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Pure Mathematics 1: Cambridge International As & a Level
Pure Mathematics 1: Cambridge International As & a Level
Sophie Goldie
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This brand new series has been written for the University of Cambridge International Examinations course for AS and A Level Mathematics (9709). This title covers the requirements of P1. The authors are experienced examiners and teachers who have written extensively at this level, so have ensured all mathematical concepts are explained using language and terminology that is appropriate for students across the world. Students are provded with clear and detailed worked examples and questions from Cambridge International past papers, so they have the opportunity for plenty of essential exam practice. Each book contains a free CDROM which features the unique 'Personal Tutor' and 'Test Yourself' digital resources that will help students revise and reinforce concepts away from the classroom: With Personal Tutor each student has access to audiovisual, stepbystep support through examstyle questions The Test Yourself interactive multiple choice questions identify weaknesses and point students in the right direction
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Year:
2012
Edition:
Pap/Cdr
Publisher:
Hodder Education
Language:
english
Pages:
312 / 322
ISBN 10:
1444146440
ISBN 13:
9781444146448
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PDF, 124.33 MB
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Bhagya Ranasinghe
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Pure Mathematics 1 Sophie Goldie Series Editor: Roger Porkess Copy  CD not yet inserted Please contact ITA office .i f you have any questions . Tel : 00861058732015 Mobile: 008613910675895 Email :zhangpei .beijing@gmail.com 5R% Zhang Pei Cambridge International A and AS Level Mathematics Pure Mathematics 1 Sophie Goldie Series Editor: Roger Porkess i7 EDUCATION HODDER AN HACHETTE UK COMPANY Questions from the Cambridge International Examinations A & AS level Mathematics papers are reproduced by permission of University of Cambridge International Examinations. Questions from the MEI A & AS level Mathematics papers are reproduced by permission of OCR. We are grateful to the following companies, institutions and individuals you have given permission to reproduce photographs in this book. page 106, ©Jack Sullivan/ Alamy; page 167, © RTimages/Fotolia; page 254, © Hunta/Fotolia; page 258, © Olga Iermolaieva/ Fotolia Every effort has been made to trace and acknowledge ownership of copyright. The publishers will be glad to make suitable arrangements with any copyright holders whom it has not been possible to contact. Hachette UK's policy is to use papers that are natural, renewable and recyclable products and made from wood grown in sustainable forests. The logging and manufacturing processes are expected to conform to the environmental regulations of the country of origin. Orders: please contact Bookpoint Ltd, 130 Milton Park, Abingdon, Oxon OX14 4SB. Telephone: (44) 01235 827720. Fax: (44) 01235 400454. Lines are open 9.005.00, Monday to Saturday, with a 24hour message answering service. Visit our website at www.hoddereducation.co.uk Much of the material in this book was published originally as part of the MEI Structured Mathematics series. It has been carefully adapted for the Cambridge International A & AS level Mathematics syllabus. The original ME! author team for Pure Mathematics comprised Catherine Berry, Bob Francis, Val Hanrahan, Terry Heard, David Martin, Jean Matthews, Berna; rd Murphy, Roger Porkess and Peter Seeker. ©ME!, 2012 First published in 2012 by Hodder Education, a Hachette UK company, 338 Euston Road London NW! 3BH Impression number Year 54 32 I 2016 2015 2014 2013 2012 All rights reserved. Apart from any use permitted under UK copyright law, no part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or held within any information storage and retrieval system, without permission in writing from the publisher or under licence from the Copyright Licensing Agency Limited. Further details of such licences (for reprographic reproduction) may be obtained from the Copyright Licensing Agency Limited, Saffron House, 610 Kirby Street, London ECIN 8TS. Cover photo by © Joy Fera/Fotolia Illustrations by Pantek Media, Maidstone, Kent Typeset in IO.Spt Minion by Pantek Media, Maidstone, Kent Printed in Dubai A catalogue record for this title is available from the British Library ISBN 978 1444 14644 8 Contents Chapter 1 Key to symbols in this book vi Introduction vii The Cambridge A & AS Level Mathematics 9709 syllabus viii Algebra Inequalities 1 6 10 12 17 20 22 25 29 34 Coordinate geometry 38 Coordinates The intersection of a line and a curve 38 39 39 41 42 46 49 56 63 70 Sequences and series 75 Definitions and notation 76 77 84 95 Background algebra Linear equations Changing the subject of a formula Quadratic equations Solving quadratic equations Equations that cannot be factorised The graphs of quadratic functions The quadratic formula Simultaneous equations Chapter 2 Plotting, sketching and drawing The gradient of a line The distance between two points The midpoint of a line joining two points The equation of a straight line Finding the equation of a line The intersection of two lines Drawing curves Chapter 3 1 Arithmetic progressions Geometric progressions Binomial expansions Chapter 4 Chapter 5 Chapter 6 Chapter 7 Functions 106 The language of functions 106 Composite functions 112 Inverse functions 115 Differentiation 123 The gradient of a curve 123 Finding the gradient of a curve 124 Finding the gradient from first principles 126 Differentiating by using standard results 131 Using differentiation 134 Tangents and normals 140 Maximum and minimum points 146 Increasing and decreasing functions 150 Points of inflection 153 The second derivative 154 Applications 160 The chain rule 167 Integration 173 Reversing differentiation 173 Finding the area under a curve 179 Area as the limit of a sum 182 Areas below the x axis 193 The area between two curves 197 The area between a curve and the y axis 202 The reverse chain rule 203 Improper integrals 206 Finding volumes by integration 208 Trigonometry 216 Trigonometry background 216 Trigonometrical functions 217 Trigonometrical functions for angles of any size 222 The sine and cosine graphs 226 The tangent graph 228 Solving equations using graphs of trigonometrical functions 229 Circular measure 235 The length of an arc of a circle 239 The area of a sector of a circle 239 Other trigonometrical functions 244 Chapter 8 The angle between two vectors 254 254 258 262 271 Answers 280 Index 310 Vectors Vectors in two dimensions Vectors in three dimensions Vector calculations Key to symbols in this book 0 This symbol means that you want to discuss a point with your teacher. If you are working on your own there are answers in the back of the book. It is important, however, that you have a go at answering the questions before looking up the answers if you are to understand the mathematics fully. This symbol invites you to join in a discussion about proof. The answers to these questions are given in the back of the book. This is a warning sign. It is used where a common mistake, misunderstanding or tricky point is being described. This is the ICT icon. It indicates where you could use a graphic calculator or a computer. Graphical calculators and computers are not permitted in any of the examinations for the Cambridge International A & AS Level Mathematics 9709 syllabus, however, so these activities are optional. This symbol and a dotted line down the righthand side of the page indicates material that you are likely to have met before. You need to be familiar with the material before you move on to develop it further. e This symbol and a dotted line down the righthand side of the page indicates material which is beyond the syllabus for the unit but which is included for completeness. Introduction This is the first of a series of books for the University of Cambridge International Examinations syllabus for Cambridge International A & AS Level Mathematics 9709. The eight chapters of this book cover the pure mathematics in AS level. The series also contains a more advanced book for pure mathematics and one each for mechanics and statistics. These books are based on the highly successful series for the Mathematics in Education and Industry (MEI) syllabus in the UK but they have been redesigned for Cambridge users; where appropriate new material has been written and the exercises contain many past Cambridge examination questions. An overview of the units making up the Cambridge International A & AS Level Mathematics 9709 syllabus is given in the diagram on the next page. Throughout the series the emphasis is on understanding the mathematics as well as routine calculations. The various exercises provide plenty of scope for practising basic techniques; they also contain many typical examination questions. An important feature of this series is the electronic support. There is an accompanying disc containing two types of Personal Tutor presentation: examinationstyle questions, in which the solutions are written out, step by step, with an accompanying verbal explanation, and test yourself questions; these are multiplechoice with explanations of the mistakes that lead to the wrong answers as well as full solutions for the correct ones. In addition, extensive online support is available via the MEI website, www.mei.org.uk. The books are written on the assumption that students have covered and understood the work in the Cambridge IGCSE syllabus. However, some of the early material is designed to provide an overlap and this is designated 'Background'. There are also places where the books show how the ideas can be taken further or where fundamental underpinning work is explored and such work is marked as 'Extension'. The original MEI author team would like to thank Sophie Goldie who has carried out the extensive task of presenting their work in a suitable form for Cambridge International students and for her many original contributions. They would also like to thank Cambridge International Examinations for their detailed advice in preparing the books and for permission to use many past examination questions. Roger Porkess Series Editor The Cambridge A & AS Level Mathematics syllabus Cambridge IGCSE Mathematics ALevel Mathematic• Algebra P1 Sherlock Holmes: 'Now the skillful workman is very careful indeed ... He will have nothing but the tools which may help him in doing his work, but of these he has a large assortment, and all in the most perfect order.' A. Conan Doy/e Cl) Background algebra Manipulating algebraic expressions You will often wish to tidy up an expression, or to rearrange it so that it is easier to read its meaning. The following examples show you how to do this. You should practise the techniques for yourself on the questions in Exercise lA. Collecting terms Very often you just need to collect like terms together, in this example those in x, those in y and those in z. 0 EXAMPLE 1. 1 What are 'like' and 'unlike' terms? Simplify the expression 2x+ 4y Sz Sx 9y+ 2z+ 4x 7y+ 8z. SOLUTION Expression= 2x+ 4x Sx+ 4y 9y7y+ 2z+ 8z Sz = 6x5x+4y16y+ =X 12y+ Sz.{ This cannot be simplified further and so it is the answer. Removing brackets Sometimes you need to remove brackets before collecting like terms together. EXAMPLE 1.2 Simplify the expression 3(2x 4y)  4(x Sy). SOLUTION Expression = 6x 12y 4x + 20y EXAMPLE 1.3 Notice (4) x (5y) = +20y Simplify x(x+ 2) (x 4). SOLUTION Expression = x2 + 2x x + 4 ..._1 '../''"..J' '""..../ EXAMPLE 1.4 Simplify a(b + c) ac. SOLUTION Expression = ab+ ac ac .._1 '../'J'..._,.'._./'...../ Factorisation It is often possible to rewrite an expression as the product of two or more numbers or expressions, its factors. This usually involves using brackets and is called factorisation. Factorisation may make an expression easier to use and neater to write, or it may help you to interpret its meaning. EXAMPLE 1.5 Factorise 12x 18y. SOLUTION Expression EXAMPLE 1.6 = 6(2x 3y) Factorise x2  2xy+ 3xz. SOLUTION Expression = x(x 2y+ 3z) Multiplication Several of the previous examples have involved multiplication of variables: cases like and a x b= ab xx x= x 2. In the next example the principles are the same but the expressions are not quite so simple. EXAMPLE 1.7 SOLUTION Expression = 3 x 4 x 5 x p 2 x p x q x q 3 x q x r x r2 = 60 X p3 X q5 X r 3 = 60p3qsr3 Fractions The rules for working with fractions in algebra are exactly the same as those used in arithmetic. EXAMPLE 1.8 SOLUTION As in arithmetic you start by finding the common denominator. For 2, 10 and 4 this is 20. Then you write each part as the equivalent fraction with 20 as its denominator, as follows . . _ lOx 4y Sz Expression  2o + 20 20 _ lOx 4y + Sz 20 EXAMPLE 1.9 x2 y2 Simplify   . y X SOLUTION . x3 y3 ExpressiOn =    xy xy x3 _ y3 = xy be left out. j 2 EXAMPLE 1.10 5yz . l"fy 3x SImpi  X  . 5y 6x SOLUTION Since the two parts of the expression are multiplied, terms may be cancelled top and bottom as in arithmetic. In this case 3, 5, x and y may all be cancelled. Expression = 'Jxr x $yz $y (jf2;t xz 2 EXAMPLE 1 .11 . l"fy (x 1? SImp 1 4.x(x _ ). 1 SOLUTION (x 1) is a common factor of both top and bottom, so may be cancelled. However, xis not a factor of the top (the numerator), so may not be cancelled. (X 1)2 Expression=    4x EXAMPLE 1. 12 . l"fy 24.x + 6 SImp I 3(4.x + 1)" SOLUTION When the numerator (top) and/or the denominator (bottom) are not factorised, first factorise them as much as possible. Then you can see whether there are any common factors which can be cancelled. 6(4.x + 1) . ExpressiOn= 3(4.x + 1) =2 EXERCISE 1A 1 Simplify the following expressions by collecting like terms. m (ii) 8x+ 3x+ 4x 6x 3p+3+5p77p9 (iii) 2k + 3m + 8n 3k 6m 5n + 2k m+ n (iv) 2a+3b4c+4a5b8c6a+2b+ 12c (v) r 2s t+ 2r 5t 6r 7t s +Ss 2t+ 4r 2 3 Factorise the following expressions. lil 4x+ By liil (iii) 72f 36g 48h (iv) (vi 12P + 144km 72kn 12a + 15b 18c p 2  pq + pr m )C Simplify the following expressions, factorising the answers where possible. 8(3x + 2y) + 4(x + 3y) liil 2(3a 4b +Se)  3(2a Sb c) (iii) 6(2p 3q + 4r) 5(2p 6q 3r) 3(p 4q + 2r) lvl w+ h)+ 3(21 w 2h) + Sw 5u6(wv)+2(3u+4wv)llu 4 Simplify the following expressions, factorising the answers where possible. 5 lil a(b+ c)+ a(b c) (iil k(m+n)m(k+n) liiil p(2q+ r+ 3s) pr s(3p+ q) (ivl x(x 2) x(x 6) + 8 (vi x(x 1) + 2(xl) x(x+ 1) Perform the following multiplications, simplifying your answers. lil 2xy x 3x2y liiil km x mn x nk (vi rs X 2stx 3tu X 4ur 5a 2bc3 x 2ab 2 x 3c livl 3pq 2rx 6p 2qrx 9pqr2 (ii) 6 Simplify the following fractions as much as possible. lil ab liil ac (ivl 2e 4f 2 (iii) £ 5x (iii) E._ X 1_ 4a2b 2ab 7 Simplify the following as much as possible. (i) (iv) b X QX f._ (ii) a c 3x 2y 8y X X 3z 5z 4x 2 2 q p 2fg X 4gh2 X 32fh3 4fh 12j3 16h s Write the following as single fractions. lil 2 + 3 (iv) 2x  3 (ii) 2x _ 5 (v) 2:::  4 2 3 + 3x 4 (iiil 3z 8 + 2z _ 5z 12 5y + 4y 8 5 9 Write the following as single fractions. (i) l + .2. (ii) l +l (iv) £. + !1. q p (v) l _l + l X X ri iii" CD m (ivl 4(1+ CD X a (iii) y b c 1+ X y 24 10 Write the following as single fractions. (i) X + 1+ X 4  1 2) (.IV1 3(2.x5+ 1)  7(x2 11 (ii) 2 (v) 2.x _ 3 X  1 5 ... 3x  5 + x  7 (Ill 1  4  6 4x + 1 + 7x  3 8 12 Simplify the following expressions. ( ') I +3 2.x + 6 X 6x12 X 2 (ivl   (ii) 6(2.x + 1)2 3(2.x + 1)5 (3x+2? (vl 6x (iii) 2x( y 3) 4 8x 2(y 3) x__L 6x + 4 Cl) Linear equations f) What is a variable? You will often need to find the value of the variable in an expression in a particular case, as in the following example. EXAMPLE 1.13 A polygon is a closed figure whose sides are straight lines. Figure 1.1 shows a sevensided polygon (a heptagon). Figure 1.1 An expression for So, the sum of the angles of a polygon with n sides, is S= 180(n2). f) How is this expression obtained? Try dividing a polygon into triangles, starting from one vertex. Find the number of sides in a polygon with an angle sum of m180° liil 1080°. SOLUTION m Substituting 180 forS gives Dividing both sides by 180 Adding 2 to both sides 180 = 180(n 2) 1 = n 2 3= n The polygon has three sides: it is a triangle. liil Substituting 1080 forS gives Dividing both sides by 180 Adding 2 to both sides 1080 = 180(n 2) 6 = n 2 8= n The polygon has eight sides: it is an octagon. Example 1.13 illustrates the process of solving an equation. An equation is formed when an expression, in this case 180(n 2), is set equal to a value, in this case 180 or 1080, or to another expression. Solving means finding the value( s) of the variable( s) in the equation. Since both sides of an equation are equal, you may do what you wish to an equation provided that you do exactly the same thing to both sides. If there is only one variable involved (liken in the above examples), you aim to get that on one side of the equation, and everything else on the other. The two examples which follow illustrate this. In both of these examples the working is given in full, step by step. In practice you would expect to omit some of these lines by tidying up as you went along. 0 A Look at the statement 5(xl) = 5x 5. What happens when you try to solve it as an equation? This is an identity and not an equation. It is true for all values of x. For example, try x= 11: 5(xl) = 5 x (111) =50; 5x 5 =55 5 =50./, or try x= 46: 5(x1) = 5 x (46 1) = 225; 5x 5 = 2305 = 225 ./, or try x = anything else and it will still be true. To distinguish an identity from an equation, the symbol= is sometimes used. Thus 5(x 1) = 5x 5. I"" :;· .. CD Ill CD ..o· .Cl c Ill = Cll P1 EXAMPLE 1.14 Solve the equation 5(x 3) = 2(x+ 6). SOLUTION 5x 15 = 2x+ 12 5x2x15 = 2x2x+ 12 3x15=12 3x15+15=12+15 3x = 27 3x 27 = 3 3 x=9 Open the brackets Subtract 2x from both sides Tidy up Add 15 to both sides Tidy up .. la .a Q) Cll ;a: Divide both sides by 3 CHECK When the answer is substituted in the original equation both sides should come out to be equal. If they are different, you have made a mistake. Lefthand side Righthand side 5(x3) 5(9 3) 2(x+ 6) 2(9 + 6) 2 X 15 30 (as required). 5X6 30 EXAMPLE 1.15 Solve the equation (x + 6) = x + (2x 5). SOLUTION Start by clearing the fractions. Since the numbers 2 and 3 appear on the bottom line, multiply through by 6 which cancels both of them. 6x Multiply both sides by 6 Tidyup Open the brackets Subtract 6x, 4x, and 18 from both sides Tidyup 6) = 6 x x+ 6 x 3(x+ 6) = 6x+ 2(2x 5) 3x+ 18 = 6x+4x10 3x6x4x = 1018 7x = 28 7x 28 7 = 7 x=4 Divide both sides by (7) CHECK Substituting x= 4 in Lefthand side 6) = x+ Righthand side + 6) 10 2 5 3 5 (as required). 5) gives: 5) EXERCISE 18 1 Solve the following equations. (i) Sa 32 = 68 (ii) 4b 6 = 3b+ 2 (iii) 2c+ 12 =Se+ 12 S(2d + 8) = 2(3d + 24) (iv) 2 (v) 3(2e1) = 6(e+ 2) + 3e (vi) 7(2 f) 3(f 4) = 10/4 (vii) Sg+ 2(g 9) = 3(2g S) + 11 (viii) 3(2h 6) 6(h+ S) = 2(4h 4) 10(h+ 4) (ix) .!_k+ .4!_ k= 36 2 (x) ius)+ (xi) i(3m+ S) + 1i(2m1) =si (xii) n+ (n+ 1) + i(n+ 2) = (ii) 3 4 Write this information in the form of an equation for a, the size in degrees of the smallest angle. Solve the equation and so find the sizes of the three angles. Miriam and Saloma are twins and their sister Rohana is 2 years older than them. The total of their ages is 32 years. m Write this information in the form of an equation for r, Rohana's age in years. (ii) What are the ages of the three girls? The length, d m, of a rectangular field is 40 m greater than the width. The perimeter of the field is 400 m. m Write this information in the form of an equation for d. liil Solve the equation and so find the area of the field. 5 Yash can buy three pencils and have 49c change, or he can buy five pencils and have 1Sc change. (i) . :oc CD n iii' ... CD Ill z= 11 The largest angle of a triangle is six times as big as the smallest. The third angle is 7S 0 • (i) m Write this information as an equation for x, the cost in cents of one pencil. liil How much money did Yash have to start with? l j P1 In a multiplechoice examination of 25 questions, four marks are given for each correct answer and two marks are deducted for each wrong answer. One mark is deducted for any question which is not attempted. A candidate attempts q questions and gets c correct. 6 7 8 (i) Write down an expression for the candidate's total mark in terms of q and c. (ii) James attempts 22 questions and scores 55 marks. Write down and solve an equation for the number of questions which James gets right. Joe buys 18 kg of potatoes. Some of these are old potatoes at 22c per kilogram, the rest are new ones at 36c per kilogram. (i) Denoting the mass of old potatoes he buys by m kg, write down an expression for the total cost ofJoe's potatoes. (iil Joe pays with a $5 note and receives 20c change. What mass of new potatoes does he buy? In 18 years' time Hussein will be five times as old as he was 2 years ago. (il Write this information in the form of an equation involving Hussein's present age, a years. (ii) How old is Hussein now? 41) Changing the subject of a formula The area of a trapezium is given by where a and b are the lengths of the parallel sides and h is the distance between them (see figure 1.2). An equation like this is often called a formula. b h a Figure 1.2 The variable A is called the subject of this formula because it only appears once on its own on the lefthand side. You often need to make one of the other variables the subject of a formula. In that case, the steps involved are just the same as those in solving an equation, as the following examples show. EXAMPLE 1.16 Make a the subject in (a+ b)h. SOLUTION It is usually easiest if you start by arranging the equation so that the variable you want to be its subject is on the lefthand side. Divide both sides by h Subtract b from both sides ==> . ::r (1) Ill c = 2A h a+ b CT ..... i' n 2A a =b ==> :I Ul Ul (a + b) h = 2A ==> I» :;· b)h =A Multiply both sides by 2 (") ::r 0 h ... I» ...0 EXAMPLE 1.17 Make T the subject in the simple interest formula I= · SOLUTION PRT =I Arrange with Ton the lefthand side EXAMPLE 1.18 Multiply both sides by 100 ==> Divide both sides by P and R ==> 100 PRT = lOOI T = lOOI PR Make x the subject in the formula v = of an oscillating point.) w.J a 2  x 2 • (This formula gives the speed SOLUTION Square both sides ==> Divide both sides by ro 2 ==> Add x2 to both sides ==> v2 Subtract 2 from both sides ==> Take the square root of both sides ==> w EXAMPLE 1.19 x= Make m the subject of the formula mv = I+ mu. (This formula gives the momentum after an impulse.) SOLUTION Collect terms in m on the lefthand side and terms without m on the other. ==> Factorise the lefthand side Divide both sides by ( v u) ==> ==> mv mu = I m(vu)=I m= _I_ vu 3 c ii' EXERCISE 1C 1 Make lil a (iil t the subject in v = u + at. 2 Make h the subject in V= l wh. 3 Make r the subject in A = 1t r2 • 4 Make lil s 5 Make h the subject in A= 2rrrh + 2rrr2. 6 Make a the subject in s = ut+ 1at2• 7 Make b the subject in h = ./ a 2 + b2 • (ii) u the subject in v2  u2 = 2as. s Make gthe subject in T= 2rr 9 10 /f. Make m the subject in E = mgh + + MakeR the subject 1 8 1mv 2 • . 2 11 Make h the subject in bh = 2A  ah. 12 Make u the subject in f = 13 Make d the subject in u 2 14 Make Vthesubjectinp 1 VM=mRT+p 2 VM. u+v  du + fd = 0. All the formulae in Exercise lC refer to real situations. Can you recognise them? Quadratic equations EXAMPLE 1.20 The length of a rectangular field is 40 m greater than its width, and its area is 6000 m 2 • Form an equation involving the length, x m, of the field. SOLUTION Since the length of the field is 40 m greater than the width, the width in m must be x 40 and the area in m 2 is x(x 40). x40 So the required equation is x(x 40) = 6000 X or x2  40x 6000 = 0. Figure 1.3 This equation, involving terms in x 2 and x as well as a constant term (i.e. a number, in this case 6000), is an example of a quadratic equation. This is in contrast to a linear equation. A linear equation in the variable x involves only terms in x and constant terms. P1 It is usual to write a quadratic equation with the righthand side equal to zero. 0 c To solve it, you first factorise the lefthand side if possible, and this requires a particular technique. Ill ... Ill Cl. c;· (1) ..c;· J:l c Ill ..= Quadratic factorisation EXAMPLE 1.21 Factorise xa + xb + ya + yb. SOLUTION xa + xb+ ya+ yb = x (a+ b)+ y (a+ b) \;{ = (x+y)(a+b) The expression is now in the form of two factors, (x+ y) and (a+ b), so this is the answer. You can see this result in terms of the area of the rectangle in figure 1.4. This can be written as the product of its length (x+ y) and its width (a+ b), or as the sum of the areas of the four smaller rectangles, xa, xb, ya and yb. X y a xa ya b xb yb Figure 1.4 The same pattern is used for quadratic factorisation, but first you need to split the middle term into two parts. This gives you four terms, which correspond to the areas of the four regions in a diagram like figure 1.4. EXAMPLE 1.22 Factorise x 2 + 7x+ 12. SOLUTION Splitting the middle term, 7x, as 4x + 3x you have x 2 + 7x+ 12 = x 2 + 4x+ 3x+ 12 = x(x+ 4) + 3(x+ 4) =(x+3)(x+4). How do you know to split the middle term, 7x, into 4x+ 3x, rather than say 5x+ 2x or 9x 2x? 3 X 3x X 4 4x 12 Figure 1.5 The numbers 4 and 3 can be added to give 7 (the middle coefficient) and multiplied to give 12 (the constant term), so these are the numbers chosen. EXAMPLE 1.23 Factorise x 2  2x 24. SOLUTION First you look for two numbers that can be added to give 2 and multiplied to give24: 6+4=2 6 X (+4) = 24. The numbers are 6 and +4 and so the middle term, 2x, is split into 6x + 4x. x 2  2x 24 = x 2  6x + 4x 24 = x(x 6) + 4(x 6) = (x+4)(x6). This example raises a number of important points. makes no difference if you write + 4x 6x instead of 6x + 4x. In that case the factorisation reads: 1 It x 2  2x 24 = x 2 + 4x 6x 24 = x(x+ 4) 6(x+ 4) = (x 6)(x+ 4) 2 3 0 c m There are other methods of quadratic factorisation. If you have already learned another way, and consistently get your answers right, then continue to use it. This method has one major advantage: it is selfchecking. In the last line but one of the solution to the example, you will see that (x + 4) appears twice. If at this point the contents of the two brackets are different, for example (x + 4) and (x 4), then something is wrong. You may have chosen the wrong numbers, or made a careless mistake, or perhaps the expression cannot be factorised. There is no point in proceeding until you have sorted out why they are different. You may check your final answer by multiplying it out to get back to the original expression. There are two common ways of setting this out. (i) Long multiplication x+4 This is x(x + 4). x 2 +4x 6x 24 2 x 2x 24 (as required) This is  6(x + 4). (iil Multiplying term by term (x + 4) ( x 6) = x 2  6x + 4x 24 = x 2  2x 24 (as required) You would not expect to draw the lines and arrows in your answers. They have been put in to help you understand where the terms have come from. EXAMPLE 1.24 Factorise x 2  20x + 100. SOLUTION x 2  20x+ 100 = x 2  10x 10x + 100 = x(x 10) lO(x 10) = (x 10)(x10) = (x10)2 ar;· Q. (clearly the same answer). Notice: ( 10) + (10) =20 ( 10) X ( 10) = + 100 . .Q c I» c;· = COl Note The expression in Example 1.24 was a perfect square. lt is helpful to be able to recognise the form of such expressions. (x+ a) 2 = x 2 + 2ax+ a 2 (x a) 2 = x 2  (in this case a= 10) 2ax+ a 2 Factorise x 2  49. EXAMPLE 1.25 x 2  49 can be written as x 2 +Ox 49. x 2 +Ox 49 = x 2  7x+ 7x 49 = x(x 7) + 7(x 7) =(x+7)(x7) Note The expression in Example 1.25 was an example of the difference of two squares which may be written in more general form as a2 G  tJ2 = (a+ b)( a b). What would help you to remember the general results from Examples 1.24 and 1.25? The previous examples have all started with the term x 2 , that is the coefficient of x 2 has been 1. This is not the case in the next example. EXAMPLE 1.26 Factorise 6x2 + x12. SOLUTION The technique for finding how to split the middle term is now adjusted. Start by multiplying the two outside numbers together: 6x(12)=72. Now look for two numbers which add to give+ 1 (the coefficient of x) and multiply to give 72 (the number found above) . (+9) (+9) + (8) = +1 X (8) = 72 Splitting the middle term gives 6x2 + 9x Sx12 = 3x(2x+ 3) 4(2x+ 3) = (3x 4)(2x+ 3) 4 is a factor of both  8x and  12. Note The method used in the earlier examples is really the same as this. it is just that in those cases the coefficient of x 2 was 1 and so multiplying the constant term by it had no effect. en 0 < :r A IQ Before starting the procedure for factorising a quadratic, you should always check that the terms do not have a common factor as for example in 2x 2  8x+ 6. .c c ....ri" ..o· I» c. I» CD This can be written as 2( x 2  4x + 3) and factorised to give 2( x 3) (x 1). .c c I» :I Ul Solving quadratic equations It is a simple matter to solve a quadratic equation once the quadratic expression has been factorised. Since the product of the two factors is zero, it follows that one or other of them must equal zero, and this gives the solution. EXAMPLE 1.27 Solve x2  40x 6000 = 0. SOLUTION => => => x 2  40x 6000 = = = (x + 60)(x lOO)= either x + 60 = or x 100 = x 2  lOOx + 60x 6000 x(x 100) + 60(x100) (x + 60)(x 100) 0 0 => x = 60 0 => x = 100 The solution is x = 60 or 100 . ...__ _ _ Q Look back to page 12. What is the length of the field? Note The solution of the equation in the example is x= 60 or 100. The roots of the equation are the values of x which satisfy the equation, in this case one root is x = 60 and the other root is x = 100. Sometimes an equation can be rewritten as a quadratic and then solved. EXAMPLE 1.28 Solve x 4  13x2 + 36 = 0 SOLUTION This is a quartic equation (its highest power of xis 4) and it isn't easy to factorise this directly. However, you can rewrite the equation as a quadratic in x2. Let y=x 2 P1 x 4  13x2 + 36 = 0 (x2 ) 2  13x2 + 36 = 0      1 y2 13y+ 36 = 0 Now you have a quadratic equation which you can factorise. (y 4)(y 9) = 0 So y= 4 or y= 9 Since y = x 2 then x 2 = 4 x = ±2 2 or x = 9 x = ±3 You may have to do some work rearranging the equation before you can solve it. EXAMPLE 1.29 Find the real roots of the equation x 2  2 = 82 • X SOLUTION You need to rearrange the equation before you can solve it. x 2  2 = _§_ xz Multiply by il: Rearrange: 2x2 = 8 2x2  8 = 0 x4 x4   This is a quadratic in x 2• You can factorise it directly, without substituting in for x 2• (x2 +2)(x2 4)=0 So this quartic equation only has two real roots . You can find out more about roots which are not real in P3. So x 2 = 2 which has no real solutions. or x 2 = 4 x = ±2 EXERCISE 10 1 2 Factorise the following expressions. (i) al+ am+ bl+ bm (ii) px+ py qx qy (iii) ur vr+ us vs (iv) m 2 + mn+ pm+ pn (v) x 2  3x+ 2x 6 (vi) y 2 + 3y+ 7y+ 21 (vii) z 2  5z+ 5z 25 (viii) q2  3q 3q + 9 (ix) 2x2 + 2x+ 3x+ 3 (x) 6v 2 +3v20v10 Multiply out the following expressions and collect like terms. (i) (a+2)(a+3) (ii) (b + 5)(b+ 7) (iii) (c 4)(c 2) (iv) (d5)(d4) (v) (e+ 6)(el) (vi) (g 3)(g+ 3) (vii) (h + 5)2 (viii) (2i  3)2 (ix) (a+b)(c+d) (x) (x+ y)(x y) 3 Factorise the following quadratic expressions. x 2 + 6x + 8 (iii) y 2 +9y+20 (v) r 2 2r15 (vii) x 2  5x 6 (ix) a 2 9 (i) x 2  6x+ 8 r 2 + 2r 15 s2  4s + 4 (iil (iv) (vi) (viii) (x) x 2 + 2x + 1 (x+ 3) 2  9 m >c CD ri .. iii' CD 4 Factorise the following expressions. 2 2x + (iii) 5x 2 + (vi 2x2 + (vii) 6x 2  (i) (ix) t l 5x+ 2 llx+ 2 14x+ 24 5x 6 2 t 2 2 c liil 2x2  5x+ 2 (ivl 5x2  llx+ 2 (vi) 4x2  49 (viiil 9x2  6x + 1 (x) 2x2  11xy+ 5f 5 Solve the following equations. lil (iiil (v) 6 x 2  llx + 24 = 0 x 2  llx+ 18 = 0 x 2  64 = 0 (iil x 2 + llx+ 24 = 0 (iv) x 2  6x + 9 = 0 Solve the following equations. 3x2  5x+ 2 = 0 liiil 3x2  5x 2 = 0 2 (v) 9x 12x+4=0 m liil 3x2 + 5x+ 2 = 0 (ivl 25x2  16 = 0 7 Solve the following equations. lil x 2  x= 20 liiil x 2 + 4 = 4x 3x2 + 5x = 4 3 15 (iv) 2x + 1 = (v) 6 X 1 = (vi) liil X X 3x + = 14 s Solve the following equations. x 4  5x2 + 4 = 0 liiil 9x4  13x2 + 4 = 0 m (v) 25x (vii) x 6  4  4x 2 9x 3 =0 +8= 0 (ivl x 4 10x2 + 9 = 0 4x4  25x2 + 36 = 0 (vi) x (viii) x (iil 6Fx + 5 = 0 Fx  6 = 0 9 Find the real roots of the following equations. liil (iv) (vi) 12 x2 1 + _!__  20 = 0 x2 x4 x2 = 1+ x 3 + 1_ = 3 x3 7 ... , 2 + 3 (VIII = Fx P1 1o Find the real roots of the equation 94 + 82 = 1. 11 The length of a rectangular field is 30 m greater than its width, w metres. 12 13 X X (i) Write down an expression for the area Am2 of the field, in terms of w. (iil The area of the field is 8800m2• Find its width and perimeter. A cylindrical tin ofheight h cm and radius rem, has surface area, including its top and bottom, Acm2• (i) Write down an expression for A in terms of r, hand n. (iil A tin of height 6 cm has surface area 54n cm2 • What is the radius of the tin? (iiil Another tin has the same diameter as height. Its surface area is 150ncm2 • What is its radius? When the first n positive integers are added together, their sum is given by 1 2n(n + 1). m Demonstrate that this result holds for the case n = 5. 14 (ii) Find the value of n for which the sum is 105. (iiil What is the smallest value of n for which the sum exceeds 1000? The shortest side AB of a rightangled triangle is x cm long. The side BC is 1 cm longer than AB and the hypotenuse, AC, is 29 cm long. Form an equation for x and solve it to find the lengths of the three sides of the triangle. Equations that cannot be factorised The method of quadratic factorisation is fme so long as the quadratic expression can be factorised, but not all of them can. In the case of x 2  6x + 2, for example, it is not possible to find two whole numbers which add to give 6 and multiply to give +2. There are other techniques available for such situations, as you will see in the next few pages. Graphical solution If an equation has a solution, you can always find an approximate value for it by drawing a graph. In the case of x 2  6x+ 2 = 0 you draw the graph of y= x 2  6x+ 2 and find where it cuts the x axis. X 0 1 2 3 4 5 6 Xl 0 1 4 9 16 25 36  6x 0 6 12 18 24 30 36 +2 +2 +2 +2 +2 +2 +2 +2 y +2 3 6 7 6 3 +2 m .o· J:l c Ill .. :I Ill :::r y n 2 I\ 1 _ _j 1\ 0 ('r1 .... Ill .. 0 j iii" Cl) I \ I 1/ $ I 1\  1/ 3 I 1\ \ 1 j \  5 r  t  6  ··· j   ! r h t t· t ·  t I \ 4 X 6 I \ 2 7 c:r Cl) j \0.3 and 0.4 \ Ill :I :I 1 t j   tr   L !J t t fJ V L. / t ·· t  . ..  . t   r  r · Figure 1.6 From figure 1.6, xis between 0.3 and 0.4 so approximately 0.35, or between 5.6 and 5.7 so approximately 5.65. Clearly the accuracy of the answer is dependent on the scale of the graph but, however large a scale you use, your answer will never be completely accurate. Completing the square If a quadratic equation has a solution, this method will give it accurately. It involves adjusting the lefthand side of the equation to make it a perfect square. The steps involved are shown in the following example. Q. P1 EXAMPLE 1.30 Solve the equation x 2  6x + 2 = 0 by completing the square. SOLUTION Subtract the constant term from both sides of the equation: ==> x 2 6x =2 Take the coefficient of x: Halve it: Square the answer: 6 ) 3 Q +9 Explain why this makes the lefthand side a perfect square. Add it to both sides of the equation: ==> x 2  6x + 9 = 2 + 9 Factorise the lefthand side. It will be found to be a perfect square: ==> (x3)2=7 Take the square root of both sides: ==> x3=± .fi ==> Using your calculator to find the value of .fi => x = 5.646 or 0.354, to 3 decimal places. The graphs of quadratic functions Look at the curve in figure 1. 7. It is the graph of y = x 2  4x + 5 and it has the characteristic shape of a quadratic; it is a parabola. Notice that: y • it has a minimum point (or vertex) at (2, 1) • it has a line of symmetry, x = 2. It is possible to find the vertex and the line of symmetry without plotting the points by using the technique of completing the square. X Figure 1.7 Rewrite the expression with the constant term moved to the right Take the coefficient of x: Divide it by 2: Square the answer: P1 4 2 +4 i ::r CD ea Add this to the lefthand part and compensate by subtracting it from the constant term on the right iil 'tl ::r Ill 2. .Q x 2 4x+4 c +54. Ill .. ...Illc. This can now be written as (x 2)2 + 1. ii' ....c .er :I n The line of symmetry is x  2 = 0 or x = 2. EXAMPLE 1.31 :I Ill Write x 2 + 5x+ 4 in completed square form. Hence state the equation of the line of symmetry and the coordinates of the vertex of the curve y = x 2 + 5x + 4. SOLUTION = x 2 +5x +4 2 x +5x+6.25 +46.25 (x+ 2.5) 2  2.25 (This is the completed square form.) The line of symmetry is x + 2.5 = 0, or x = 2.5. The vertex is (2.5, 2.25). y X Figure 1.8 P1 A For this method, the coefficient of x 2 must be 1. To use it on, say, 2x 2 + 6x+ 5, you must write it as 2(x 2 + 3x + 2.5) and then work with x 2 + 3x + 2.5. In completed square form, it is 2(x + 1.5)2 + 0.5. Similarly treat  x 2 + 6x + 5 as 1 (x 2  6x 5) and work with x 2  6x 5. In completed square form it is 1(x  3)2 + 14. Completing the square is an important technique. Knowing the symmetry and least (or greatest) value of a quadratic function will often give you valuable information about the situation it is modelling. EXERCISE 1E 1 For each of the following equations: (a) write it in completed square form (b) hence write down the equation of the line of symmetry and the coordinates of the vertex (c) sketch the curve. (i) y= x 2 +4x+ 9 (ii) (iii) y=x 2 + 4x+ 3 y= x 2 + 6x1 y= x 2 + x+ 2 (iv) lvl (vii) (ix) 2 y=x 2 1  (viii) y= x 2  4x+ 9 y= x 2 4x+ 3 y= x 2 10x y = x 2  3x7 (x) y= x 2 + 0.1x+ 0.03 (vi) Write the following as quadratic expressions in descending powers of x. lil (x+2)23 liil (x+ 4) 2  4 (iii) (x 1)2 + 2 (iv) (x10) 2 +12 lv> (x (vi) (x + 0.1) 2 + 0.99 + 3 Write the following in completed square form. (ii) 3x 2 18x27 (iv) 2x 2  2x 2 (v) 2x 2 +4x+ 6 x 2  2x+ 5 5x 2  lOx+ 7 (vi) (vii) 3x 2 12x (viii) 4x 2 4x4 8x 2 + 24x 2 (i) (iii) The curves below all have equations of the form y = x 2 + bx + c. 4 In each case find the values of band c. (i) (ii) y y l :::r CD .Q c ..... .3 Ill c. Ill ii' (3, l) X 0 c i» (l ,  I) X (iv) (iii) y y V (3, 2) (4, 0) 5 X X Solve the following equations by completing the square. m x 2  6x + 3 = 0 (iil x 2  8x l = 0 (iiil x 2  3x + l = 0 (ivl 2x 2  6x + l = 0 (vi Sx 2 + 4x 2 = 0 The quadratic formula Completing the square is a powerful method because it can be used on any quadratic equation. However it is seldom used to solve an equation in practice because it can be generalised to give a formula which is used instead. The derivation of this follows exactly the same steps. To solve a general quadratic equation ax2 + bx+ c= 0 by completing the square: First divide both sides by a: x2 + b: + = 0. Subtract the constant term from both sides of the equation: x2+bx a = _f a Take the coefficient of x: +Q P1 a Halve it: Square the answer: Add it to both sides of the equation: ==>xz+bx+£_=£__f. a 4a 2 4a 2 a Factorise the lefthand side and tidy up the righthand side: ==> (x + jz_)2 = b2  4ac 4a2 2a Take the square root of both sides: ==> x + jz_ = + b2  4ac 2a ==> X = b ± 2a 4ac .....;:...=....:.....:__=.cc_ 2a This important result, known as the quadratic formula, has significance beyond the solution of awkward quadratic equations, as you will see later. The next two examples, however, demonstrate its use as a tool for solving equations. EXAMPLE 1.32 Use the quadratic formula to solve 3x 2  6x+ 2 = 0. SOLUTION Comparing this to the form ax 2 + bx + c = 0 gives a= 3, b=6 and c= 2. Substituting these values in the formula x = b ± 2a 4ac . g1ves x = 6 = 0.423 or 1.577 (to 3 d.p.). EXAMPLE 1.33 Solve x 2  2x+ 2 = 0. SOLUTION The first thing to notice is that this cannot be factorised. The only two whole numbers which multiply to give 2 are 2 and 1 (or 2 and 1) and they cannot be added to get 2. Comparing x 2 2x + 2 to the form ax 2 + bx + c = 0 gives a = 1, b = 2 and c = 2. . . h . Su bstltutmg t ese va1ues m x . gtves 2 = b+.Jb  2a 4ac 2 ::r .a 2 CD 1: .... I» Trying to find the square root of a negative number creates problems. A positive number multiplied by itself is positive: +2 x +2 = +4. A negative number multiplied by itself is also positive: 2 x 2 = +4. Since can be neither positive nor negative, no such number exists, and so you can find no real solution. a. I» ...ti" 0 .J=4 3 1: iii Note lt is not quite true to say that a negative number has no square root. Certainly it has none among the real numbers but mathematicians have invented an imaginary number, denoted by i, with the property that i2 = 1. Numbers like 1 + i and 1  i (which are in fact the solutions of the equation above) are called complex numbers. Complex numbers are extremely useful in both pure and applied mathematics; they are covered in P3. To return to the problem of solving the equation x 2  2x+ 2 = 0, look what happens if you draw the graph of y = x 2  2x + 2. The table of values is given below and the graph is shown in figure 1.9. As you can see, the graph does not cut the x axis and so there is indeed no real solution to this equation. y 1 0 1 2 3 x2 +1 0 +1 +4 +9 2x +2 0 2 4 6 +2 +2 +2 +2 +2 +2 y +5 +2 +1 +2 +5 X   X Figure 1.9 P1 The part of the quadratic formula which determines whether or not there are real roots is the part under the square root sign. This is called the discriminant. X = b ± .Jb2  4ac ======2a If b2  4ac > 0, the equation has two real roots (see figure 1.10). X Figure 1.10 If b2  4ac < 0, the equation has no real roots (see figure 1.11). X Figure 1.11 If b2  4ac= 0, the equation has one repeated root (see figure 1.12). X Figure 1.12 EXERCISE 1F 1 Use the quadratic formula to solve the following equations, where possible. m (ii) x 2 +2x+4=0 (iiil x 2  Sx19 = 0 (iv) 5x2  3x + 4 = 0 3x2 + 2x 4 = 0 (vi) x 2  12 = 0 (v) 2 x 2 + 8x + 5 = 0 (11 3" Find the value of the discriminant and use it to find the number of real roots for each of the following equations. ;; m x 2  3x + 4 = 0 (iil x 2  3x 4 = 0 Ul (iii) 4x2 (iv) 3x2 + 8 = 0 (vi) x 2 + lOx+ 25 = 0 (v)  3x = 0 3x2 + 4x + 1 = 0 c Ill :I CD 0 c CD .o· ..Q c Ill :I Ul 3 Show that the equation ax2 + bx a= 0 has real roots for all values of a and b. 4 Find the value(s) of k for which these equations have one repeated root. m x 2  2x + k = 0 (iiil kx2 + 3x 4 = 0 (v) 5 (iil 3x2  6x+ k= 0 (iv) 2x 2 + kx + 8 = 0 3x2 + 2kx 3k= 0 The height h metres of a ball at time t seconds after it is thrown up in the air is given by the expression h = 1 +1St 5t2 . m Find the times at which the height is 11 m. (iil Use your calculator to find the time at which the ball hits the ground. (iii) What is the greatest height the ball reaches? Simultaneous equations There are many situations which can only be described mathematically in terms of more than one variable. When you need to find the values of the variables in such situations, you need to solve two or more equations simultaneously (i.e. at the same time). Such equations are called simultaneous equations. If you need to find values of two variables, you will need to solve two simultaneous equations; if three variables, then three equations, and so on. The work here is confined to solving two equations to fmd the values of two variables, but most of the methods can be extended to more variables if required. Linear simultaneous equations EXAMPLE 1.34 At a poultry farm, six hens and one duck cost $40, while four hens and three ducks cost $36. What is the cost of each type of bird? SOLUTION Let the cost of one hen be $hand the cost of one duck be $d. Then the information given can be written as: 6h+ d = 40 4h+3d = 36. CD @ There are several methods of solving this pair of equations. Method 1: Elimination CD by 3 => 18h+ 3d= => 4h+3d = Subtracting => 14h Dividing both sides by 14 => h Substituting h = 6 in equation CD gives 36+ d = d = => Multiplying equation Leaving equation @ 120 36 84 6 40 4 Therefore a hen costs $6 and a duck $4. Note 1 The first step was to multiply equation CD by 3 so that there would be a term 3d in both equations. This meant that when equation @was subtracted, the variable dwas eliminated and so it was possible to find the value of h. 2 The value h = 6 was substituted in equation CD but it could equally well have been substituted in the other equation. Check for yourself that this too gives the answer d= 4. Before looking at other methods for solving this pair of equations, here is another example. EXAMPLE 1.35 Solve 3x+ 5y= 12 2x 6y= 20 SOLUTION CD x6 @x5 Adding Giving Substituting x = 1 in equation CD Adding 3 to each side Dividing by 5 Therefore x= 1, y= 3. => => => 18x+ 30y 10x 30y 72 100 28 1 28x X => => = = 3 + 5y 5y = 12 = 15 y= 3 Note In this example, both equations were multiplied, the first by 6 to give +30y and the second by 5 to give 30y. Because one of these terms was positive and the other negative, it was necessary to add rather than subtract in order to eliminate y. Ill 3" Returning now to the pair of equations giving the prices of hens and ducks, I» :I CD CD 6h+ d = 40 4h+3d = 36 c ;:; 0 c Cll @ CD J:l . c I» here are two alternative methods of solving them. c;· :I Cll Method 2: Substitution The equation 6h + d = 40 is rearranged to make d its subject: d=406h . This expression for d is now substituted in the other equation, 4h + 3d= 36, giving => => => 4h+ 3(40 6h) = 36 4h+ 120 18h = 36 l4h = 84 h= 6 Substituting for h in d = 40 6h gives d = 40 36 = 4. Therefore a hen costs $6 and a duck $4 (the same answer as before, of course) . Method 3: Intersection of the graphs of the equations Figure 1.13 shows the graphs of the two equations, 6h + d = 40 and 4h +3d= 36. As you can see, they intersect at the solution, h = 6 and d = 4. d 10 I \ 9 '\ 8 7 6  r\ 36 3d \ 2 0 Figure 1.13  \ 4 3 I r.  f  r\.  5 1 f r 2  · t r  3 [\  5 6   r \ I 4 ±' 7 8 9 10 h Nonlinear simultaneous equations P1 The simultaneous equations in the examples so far have all been linear, that is their graphs have been straight lines. A linear equation in, say, x and y contains only terms in x and y and a constant term. So 7x + 2y = 11 is linear but 7x 2 + 2y= 11 is not linear, since it contains a term in x 2 • You can solve a pair of simultaneous equations, one of which is linear and the other not, using the substitution method. This is shown in the next example. Solve EXAMPLE 1.36 CD x+ 2y= 7 x 2 + y 2 = 10 @ SOLUTION Rearranging equation CD gives x= 7 2y. Substituting for x in equation@: (7  2y)2 + y 2 = 10 Multiplying out the (7 2y) x (7 2y) gives 49 14y 14y+ 4y2 = 49 28y+ 4y2 , so the equation is 49 28y+ 4y2 + y 2 10. This is rearranged to give 5y2  28y+ 39 sy2 1sy 13y+ 39 Sy(y 3) 13(y 3) (Sy 13)(y 3) :::::} :::::} :::::} Either Or Sy13=0 y3=0 y = 2.6 y =3 :::::} :::::} A quadratic iny which you can now solve using factorisation or the formula. y = 2.6 y = 3 :::::} :::::} Substituting in equation 0 0 0 0 CD, x+ 2y= 7: 1.8 x= 1 X= The solution is either x = 1.8, y = 2.6 or x = 1, y = 3. A Always substitute into the linear equation. Substituting in the quadratic will give you extra answers which are not correct. EXERCISE 1G 1 Solve the following pairs of simultaneous equations. (i) 2x+ 3y= 8 3x+ 2y= 7 (ii) x+4y= 16 3x+ 5y=20 (iiil 7x+ y= 15 4x+ 3y= 11 (iv) Sx 2y= 3 (v) 8x 3y= 21 5x+ y= 16 (vi) 8x+ y= 32 7x 9y=28 (viii) 3u2v= 17 5u3v=28 (ixl 41 3m= 2 5l7m=9 x+ 4y= 5 4x+ 3y= 5 2x 6y=5 (viil 2 3 A student wishes to spend exactly $10 at a secondhand bookshop. All the paperbacks are one price, all the hardbacks another. She can buy five paperbacks and eight hardbacks. Alternatively she can buy ten paperbacks and six hardbacks. m Write this information as a pair of simultaneous equations. (iil Solve your equations to find the cost of each type of book. The cost of a pear is Se greater than that of an apple. Eight apples and nine pears cost $1.64. (i) Write this information as a pair of simultaneous equations. (iil Solve your equations to find the cost of each type of fruit. 4 A car journey of 380 km lasts 4 hours. Part of this is on a motorway at an average speed of 110 kmh 1, the rest on country roads at an average speed of 70 kmh 1. 5 m Write this information as a pair of simultaneous equations. (ii) Solve your equations to find how many kilometres of the journey is spent on each type of road. Solve the following pairs of simultaneous equations. m (iil x 2 2y 2 =8 x+ 2y= 8 k2 + km = 8 m=k6 (v) t12  k(k m) = 12 k(k+ m)= 60 tviiil x2+y2=10 x+y=4 (iv) (viil t 22 = 75 tl = 2t2 P/  P/ = 0 PI+ p2 = 2 (iiil (vi) 2x 2 + 3y = 12 x y =1 p + q+ 5 =0 p2 = q2 + 5 P1 6 The diagram shows the net of a cylindrical container of radius r cm and height hem. The full width of the metal sheet from which the container is made is 1 m, and the shaded area is waste. The surface area of the container is 1400n: cm 2 • h !m 7 (i) Write down a pair of simultaneous equations for rand h. (ii) Find the volume of the container, giving your answers in terms of n:. (There are two possible answers.) A large window consists of six square panes of glass as shown. Each pane is x m by x m, and all the dividing wood is y m wide. X y (i) Write down the total area of the whole window in terms of x and y. (iil Show that the total area of the dividing wood is 7xy + 2y 2 • (iii) The total area of glass is 1.5 m 2, and the total area of dividing wood is 1m2 • Find x, and hence find an equation for y and solve it. [MEI] Inequalities Not all algebraic statements involve the equals sign and it is just as important to be able to handle algebraic inequalities as it is to solve algebraic equations. The solution to an inequality is a range of possible values, not specific value(s) as in the case of an equation. Linear inequalities A The methods for linear inequalities are much the same as those for equations but you must be careful when multiplying or dividing through an inequality by a negative number. :i CD .Q c !. ;::;: ;· Ill Take for example the following statement: Multiply both sides by 1 A 5 > 3 is true 5 > 3 is false. It is actually the case that multiplying or dividing by a negative number reverses the inequality, but you may prefer to avoid the difficulty, as shown in the examples below. EXAMPLE 1.37 Solve Sx 3 o% 2x 15. SOLUTION Add 3 to, and subtract 2x from, both sides Tidy up Divide both sides by 3 => Sx 2x ,;;;: 15 + 3 => 3x o% 12 X o% 4 Note Since there was no need to multiply or divide both sides by a negative number, no problems arose in this example. EXAMPLE 1.38 Solve 2y+ 6 > 7y+ 11. SOLUTION Subtract 6 and 7y from both sides Tidy up :::::} 2y7y > 116 :::::} Add 5y to both sides and subtract 5 :::::} Divide both sides by +5 1 > y Note that logically 1 > y is the same as y < 1, so the solution is y < 1. Quadratic inequalities P1 EXAMPLE 1.39 Solve x2  4x+ 3 > 0 (il liil x2  4x+ 3 :os; 0. SOLUTION The graph of y = x2  4x + 3 is shown in figure 1.14 with the green parts of the x axis corresponding to the solutions to the two parts of the question. (i) You want the values of x for which y > 0, which that is where the curve is above the x axis. (ii) You want the values of x for y :os; 0, that is where the curve crosses or is below the x axis. Here the end points are not included in the inequality so you draw open circles: 0 y I 3 r2  i\  I I fl0 t1 V X I X I Figure 1.14 The solution is x < 1 or x > 3. EXAMPLE 1.40 The solution is x 1 and x usually witten 1 :os; x :os; 3. Find the set of values of k for which x2 + kx + 4 = 0 has real roots. SOLUTION A quadratic equation, ax2 + bx + c = 0, has real roots if b2  4ac So x2 + kx + 4 = 0 has real roots if k2  4 x 4 x 1 So the set of values is k 4 and k 0. 0. 4. Take care: ( 5) 2 = 25 and (3)2 = 9, so k must be less than or equal to 4. :os; 3, EXERCISE 1H 1 2 Solve the following inequalities. P1 m Sa+ 6 > 2a + 24 (ii) 3b 5 ,::;; b1 (iiil 4(cl) > 3(c 2) (iv) d 3(d + 2) (v) !e + 3!2 < e 2 (vi) J 2f 3 < 4(1 +f) (vii) 5(2 3g) + g (viii) 3(h + 2) 2(h 4) > 7(h + 2) 8(2g 4) 2(1 + 2d) Solve the following inequalities by sketching the curves of the functions involved. (i) p2  5p+4 < 0 (ii) p2  5p+4 (iii) x 2 + 3x+ 2 0 (iv) (v) y 2  2y 3 > 0 (vi) x 2 + 3x > 2 z(z1) 20 (viii) y(y 2) > 8 (x) 2y 2  lly 6 (xii) 10y 2 > y+ 3 (vii) q2 (ix) (xi) 4q + 4 > 0 3x 2 + 5x 2 < 0 4x 3 x 2 0 0 3 Find the set of values of k for which each of these equations has two real roots. (i) (iiil 4 2x 2  3x + k = 0 5x 2 + kx+ 5 = 0 =0 (iil (iv) 3x 2 + 2kx+ k= 0 Find the set of values of k for which each of these equations has no real roots. (i) x 2  6x + k = 0 liiil 4x 2  (iil kx+ 4 = 0 (iv) I<EY POINTS 1 The quadratic formula for solving ax2 + bx + c = 0 is ± .Jb  4ac = b =='=="'2 X 2a where b2  4ac is called the discriminant. If b2  4ac > 0, the equation has two real roots. If b2  4ac = 0, the equation has one repeated root. If b2  4ac < 0, the equation has no real roots. 2 To solve a pair of simultaneous equations where one equation is nonlinear: • • • • first make x or y the subject of the linear equation then substitute this rearranged equation for x or y in the nonlinear equation solve to find y or x substitute back into the linear equation to find pairs of solutions. 3 Linear inequalities are dealt with like equations but if you multiply or divide by a negative number you must reverse the inequality sign. 4 When solving a quadratic inequality it is advisable to sketch the graph. ... ::t Coordinate geometry .. A place for everything, and everything in its place CD E Samuel Smiles 0 CD .. CD Cll c .. :c 0 0 (J Coordinates Coordinates are a means of describing a position relative to some fixed point, or origin. In two dimensions you need two pieces of information; in three dimensions, you need three pieces of information. In the Cartesian system (named after Rene Descartes), position is given in perpendicular directions: x, y in two dimensions; x, y, z in three dimensions (see figure 2.1). This chapter concentrates exclusively on two dimensions. z 5 (3 , 4, 5) 4 y 3 3 5 2 3 ··· · ··········· · ·· (3, 2) 2 2 1 0 1 Figure 2.1 2 4 X 3 .· · 4 •• •• 3 y Plotting, sketching and drawing In two dimensions, the coordinates of points are often marked on paper and joined up to form lines or curves. A number of words are used to describe this process. P1 l :::r Plot (a line or curve) means mark the points and join them up as accurately as you can. You would expect to do this on graph paper and be prepared to read information from the graph. Cll cc ii! Cl. ...... Cij" :I 0 Sketch means mark points in approximately the right positions and join them up in the right general shape. You would not expect to use graph paper for a sketch and would not read precise information from one. You would however mark on the coordinates of important points, like intersections with the x and y axes and points at which the curve changes direction. Draw means that you are to use a level of accuracy appropriate to the circumstances, and this could be anything between a rough sketch and a very accurately plotted graph. The gradient of a line In everyday English, the word line is used to mean a straight line or a curve. In mathematics, it is usually understood to mean a straight line. If you know the coordinates of any two points on a line, then you can draw the line. The slope of a line is measured by its gradient. It is often denoted by the letter m. In figure 2.2, A and B are two points on the line. The gradient of the line AB is given by the increase in the y coordinate from A to B divided by the increase in the x coordinate from A to B. y AL B(6, 7) (2, 4) 0 Figure 2.2 X I» :;· Cll In general, when A is the point (x 1, y 1) and B is the point (x 2 , yz), the gradient is Yz Yt m=. X2  x1 When the same scale is used on both axes, m= tan e (see figure 2.2) . Figure 2.3 shows four lines. Looking at each one from left to right: line A goes uphill and its gradient is positive; line B goes downhill and its gradient is negative. Line C is horizontal and its gradient is 0; the vertical line D has an infinite gradient. .. Q) E 0 Q) .. Cl Q) ea c :s.. 0 0 (.) 2 0 3 4 5 6 7 8 X Figure 2.3 ACTIVITY 2.1 On each line in figure 2.3, take any two points and call them (xl' y1) and (x 2, y2 ). Substitute the values of xl' yl' x 2 and y2 in the formula m= Yz  Y1 Xz XI and so find the gradient. G Does it matter which point you call (x1, y 1) and which (x2 , y2 )? Parallel and perpendicular lines If you know the gradients m 1 and m 2 of two lines, you can tell at once if they are either parallel or perpendicular see figure 2.4. Figure 2.4 parallel lines: m 1 =m 2 perpendicular lines: m 1m 2 = I Lines which are parallel have the same slope and so m 1 = m2. If the lines are perpendicular, m 1m2 = 1. You can see why this is so in the activities below. I ACTIVITY 2 .2 ACTIVITY 2.3 Draw the line L 1 joining (0, 2) to (4, 4), and draw another line l 2 perpendicular to L 1• Find the gradients m 1 and m2 of these two lines and show that m 1 m2 = 1. The lines AB and BC in figure 2.5 are equal in length and perpendicular. By showing that triangles ABE and BCD are congruent prove that the gradients m 1 and m2 must satisfy m1 m2 = 1. P1 4 :r CD a. !'Ill :I n CD er ::!: . . CD CD :I y B ::!: 0 \.:::::_.. "Cl 0 B gradient m 1 :;· Ul gradient m 2 A B I E Dl c 0 X Figure 2.5 A Lines for which m 1m2 = 1 will only look perpendicular if the same scale has been used for both axes. The distance between two points When the coordinates of two points are known, the distance between them can be calculated using Pythagoras' theorem, as shown in figure 2.6. y A LJC (2, 4) 0 Figure 2.6 X This method can be generalised to find the distance between any two points, A(xl' y 1 ) and B(x 2, y2 ), as in figure 2.7. y .. Ql E ., 0 Q) !Ill c 0 0 u 0 X Figure 2.7 The midpo·nt of a line joining two oin s Look at the line joining the points A(2, 1) and B(8, 5) in figure 2.8. The point M(5, 3) is the midpoint of AB. y 5 4 3 012345678 X Figure 2.8 Notice that the coordinates of M are the means of the coordinates of A and B. 5= i(2 + 8); 3= iO + 5). This result can be generalised as follows. For any two points A(xl' y 1) and B(x 2 , y 2 ), the coordinates of the midpoint of AB are the means of the coordinates of A and B so the midpoint is ( Xi+ Xz Y1 + Yz) 2 ' 2 . EXAMPLE 2.1 A and Bare the points (2, 5) and (6, 3) respectively (see figure 2.9). Find: (i) (iil (iiil (ivl P1 the gradient of AB the length of AB the midpoint of AB the gradient of a line perpendicular to AB. .... ::r 3 a: ,; SOLUTION ..:;·.... 0 Taking A(2, 5) as the point (x 1, y 1), and B(6, 3) as the point (x 2 , y 2 ) gives x 1 = 2, y 1 =5,x2 =6,y 2 =3. 0 Ill y (i) Gradient= y2  y, :r o·:;· A(2, 5) :;· . cc _35_ 1 622 liil Length AB= B(6, 3) x,)l + (y 2  ..:;· Yl Ill 0 = 4= = 0 'C 0 J2o X Figure 2.9 .. M1dpomt . . = (xl +2x2, Y1 +2 Y2) hi•l = (2; 6) 5 ; (ivl 3) = (4, 4) Gradient of AB= m 1 = If m2 is the gradient of a line perpendicular to AB, then m 1 m 2 = 1 ::::::} 1 = 1 m 2 =2. EXAMPLE 2.2 Using two different methods, show that the lines joining P(2, 7), Q(3, 2) and R(O, 5) form a rightangled triangle (see figure 2.10). R(O, 5) 5 SOLUTION 4 Method 1 75 Gradient of RP 20   1 Gradient ofRQ = 6 25 _ = 1 3 0 ::::::} Product of gradients= 1 x (1) = 1 ::::::} Sides RP and RQ are at right angles. 31 +" 21 + + 0 [ Figure 2.10 2 3 4 X Method2 P1 Pythagoras' theorem states that for a rightangled triangle whose hypotenuse has length a and whose other sides have lengths band c, a 2 = b2 + c2. Conversely, if you can show that a 2 = b2 + c2 for a triangle with sides of lengths a, b, and c, then the triangle has a right angle and the side oflength a is the hypotenuse. ; E 0 Ill 01 Ill Ill .. This is the basis for the alternative proof, in which you use length 2 = (x 2  x 1) 2 + (y 2  Y1V c . :c 0 PQ 2 = (3 2) 2 + (2 7) 2 = 1 + 25 = 26 0 (J RP 2 = (2 0) 2 + (7 5) 2 = 4 + 4 = 8 RQ 2 = (3 0) 2 + (2 5) 2 = 9 + 9 = 18 PQ2 = RP2 + RQ2 Since 26 = 8 + 18, :::::} Sides RP and RQ are at right angles. EXERCISE 2A 1 For the following pairs of points A and B, calculate: (a) the gradient of the line AB (bl the midpoint of the line joining A to B (cl the distance AB (d) the gradient of the line perpendicular to AB. m A(O, 1) !iiil A( 6, 3) (v) A(4, 3) B(2, 3) !iil A(3,2) B(4,1) B(6,3) !ivl A(5, 2) B(2, 8) B(2,0) !vil A(l, 4) B(l, 2) 2 The line joining the point P(3, 4) to Q(q, 0) has a gradient of2. Find the value of q. 3 The three points X(2, 1), Y(8, y) and Z(ll, 2) are collinear (i.e. they lie on the same straight line) . Find the value of y. 4 The points A, B, C and D have coordinates (1, 2), (7, 5), (9, 8) and (3, 5). m Find the gradients of the lines AB, BC, CD and DA. (ii) What do these gradients tell you about the quadrilateral ABCD? (iiil Draw a diagram to check your answer to part (ii) . 5 The points A, Band C have coordinates (2, 1), (b, 3) and (5, 5), where b > 3 and LABC = 90°. Find: m the value of b (iil the lengths of AB and BC (iiil the area of triangle ABC. 6 The triangle PQR has vertices P(8, 6), Q(O, 2) and R(2, r). Find the values of r when the triangle: lil has a right angle at P liil has a right angle at Q liiil has a right angle at R livl is isosceles with RQ =RP. 7 The points A, B, and C have coordinates (4, 2), (7, 4) and (3, 1). m Draw the triangle ABC. liil Show by calculation that the triangle ABC is isosceles and name the two equal sides. liiil Find the midpoint of the third side. (ivl By calculating appropriate lengths, calculate the area of the triangle ABC. 8 For the points P(.x, y), and Q(3.x, Sy), find in terms of x and y: lil the gradient of the line PQ liil the midpoint of the line PQ liiil the length of the line PQ. 9 A quadrilateral has vertices A(O, 0), B(O, 3), C(6, 6) and D(12, 6). lil Draw the quadrilateral. liil Show by calculation that it is a trapezium. liiil Find the coordinates of E when EBCD is a parallelogram. 10 Three points A, Band C have coordinates (1, 3), (3, 5) and (1, y). Find the values of y when: 1i1 AB=AC liil AC= BC liiil AB is perpendicular to BC livl A, B and C are collinear. 11 The diagonals of a rhombus bisect each other at 90°, and conversely, when two lines bisect each other at 90°, the quadrilateral formed by joining the end points of the lines is a rhombus. Use the converse result to show that the points with coordinates ( 1, 2), (8, 2), (7, 6) and (0, 10) are the vertices of a rhombus, and find its area. The equation of a straight line The word straight means going in a constant direction, that is with fixed gradient. This fact allows you to find the equation of a straight line from first principles. ti E EXAMPLE 2.3 Find the equation of the straight line with gradient 2 through the point (0, 5). 0 CD .. 01 CD 111 SOLUTION c .. y :c 0 0 (.) X Figure 2.11 Take a general point (.x, y) on the line, as shown in figure 2.11. The gradient of the line joining (0, 5) to (.x, y) is given by . gradient= y(5) x0 y+5 = . x Since we are told that the gradient of the line is 2, this gives y+5=2 X y=2x5. Since (.x, y) is a general point on the line, this holds for any point on the line and is therefore the equation of the line. The example above can easily be generalised (see page 50) to give the result that the equation of the line with gradient m cutting they axis at the point (0, c) is y= mx+ c. (In the example above, m is 2 and c is 5.) This is a wellknown standard form for the equation of a straight line. Drawing a line, given its equation There are several standard forms for the equation of a straight line, as shown in figure 2.12. When you need to draw the graph of a straight line, given its equation, the first thing to do is to look carefully at the form of the equation and see if you can recognise it. 4 ::r CD CD J:l ..a· c Ill :I (a) Equations of the form x= a (b) Equations of the form y= b y y .... 0 ... Ill Ill Ill x= 3 =·..::r (0, 2) l       y= 2 :I CD 0 (3 , 0) 0 X (c) Equations of the form y = mx X (d) Equations of the form y = mx + c y y (0, I) X (e) Equations of the form (0,  J) y=tx+ I px + qy + r = 0 Figure 2.12 (a), {b): Lines parallel to the axes Lines parallel to the x axis have the form y = constant, those parallel to they axis the form x = constant. Such lines are easily recognised and drawn. (c), fd): Equations of the form y = mx + c The line y = mx + c crosses the y axis at the point (0, c) and has gradient m. If c = 0, it goes through the origin. In either case you know one point and can complete the line either by fmding one more point, for example by substituting x = 1, or by following the gradient (e.g. 1 along and 2 up for gradient 2). ; E 0 . Ill (e): Equations of the form px + qy + r en Ill «< =0 In the case of a line given in this form, like 2x + 3y 6 = 0, you can either rearrange it in the form y= mx+ c (in this example y = + 2), or you can find the coordinates of two points that lie on it. Putting x = 0 gives the point where it crosses they axis, (0, 2), and putting y= 0 gives its intersection with the x axis, (3, 0). c 0 0 u EXAMPLE 2.4 Sketch the lines x = 5, y = 0 and y = x on the same axes. Describe the triangle formed by these lines. SOLUTION The line x= 5 is parallel to they axis and passes through (5, 0). The line y = 0 is the x axis. The line y = x has gradient 1 and goes through the origin. y Figure 2.13 The triangle obtained is an isosceles rightangled triangle, since OA =AB= 5 units, and LOAB = 90°. EXA PLE 2.5 Draw y = x 1 and 3x + 4y = 24 on the same axes. SOLUTION The line y = x 1 has gradient 1 and passes through the point (0, 1). Substituting y= 0 gives x= 1, so the line also passes through (1, 0). Find two points on the line 3x + 4y = 24. Substituting x = 0 gives 4y = 24 so y=6. Substituting y =0 so x=8. g1ves 3x =24 The line passes through (0, 6) and (8, 0). P1 "11 :;· Q, . i:i' CD ':1' CD CD ..s· .Cl c I» X :I ... 0 I» :;· Figure 2.14 EXERCISE 28 1 Sketch the following lines. (ii) X= (iv) y=2 y=3x (v) y= 3x+ 5 (vi) y=2x y=x4 (viil y=x+4 (viii) y= (ix) y = 2x + (x) y=4x+8 (xi) y= 4x 8 (xii) y= x+ 1 •.• , ( XIII Y =2 (xiv) y= 1 2x (xv) 3x 2y= 6 (xvil 2x+ 5y= 10 x+ 3y 6 = 0 (xvii) 2x+ y 3 = 0 y= 2 x (xviii) 2y= 5x 4 (i) (xix) 2 CD 1 x 2 (xx) 5 (iiil 2 By calculating the gradients of the following pairs of lines, state whether they are parallel, perpendicular or neither. (i) y=4 (iii) (v) 2x+ y= 1 3x y+ 2 = 0 (vii) x+2y1=0 (ix) y=x2 (xi) x+3y2=0 x=2 x 2y= 1 3x+ y= 0 x+2y+ 1 =0 x+y=6 y= 3x+ 2 (viii) y=3x y= 2x+ 3 2x+ 3y= 4 y= 2x1 (x) y=42x (xii) y=2x (ii) (iv) (vi) x=3y 4x y+ 1 = 0 2y= 3x 2 2x y+3=0 x+ 2y= 8 4x+ 2y= 5 Finding the equation of a line The simplest way to find the equation of a straight line depends on what information you have been given. (i) Given the gradient, m, and the coordinates (x1 , y 1 ) of one point on the line Take a general point (x, y) on the line, as shown in figure 2.15. .. Cl) E 0 Cl) .. en Cl) Ill c . :c 0 0 (J X Figure 2.15 The gradient, m, of the line joining (xl' y 1 ) to (x, y) is given by yy m=1 x  x1 => yy1 =m (xx 1 ). This is a very useful form of the equation of a straight line. Two positions of the point (x1, y 1) lead to particularly important forms of the equation. (a) When the given point (xl' y1) is the point (0, c), where the line crosses the y axis, the equation takes the familiar form y= mx+ c as shown in figure 2.16. (b) When the given point (xl' y 1) is the origin, the equation takes the form y=mx as shown in figure 2.17. y Figure 2.16 Figure 2.17 Find the equation of the line with gradient 3 which passes through the point (2, 4). EXAMPLE 2.6 P1 SOLUTION Using y  y 1 = m(x x 1) :::::} y (4) = 3(x 2) :::::} y+ 4 = 3x 6 :::::} , :;· Cl. . :r cc y=3x10. ::r CD CD ..o· (ii) Given two points, (x1 , y 1 ) and (x2 , y2 ) .Cl The two points are used to find the gradient: :I c m y ...m 0 :;CD This value of m is then substituted in the equation y y 1 =m (x x 1). 0 This gives y YI = (Y2YI)(x xJ x x 2  X Figure 2.18 1 Rearranging the equation gives Y Y1 = x X1 or Y Y1 = Y2 Y1 Find the equation of the line joining (2, 4) to (5, 3). EXAMPLE 2.7 SOLUTION Taking (xi' y 1) to be (2, 4) and (x 2, y2 ) to be (5, 3), and substituting the values in y YJ = X X] Y2 Y1 x2 X1 . y4 x2 gives 3  4 = 5  2. y This can be simplified to x+ 3y 14 = 0. f) Show that the equation of the line in figure 2.19 can be written X Figure 2.19 Different techniques to solve problems P1 ; The following examples illustrate the different techniques and show how these can be used to solve a problem. EXAMPLE 2.8 Find the equations of the lines (a) (e) in figure 2.20. E 0 y Gl .. Cl Gl Ill c (c) 0 0 u  1\ \ v V, J 4_ 1_1 v V "I 5   (b) I I ') i / 1\ V I + l\ v 24..0 I • I " , I 2 I I I 4 6 t 8 I X · I 1\ \ .___ I >1\ __(___._... ! I I I Figure 2.20 SOLUTION Line (a) passes through (0, 2) and has gradient 1 :::::} equation of (a) is y= x+ 2. Line (b) is parallel to the x axis and passes through (0, 4) :::::} equation of (b) is y = 4. Line (c) is parallel to they axis and passes through (3, 0) :::::} equation of (c) is x = 3. Line (d) passes through (0, 0) and has gradient 2 :::::} equation of (d) is y = 2x. Line (e) passes through (0, 1) and has :::::} equation of (e) is y = 1. This can be rearranged to give x+ 5y+ 5 = 0. EXAMPLE 2.9 Two sides of a parallelogram are the lines 2y = x + 12 and y = 4x 10. Sketch these lines on the same diagram. The origin is a vertex of the parallelogram. Complete the sketch of the parallelogram and find the equations of the other two sides. SOLUTION The line 2y= x+ 12 has (since dividing by 2 gives y = and passes through the point (0, 6) + 6). The line y= 4x 10 has gradient 4 and passes through the point (0, 10). P1 ., ;· a. :r .. ..o· IQ y ::r CD CD J:l c Ill = Ill ;· CD X Figure 2.21 The other two sides are lines with i.e. y= EXAMPLE 2.10 and 4 which pass through (0, 0), and y= 4x. Find the equation of the perpendicular bisector of the line joining P( 4, 5) to Q(2, 3). SOLUTION y 0 Figure 2.22 X The gradient of the line PQ is P1 35 2(4) 2 6 1 = 3 and so the gradient of the perpendicular bisector is +3. ... The perpendicular bisector passes throught the midpoint, R, of the line PQ. The coordinates of R are Q) E 0 Q) D) Cll ( c: 0 2+(4) 3+5). 2 ' 2 1.e. (1 4) ' · Using y y1 = m(x x 1), the equation of the perpendicular bisector is y4=3(x(1)) y 4 = 3x+ 3 0 0 y=3x+7. EXERCISE 2C 1 Find the equations of the lines (i)  (x) in the diagrams below. y X y X 2 Find the equations of the following lines. (i) P1 parallel toy= 2x and passing through (1, 5) (iil parallel toy= 3x 1 and passing through (0, 0) (iiil parallel to 2x + y 3 = 0 and passing through (4, 5) m .. (ivl parallel to 3x y 1 = 0 and passing through (4, 2) >C CD () (vi parallel to 2x+ 3y= 4 and passing through (2, 2) (vi) 3 iii' CD parallel to 2x y 8 = 0 and passing through (1, 5) N n Find the equations of the following lines. (i) perpendicular toy= 3x and passing through (0, 0) y = 2x + 3 and passing through (2, 1) (iiil perpendicular to 2x+ y= 4 and passing through (3, 1) (ivl perpendicular to 2y = x + 5 and passing through (1, 4) (vi perpendicular to 2x + 3y = 4 and passing through (5, 1) (vil perpendicular to 4x y+ 1 = 0 and passing through (0, 6) (iil perpendicular to 4 Find the equations of the line AB in each of the following cases. lil A(O, 0) liiil A(2, 7) (v) 5 A(2, 4) B(4,3) liil A(2, 1) B(2, 3) (iv) A(3, 5) B(5, 3) (vi) A(4, 2) B(3, 0) B(5, 1) B(3, 2) Triangle ABC has an angle of 90° at B. Point A is on they axis, AB is part of the line x 2y+ 8 = 0 and C is the point (6, 2). (i) Sketch the triangle. (iil Find the equations of AC and BC. (iiil Find the lengths of AB and BC and hence find the area of the triangle. (iv) Using your answer to part (iiil, find the length of the perpendicular from B toAC. 6 A median of a triangle is a line joining one of the vertices to the midpoint of the opposite side. In a triangle OAB, 0 is at the origin, A is the point (0, 6) and B is the point (6, 0). (i) Sketch the triangle. (iil Find the equations of the three medians of the triangle. (iiil Show that the point (2, 2) lies on all three medians. (This shows that the medians of this triangle are concurrent.) 7 A quadrilateral ABCD has its vertices at the points (0, 0), (12, 5), (0, 10) and (6, 8) respectively. (i) Sketch the quadrilateral. (iil Find the gradient of each side. (iiil Find the length of each side. (iv) Find the equation of each side. (vi Find the area of the quadrilateral. P1 The intersection of two lines The intersection of any two curves (or lines) can be found by solving their equations simultaneously. In the case of two distinct lines, there are two possibilities: ti E 0 Ql .. Cl Ql ea c (il they are parallel (ii) they intersect at a single point. 0 0 (J EXAMPLE 2.11 Sketch the lines x+ 2y= 1 and 2x+ 3y= 4 on the same axes, and find the coordinates of the point where they intersect. SOLUTION The line x+ 2y= 1 passes through ( 0, The line 2x+ 3y= 4 passes through ( 0, (1, 0). j) and (2, 0) . y X Figure 2.23 CD: x+ 2y= 1 CD:x 2: 2x+ 4y= 2 (i): 2x+ 3y= 4 (i): 2x+3y=4 Subtract: y= 2. Substitutingy=2 in CD: x4=1 ::::} X= 5. The coordinates of the point of intersection are ( 5, 2) . EXAMPLE 2.12 Find the coordinates of the vertices of the triangle whose sides have the equations x+ y= 4, 2x y= 8 and x+ 2y= 1. SOLUTION A sketch will be helpful, so first find where each line crosses the axes. ::r CD x+ y= 4 crosses the axes at (0, 4) and (4, 0). :r t . (I) (I) @ 2x y = 8 crosses the axes at (0, 8) and (4, 0). ® x + 2y = 1 crosses the axes at ( 0, (I) !1 o·:I and (1, 0). .....:e 0 y 0 :r (I) Cll X Figure 2.24 Since two lines pass through the point (4, 0) this is clearly one of the vertices. It has been labelled A on figure 2.24. Point B is found by solving @and ®simultaneously: @x2: ®: Add 4x 2y= 16 x+2y=1 5x =15 so x=3. Substituting x= 3 in@ gives y= 2, soB is the point (3, 2). Point C is found by solving CD and ®simultaneously: CD: ®: Subtract x+ y=4 x+2y=1 y = 5 so y = 5. Substituting y = 5 in CD gives x = 9, so C is the point (9, 5). P1 0 The line l has equation 2x y = 4 and the line m has equation y = 2x 3. What can you say about the intersection of these two lines? ; E 0 Gl .. Historical note en Rene Descartes was born near Tours in France in 1596. At the age of eight he was sent to a Jesuit boarding school where, because of his frail health, he was allowed to Gl Ill stay in bed until late in the morning. This habit stayed with him for the rest of his life c :s.. and he claimed that he was at his most productive before getting up . 0 0 u After leaving school he studied mathematics in Paris before becoming in turn a soldier, traveller and optical instrument maker. Eventually he settled in Holland where he devoted his time to mathematics, science and philosophy, and wrote a number of books on these subjects. In an appendix, entitled La Geometrie, to one of his books, Descartes made the contribution to coordinate geometry for which he is particularly remembered. In 1649 he left Holland for Sweden at the invitation of Queen Christina but died there, of a lung infection, the following year. EXERCISE 20 1 m Find the vertices of the triangle ABC whose sides are given by the lines AB: x 2y= 1, BC: 7x+ 6y= 53 and AC: 9x+ 2y= 11. liil Show that the triangle is isosceles. 2 Two sides of a parallelogram are formed by parts of the lines 2x y = 9 and x 2y=9. (i) Show these two lines on a graph. (iil Find the coordinates of the vertex where they intersect. Another vertex of the parallelogram is the point (2, 1). liiil Find the equations of the other two sides of the parallelogram. (ivl Find the coordinates of the other two vertices. 3 A(O, 1), B(l, 4), C(4, 3) and D(3, 0) are the vertices of a quadrilateral ABCD. 4 (i) Find the equations of the diagonals AC and BD. (iil Show that the diagonals AC and BD bisect each other at right angles. (iiil Find the lengths of AC and BD. (iv) What type of quadrilateral is ABCD? The line with equation Sx + y = 20 meets the x axis at A and the line with equation x + 2y = 22 meets the y axis at B. The two lines intersect at a point C. m Sketch the two lines on the same diagram. (iil Calculate the coordinates of A, B and C. (iiil Calculate the area of triangle OBC where 0 is the origin. (iv) Find the coordinates of the point E such that ABEC is a parallelogram. 5 A median of a triangle is a line joining a vertex to the midpoint of the P1 opposite side. In any triangle, the three medians meet at a point. The centroid of a triangle is at the point of intersection of the medians. Find the coordinates of the centroid for each triangle shown. m (i) .. >c et (iil y n iij' et (0, 12) N c y (0, 9) 6 You are given the coordinates of the four points A(6, 2), B(2, 4), C( 6, 2) and D( 2, 4). (i) Calculate the gradients of the lines AB, CB, DC and DA. Hence describe the shape of the figure ABCD. (ii) Show that the equation of the line DAis 4y 3x = 10 and find the length DA. (iiil Calculate the gradient of a line which is perpendicular to DA and hence find the equation of the line l through B which is perpendicular to DA. (ivl Calculate the coordinates of the point P where l meets DA. (vi Calculate the area of the figure ABCD. 7 [MEI) The diagram shows a triangle whose vertices are A( 2, 1), B(l, 7) and C(3, 1). The point L is the foot of the perpendicular from A to BC, and M is the foot of the perpendicular from B to AC. (il Find the gradient of the line BC. y (iil Find the equation of the line AL. B(l , 7) (iiil Write down the equation of the line BM. A ( 2, I) X The lines AL and BM meet at H. ti E 0 . Ill Cl Ill livl Find the coordinates of H. (v) Show that CH is perpendicular to AB. (vi) Find the area of the triangle BLH. [MEI] 8 The diagram shows a rectangle ABCD. The point A is (0, 2) and C is ( 12, 14). The diagonal BD is parallel to the x axis. ea c .. :c 0 C(l2, 14) 0 (.) X lil Explain why the y coordinate of D is 6. The x coordinate of D is h. liil Express the gradients of AD and CD in terms of h. liiil Calculate the x coordinates of D and B. (ivl Calculate the area of the rectangle ABCD. [Cambridge AS & A Level Mathematics 9709, Paper 12 Q9 November 2009] 9 The diagram shows a rhombus ABCD. The points Band D have coordinates (2, 10) and (6, 2) respectively, and A lies on the x axis. The midpoint of BD is M. Find, by calculation, the coordinates of each of M, A and C. y c A 0 X [Cambridge AS & A Level Mathematics 9709, Paper 1 QS June 2005] 10 Three points have coordinates A(2, 6), B(8, 10) and C(6, 0). The perpendicular bisector of AB meets the line BC at D. Find (i) the equation of the perpendicular bisector of AB in the form ax + by= c lii) the coordinates of D. m [Cambridge AS & A Level Mathematics 9709, Paper 1 Q7 November 2005] 11 The diagram shows a rectangle ABCD. The point A is (2, 14), B is (2, 8) and C lies on the x axis. y A(2, 14) B( 2, 8) D c 0 X Find (i) the equation of BC. liil the coordinates of C and D. [Cambridge AS & A Level Mathematics 9709, Paper 1 Q6 June 2007] 12 The three points A(3, 8), B(6, 2) and C(lO, 2) are shown in the diagram. The point D is such that the line DA is perpendicular to AB and DC is parallel to AB. Calculate the coordinates of D. y D A(3 , 8) B(6, 2) 0 P1 C(IO, 2) X [Cambridge AS & A Level Mathematics 9709, Paper 1 Q6 November 2007] CD c; iii' CD N c P1 13 In the diagram, the points A and C lie on the x and y axes respectively and the equation of AC is 2y+ x= 16. The point B has coordinates (2, 2). The perpendicular from B to AC meets AC at the point X. y .. .. c Q) E 0 Q) en Q) «< c :c.. 0 0 0 A 0 X lil Find the coordinates of X. The point D is such that the quadrilateral ABCD has AC as a line of symmetry. (ii) Find the coordinates of D. (iiil Find, correct to 1 decimal place, the perimeter of ABCD. [Cambridge AS & A Level Mathematics 9709, Paper 1 Q11 June 2008] 14 The diagram shows points A, Band C lying on the line 2y= x+ 4. The point A lies on they axis and AB= BC. The line from D(lO, 3) to B is perpendicular to AC. Calculate the coordinates of B and C. y 0 X [Cambridge AS & A Level Mathematics 9709, Paper 1 QS June 2009] e rawing curves You can always plot a curve, point by point, if you know its equation. Often, however, all you need is a general idea of its shape and a sketch is quite sufficient. Figures 2.25 and 2.26 show some common curves of the form y = x" for n = 1, 2, 3 and 4 and y = _l_ for n = 1 and 2. = x" for n = 1, 2, 3 and 4 Ill y=x X (bl n = 2, y n = l,y=x = x2 X X (cl n = 3,y=x 3 (dl n = 4, y = x 4 Figure 2.25 G ill CD y (al c :E :r ea () c x" Curves of the form y P1 How are the curves for even values of n different from those for odd values of n? Stationary points A turning point is a place where a curve changes from increasing (curve going up) to decreasing (curve going down), or vice versa. A turning point may be described as a maximum (change from increasing to decreasing) or a minimum (change from decreasing to increasing). Turning points are examples of stationary points, where the gradient is zero. In general, the curve of a polynomial of order n has up to n 1 turning points as shown in figure 2.26. a maximum point y . . stationary point. Gl E C) X Gl Cl a minimum point Gl ea c: :s.. C) 0 0 \ y X 4, Figure 2.26 There are some polynomials for which not all the stationary points materialise, as in the case of y= x 4  4x 3 + 5x 2 (whose curve is shown in figure 2.27). To be accurate, you say that the curve of a polynomial of order n has at most n 1 stationary points. Figure 2.27 Behaviour for large x (positive and negative) What can you say about the value of a polynomial for large positive values and large negative values of x? As an example, look at f(x) = x 3 + 2x2 + 3x+ 9, and take 1000 as a large number. f(lOOO) = 1000000000 + 2000000 + 3000 + 9 = 1 002 003 009 Similarly, f( 1000) = 1000 000 000 + 2 000 000 3000 + 9 = 998002 991. Note 1 The term x 3 makes by far the largest contribution to the answers. lt is the dominant term. For a polynomial of order n, the term in xn is dominant as X7 ± oo. 2 In both cases the answers are extremely large numbers. You will probably have noticed already that away from their turning points, polynomial curves quickly disappear off the top or bottom of the page. For all polynomials as x 7 ±oo, either f(x) 7 +oo or f(x) 7 oo. When investigating the behaviour of a polynomial of order n as x 7 ± oo, you need to look at the term in xn and ask two questions. m (ii) Is n even or odd? Is the coefficient of x" positive or negative? According to the answers, the curve will have one of the four types of shape illustrated in figure 2.28. Intersections with the x and y axes The constant term in the polynomial gives the value of y where the curve intersects they axis. So y = x 8 + 5x 6 + 17 x 3 + 23 crosses they axis at the point (0, 23) . Similarly, y = x 3 + x crosses the y axis at (0, 0), the origin, since the constant term is zero. When the polynomial is given, or known, in factorised form you can see at once where it crosses the x axis. The curve y = ( x 2 Xx 8 Xx 9), for example, crosses the x axis at x = 2, x = 8 and x = 9. Each of these values makes one of the brackets equal to zero, and soy= 0. P1 n even n odd P1 ... coefficient of x" positive Ill E ,,  0 Ill r::n ...ea Ill \ \ ,,, ... \ c .. :s 0 0 u  ,_/,,'...,,/ ,.. \ I coefficient of x" negative Figure 2.28 EXAMPLE 2.13 Sketch the curve y= x 3  3x2  x+ 3 = (x+ 1) (x1) (x 3). y SOLUTION I ,\J '" y=il 3x2+x+ 3 I 2 4 X Figure 2.29 Since the polynomial is of order 3, the curve has up to two stationary points. The term in x 3 has a positive coefficient (+1) and 3 is an odd number, so the general shape is as shown on the left of figure 2.29. The actual equation y= x 3  3x2  x+ 3 = (x+ 1) (x 1) (x3) tells you that the curve: crosses they axis at (0, 3) crosses the x axis at (1, 0), (1, O) and (3, 0). This is enough information to sketch the curve (see the right of figure 2.29). In this example the polynomial x 3  3x2  x + 3 has three factors, (x + 1), (x 1) and (x 3). Each of these corresponds to an intersection with the x axis, and to a root of the equation x 3  3x2  x + 3 = 0. Clearly a cubic polynomial cannot have more than three factors of this type, since the highest power of xis 3. A cubic polynomial may, however, cross the x axis fewer than three times, as in the case of f(x) = x 3  x 2  4x + 6 (see figure 2.30). X Figure 2.30 Note This illustrates an important result. If f(x) is a polynomial of degree n, the curve with equation v= f(x) crosses the x axis at most n times, and the equation f(x) = 0 has at most n roots. An important case occurs when the polynomial function has one or more repeated factors, as in figure 2.31. In such cases the curves touch the x axis at points corresponding to the repeated roots. f(x) f(x) 0 f(x)=(x  l)(x  3f Figure 2.31 X X f(x) = x2(x  4f P1 P1 Sketch the following curves, marking clearly the values of x and y where they cross the coordinate axes. EXERCISE 2E . . Gl E 0 1 y= x(x 3)(x+ 4) 2 y= (x+ 1)(2x 5)(x 4) 3 y= (5 x)(xl)(x+ 3) 4 y= x 2 (x 3) 5 y= (x+ 1) 2 (2 x) 6 y= (3x 4)(4x 3)2 7 y= (x+ 2) 2 (x 4) 2 8 y= (x 3) 2 (4 + x) 2 9 Suggest an equation for this curve. Gl Cl Gl «< c .. :s 0 0 u y 3 Q X What happens to the curve of a polynomial if it has a factor of the form (x a)3? Or (x a) 4 ? 1 Curves of the form y =;,;(for x :1 0) y y X X (a) n = l,y= 1 x (b) n = 1 2, y = x2 Figure 2.32 The curves for n = 3, 5, .. . are not unlike that for n = 1, those for n = 4, 6, ... are like that for n = 2. In all cases the point x = 0 is excluded because is undefined. t An important feature of these curves is that they approach both the x and the y axes ever more closely but never actually reach them. These lines are described as asymptotes to the curves. Asymptotes may be vertical (e.g. they axis), horizontal, or lie at an angle, when they are called oblique. Asymptotes are usually marked on graphs as dotted lines but in the cases above the lines are already there, being coordinate axes. The curves have different branches which never meet. A curve with different branches is said to be discontinuous, whereas one with no breaks, like y= x 2 , is continuous. P1 c iil :I IQ n c Ill The circle You are of course familiar with the circle, and have probably done calculations involving its area and circumference. In this section you are introduced to the equation of a circle. The circle is defined as the locus of all the points in a plane which are at a fixed distance (the radius) from a given point (the centre). (Locus means path.) As you have seen, the length of a line joining (xl' y 1) to (x 2, y 2 ) is given by length= x/ + (y2  YY. This is used to derive the equation of a circle. In the case of a circle of radius 3, with its centre at the origin, any point (x, y) on the circumference is distance 3 from the origin. Since the distance of x, ) from (0, 0) is given by W + (y 0)2, this means that (x W + (y W = 3 or 2 2 x + y = 9 and this is the equation of the circle. This circle is shown in figure 2.33. y X Figure 2.33 These results can be generalised to give the equation of a circle centre (0, 0), radius r as follows: P1 .. The intersection of a line and a curve When a line and a curve are in the same plane, there are three possible situations. m All points of intersection are distinct (see figure 2.34). Gl E y 0 . Gl Cl Gl Ill c :s.. 0 0 (J X Figure 2.34 (ii) The line is a tangent to the curve at one (or more) point(s) (see figure 2.35). In this case, each point of contact corresponds to two (or more) coincident points of intersection. It is possible that the tangent will also intersect the curve somewhere else. Figure 2.35 (iiil The line and the curve do not meet (see figure 2.36). The coordinates of the point of intersection can be found by solving the two equations simultaneously. If you obtain an equation with no real roots, the conclusion is that there is no point of intersection. P1 i ::r ... .o· y Cl) :;· Cl) Ul Cl) n :I ... C) I» :;· Cl) X I» :I Cl. I» . n c < Cl) Figure 2.36 The equation of the straight line is, of course, linear and that of the curve nonlinear. The examples which follow remind you how to solve such pairs of equations. EXAMPLE 2.14 Find the coordinates of the two points where the line y 3x = 2 intersects the curve y = 2:il. SOLUTION First sketch the line and the curve. Figure 2.37 You can find where the line and curve intersect by solving the simultaneous equations: P1 CD y 3x = 2 and y =2i2 Make y the subject of CD: y = 3x + 2 Substitute ® into @ : y = 2:x? . Ill E 0 Ill Cl Ill Ill .. @ ® 3x+2 =2i2 2:x?3x2 =0 c . :s (2x+ l)(x 2) = 0 0 0 (.) coordinates ofthe x=2orx=! 2 Substitute into the linear equation, y = 3x + 2, to fmd the corresponding y coordinates. 2 2 So the coordinates of the points of intersection are (2, 8) EXAMPLE 2.15 (i) Find the value of k for which the line 2y = x + k forms a tangent to the curve l=2x. liil Hence, for this value of k, find the coordinates of the point where the line 2y = x + k meets the curve. SOLUTION lil You can find where the line forms a tangent to the curve by solving the simultaneous equations: 2y = x+ k y 2 =2x and CD @ When you eliminate either x or y between the equations you will be left with a quadratic equation. A tangent meets the curve at just one point and so you need to find the value of k which gives you just one repeated root for the quadratic equation. Make x the subject of CD: Substitute ®into@: => x=2yk I =2x y2 ® = 2(2y k) .r =4y2k .r 4y+ 2k = o G) You can use the discriminant, b2  4ac, to find the value of k such that the equation has one repeated root. The condition is b2  4ac = 0 y 2 4y+2k=O b2  4ac = 0 a=1,b=4andc=2k (4)2 4 x 1 x 2k = 0 16Sk=O k=2 So the line 2y = x + 2 forms a tangent to the curve y 2 = 2x. {ii) You have already started to solve the equations 2y = x + 2 and y 2 = 2x in part (i). Look at equation @ : y2  4y + 2k = 0 You know from part mthat k = 2 so you can solve the quadratic to find y. y 2  4y+ 4 = 0 (y 2)(y 2) = 0 y=2 Notice that this is a repeated root so the line is a tangent to the curve. Now substitute y = 2 into the equation of the line to find the x coordinate. Wheny= 2: 2y= x+ 2 x=2 4=x+ 2 So the tangent meets the curve at the point (2, 2). EXERCISE 2F 1 Show that the line y = 3x + 1 crosses the curve y = x 2 + 3 at ( 1, 4) and find the coordinates of the other point of intersection. 2 (i) (ii) 3 m Find the coordinates of the points of intersection of the line y = 2x and the curve y= x 2 + 6x 5. (ii) 4 5 Find the coordinates of the points A and B where the line y = 2x 1 cuts the curve y = x 2  4. Find the distance AB. Show also that the line y = 2x does not cross the curve y = x 2 + 6x + 5. The line 3y = 5 x intersects the curve between the two points. 2y = x at two points. Find the distance The equation of a curve is xy = 8 and the equation of a line is 2x + y = k, where k is a constant. Find the values of k for which the line forms a tangent to the curve. 6 Find the value of the constant curve y2 = 4x. c for which the line y = 4x + c is a tangent to the P1 m .. >c CD n iii' CD .,N 7 P1 The equation of a curve is xy = 10 and the equation of a line l is 2x + y = q, where q is a number. m In the case where q = 9, find the coordinates of the points of intersection of l and the curve. (iil Find the set of values of q for which l does not intersect the curve. ; E 0 8 The curve y2 = 12x intersects the line . G) 1:11 3y = 4x + 6 at two points. Find the distance between the two points. G) ea c [Cambridge AS & A Level Mathematics 9709, Paper I QS June 2006] 0 0 9 Determine the set of values of the constant (.) k for which the line y = 4x + k does not intersect the curve y = i2. [Cambridge AS & A Level Mathematics 9709, Paper I QI November 2007] 1o Find the set of values of k for which the line y = kx 4 intersects the curve y = i2 2x at two distinct points. [Cambridge AS & A Level Mathematics 9709, Paper I Q2 June 2009] KEY POINTS 1 The gradient of the straight line joining the points (xi, y1 ) and (x2 , y2 ) is given by y y Xz X. 2 1 gradient =  . when the same scale is used on both axes, m= tanO. 2 Two lines are parallel when their gradients are equal. 3 Two lines are perpendicular when the product of their gradients is 1 . 4 When the points A and B have coordinates (xi, y1 ) and (X2, y2 ) respectively, then the distance AB is xi+ (y 2  Yi the midpoint of the line AB is 5 (X. : Xz, Yt : Yz J The equation of a straight line may take any of the following forms: • • • • • • line parallel to the y axis: x = a line parallel to the x axis: y = b line through the origin with gradient m: y = mx line through (0, c) with gradient m: y = mx + c line through (x 1, y 1) with gradient m: y y 1 = m(x x 1) line through (x 1, y 1) and (x 2 , yz): Y Y1 Yz Y1 x X. Y Yt or Xz X. x X. Yz Y1 Xz X. Sequences and series P1 en Population, when unchecked, increases in a geometrical ratio. Subsistence increases only in an arithmetical ratio. A slight acquaintance with numbers will show the immensity of the first .zc CD :I n CD Ul I» power in comparison with the second. :I Q. Thomas Malthus (1798) ..a;· Ul CD Ul FISH&CHIPS Sunday 128 Monday closed Tuesday to Friday 1210 Saturday 1111 ASIAN SAVINGS DOUBLE your$$ every 10 years 0 Each of the following sequences is related to one of the pictures above. (i) (ii) (iii) (iv) 5000, 10000, 20000,40000, ... . 8, 0, 10, 10, 10, 10, 12, 8, 0, ... . 5, 3.5, 0, 3.5, 5, 3.5, 0, 3.5, 5, 3.5, .... 20, 40, 60, 80, 100, .... Identify which sequence goes with which picture. (bl Give the next few numbers in each sequence. (cl Describe the pattern of the numbers in each case. (d) Decide whether the sequence will go on for ever, or come to a stop. tal P1 Definitions and notation A sequence is a set of numbers in a given order, like 1 1 1 1 2' 4' 8' 16' · · ·· Ill G) 'i: G) Ill c IQ Ill Each of these numbers is called a term of the sequence. When writing the terms of a sequence algebraically, it is usual to denote the position of any term in the sequence by a subscript, so that a general sequence might be written: G) u ul' u2 , u 3, ... , with general term uk. cG) :I c:r G) Cl) i, For the sequence above, the first term is u1 = the second term is u2 = and soon. When the terms of a sequence are added together, like 1 1 1 1 2+4+8+16+ ... the resulting sum is called a series. The process of adding the terms together is called summation and indicated by the symbol (the Greek letter sigm a), with the position of the first and last terms involved given as limits. L k=S 5 So u 1 + u 2 + u 3 + u4 + u 5 is written I,uk or I,uk. k=l k=l In cases like this one, where there is no possibility of confusion, the sum would 5 normally be written more simply as I,uk. 1 If all the terms were to be summed, it would usually be denoted even more simply, as I,uk, or even I,uk. k A sequence may have an infinite number of terms, in which case it is called an infinite sequence. The corresponding series is called an infinite series. In mathematics, although the word series can describe the sum of the terms of any sequence, it is usually used only when summing the sequence provides some useful or interesting overall result. For example: (1 + x) 5 = 1 + Sx+ 10x 2 + 10x3 + 5x4 + x 5 infinite number of terms. The phrase 'sum of a sequence' is often used to mean the sum of the terms of a sequence (i.e. the series). Arithmetic progressions ___ o / / / / / / / ./ ./ ./ ./ / / / /   P1  . l> ::+ ::r 3 So ii' .. 'C SCORECARD · . 0 OUT 4 1.. 5 10 3 HOME 3 5 7 I 2 4 8 6 4 5 4 5 7 cc Ill Ill c;· :I Ill Figure 3.1 Any ordered set of numbers, like the scores of this golfer on an 18hole round (see figure 3.1) form a sequence. In mathematics, we are particularly interested in those which have a welldefined pattern, often in the form of an algebraic formula linking the terms. The sequences you met at the start of this chapter show various types of pattern. A sequence in which the terms increase by the addition of a fixed amount (or decrease by the subtraction of a fixed amount), is described as arithmetic. The increase from one term to the next is called the common difference. Thus the sequence 5 8 11 14 ... is arithmetic with +3 +3 +3 common difference 3. This sequence can be written algebraically as uk= 2 + 3kfor k= 1, 2, 3, ... When k = 1, u1 = 2 + 3 = 5 k = 2, u 2 = 2 + 6 = 8 k= 3, u3 = 2 + 9 = 11 This version has the advantage that the righthand side begins with the first term of the sequence. and so on. (An equivalent way of writing this is uk = 5 + 3(k 1) for k = 1, 2, 3, .... ) As successive terms of an arithmetic sequence increase (or decrease) by a fixed amount called the common difference, d, you can define each term in the sequence in relation to the previous term: When the terms of an arithmetic sequence are added together, the sum is called an arithmetic progression, often abbreviated to A.P. An alternative name is an arithmetic series. Notation When describing arithmetic progressions and sequences in this book, the following conventions will be used: • first term, u 1 = a Ill Ql ·;:: Ql Ill • number of terms = n '1:1 c Ill • last term, un = l Ill Ql u c • common difference = d Ql :I er Ql en • the general term, uk, is that in position k (i.e. the kth term) . Thus in the arithmetic sequence 5, 7, 9, 11, 13, 15, 17, a= 5, l = 17, d = 2 and n = 7. The terms are formed as follows. u 1 =a =5 u2 =a+d u 3 = a+ 2d u 4 =a+ 3d u5 =a+ 4d u6 =a+ 5d u7 =a+ 6d =5+2 =7 = 5+ 2x 2 = 9 = 5 + 3 X 2 = 11 = 5 + 4 X 2 = 13 = 5 + 5 x 2 = 15 = 5 + 6 X 2 = 17 You can see that any term is given by the first term plus a number of differences. The number of differences is, in each case, one less than the number of the term. You can express this mathematically as For the last term, this becomes l= a+ (nl)d. These are both general formulae which apply to any arithmetic sequence. EXAMPLE 3.1 Find the 17th term in the arithmetic sequence 12, 9, 6, .... SOLUTION In this case a= 12 and d = 3. Using uk =a+ (kl)d, you obtain u17 = 12 + (17 1) x ( 3) = 1248 =36. The 17th term is 36. EXAMPLE 3.2 How many terms are there in the sequence 11, 15, 19, ... , 643? P1 SOLUTION This is an arithmetic sequence with first term a common difference d = 4. = 11, last term l = 643 and .. :r> :+ ::r l = a+ (n l)d, you have 643 = 11 + 4(n 1) ==> 4n = 643  11 + 4 ==> n = 159. Using the result There are 159 terms. I  a d n=   +1 This gives the number of terms in an A.P. directly if you know the first term, the last term and the common difference. The sum of the terms of an arithmetic progression When Carl Friederich Gauss (17771855) was at school he was always quick to answer mathematics questions. One day his teacher, hoping for half an hour of peace and quiet, told his class to add up all the whole numbers from 1 to 100. Almost at once the 10yearold Gauss announced that he had done it and that the answer was 5050. Gauss had not of course added the terms one by one. Instead he wrote the series down twice, once in the given order and once backwards, and added the two together: 0 0 0 0 0 0 + 98 + 99 + 100 + 3 + 2 + 1. Adding, 25 =