both equal to 10, then nd the values of a and c. $¬}1i7%'I'i¬}i§ Thegivenequation is axz- 5x+ C= 0. Sum of the roots, 2: a 10, _-=>a = 3 Product oftheroots, %= 10 2; c=10a=10><%=5 Hence, a=~%~ and c=5 Aégebm If a/andijB are theroots ofax;+bx+cB:Q,then many expressions aand Blike a2+ B3252, iazL... etc., can beevaluated using thevalues ofa +Banda/B.BL7 Letussiwrite some iresultsinvolving aandB x V ;B (11);a2+2i=[(a+B)2--2qi,9] BB B V V i LB (iii)iii0125-B2j=i (Q! 9:; 5)(qa..}5)i=j (ai+eB)[ ] only ifqaZBi av) a3+z?=3~sa5 (v)j 03%-Be3i=(ii"3)3+5"5(¢B)iiis; (vi)Vof+6*#(a/2+ /f)2-q2cizi2,5= [(a+i[3)2 -201312 +2(a/B)2 (vii)Bct?-ii£>"é(cv+)(d+-B)(0f+B2) Example 3.50 If a andB aretherootsof theequation 2x2 3x 1 = O,nd thevaluesof (i) a2 +E2 (ii) %+5- (iii)0:Bifa>B (iv)(F2 +7) (V)(a+ +6) (vi)a4 +[34 (vii)%3 +g »,§;,s£a:£§{2s/z Given equationis 2x2- 3x - 1 = 0 Letthegivenequation bewrittenasaxz+ bx+ c = 0 Then, a = 2, b = ~ 3, c = -- 1. Given at and B are the roots of the equation. _ ---3 a+B=5=(2)=%andaB=~~% (1) a/2+B2=(0z+B)2 2ag=(g)2 2( 1)~ 701/9Std. Mat/yeimzlicx C?2_+_B2 _a3+§ _ (a+B)3""3aB(a+B)_%+%_ [3 a (V) ab ab --T1 4 B) 0!+1)(1 a-+0! _([> _(1+ab)2_(1'i")_ 1 ab ____1_ 2 2 (vi) a/4+ B = (of+ [952 - 2a2[)2 _(143>2 2( i>2_(11669 i>11661' (Vii)cf 161_ B+a[)3_a4+[)4_<161>(2>_ ab 16 1 8 Example 3.5} Formthequadratic equation whoserootsare7 + /3 and7 - x/§ . Sm:s;f:°wz Given roots are 7 + «/§ and 7 -- «/§. -. Sumofthe roots = 7 + /5 + 7 -- £3 =14. Product ofroots= (7+ /§)(7 - /3) = (7)2- (/§)2 = 49-3 = 46. Therequired equation is x2-- (sumof theroots)x+ (product of theroots)= 0 Thus,therequired equation is x2-~14x+ 46 = 0 Example 3.52 If aand B are the roots of the equation 3x2 4x +1=0,form aquadratic equation whose roots are 6% and %. 2 Sfigifilt Since01,3aretherootsof theequation 3x2 - 4x + 1=0, wehave a+B=~%-, a[)=:1%~ 2 3 Now, for the required equation, the sum ofthe roots =(-9+ =a+B3 [3 ct ab 4 3 ... .1. A _ (a+b)3--30zl>(w+b) _ (3') 3X3X3 _ 28 ab a2 -67 [92=ab=31 Also,product oftheroots= -?~ L 9 3 Therequired equation is x2- 2-8-x + -1-= O or 9x2- 28x+ 3 = 0 9 3 Aégebm Find the sum and the product of the roots of the following equations. (1) x2-6x+5 = 0 (11) kxz+ rx +pk = 0 ( iii ) 3x2-- 5x = 0 (iv) 8x2-- 25= 0 Form a quadratic equation whose roots are (i) 3,4 (ll)3+\/7,3/7 (in) 4+/7 2 , 4-/7 2 If a andB aretherootsof theequation 3x2-- 5x+ 2 = O, thennd thevaluesof . Q !_ .. (1) B+ a _ (11) at B 10. Q3 5: (111)3 + Q If 0!andB aretherootsof theequation 3x2~6x + 4 = 0,nd thevalueof a/2+ B2. If a/,B aretherootsof 2x2- 3x- 5=0,formaequationwhose rootsareazandB2. If a/, B aretherootsof x2- 3x + 2 = 0, forma quadratic equation whoserootsare C?and B . If a andB aretherootsof x2- 3x l=0, thenformaquadratic equation roots are Lzandi. ll.whose oz If C?andB aretherootsof theequation3x2- 6x+ 1 = 0, formanequation whose roots are 12. (1)-2? (ii) M5,B20 (iii) 2a+B,26+a Find a quadratic equation whose roots are the reciprocal of the roots of the equation 4x2-- 3x - 1 = 0. If onerootof theequation 3x2+ kx- 81 = 0 is thesquare of theother,nd k. If onerootof theequation 2x2- ax+ 64 = 0 is twicetheother,thennd thevalueof a If a/andB aretherootsof 5x2-px +1 = 0 anda B = 1,thenndp. Exercise Choose the correct answer. If onezeroof thepolynomial p(x) = (k+4)x2+l3x-I-3k isreciprocal oftheother, then k is equal to (A) 2 (B) 3 (C) 4 (D) 5 Thesumof twozerosof thepolynomial f(x) = 2x2+ (p+ 3)x+ 5 is zero,thenthe value of p is (A) 3 (B) 4 (C) -3 (D) 4 Theremainder whenx2~2x + 7 isdivided byx+4is (A) 28 (B) 29 (C) 30 (D) 31 Thequotient whenx3- 5x2+ 7x~4 isdivided byxl is (A)x2+4x+3 (B) x24x+3 (C) x2-4x-3 (D) x2+4x3 TheGCDof(x3 + 1) and x4A 1 is 10. ll. (A) x31 (B) x3+1 (c) x+l (D) x-1 TheGCDof x2-- Zxy+ yzandx4- y4is (A) 1 (B) x+y (C) x~y (D)x2- y2 TheLCMof x3 a3and(x a)2is (A) (x3- a3)(x + a) (B) (x3~as)(x ~-(1)2 (C) (x ~a)2(x2 + ax+ (12) (D) (x + a)2(x2+ ax+ a2) TheLCMof ak,ak+3,ak+5wherekeN is (A) ak *9 (c) ak *6 (B) a" (D) ak * 5 2 12. The lowest form ofthe rational expression L2 is x x -~ 3 x + 3 x + 2 (A x+3 3 303) x-3 13. -- x ~ (C)x~3 x - 3 (D x+2 If3+2and a3 " 23are the two rational expressions, then their product is " Cl + 2 2 2 2 2 2 2 2 (A) _6.l.§..'i.._6.l.b..:*.'..b..2._ (B)a+ab+b .6.l.§..:"...¬£(.2.:l.'....b..2. (C)a+ab+b £L2..:;.¬(2..::...b§. (D)a~ab-b .6.l.i:..l:..6.l.b...'J aab+b 2 14. On dividing xx;is by (A) (x 5)(x3) 3 15. 16. 17. isequal to (B) (x 5)(x+3) (C) (x +5)(x3) (D) (x +5)(x+3) 3 If aC:bisadded withb?a,then the new expression is (A) a2+ ab+ b2 (B) a2- ab + b2 (c) a3+ b3 (D) a3 b3 Thesquare rootof 49(x2 Zxy+ y2)2is (A) 71x yl (B) 7(x+ y)(x-~y) (C) 7(x+ y)2 (D) 7(x- y)2 Thesquare rootof x2+ y2+ z2- Zxy+ 2yz- Zzx (A) |x+)-zl (B) Ix--y+zl (C) |x+y+z| iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii (D) Ix-y-zl Wlif/$23)?) 18. Thesquare rootof l2lx4y8z6(l - m)2is (A) llx2y4z4|l-ml (B) llx4y4lz3(l-m)[ (C) 1lx2y4z6]l-ml (D) 11x2y4[z3(l-m)| 19. If axz+ bx+ c = 0 hasequalroots,thenc is equal L2 b_2 L2 _._b_2 (A) 2a 20. (B) 4a (D) 4a If x2+ 5kx+ 16= Ohasnorealroots,then (A)k>§5 21. (C) 2a (B)k>-§ 5 (C) -§a2+B2=;l¢ aB=§ a+6=§ (D);};+}«=-c~ If a and B arethe roots of axz+ bx + c = 0, then oneof the quadraticequationswhose roots are %and %,is (A) ax2+bx+c=0 (B) bx2+ax+c=0 (C) cx2+bx+a=0 (D) cx2+ax+b=0 25. Letb = a + c . Thentheequation axz+ bx+ c = 0 hasequlroots,if (A)a=c (B)a=c (C) a=2c (D) a=2c _Pdi1its_:'5to§_Réiev1nbé1*i" El A set of nite number of linear equations in two variables x and y is called a system of linear equations in x and y. Such a system is also called simultaneous equations. Eliminating one of the variables rst and then solving a system is called method of elimination. The followingarrowdiagramhelpsus Verymuchto applythe methodof cross multiplication in solvingalx + bly + cl = O, azx+ by + c2= O. A realnumber k is saidto bea zeroof a polynomial p(x), if p(k)= 0. Oz/9Std. Mat/mrmlicx The basic relationshipsbetweenzeros and coefcients of a quadratic polynomial p(x) =axz +bx+care Sum of zeros _b_ coefficient ofx , a Product of zeros C a coefficient of x2 Constant term2 coefficient of x (i) For any polynomial p(x), x = a is zero if and only if p(a) ==0. (ii) x a is a factor for p(x) if and only if p(a) = 0. GCD of two or more algebraic expressions is the expression of highest degree which divides each of them without remainder. LCM of two or more algebraic expressions is the expression of lowest degreewhich is divisible by each of them without remainder. The product of LCM and GCD of any two polynomials is equal to the product of the two polynomials. Let a E R be a non negativereal number.A squareroot of a, is a real numberb such thatb2= a. Thesquare rootof a isdenoted by 2 a or x/;. A quadratic equation in thevariablex is of theformaxz+ bx+ c = 0, wherea,b,c are real numbers and a 79 O. A quadratic equation can be solved by (i) the method of factorization (ii) the method of completing square (iii) using a quadratic formula. Therootsof a quadratic equation axz+ bx+ c = Oaregivenby -bJ_rx/b24ac 2a providedb2- 4acZ 0. A quadratic equation axz+ bx+ c = 0 has (i) twodistinctrealrootsif b2- 4516> 0 (ii) twoequalrootsif b2- 4ac = O,and (iii) norealrootsif b2-~4ac< 0 Do you know? L N V LFern1a.ts lasttheorem: Theequation x" + y" = z"hasnointeger solution whenn > 2. Fermat wrote,i I havediscovered a truelyremarkable proofwhichthismargin is toosmall to contain . Noonewasabletosolve thisforover300) years untilBritish mathematician Andrew Wiles solved it in1994. Interestingly hecame toknow about this problem inhiscitylibrary when hewasahighschool student. Aégebm MATRICES Number;plate,andcombzhafion - I/yeibreemterrerling but dirtimt .5]?/9676! of thoughtto which all mathematical zk/ea; admitof beingreferred .S_'y/1/erter 4.1 Introduction In this chapter we are going to discuss an important Ivlatriceso mathematical object called MATRIX. .puca§m.o£:.pe Here, we shall introduce matrices and study the basics of matrix algebra. Matrices were formulated and developed as a concept during 18th and 19th centuries. In the beginning, their development was due to transformation of geometric objects and solution of linear equations. However matrices are now one of the most powerful tools in mathematics. Matrices are useful because they enable us to consider an array of many numbers as a single object and perform calculations with these symbols in a very compact form. The mathematical shorthand thus obtained is very elegant and powerful and is _1i81i4"1.897> '....Y1i'es¢<%r"i suitable for various practical problems. I The term Matrix for arrangement of numbers, was introduced in 1850 by JamesJoseph Sylvester. Matrix is the Latin word for womb, and it retains that sense in English. It can also mean more generally any place in which something Ip ért/yeéejiarrziebmeiétériz x.?He7 is formed or produced. -~defe7;772zin'ed it we Now let us consider the following system of linear I if equationsin x andy: 3x--2y = 4 2x+5y = 9 (1) (2) We already know how to get the solution (2, 1) of this system by the method of elimination (also known as Gaussian Elimination method), where only the coefcients are used and not the variables. The same method can easily be executed and the solution can thus be obtained using matrix algebra. 4.2 Formation of matrices Let us consider some examples of the ways that matrices can arise. Kumar has 10 pens. We may express it as (10), with the understanding that the number inside ( ) is the number of pens that Kumar has. Now, if Kumar has 10 pens and 7 pencils, we may express it as (10 7) with the understanding that the rst number inside ( ) is the number of pens while the other one is the number of pencils. Look at the following information : Pens and Pencils owned by Kumar and his friends Raju and Gopu are as given below. Kumar has Raju has Gopu has 10 pens and 7 pencils 8 pens and 4 pencils 6 pens and 5 pencils This can be arranged in tabular form as follows: This can be expressed in a rectangular array where the entries denote the number of respective items. 10 7 has 2rows and 3columns. So, the order ofAis2X3. yam n matrix, thefirstletter Inalways denotes thenumber ofrows andthe second letter11 alwaysdenoites thenumber of columns. 4.3 Types of matrices Let us learn certain types of matrices. (i) Row matrix A matrix is said to be a row matrix if it has only one row. A row matrix is also called as a row vector. For example, A = (5 3 4 1) andB = (-3 0 5 ) are row matricesoforders 1 X 4 and 1 X 3 respectively. Ingeneral, A = (alj)1Xn isarowmatrix oforder1Xn. (ii) Column matrix A matrix is said to be a column matrix if it has only one column. It is also called as a columnvector. 1 respectively. 5 For example, A=<3>and B= 2 are column matrices oforders 2X1and 3X1 In general,A = [a ljimx1is a columnmatrixof orderm X 1. (iii) Square matrix A matrix in which the number of rows and the number of columns are equal is said to be a square matrix. For example, 3 0 2 7 6 1 A=<:1;31 >and B= 1 5- 7 are square matrices oforders 2and 3respectively. Ingeneral, A = [alj]X isasquare matrix oforder m. Theelements al1,a22 ,a33,---,amarecalledprincipalor leadingdiagonalelements of the square matrix A. (iv) Diagonal matrix A square matrix in which all the elements above and below the leading diagonal are equal to zero, is called a diagonal matrix. For example, 3 0 0 0 0 1 A=<(5) E>and B= 0 0 0 arediagonal matrices oforders 2 and 3 respectively. Ingeneral, A = [alj]X issaidtobeadiagonal matrix if alj= 0 foralli 7% j. (V) Scalar matrix A diagonal matrix in which all the elements along the leading diagonal are equal to a non-zero constant is called a scalar matrix. For example, 7 0 O A=<3 E>and B= 0 7 0 are scalar matrices oforders 2and 3respectively. 007 h _¢ , Ingeneral, A=[a_,] issaid tobe ascalar matrix ifa_. ={OW enJ, U mxm where k is a constant. (Vi) U k, when l :1 Unit matrix A diagonal matrix in which all the leading diagonal entries are l is called a unit matrix.A unit matrix of ordern is denotedby In. For example, 1 0 0 I2=<(1) (1) >and I3= O1 0 are unit matrices oforders 2and 3respectively. 0 O 1 . Ingeneral asquare matrix A=(a ) isaunit matrix if a,,= 01ifIf i;éj i =j zjnxn A unitmatrixisalsocalled Every unitma,tn'xis clearly ascalar tomultiplicationy, A A L V withrespect However ascalar need notbeaunity Aunit matrixplaysytherole ofthenumber lin (Vii) . A AA A AA Null matrix or Zero-matrix A matrix is said to be a null matrix or zero-matrix if each of its elements is zero. It is denoted by 0. For example, 0:; (3 andO=arenullmatricesoforder 2x3and 2x2. :iy(i) Azero-matrix need notbea,square matrix; (ii) Zero.-rniatrix plays therole Anumber zeroinnumbers.i5(iii) Amatrix does notchange [ifthezeromatrix ofsame, A order is addedtoitor subtractedffrom it. (viii) A A it A A A Transpose of a matrix §}z3}*i:s§z§e;s The transpose of a matrix A is obtained by interchanging rows and columns of thematrixA andit is denoted by AT(readasA transpose). Forexample, 13 if A=<§31 then AT:2 4 In general,if A = [a_ 5 then 6 lJ']m>-\I 4.3 Let A = [cu] U Find (i) theorderof thematrix(ii) theelements an andam (iii) thepositionof theelement 2. Safzlszigm(i) Since the matrix A has 4 rows and 3 columns, A is of order 4 X 3. Example 4.4 Constructa 2 X 3 matrix A = [aU]whose elements aregiven byalj= |2i-- 3j| tksfézsfisssr In general a 2 X 3 matrix is given by A=are not equal as the orders of the . . matrices are different. 6 3 1 5 3 Also <1 2>7é< 1 8>, since some ofthe corresponding elements are not equal. 85 25 Example 4.6 Fdh 1 f,xyandz1<5 f 91 >=<5 yl > intevaueso . . 5 4 3 5 z lbfzsissyzAs the given matrices are equal, their corresponding elements must be equal. Comparing the corresponding elements, we get x = 3, y = 9 and z = 4. Example 4.7 6 2 Solve: =131 +4y ) - x r§:;Z:zI:r)sz Since the matrices are equal, the corresponding elements are equal. Comparing the corresponding elements, we get y = 6 2x and 3x = 31 + 4y. Using y = 6 2x in the other equation, we get 3x = 31 + 4(6 - 2x) 3x = 31 + 24 ~ 8x x = 5 and hence y = 6 -~-2(5): Thus, x=5andy (ii) Multiplication -4. =- 4. of a matrix by a scalar Fora givenmatrixA = [a..1 V B= ll mxn 11mxn anda scalar(realnumber)k, wedenea newmatrix , whereb., = ka., forall 1' andj. 1] ,lJ' Thus, the matrix B is obtained by multiplying each entry of A by the scalar k and written as B = kA. This multiplication is called scalar multiplication. Forexample,ifA=<:ll S;>thenkA=k<:ll 1: 11:]: Example 4.8 -1 2 4 lfA<3 6_5)thennd3A .§:;§5:£§s;sz The matrix 3A is obtained by multiplying every element of A by 3. -1 3 2 4 6 -5 3A=3( _3(-1) 3(3) 3(2) 3(6) 3(4) 3(--5) __-3 " 9 6 12 18 --15 (iii) Addition of matrices Matrices A and B given below show the marks obtained by 3 boys and 3 girls in the subjects Mathematics and Science respectively. Mathematics Science A__<45 81>Boys B_<51 807090>Boys " 30 72 90 65 Girls " 42 85 Girls OI/9Std Mat/yelrmlicr To find the total marks obtained by each student, we shall add the corresponding entries of A and B. We write A+B=<45 7281>+<5180 90> 45+51 72+80 81+9O<96152171) 30 90 65 30+42 42 90+85 85 70 65+70 72 175 135 The nal matrix shows that the rst boy scores a total of 96 marks in Mathematics and Science. Similarly, the last girl scores a total of 135 marks in Mathematics and Science. Hence, we observe that the sum of two matrices of same order is a matrix obtained by adding the corresponding entries of the given matrices. L IfA: [alj]Xand B:4 Xare two matrices ofthesame order, thenthe addition iofAandB isamatrix C=,[clj] ,iwhere clj=alj+bi],foralli andj. A M Lit Note that the operation ofaddition onmatrices isdened asfornumbers. The addition of twomatrices A andB isdenoted byA+B.Additionisnotdenedformatrices of differentorders. Example 4.9 832 11 LetA=<591)andB=<3 O Ss;§z;siwa Since A is order of 2 X 3 and B is of order 2 X 2, addition of matrices A and B is not possible. ). FindA+B ifitexists. Example 4.10 5 6-2 3 3 -~14 7 IfA<1 0 4 2>andB<2 8 2 3),thenndA+B. .»S}<3f;§Iism Since A and B are of the same order 2 X 4 , addition of A and B is dened. 56--23>+<3-147> So, A+B=< 10 4 2 2 8 2 3 <5+3 6~1-2+43+7) 1+ 2 Thus, A +B = O+ 8 4 + 2 2 + 3 <85210) 3 8 6 5 (iv) Negative of a matrix is denoted by A Thenegativeof a matrixA = [a. Thatis, -A = [b5] mm ,where b..U = - a..U for all i andj. and is dened as -A = ( 1)A. (V) Subtraction of matrices If A = [alj] X andB = X aretwo matricesof thesameorder,thenthe subtraction A B isdened asA --B = A + (- 1)B.Thatis,A -~B = where clj= alj- bi]. for alli andj. Example 4.11 Matrix A shows the weight of four boys and four girls in kg at the beginning of a diet programme to lose weight. Matrix B shows the corresponding weights after the diet programme. A_<35 40 28 45>Boys B__ <32 35 27 41 >Boys 42 38 41 30 Girls 40 30 34 27 Girls Find the weight loss of the Boys and Girls. 35402845)__<32 352741> 42 38 41 30 40 30 34 27 SiiiiififWeight loss matrix A~B__(3514> =< 2873' 4.5 Properties of matrix addition (i) Matrix addition is commutative If A and B are any two matrices of same order, then A+B = B+A (ii) Matrix addition is associative 1fA, B and C are any three matrices of same order, then A + (B + C) = (A + B) + C (iii) Existence of additive identity Null or zero matrix is the additive identity for matrix addition. If A is a matrix of order m X n, thenA + O = O + A =A, where O is the null matrix of order m>_.-. <153 > 3. HA =(S2)~(;_f ),then ndthe additive inverse ofA. 5. IfA(5____9)andB<__1__3)nd6A~~3B 4-2 8 2 6. Find aandb 1fa<§)+b('"i >=<150 7. FindXandYif 2X+3Y=and3X+2Y=<__? 8. Solveforxandyif(x2)+3(2x)=< 9). 2 ..... y 9 I . £4: 3 2 "Y 142 4 114: 12 1-2 0 Verifythat 11. 0 0 <51) (2 3>andO<00>ten B: = verify: (1) A+B=B+A 10. 4 3 (11) A+(-~A)=O=:(-A)+A. 2 o 4 3 ,3: 6 2 8 2 2 4 6 A+(B+C)= h andC= 1 23 5 0 1..1 2 ,then 1 (A+B)+C. An electronic company records each type of entertainment device sold at three of their branch stores so that they can monitor their purchases of supplies. The sales in two weeks are shown in the following spreadsheets. Cniriayers pg Find the sum of the items sold out in two weeks using matrix addition. 12. The fees structure for one-day admission to a swimming pool is as follows: Before2.00pm. After 2.00p.m. 2 Before2.00pm. After 2.00 p.m. Write the matrix that represents the additional cost for non-membership. 4.6 Multiplication of matrices Suppose that Selvi wants to buy 3 pens and 2 pencils, while Meena needs 4 pens and 5 pencils. Eachpen and pencil cost ?l0 and ?5respectively. How much money doeseach need to spend? Clearly, Since 3 X 10 + 2 X 5 = 40, Selvi needs ? 40. Since 4 X 10 + 5 X 5 = 65, Meena needs 3 65. We can also do this using matrix multiplication. Let us write the above information Requirements as follows: Price (in ?) Money Needed(in ?) Se1vi(3 2) (10) <3><10+2><5)_<40 Meena 4 5 5 4><5 65 Suppose the cost of each pen and pencil in another shop are ?8 and 34 respectively. The money required by Selvi and Meena will be 3 X 8 + 2x 4 = ?32 and 4 X 8 + 5 X 4= ?' 52 . The aboveinformation canbe representedas Requirements Selvi 3 Price (in ?) 2 8 Money Needed(in ?) 3><8+2><4 32 Meena<4 5) <4) <4><8+5><4>=<52> Now, the above information in both the cases can be combined in matrix form as shown below. Requirements Price (in ?) Selvi<3 2) (108) Meena 4 5 5 4 Money needed(in ?) 3><5 3><8+2><4 :<4032) 4><5 4><8+5><4 65 52 From the above example, we observe that multiplication of two matrices is possible if the number of columns in the rst matrix is equal to the number of rows in the second matrix. Further, for getting the elements of the product matrix, we take rows of the rst matrix and columns of the second matrix, multiply them element-wise and sum it. The following simple example illustrates how to get the elements of the product matrix when the product is dened. LetA=and B=<2M: Then the product ofABisdefined and is given by AB :: 2 -l 3~9 <3 4>< 5 7> Step E : Multiply the numbers in the first row of A by the numbers in the first column of B, add the products, and put the result in the rst row and rst column of AB. (3~41)79>=(2(3)+(-1)5 > fétep2: Follow the same procedure as in step 1, using the rst row of A and second column of B. Write 2 ~l the result in the rst 3 -9 row and second column 2(3)+(--l)5 2(~i9)+(~l)7 <3457: >( > ( Step 3: of AB. l Follow the same procedure with the second row of A and rst column of B. Write the result in the second row and rst column of AB. @--D «-3) =(233/)3))++(;-(l5))5 2(~ 9)+(~D7) Step 4: The procedure is the same for the numbers in the second row of A and second columnofB. <2 -1><3 «9>__ 2(3)+(-1)5 2(--9)+(~1)7 3 t top 5: 4 5 7 ' 3(3)+4(5) 3(9)+4(7) Simplify to get the product matrix AB 2(3)+(1)5 2(~9)+(~1)7 ~25) 3(3)-+4(5) 3(~«9)+4(7) "__<1 thentheproduct matrix ABisdened andis of IfA#[a_] iandB=[b3,i ti mxn li]n>

'<13> S::f:1f§sm Given that <43 2 8> 5 > y =< 13 3x+2y <138> 4x + 5y :2 2 Equating the corresponding elements, we get 3x+2y=8 and =>3x+2y~8=O and 4x+5y =13 4x+5y-l3=O. Solving the equations by the method of cross multiplication, we get x y 5 l -8 3 -13 4 x _ Y ""3-26+4o Thus, 5 _ -32+39 x=2, 1 x 15-8 """ _}_l 14 7 7 y=l Example 4.14 1£A=<" b>and1=(1 0>,thenshowthatA2(a+d)A=(bc~ c d 2 0 1 2 §§£}§§»§f§s?§§ Consider A2 = A > =(a2+bc ab+bd) C d C 61 ac + cd be + d2 Now,(a+d)A= (a+d)("I) c d ,C=<:53 :)andD=<(1) Then, (i) AB 75BA (ii) AD = 0, however, A andD are not zero-matrices and (iii) AB = AC, but B ¢ C. Let us see some properties of matrix multiplication through examples. (i) Matrix multiplication is not commutative in general If A and B are two matrices and if AB and BA both are dened, it is not necessary that AB = BA. Example 4.15 8"'7 9-3 2 If A = -2 4 and B = 0 3 , then nd AB and BA if they exist. 6 -1 -5 S%.s§£§zigm The matrix A is of order 3 X 2 and B is of order 2 X 3. Thus, both the products AB and BA are dened. 8-7 Now,AB=<-.2 4)<93 2) =(-18+24 6-4--4-20)< 6 2-24) BA=<: _§ __§)(..(2) .3i>=(;g __g§). (Notethat 0 3 72--42 .. 24+7 0+18 Similarly, (ii) 6~1-~5 Matrix multiplication 16+35 O--3 30 0--15 8 _7 18 -17 ~3 51 -15 is always associative For any three matrices A, B and C, we have (AB)C = A(BC), whenever both sides of the equality are dened. (iii) Matrix multiplication is distributive over addition For any three matrices A, B and C, we have (i) A(B + C) = AB + AC (ii) (A + B)C = AC + BC , whenever both sides of equality are dened. Example 4.16 3 2 --2 5 1 1 . IfA=(_.1 4>,B=< 6 7>andC=(~_5 3)Ver1fythatA(B+C Saaigstfrm Now, B+C=<+<___5 3)=+< 3) AB + AC 2+24 -5+28 -1-20 1+12 6 29>+ -7 9 < < 26 From (iv) and 23 ;, we have -2111 A(B + C) = AB + AC. Existence of multiplicative identity In ordinary algebra we have the number 1, which has the property that its product with any number is the number itself. We now introduce an analogous concept in matrix algebra. For any square matrix A of order n, we have AI = [A = A, where I is the unit matrix of order 11.Hence, I is known as the identity matrix under multiplication. Example 4.17 lfA=< ; __g),then verify AI=IA =A,wherelis the unit matrix oforder 2. .§r;f£;iae2s/2 1 3 1 0 _ 1+0 0+3 1+0 3+0 __ 1 3 1 3 _ N°W AI'i9 6)<01>_(9+0 o~6)'i9 6)A A1s°IA"<0 1>i9 6)"0+90~6 9-6 A 1 Hence (v) AI = [A 0 1 3 = A. Existence of multiplicative inverse If A is a square matrix of order n, and if there exists a square matrix B of the same order n, such that AB = BA = I, where I is the unit matrix of order n, then B is called the multiplicative inverse matrixofA andit isdenoted by A-1. 2y3 2 not have multiplicative inverses. 2(Do Some ofthesquare matrices like <4 6 )do (ii) If B isthemultiplicative inverse of/1,thenA isthemultiplicativeinverse ofB. (iii) If multiplicative inverse ofasquare matrix exists, thenit.isunique. Example 4.18 Prove that and ( 1 3)are multiplicative inverses toeach other. 2 OI/9Sid. Mal/aemalicx ._. f5%;fz£I:'mz Now,<3 5)(2~5>=(6-"5 m15+15)=<(1) (1)>=I 12 --1 3 2--2 --5+6 Also 2 -5 <35>__ 6-5 -5+6 10-10 __(10>__I -1312"-3+3 oi "' The given matrices are inverses to each other under matrix multiplication. (Vi) Reversal law for transpose of matrices If A andB aretwomatrices andif AB is dened, then (AB)T= BTAT. Example 4.19 IfA=( 4)andB=(1 3-6),then verifythat(AB)T $£?§!i§§(3§E Now, AB (4)(1 3-6) =<4 12 -24) Thus, (AB)T (--61215 > Now, BTA=( 3)(-2 45) 2 -2 --2 5 5 -2 4 12 - 24 --6 15 12 ~30 5 - 30 1 -6 2 4 5 ~6 12 15 12 - 24 - 30 Fromi : andAI? :, weget (AB)T= BTAT. 1. Determine whether the product of the matrices is dened in each case. If so, state the order of the product. (i) AB,whereA = [alj]4X3 , B = [blj]3X2 (ii) PQ,whereP = [pl_J_]4X3 , Q= [qlj]4x3 (iii) MN,whereM = [mlj]3x1 , N = [nlj]1X5(iv) RS,whereR= [rl.J.]2X2 , S= [slj]2X2 2. Find the product of the matrices, if exists, <21><:> <: f><;;> (iii) (31 3 "g)(_E(iv) (__§)(2 7) A fruit vendor sells fruits from his shop. Selling prices of Apple, Mango and Orange are ? 20, ? 10 and ? 5 eachrespectively. The salesin three daysare given below .5 II le 4 an0" .p"5OAp :4" V e360. a- 7330. r 10 60. Write the matrix indicating the total amount collected on each day and hence nd the total amount collected from selling of all three fruits combined. 1 2 x 0 x 0 Findtevauesoxandyi <33><0 y) <90) 'h l f 'f = . IfA=<5 3),X= and C=(-.5)andifAX=C, thenndthevalu 7 5 y -11 ofxandy. IfA =<:"31)thenshowthatA2 ..4A+ 512 =0. IfA=<4 0>and B=<3 2>then ndAB and BA. Are they equal? 3 2 3 0 0 10. 1£A=,B=<1)andC=(2 1)verify(AB)C= 2 1fA=<5 2>and B=(_2 _1)verifythat(AB)T=B A. Prove that A=<3§>and B=(*3E)are inverses toeach other under matrix TT ll. 73 12. 11 multiplication. 13. Solve (x1)(*: IfA=(1"4)andB=('"1 6),thenprovethat(A+B)27 -2 3 3 --2 IfA <76>,B<09>andC <4 6),nd(+ )Cand C+C 3 3 8 77 70f/9 Std. = Mal/yemalicr = Is (A+B)C=AC+BC? 7 = 2-3 A B A B. Choose 1. the correct answer. Which one of the following statements is not true? (A) A scalar matrix is a square matrix (B) A diagonal matrix is a square matrix (C) A scalar matrix is a diagonal matrix (D) A diagonal matrix is a scalar matrix. 2. Matrix A = [(1. is a square matrix lj]m>n if (C)m=1 (D)m=n 3. If( y+1 2~_ 3x)(8 8 >then the values ofxand yrespectively are 3x + 7 5 (A)2,7 _ 1 y-2 (B)L,7 . (c)%,-% (D) 2,7 -1 4.IfA=(1-2 3)andB=< 2>thenA+B (A) (O 0O) (B) =4<2 1>thenthevalue ofxis1 x 8 12 (A) 1 7. (B) 2 (C) -4- (D) 3 (D) 4 IfA is of order 3 X 4 and B is of order 4 X 3, then the order ofBA is (A) 3 X 3 (B) 4 X 4 (C) 4 X 3 (D) not dened 8. IfA><<(1) ;>=(l 2),thentheorderofAis (A)2><1 9. (B)2><2 (C)1><2 (D)3><2 If A and B are square matrices such that AB = I and BA = I , then B is (A) Unit matrix (B) Null matrix (C) Multiplicative inverse matrix of A (D) A 10.If<: ? x>=<31 ),then the values ofxandy respectively, are 3 (A)2,0 (B)0,2 (C)0,--2 (D)1,l 11. IfA=(§m:)andA+B=O,thenBis (..i1f)(":mi)(:§:i) <é£> IfA=(::§>,thenA2is «~><:::><1:::> «»<:::> «»(:::> A is of order m X n and B is of order p X q, addition of A and B is possible only if (A)m=P 14. (B)n=q (C)n=P (D)m=p,n=q IfissuchthatA2=I,then (A)1+a2+[>7=0 (B)1--0z2+[>7=0 (C)1~0z2--37:0 (D)1+0z2-[)7=O If/l=[a_.]lJ2x2 anda_.=i+j,thenA= :1 1 2 2 (3 4) A 0% 4) (O1 (A) -1,0,0, 18. 3 2 3 ((4 5) 4 5 W6 7) _(1) >,thentheValues ofa,b,canddrespect -1 (B) 1, 0,0,1 (C) --1,0,1,0 (D) 1, 0, 0, 0 IfA=7 3 (B)-7 (0%; (me 2 20. Which one of the following is true for any two square matrices A and B of same order?. (A) (AB)T= ATBT (B) (ATB)T= ATBT (C) (AB)T= BA (D) (AB)T= BTTA 70¢/9Std. Mal/mmzlitr TI"3§Y3s A matrix is a rectangular array of numbers. A matrix having m rows and 11columns, is of the order m X n. A = [alj]mxn isarowmatrixifm = l. A = [a__] is a columnmatrixifn = l. l] m>30yMy x = 20. Thus, 2y '. The position ofthe relay tower is at P(20, 10). Taking the above problem as a model, we shall derive the general section formula. Let A (x1,y1)andB(x2,yz)betwodistinctpointssuchthatapointP(x,y) dividesAB internally inthe ratio 1:m.That is,%g=72From the Fig. 5.2, we get AF=CD=OD-OC=x-xl PG-DE=OE-OD=x2-~x Also, PF:-PD-FD=y--yl BG=BE-GE=y2y Now, AAFP and APGB are similar. (Refer chapter 6, section 6.3) Thus AFEAP PGBG_PB_m £_L PG_m => x~x1:L x2--x E and 2: m ==> mx-mx1=lx2~~lx lx+mx=lx2+mx1 =L BG m y~y1:L yz--y m =2» my~-my1=ly2-ly ly+my=ly2+my1 Coordimzte Geometijl Thus, the pointP which divides the line segmentjoining the two points A A(ix1, yl)andB(x2i,y2) internally intheratio1:mis B g A1P 2 2 ' 2 A 2 Themidpointof thelinesegment joiningthepoints A A Q _ x +x y,+y A(px1a}1)ia,1V1dpB,(X2i}2)p1SV([l2 12 (iii) Centroid of a triangle Consider a AABCwhoseverticesareA(x1,y1), B(x2,y2)andC(x3,y3). LetAD, BE and CF be the medians of the AABC. We know that the medians of a triangle are concurrent and the point of concurrency is the centroid. Let G(x , y) be the centroid of AABC. Now the midpoint ofBC isD<22 3,y22y3> X +96 + Bytheproperty oftriangle, thecentroid Gdivides the B median AD internally in the ratio 2 : l By section formula, the centroid G(x,y) G 2(x2+x3) 2 2+1 +109) 2?+1) 2+1 _G 3 0;/9Std. Mat/mmzlicx 3 eThecentroid of thetriangle whose vertices areL M, _ Midpoint ofthe line ségment joining theA(3 0) Fig(x5:) B(14) points (3,0) and (- 1,4) is M(x,y)=(3'2;1,0§4) =M(1,2). Example 5.2 Find the point which divides the line segment joining the points (3 , 5) and (8 , 10) internally in the ratio 2 : 3. Sz:1sm'o:z Let A(3 ,5)and B(8,10) be the given points. 2 3 Let the point P(x,y) divide the line AB A(3, 5) internally in the ratio 2 :3. P(x, y) B(8,l0) Fig. 5.5 By section formula P(xy) P:33(5>) - ,,(5 7) Example 5.3 In what ratio does the point P(2 A( 3, 5) and B ( 4, , 3) divide the line segment joining the points 9) internally? Sz2§m§0:z Given points are A(-~3 , 5) and B(4 , - 9). Let P ( 2 , 3) divide AB internally in the ratio l : m I Bythesection formula, P< lxz +mxl, ly2 +m)1)%P(2 3) l + m l + m in A('_35) P(_2,3) Fig. 5.6 B' (4,_9) _> :_____2 6l=m _l_=_1_ m 6 i.e., l:m=l:6 Hence P divides AB internally in the ratio 1 : 6 (1) jin nthéiaboveex-ample; am: imiayi getthe ratio by? equating y-coogdi"na:¢s a1s¢;y 2 i(ii)_y Thengratiosniobitained jequatyinigf2C§Coordiinates gbyfequiating yy-Coordinates are gt[sameonlyivvhenithethree pogintsareicollinear.A A f TP' P 2 (iiijfi :1*fy7a point idivides the lime(segment!itemaiiyimthe ratio Ithai positive.;or If apomt diviydes;thfe1ignelnsegrnentjexternallyfinjtheratio 2;m,t11éir1f,LfQ is? ' t Example 5.4 Find the points of trisection of the line segmentjoining (4, - 1) and (~-2, - 3). Seiuzém-2Let A(4, 1) and B( 2, 3) be the given points. Let P(x,y) and Q(a,b) be the points of . . tr1sect1on of AB sothatAP= PQ= QB -Tome A(4,-1) Hence P divides AB internally in the ratio122andQdivides ABinternally D Fig5;,- 1 14(4) _1) Q B(2,-3) 2 P in the ratio 2 : l B(_2 _3) F3Q-5-8 Bythesection formula, therequired points are 2 l """:"'"""""""'°"""' Pi1(-2)+2(4) 1+2 1(-3)+2(~1) 1+2 land M1) Fig.5.9QB('2" 2(-2) 1(4) 2(--3) 1(-1) Q< 2++1 2:1 > :> P(x,y)=P(-23+8,~-33-2) andQ(a,b)=Q<~43+4,63-1) =P<2»§> =Q» Note that Q is the midpoint of PB and P is the midpoint of A Q. Example 5.5 Find the centroid of the triangle whose Vertices are A(4, -6), B(3,2) and C(5, 2). ..§(§3:/zzim: The centroid G(x , y) of a triangle whose Vertices are (x1,y1) , (x2,y2)and(x3,y3)is givenby A(4,-6) G(x,y)_G _ 3(3g-2) D_ = G(4, .. 2)_ C(5,2) Fig.5.10 Example 5.6 If (7, 3),(6, 1),(8, 2) and (p, 4) are the Verticesof a parallelogramtaken in order,then nd the value of p. ;§:;imi:mLet the Verticesof the parallelogrambe A(7, 3),B(6, 1) ,C(8, 2) and D(p, 4). We know that the diagonals of a parallelogram bisect each other. The midpoints of the diagonalAC D(p,4) andthediagonal BD coincide. C(89) Hence(7+8,3+2)_<6+I9,l+4> "5 (6:p3> A03)Fig. 5.11 M651) Equating the x-coordinates, we get, 6 + p : _1_'5_ 2 2 '. p =9 Example 5.7 If C is the midpoint of the line segment joining A(4 , 0) and B(0 , 6) and if 0 is the origin, then show that C is equidistant from all the Vertices of A OAB. mmmm'mmm@mmmmBm<14:Q 2 Weknowthatthedistance between P(x,,y1) andQ(x,,y,) is (x, x2)2+ (y, yz) . Distancebetween O(0,0) and C (2, 3) is 0C=/@~0Y+B~0Y=¢B. Coordimzte Geometijl DistancebetweenA(4, 0) and C(2, 3), AC:/Q4¥+@0Y=¢4+9=¢m DistancebetweenB (0,6) and C(2,3), BC:/Q~OY+B6Y=/4+9=/B 0C=AC=BC Fig. 5.12 The point C is equidistant from all the vertices of the AOAB. «ihe.inidj3.ointI [Cofth¬i¢'1i1i37P -an +y3)(x3 - x.) : %{x1y2 x2y2 +x1y1"x2y1+x3y1" x1y1+x3y3 "xiys-x3y2 +x2y2 '" x3y3 +x2y3} Area ofthe AABC is%{x1(y2 - 323) +x2(y3 - y1)+ x3(y1 y2)}.sq.units. I _ _ _ _ _,,,,,,, . then the area ofthe AABC is%{x1(y2 - 323) +x2 (y3 - yl)+x3(y1 - y2)}.sq.units. B The areaof thetrianglelcanalsobewrittenas V . i L. "2{71pY2_'f xiy3.+ xzy3 Wxzyid"x3y1 '5 ,3y2} Sqfun1ts' (or)A,ée{(x1y2p¥+fpxélyé+x3y1),#+ei(xé,y1i x1y3)}sq.units The following pictorial representation helps us to write the above formula very easily. Takethe verticesA(x1,y1),B(x2,y2)andC(x3,y3)of AABCin counterclockwise direction and write 2 them column-wise as shown below. vx32m>__,..v xl} andx3y1asshownin thedarkarrows. Add the diagonalproductsxlyz, x2323 Coordimzte Geomefal Alsoaddtheproductsxzyl, x3y2andx1y3asshownin thedottedarrowsandthensubtract the latter from the former toget the expression %{(x1y2 +x2y3+x3y1) - (xzyl +963372 +961373)} Tond theareaof atriangle,thefollowingstepsmaybeuseful. (i) Plot thepointsin a roughdiagram.T i (ii) Take thevertices in counter clockwise direction. Otherwise theformulagives a negative value. (iii)Use the forrnula. area ofthe AABC =:{(x1 y2x2323 +x3y1i) g-(xzyl +x3y2 +x1y3)} 5.4 Collinearity of three points Three or more points in a plane are said to be collinear, if they lie on the samestraight line. In otherwords,threepointsA(x1,y1), B(x2,y2)andC(x3,y3)arecollinearif anyoneof these points lies on the straight line joining the other two points. Suppose thatthethreepointsA(x1,y1), B(x2,y2) andC(x3,y3)arecollinear. Thenthey cannot form a triangle. Hence the area of the AABC is zero. 1 _ Le" §{(x1y2 + x2y3 + x3y1)(xzyi+ x3y2 + x1y3)} _0 2: x1y2+x2y3+x3y1:x2y1+x3y2+x1y3 One can prove that the converse is also true. Hence the area of AABC is zero if and only if the points A, B and C are collinear. 5.5 Area of the Quadrilateral LetA(x1,y1) ,B(x2,y2),C(x3,y3) andD(x4,y4)betheverticesofa quadrilateral ABCD. Now the area of the quadrilateral ABCD = area of the AABD +area of the ABCD __ 1 " §{(x1y2+ x2y4+ x4y1)"(x2y1+ x4y2+ x1y4)} 1 +§{(x2y3 + x3y4 + x4y2)(x3y2 + x4y3 + x2y4)} Area of the quadrilateral ABCD Y _l _§{(x1y2 +x2y3 +x3y4 +x4y1)"(x2y1+ x3y2 +x4y3 +x1y4)} or 1 . §{(x1 x3)(y2 H"y4)" (xz " x4)(y1"y3)}Squmts The following pictorial representation helps us to writetheaboveformulaveryeasily. 701/9Std. Mai/éidliff Fig514 Takethe verticesA(x1,y1),B(x2,y2 ),C(x3,y3)andD(x4,y4) in counterclockwise direction and write them column-wise as shown below. Follow the same technique as we did in the case of nding the area of a triangle. This helps us to get the required expression 1 '2"{(x1y2 +x2y3+x3y4+x4y1)"'(x2y1+x3y2 + x4y3 +x1y4)}' Example 5.8 y . . . 3(-3» 4) Find the area of the triangle whose vertices are ; (1,2),(-3 , 4),and(5 ,6). \:I~A(1,2) 3(;fzs¥f:;§8 Plot the points in a rough diagram and take them in order. Let the verticesbe A(l , 2), B( 3 , 4) and C (-5, -6). Now the area of AABC is 'iC(5, %) _ l _ 'f{(x1y2 +x2y3+x3y1)(x2y1+x3y2 +x1y3)} = 12{(4+18-10)-(-6-20-6)} =%{12 +32}=22.sq.units Example 5.9 If the area of the AABC is 68 sq.units and the vertices are A(6 ,7), B(4 , 1) and C(a , -9) taken in order, then nd the Value of a. §¥?££v£§¢.§fi Area of AABC is %{(6 +36 +7a) --(-28 +a- 54)}: 68 use :if C, 6} ==>(42 + 7a)~ (a - 82) = 136 =2 6a = 12 a = 2 Example 5.10 Show that the points A(2 , 3), B(4 , 0) and C(6, 3) are collinear. .§e;!::§ir;rzz Area of the AABC is =%{(0 - 12 +18) -(12 +0-6)}use :i{2 6 The given points are collinear. Coordimzte Geometijl Example 5.11 lfP(x , y) is any point on the line segmentjoining the points (a , 0) and (0, 19),then , prove that -2+-351 =1,where a,1)¢0. .,§mzi=3§mz Now the points (x, y), (a ,0) and (0, b) are collinear. '. The area of the triangle formed by them is zero. =_.>abbxay=O use:l{ bx + ay = ab cl} Dividing by ab on both sides, we get, E-+y=l, a 19 where a,b7é0 Example 5.12 Find the area of the quadrilateral formed by the points (49 -2): (_ 3: -5): (39 and 9 .S})§mi:m Let us plot the points roughly and take the Vertices in counter clock-wise Let the vertices A( 4, direction. be 2), B( 3, 5), C(3, 2) and D(2, 3). Area of the quadrilateral ABCD _ l _ §{(x1y2+x2y3+x3y4+ x4y1)"'(x2y1+x3y2 +x4y3+ x1y4)}' =%{(20+6+9-4)-~(6-15--4-12)} 1 : N31+ 25}=28sq.units. l. W3 2 ...2-«-"- 5 Find the area of the triangle formed by the points (O90): (3: and (09 (-47 -5): (49 2. LM (57 2): (39 -5) and (-53 -1) and (-19 -6) Vertices of the triangles taken in order and their areas are given below. In each of the following nd the value of a. Vertices (i) (0,0), (4, a), (6, 4) Area (in sq. units) 17 (ii) (a, a), (4, 5), (6,l) 9 (iii) (a, -3), (3, a), (l,5) 12 Oz/9Std. Mat/mmlics 3. Determine if the following set of points are collinear or not. (i) (4, 3), (1, 2) and (-2, 1) (iii) 4. (ii) (-2, -2), (-6, -2) and (-2, 2) 3),(6, -2)and (-3,4) In each of the following, nd the value of k for which the given points are collinear. (i) (k, -1), (2, 1) and (4, 5) (ii) (2, - 5), (3, 4 )and (9, k) (iii) (k, k), (2, 3) and(4, - 1) 5. Find the area of the quadrilateral whose vertices are (i) (6, 9), (7, 4), (4, 2) and(3,7) (iii) ( 4, 5), (O,7), (5, - 5) and(- 4, -- 2) 6. (ii) (-- 3, 4), (~ 5, ~ 6), (4, -- 1) and(1, 2) If the three points (h, 0), (a, b) and (O,k) lie on a straight line, then using the areaof the triangle formula, show that %+%= 1,where h,k75 O. 7. Find the area of the triangle formed by joining the midpoints of the sides of a triangle whoseVerticesare (0, ~ 1), (2, 1) and(0,3). Find the ratio of this areato the area of the given triangle. 5.6 Straight Lines 5.6.1 Angle of Inclination Let a straight line I intersect the x-axis at A. The angle between the positive x-axis and the line I, measured in counter clockwise direction is called the angle of inclination of the straight line l . Rmarks 5.6.2 Slope of a straight line If 6is the angleof inclination of a non-vertical straightline I, then tan6' is calledthe Slope or Gradient of the line and is denoted by m. '. Theslopeofthe straightline, m = tan6' for 0°S 6S 180°,67590° Coordinate GeometijlIp 5.6.3 Slope of a straight line when any two points on the line are given Let A(xLy1 ) andB(x2,yz)beanytwopointsonthestraight linel whose angleof inclination is 6'. Here, 0° 5 6 5 180°, 6' 7e90° Let the straight line AB intersect the x-axis at C. Now, the slopeof the line l is m = tan 6' Draw AD and BE perpendicular to x-axis and draw the perpendicular AF line from A to BE. From the gure, we have AF=DE = OE-OD=x2~x1 and BF=BE--EF=BE-AD=y2~y1 Also, we observethat A = /l7%AB= 6 In the right angled AABF , we have __BF_)"'} . tan6-x:__x: Lfxlyéxz From 2and =;_i'J %, weget theslope, in=yz -yl x2- x1 7 m=x2x1 y2y1 =y1_y2 wherex ¢x as67é90°. x1x2 1 2 4% Theslope ofthestraight linejoining thepoints (xliy1[)pand (x2, y2)isalso interpreted as m_ Y2" 371 A_ change inycoordinates A _ x2x1 101/9Std Mm/mrzaiicx change inx coordinates ' 5.6.4 Condition for parallel lines in terms of their slopes Consider parallellinesZ1 andZ2whoseanglesof inclinationare61and6'2 andslopes aremlandm2respectively. J Since Z1 andZ2areparallel,theanglesof inclinations 6'1 and6'2 areequal. tan61=tan62 ==> m1= m2 If two non-vertical straight lines are parallel, then their slopes are equal. The converse is also true. i.e., if the slopes of two lines are equal, then the straight lines are parallel. 5.6.5 Condition for perpendicular lines in terms of their slopes Let l1andZ2betwoperpendicular straightlinespassing throughthepointsA(x1,y1) andB(x2,y2) respectively. Let ml and m2be their slopes. 2 Let C(x3,323) betheirpointof intersection. Theslope ofthestraight linel ism= y3 yl 1 1 x3" xl . . . V2 The slope ofthe straight line Z2 is m2 =Y3"' x3 --x2 0 In the right angled AABC, we have Fig. 5.20 AB2 = AC2+ BC2 2 (x2 x1)2+ (y2"y1)2 = (x3_ x1)2+ (y3*y1)2+ (x3_x2)2+ (3)3" y2)2 =:>(x2x3+x3-x1)2+(y2"'y3+y3"y1)2 = (x3"' x1)2+ (ys"" y1)2+ (x3"' x2)2+ (ya'" y2)2 2:)(x2-x3)2 +(x3 "x1)2 +2(x2 "x3)(x3 ~x1) +(Y2 " y3)2 +(y3 ''"y1)2 +2(y2 _y3)(y3 'y1) = (x3-x.>2+2+ (y2y3)(y3"y1)= '"(x2x3)(x3"x1) (y)() x3--xl x3x2 ==> m1m2=~1orm1=mL 2 if two non~vertical straightlineswith slopesin! andm2H areperpendicular, then m}m2=l. On the other hand, if mlm2= l , then the two straight lines are perpendicular. CoordinateGeometijl Thestraightlinesix-axisandy-axisareperpendicular to eachother.But,thecondition mlmz == -91isnottruebecause theslopeof thex-axis iszeroandtheslope ofthey-axis is not dened. Example 5.13 Find the angle of inclination of the straight line whose slope is ;_ J5 £§g2§;:4zf{m If 6 is the angle of inclination of the line, then the slopeof the line is m = tan6' where 0° S 6Sl80° , 6' 7é90°. tan6=-x/1-? 2:) 61:30) Example 5.14 Find the slope of the straight line whose angle of inclination is 45°. SrizsrzwrIf 6 is the angle of inclination of the line, then the slopeof the line is m = tan6 Given that m = tan45° ==> m = 1. Example 5.15 Find the slopeof the straight line passingthrough the points (3, - 2) and (- 1,4). §£3!£§g§{?§§ Slopeof thestraightlinepassingthroughthepoints(xl , yl) and(x2, yz) is givenby m = y2my1 x2 - xl Slope of the straight line passing through the points (3 , -2) and (l m Z , 4) is 4 + 2 2 __3_ ~1~3 2' Example 5.16 Using the concept of slope, show that the points A(5, -2), B(4, l) and C(l, 2) are collinear. .§?s:£2m};:r Slope ofthe line joining the points (x1,y1) and (x2, y2) isgiven bym= : Slope ofthe line AB joining the points A(5, -2)andB(4 -~1)ism1 = F: = l Slope ofthe line BCjoining thepoints B(4,l)andC(l,2)is m2 = =1 Thus, slope of AB = slope of BC. Also, B is the common point. Hence, the points A , B and C are collinear. Oz/9Std. Mai/éidliff Example 5.17 Using the concept of slope, show that the points (-2 , -1), (4 , 0), (3 , 3) and ( 3 , 2) taken in order form a parallelogram. Siifffiii Let A( 2 , 1), B(4 , 0), C(3 , 3) and D( 3 , 2) be the given points taken in order. Nowtheslope of AB= 0+ 1 = SlopeofCD= 2"? = Slope of AB = slope of CD Hence, AB is parallel to CD. Now theslope ofBC=%:'_g = : 2+l___ SlopeofAD ___3+2 3 Slope ofBC= Hence, From and slope ofAD BC is parallel to AD. 2,we see that opposite sides of quadrilateral ABCD are parallel '. ABCD is a parallelogram. Example 5.18 The Vertices of a AABC are A(l , 2), B(4 , 5) and C(O , 1). Find the slopes of the altitudes of the triangle. §§:s§ss£§e.e Let AD, BE and CF be the altitudes of a AABC. 140,7-3 slope ofBC:1-*50+4 :_ 1 Since the altitude AD is perpendicular to BC, slopeofAD = l mlmz= -1 Fig. 5.22 1 °~ 2 = slope ofAC: 0 __1 Thus, slope ofBE = 1 BE _LAC Also, slope ofAB= __Z:% =-% slope ofCF=% CF_L AB Coordimzte Geometijl Find theangle ofinclination ofthestraightline whose slope is (1)1 (ii) /3 (iii) 0 Find the slope of the straight line whose angle of inclination is (i) 30° (ii) 60° (iii) 90° Find the slope of the straight line passing through the points (i) (3 , -2) and (7 , 2) (ii) (2 , -4) and origin (iii) (1+,/§,2) and(3 +/3,4) Find the angle of inclination of the line passing through the points (1)(1,2) and(2,3) (iii) (a , b) and (-a, (ii) (3,) and(0,0) b) Find the slope of the line which passesthrough the origin and the midpoint of the line segmentjoining the points (0 , - 4) and (8 , 0). The side AB of a squareABCD is parallel to x-axis . Find the (i) slope of AB (ii) slope of BC (iii) slope of the diagonal AC The side BC of an equilateral AABC is parallel to x-axis. Find the slope of AB and the slope of BC. Using the conceptof slope,showthat eachof the following setof points arecollinear. (i) (2 , 3), (3 , - 1) and (4 , -5) (ii) (4 , 1), (-2 , -3) and (- 5, -5) (iii) (4 ,4), (-2 , 6)aI1d(1 , 5) Ifthepoints (a,l),(1,2)and (0,b+l)are collinear, then show that %+%=l. 10. The line joining the points A(-2 , 3) and B(a , 5) is parallel to the line joining the points C(O , 5) and D(2 , 1). Find the Value of a. ll. The line joining the points A(0, 5) and B(4, 2) is perpendicular to the line joining the points C( 1, -2) and D(5, 1)).Find the value of b. l2. The Vertices of AABC are A(l, 8), B(- 2, 4), C(8, - 5). HM and Nare the midpoints of AB and AC respectively, find the slope of MN and hence Verify that MN is parallel to BC. l3. A triangle has Vertices at (6 , 7), (2 , -9) and (-4 , 1). Find the slopes of its medians. The Vertices ofa AABC are A(- 5 , 7), B(-4 , - 5) and C(4 , 5). Find the slopes of the altitudes of the triangle. 701/9Std. Mat/yemalics 15. Using the concept of slope, show that the vertices (1 , 2), (-2 , 2), (-4 , -3) and ( l, 16. 3) taken in order form a parallelogram. Show that the opposite sides of a quadrilateral with Vertices A(2 ,4), B(5 , 1), C(6 , 4) and D( l, 1) taken in order are parallel. 5.6.6 Equation of a straight line Let L be a straight line in the plane. A rst degree equation px + qy + r = 0 in the Variables x and y is satised by the x-coordinate and y-coordinate of any point on the line L and any values of x and y that satisfy this equation will be the coordinates of a point on the line L. Hence this equation is called the equation of the straight line L. We want to describe this line L algebraically. That is, we want to describe L by an algebraic equation. Now L is in any one of the following forms: (i) horizontal line (i) (ii) vertical iine Horizontal line: Then either L is x-axis Case (a) (iii) neither vertical nor horizontal Let L be a horizontal line. or L is a horizontal line other than x-axis. If L is x axis, then a point (x, y) lies on L if and only if y = 0 and x can be any real number. Thus, y = 0 describes x axis. The equation of x-axis is y = 0 Case (b) L is a horizontal line other than x-axis. That is, L is parallel to x-axis. Now, a point (x, y) lies on L if and only if the y-coordinate must remain a constant and x can be any real number. The equation of a straight line parallel to x-axis is y = k, where k is a constant. Note that if k > 0, then L lies above x-axis and if k < 0, then L lies below x- axis. If k = 0, then L is nothing but the x-axis. (ii) Vertical line: Let L be a vertical line. Then either L is y-axis or L is a Vertical line other than y-axis. Coordimzte Geometijl Case (a) If L is y-axis, then a point (x, y) in the plane lies on L if and only if x = 0 and y can be any real number. Thus x = 0 describes y axis. The equation ofy-axis is x = 0 Case (b) If L is a Vertical line other than y-axis, then it is parallel to y-axis. Now a point (x, y) lies on L if and only if x-coordinate must remain constant and y can be any real number. The equation of a straight line parallel to y-axis is x = c, where Note that C is a constant. if c > 0, then L lies to the right y-axis and if c < 0, then L lies to the left of y-axis. If c = 0, then L is nothing but the y-axis. (iii) Neither vertical nor horizontal: Let L be neither Vertical nor horizontal. In this casehow do we describeL by an equation?Let (9denotethe angleof inclination. Observethat if we know this :9 and a point on L, then we can easily describeL. LSlope mofatnon-yertical lineLcan becalculated using .' L (i) i M m : tan9 if weknowtheangleof inclination6. d yz __yl (mm: x2-Lxl , . . L . . . ifwe xI . knowtwo , distinct , , points , _ V V K1 xK2,y2 , onL. (iii) m:=0 if andonlyifL is horizontal. Uis Now considerthe casewhere L is not a Vertical line and derive the equationof a straight line in the following forms: (a) SlopePoint form 03) Two-Points form (C)Slopeintercept form (d)intercepts form (a) Siope-=-Point form Letinbethe slope ofLand Q(x1 ,yl)beapoint onL. L 1) J) LetP(x, y) beanarbitrary pointonL other thanQ.Then,wehave VT) _. O F 526 m=x-x:<:>m(x"'x1)"y"'y1 *9 A Thus,theequationof a straightline with slopem andpassingthrough (xl , yl) is y ~y1= m(x~x1)forallpoints(x , )2) on L. 701/7Std. Mat/yeirzalics (b) Two-Points form y Suppose thattwo distinctpoints (xl,yl),(x2 ,yz)are given ona yl) non-Vertical lineL. J) To nd the equation of L, we nd the slope ofL rstand then usei 2. The slope of L is yz -- yl m= x2 ~-xl L B(x2,y2) 0 , wherex275x1asL is non-vertical. Now, the formula (1) gives y-~y1 yzyi (C) xx1 xznxi x~x1=y--yl xzgxi for all points(x , y) onL y2-yl Slope-Intercept form Suppose that m is the slope of L and c is the y-intercept of L. Since0 is they-intercept,thepoint (0 , C) lies on L. Now using Lf:with (xi , y1)=(0 , c) we obtain,y c = m(x => y = mx + c _ O) for all points(x , y) onL. Fig. 5.27 Thus, y = mx + c is the equation of straight line in the Sl0peInter(:epi; form. CoordinateGeomefijl (d) intercepts form Suppose that the straight line L makes non-zero intercepts a and b on the x-axis and on the y-axis respectively. '. The straight line cuts the x-axis at A(a, O) and the y-axis at B(0, 1)) The slope ofAB ism=-S-. Now gives, y 0 =--%(x- a) ==> ay = -bx + ab bx + ay = ab Divide byabtoget-3+-2:=1 Equation of a straight line having xintercept a and yintercept b is %+%=1 forallpoints (x,y) onL (i) If the lineLwithtslopem, rnakes,;c;intercept pdythentheequation ofthe is y m. The pstraight liney=ly passeps,,through both xjand y-intercepts are zeroform750).I T (iii)Equatiyonsi (1), (2)and (4)can besimplied tosiope-intemep: romgivenby (3);T (iv)Each equation (), (2),(3 and (4)rcaI.1bye lrevvritten, inthe; form t Tispx+,qy+ r :40,forallcpoints (xV,32) onL,which iscalled thege::§ra1 formpf equation ofastraight line. , Example 5.19 Find the equations of the straight lines parallel to the coordinate axes and passing through the point (3, - 4). ,§};fm§¢mLet L and L be the straightlines passingthroughthe point (3, - 4) andparallelto x-axis andy-axis respectively. The y-coordinate of every point on the line L is 4. Hence, the equation of the line L is y = - 4 Similarly, the x-coordinate of every point on the straight line L is 3 Hence,the equationof the line L is x = 3. 701/7Std. Mat/mizalicx L Example 5.2% Find the equation of straight line whose angle of inclination is 45° and y-intercept is tlziswiw Slopeof the line, m = tan6' = tan 45° = 1 y-intercept is c= %~ By the slope-intercept form, the equation of the straight line is y = mx + c y_x+g ___> y_ 5x5+ 2 The equation of the straight line is 5x - 5y + 2 = 0 Example 5.21 Find the equation ofthe straight line passing through the point ( 2,3)withslope ,3?:),¬s:fims Given that theslope m=%and apoint (xl,yl)=(--2,3) By slope-point formula, the equation of the straight line is y"y1= m(x"x1) ==¢y- 3= Thus, x +2) 332+ 1l= O is the required equation. Example 5.22 Find the equationof the straight line passingthrough the points ( 1,1) and (2, 4). Saixgfimz LetA(x1, yl) andB(x2, yz)bethegivenpoints. Herex1-- 1, y1= 1 and x2 = 2, yz =-~4. Using two-points formula, the equation of the straight line is y yl : x - xl y2 T y1 i )" 1 --4 - 1 x2 _ xl Z x+ 1 2 + 1 _..> 3y-3=-5x-5 Hence, 5x + 3y + 2 = 0 is the required equation of the straight line. Example 5.23 The Vertices ofa AABC are A(2, 1), B(2, 3) and C(4, 5). Find the equation of the median through the Vertex A. Coordimzte Geometijl ,§§2£:s2'i:m Median is a straight line joining a vertex and the midpoint of the opposite side. LetD bethemidpoint ofBC. /{(2:1) Midpointof BCisD(2 +4, 3+5) =D(1, 4) 2 2 Now the equation of the median AD is ij _. 1 __ = f __2? =(2,1)and =(1,4) 3" 1 Z x ~ 2 3 --1 ' 3x+y-~7=0 istherequired equation. .. .. ._ B(23) ID Fig. 5.30 C(4, 5) Example 5.24 Ifthex-intercept and y-intercept ofastraight lineare~§and %respectively, then nd the equation of the straight line. $£)§£§§i<)§? Given that x-intercept ofthestraight line,a=% :1 and the y-intercept of the straight line, b 4 Using intercept form, the equation of the straight line is x+)_1___>x+)_1 a b 2 1 3 4 ==>»..3_& 2 + 2!. 3 Hence, : 1 9x + 8y 6 = 0 is the required equation. Example 5.25 Find the equations of the straight lines each passing through the point (6, -2) and whose sum of the intercepts is 5. S};izs§§s:§2 Let a and b be the x-intercept and y-intercept of the required straight line respectively. Given that sum of the intercepts, a + b = 5 ==> b = 5 - a Now, the equation of the straight line in the intercept form is x a+by_ 1"» 2:) Thus, x 3 a+5-~a 1 (5~-a)x+ay: a(5-a) (5-a)x+ay=a(5-a) Since the straight line given by 701/9Std. Mat/mmlics _ passesthrough (6, 2), we get, (5-a)6+a(==> That is, 2) =a(5-a) a2-l3a+30=O. (a-3)(a- 10) =0 a=3ora= 10 Whena=3, :{:==>(5-3)x+3y=3(5-3) ==>2x+3y=6 Whena= 10, ==>(5--l0)x+l0y => That is, =l0(5-10) -5x+lOy=50 x-2y- 10 = 0. Hence, 2x + 3y = 6 and x - 2y - 10 = 0 are the equations of required straight lines. 1. Write the equations of the straight lines parallel to x- axis which are at a distance of 5 units from the x-axis. 2. Find the equations of the straight lines parallel to the coordinate axes and passing through the point (-5,2). 3. 4. Find the equation of a straight line whose (i) slope is -3 and y-intercept is 4. (ii) angle of inclination is 60° andy-intercept is 3. Find the equation of the line intersecting the y- axis at a distance of 3 units above the origin and tan6 =%,where 6istheangle ofinclination. 5. Find the slope and y-intercept of the line whose equation is (i)y=x+l 6. (ii)5x=3y (iii)4x-2y+l=O (iv) lOx+15y+6=0 Find the equation of the straight line whose (i) slope is -4 and passing through (1, 2) .2. (ii) 7. slope is 3and passing through (5, -4) Find the equation of the straight line which passes through the midpoint of the line segmentjoining (4, 2) and (3, 1) whoseangleof inclination is 30°. 8. Find the equation of the straight line passing through the points (i) (-2, 5) and (3, 6) 9. (ii) (0, -6) and (-8, 2) Find the equation of the median from the Vertex R in a APQR with Vertices at P(l, -3), Q(2, 5) and R(3, 4). 10. By using the concept of the equation of the straight line, prove that the given three points are collinear. (1) (4,2), (7, 5) and (9, 7) ll. Find the equation of the straight line whose x and y-intercepts on the axes are given by (1) 2and 3 12. (11) (1, 4), (3, -2) and (-3, 16) .. 1 3 (11) -3- and-2~ (111) %and -14 Find the x and y intercepts of the straight line (i)5x+3y-15:0 (ii)2x-y+l6:=0(iii)3x+10y+4=0 13. Find the equation of the straight line passing through the point (3, 4) and has intercepts which are in the ratio 3 : 2. l4. Find the equation of the straight lines passing through the point (2, 2) and the sum of the intercepts is 9. 15. Find the equation of the straight line passing through the point (5, -3) and whose intercepts on the axes are equal in magnitude but opposite in sign. 16. Find the equation of the line passing through the point (9, -l) and having its x-intercept thrice as its y-intercept. 17. A straight line cuts the coordinate axes atA and B. If the midpoint of AB is (3, 2), then nd the equation of AB. 18. Find the equation of the line passing through (22, -6) and having intercept on x-axis exceeds the intercept on y-axis by 5. 19. If A(3, 6) and C(l, 2) are two vertices of a rhombus ABCD, then nd the equation of straight line that lies along the diagonal BD. 20. Findtheequation ofthelinewhose gradient is12 andwhichpasses through P,where P divides the line segmentjoining A(2, 6) and B (3, -4) in the ratio 2 : 3 internally. 5.7 General Form of Equation of a straight line We have already pointed out that different forms of the equation of a straight line may be converted into the standard form ax + by + C = 0, where a , b and C are real constants such that either a ¢ 0 or b ¢ 0 . Now let us nd out (i) the slope of ax + by + C = 0 (ii) the equation of a straight line parallel to ax + by + C = O the equation of a straight line perpendicular to ax + by + C = O and the point of intersection of two intersecting straight lines. oZif§7fiZ;;;32Z;§Z} WWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW W (i) The general form of the equation of a straight line is ax + by + c = 0. Theaboveequationis rewrittenas y = -- ix b Comparing -~ 9, b 75O b 7;with the slope-intercept form y = mx + k, we get, slope,m = Q andthey-intercept= 9 b b For the equation ax + by + c = 0, we have coefficient ofx slope in = (ii) constant term ofy' Coefficient ofy andthey-intercept is coefficient Equation of a line parallel to the line ax + by + c = 0. We know that two straight lines are parallel if and only if their slopes are equal. Hence the equations of all lines parallel to the line ax + by + c = 0 are of the form ax + by + k = 0, for different values of k. (iii) Equation of a line perpendicular to the line ax + by + c =-.G We know that two non-Vertical lines are perpendicular if and only if the product of their slopes is -1. Hence the equations of all lines perpendicular to the line ax + by + c = 0 are bx - ay + k = 0, for different values of k. y,g;; ,Two[strapight:1yinesi.a1xp.+b:,yi+ i_-o,.and lazx + b2ye+l"¢,;.+_-g 0,where the; and T T _(mice are perpenaicuiarir andoayrr .ia1.ta2j+b;b; 0 (iv) The point of intersection of two straight lines If two straight lines are not parallel, then they will intersect at a point. This point lies on both the straight lines. Hence, the point of intersection is obtained by solving the given two equations. Example 5.26 Show that the straight lines 3x + 2y - 12 = 0 and 6x + 4y + 8 = 0 are parallel. Sasissriarzz Slope ofthe straight line3x+2y 12=0ismr gi Similarly, theslope ofthe line 6x+4y+8=0is m2 =---261=--3 m1= mg. Hence, the two straight lines are parallel. Coordimzte Geometijl 3 Example 5.27 Prove that the straight lines x + 2y + 1 = 0 and 2x - y + 5 = 0 are perpendicular to each other. .;§£3!z:§§5)§3 Slope ofthe straight linex+2y+1=0isml Slope ofthe straight line 2x- y+5=0 ism2 Product oftheslopes $3: : :3 2 mlmz =-~£>< 2= 1 The two straight lines are perpendicular. Example 5.28 Find the equation of the straight line parallel to the line x - 8y + 13 = 0 and passing through the point (2, 5). §§g.»Ez:s?izm Equation of the straight line parallel to x - 8y + 13 = 0 is x - 8y + k = 0 Since it passesthrough the point (2, 5) 2--8(5)+k = 0 ==> k = 38 Equation of the required straight line is x ~-8y + 38 = 0 Example 5.29 The Vertices of AABC are A(2, 1), B(6, -1) and C(4, 11). Find the equation of the straight line along the altitude from the Vertex A. Széitsifem Slope ofBC=141j_'61 =6 Since the line ADisperpendicular totheline BC, slope ofAD=% A(2,l) Equation ofAD is y - yl = m(x -~x1) 3 1 ax 2) M 6} 6 "X 2 B(6,-1) Equation of the required straight line is x - 6y + 4 = 0 C(4,11) Fig 5-31 1. Find theslope ofthestraight line (i)3x+4y--6:0 (ii)y=7x+6 (iii)4x=5y+3. 2. Show that the straight lines x + 2y + 1 = 0 and 3x + 6y + 2 = 0 are parallel. 3. Show that the straight lines 3x ~-5y + 7 = 0 and 15x + 9y + 4 ==0 are perpendicular. 4. Ifthestraight lines %- xp andax +5= 3yare parallel, then nda. 5. Find the Value of a if the straight lines 5x - 2y - 9 = 0 and ay + 2x - 11 = 0 are perpendicular to each other. Oz/9Std. Mai/éidliff Find the values of p for which the straight lines 8px+(2 - 3p)y+ 1 = 0 and px + 8y 7 = 0 are perpendicular to each other. If the straight line passing through the points (h, 3) and (4, 1) intersectsthe line 7x - 9y - 19 = 0 at right angle, then nd the Value of h. Find the equation of the straight line parallel to the line 3x ~ y + 7 = O and passing through the point (1, 2). Find the equation of the straight line perpendicular to the straight line x - 2y + 3 = 0 and passing through the point (1, 2). 10. Find the equation of the perpendicular bisector of the straight line segmentjoining the points (3, 4) and ( 1, 2). 11. Find the equation of the straight line passing through the point of intersection of the lines 2x + y -~ 3 = O and 5x + y -~ 6 = 0 and parallel to the line joining the points (1, 2) and (2, 1). 12. Find the equation of the straight line which passesthrough the point of intersection of the straight lines 5x - 6y ==1 and 3x + 2y + 5 ==0 and is perpendicular to the straight line 3x 5y+11= 13. 0. Find the equation of the straight line joining the point of intersection of the lines 3x - y + 9 = 0 and x + 2y = 4 and the pointofintersection ofthe lines 2x + y - 4 = 0 andx-2y+3== 14. 0. lfthe Vertices ofa AABC are A(2, 4), B(3, 3) and C( 1, 5). Find the equation of the straight line along the altitude from the Vertex B. 15. If the Vertices ofa AABC are A( 4,4 ), B(8 ,4) and C(8,l0). Find the equation of the straight line along the median from the Vertex A. 16. Find the coordinates of the foot of the perpendicular from the origin on the straight line 3x + 2y = 13. 17. If x + 2y = 7 and 2x + y = 8 are the equations of the lines of two diameters of a circle, nd the radius of the circle if the point (0, 18. 2) lies on the circle. Find the equation of the straight line segment whose end points are the point of intersection of the straight lines 2x - 3y + 4 : 0, x ~ 2y + 3 = 0 and the midpoint of the line joining the points (3, 2) and ( 5, 8). 19. In an isosceles APQR, PQ = PR. The base QR lies on the x-axis, P lies on the y- axis and 2x - 3y + 9 = 0 is the equationof PQ. Find the equationof the straightline alongPR. """""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""" WE?;?§};2?;""5§Z;2Zf§ Exercise5.6 Choose 1. the correct answer The midpoint of the line joining (a, - b) and (3a, 5b) is (A) (- a, 2b) (B) (2a, 4b) (C) (261,2b) (D) (- 61,~ 319) The point P which divides the line segmentjoining the points A(1, - 3) and B( 3, 9) internally in the ratio 1:3 is (A)(2.1) (B)(0,0) (C)(3-,2) (B)(1,~2) If the line segmentjoining the points A(3, 4) and B(14, 3) meetsthe x-axis at P, then the ratio in which P divides the segment AB is (A)4:3 (B)3:4 (C)2:3 (D)4:1 The centroid ofthe triangle with Verticesat (- 2, - 5), (- 2,12) and (10, ~~1) is (A) (6, 6) (B) (4, 4) (C) (3,3) (D) (2, 2) If (1, 2), (4, 6), (x, 6) and (3, 2) arethe Verticesof a parallelogramtakenin order,then the Value of x is (A) 6 (B) 2 (C) 1 (D) 3 Area of the triangle formed by the points (0,0), (2,0) and (0,2) is (A) 1 sq. units (B) 2 sq. units (C) 4 sq. units (D) 8 sq. units Area of the quadrilateralformed by the points (1,1), (0,1), (0,0) and (1,0) is (A) 3 sq. units (B) 2 sq. units (C) 4 sq. units (D) 1 sq. units The angle of inclination of a straight line parallel to x-axis is equal to (A) 0° (B) 60° (C) 45° (D) 90° Slope oftheline joining thepoints (3,- 2)and (-1,a)is-3-,thentheValue ofa is equal to (A) 1 10. (B) 2 (C) 3 (D) 4 Slope of the straight line which is perpendicular to the straight line joining the points (- 2, 6) and (4, 8) is equalto (A)§ 11. (A) (- 1,7) 12. (B) (7,1) (B)§ y 2 = 0 and 2x + y - 9 = O is (C) (1,7) (D) (-* 1,"' 7) (B) (0,4) (C) (3,4) (D) (0, -* 4) The slope of the straight line 7y ~ 2x = 11 is equal to -1 (A) 2 14. (c)-3 The straight line 4x + 3y - 12 = 0 intersects the y axis at (A) (3,0) 13. (B)3 The point of intersection of the straight lines 9x 1 (B) 2 2 (c) 7 ,2 (D) 7 The equation of a straight line passingthrough the point (2 , -7) and parallel to x-axis is (A)x=2 (B)x=7 (C)y=~7 (D)y=2 7'"oi"2},_E"2Zz.i"E("4";",;,"",?Z,"£-Z} WWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW W 15. The x and y-intercepts of the line 2x ~ 3y + 6 = 0, respectively are 16. The centre of a circle is ( 6, 4). If one end of the diameter of the circle is at ( l2, 8), then the other end is at (A) (-18) 12) 17. (B) ( 9, 6) (C) (- 3) 2) (D) (0, 0) The equation of the straight line passing through the origin and perpendicular to the straight line 2x + 3y -- 7 = 0 is (A)2x+3y=0 18. (B)x+2=O (C)y+5=0 (B) 4 (C) (B) 4 (D) 8 (D) - 3 The equation of a straight line having slope 3 and y-intercept 4 is (B)3x+y-4:0 (C)3x~y+4=0 (D)3x+y+4=O The point of intersection of the straight lines y = 0 and x = 4 is (A) (0) "' 4) 23. - 4 (C) 5 (A)3x-y-4=O 22. (D)y-5=0 If a straight line y = 2x + k passesthrough the point (1, 2), then the value of k is equal to (A) 0 21. (D)y-5:0 If the points(2, 5), (4, 6) and (a, a) arecollinear,thenthe valueof a is equalto (A) - 8 20. (C)y+5=O The equationof a straightline parallelto y-axis andpassingthroughthe point (~-2, 5) is (A)x2=0 19. (B)3x-2y=0 (B) (- 4: 0) (C) (0, 4) (D) (4, 0) The value of k if the straight lines 3x + 6y + 7 = Oand 2x + ky = 5 are perpendicular is (A)1 (B)A (C)2 (D)§ P,o,i:1tsnto. Remember 7 III The distance between P(x1,y1) and Q(x2,is ,/(x2 - x1)2 +(yz-y1)2 CI Thepoint Pwhich divides thelinesegment joining thepoints A(x1, yl)andB(x2, A inthe +mxllyz +myl ) internally ratio :mis Cextrenally inthe ratio lzm is< ,plyz y1+y2) CIA Midpoint ofthe line segment joining the points (x1, y1) and (x2, y2) is