International Examinations. Advanced Level Mathematics Pure Mathematics 1 Hugh Neill and Douglas Quadling

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1 International Eaminations Advanced Level Mathematics Pure Mathematics Hugh Neill and Douglas Quadling

2 PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE The Pitt Building, Trumpington Street, Cambridge, United Kingdom CAMBRIDGE UNIVERSITY PRESS The Edinburgh Building, Cambridge CB RU, UK 40 West 0th Street, New York, NY 00-4, USA 477 Williamstown Road, Port Melbourne, VIC 307, Australia Ruiz de Alarcón 3, 804 Madrid, Spain Dock House, The Waterfront, Cape Town 800, South Africa Cambridge Universit Press 00 This book is in copright Subject to statutor eception and to the provisions of relevant collective licensing agreements, no reproduction of an part ma take place without the written permission of Cambridge Universit Press First published 00 Printed in the United Kingdom at the Universit Press, Cambridge Tpefaces Times, Helvetica Sstems Microsoft Word, MathTpe A catalogue record for this book is available from the British Librar ISBN paperback Cover image: James L Amos/CORBIS

3 Contents Introduction iv Coordinates, points and lines Surds and indices 7 3 Functions and graphs 3 4 Quadratics 5 5 Inequalities 65 Revision eercise 73 6 Differentiation 75 7 Applications of differentiation 95 8 Sequences 4 9 The binomial theorem 8 0 Trigonometr 38 Combining and inverting functions 56 Etending differentiation 74 Revision eercise 87 3 Vectors 90 4 Geometric sequences 0 5 Second derivatives 5 6 Integration 36 7 Volume of revolution 58 8 Radians 64 Revision eercise 3 79 Practice eaminations 83 Answers 88 Inde 33

4 Coordinates, points and lines This chapter uses coordinates to describe points and lines in two dimensions When ou have completed it, ou should be able to find the distance between two points find the mid-point of a line segment, given the coordinates of its end points find the gradient of a line segment, given the coordinates of its end points find the equation of a line though a given point with a given gradient find the equation of the line joining two points recognise lines from different forms of their equations find the point of intersection of two lines tell from their gradients whether two lines are parallel or perpendicular The distance between two points When ou choose an origin, draw an -ais to the right on the page and a -ais up the page and choose scales along the aes, ou are setting up a coordinate sstem The coordinates of this sstem are called cartesian coordinates after the French mathematician René Descartes, who lived in the 7th centur In Fig, two points A and B have cartesian coordinates ( 4, 3) and ( 0, 7) The part of the line AB which lies between A and B is called a line segment The length of the line segment is the distance between the points A third point C has been added to Fig to form a right-angled triangle You can see that C has the same -coordinate as B and the same -coordinate as A ; that is, C has coordinates ( 0, 3) ( 4,3) A Fig 6 C B ( 0,7) 4 ( 0,3) It is eas to see that AC has length 0 4 = 6, and CB has length 7 3 = 4 Using Pthagoras theorem in triangle ABC shows that the length of the line segment AB is ( 0 4) + ( 7 3) = = = 5 You can use our calculator to give this as 7, if ou need to, but often it is better to leave the answer as 5 The idea of coordinate geometr is to use algebra so that ou can do calculations like this when A and B are an points, and not just the particular points in Fig It often helps to use a notation which shows at a glance which point a coordinate refers to One wa of doing this is with suffies, calling the coordinates of the first point (, ), and

5 PURE MATHEMATICS the coordinates of the second point (, ) Thus, for eample, stands for the -coordinate of the first point Fig shows this general triangle You can see that C now has coordinates (, ), and that AC = and CB = Pthagoras theorem now gives ( ) + ( ) AB = (, ) A C B (, ) (, ) An advantage of using algebra is that this formula works whatever the shape and position of the triangle In Fig 3, the coordinates of Fig A are negative, and in Fig 4 the line slopes downhill rather than uphill as ou move from left to right Use Figs 3 and 4 to work out for ourself the length of AB in each case You can then use the formula to check our answers B( 3,5) A(,5) C A(, ) C B( 6,) Fig 3 Fig 4 In Fig 3, so = 3 ( )= 3+ = 5 and = 5 ( )= 5+ = 6, ( ) + ( ) = + = + = AB = 3 ( ) 5 ( ) And in Fig 4, so = 6 = 5 and = 5= 3, AB = ( 6 ) + ( 5) = 5 + ( 3) = 5+ 9 = 34 Also, it doesn t matter which wa round ou label the points A and B If ou think of B as the first point (, ) and A as the second point (, ), the formula doesn t change For Fig, it would give BA = ( ) + ( ) = + = ( 4 0) + ( 3 7) = , as before

6 CHAPTER : COORDINATES, POINTS AND LINES 3 The distance between the points, line segment joining them) is ( ) + ( ) ( ) and (, ) (or the length of the The mid-point of a line segment You can also use coordinates to find the mid-point of a line segment Fig 5 shows the same line segment as in Fig, but with the mid-point M added The line through M parallel to the -ais meets AC at D Then the lengths of the sides of the triangle ADM are half of those of triangle ACB, so that ( )= ( )= AD = AC = , DM = CB = ( 7 3)= ( 4)= The -coordinate of M is the same as the -coordinate of D, which is 4+ AD = 4+ ( 0 4)= 4+ 3= 7 ( 4,3) A M D C B ( 0,7) ( 0,3) The -coordinate of M is 3+ DM = 3+ ( 7 3)= 3+ = 5 So the mid-point M has coordinates ( 7, 5) Fig 5 In Fig 6 points M and D have been added in the same wa to Fig Eactl as before, ( ) = = ( ) AD = AC = DM CB, So the -coordinate of M is + AD= + ( )= + = + = ( + ) The -coordinate of M is M B (, ) (, ) (, ) A D C + DM = + ( )= + = + = ( + ) Fig 6 The mid-point of the line segment joining, ( ( ) ( + )) +, ( ) and (, ) has coordinates

7 4 PURE MATHEMATICS Now that ou have an algebraic form for the coordinates of the mid-point M ou can use it for an two points For eample, for Fig 3 the mid-point of AB is ( ( ) ( ))= () ( ) ( )= ( ) ( )+ 3, ( )+ 5, 4, And for Fig 4 it is ( ( + 6), ( 5+ ) )= ( ( 7), ( 7) )= ( 3, 3 ) Again, it doesn t matter which ou call the first point and which the second In Fig 5, if ou take (, ) as ( 0, 7) and (, ) as ( 4, 3), ou find that the mid-point is ( ( 0 + 4), ( 7 + 3) 7 5 )= (, ), as before 3 The gradient of a line segment The gradient of a line is a measure of its steepness The steeper the line, the larger the gradient Unlike the distance and the mid-point, the gradient is a propert of the whole line, not just of a particular line segment If ou take an two points on the line and find the increases in the - and -coordinates as ou go from one to the other, as in Fig 7, then the value of the fraction -step -step is the same whichever points ou choose This is the gradient of the line In Fig on page the -step and -step are and, so that: -step Fig 7 -step The gradient of the line joining, ( ) to (, ) is This formula applies whether the coordinates are positive or negative In Fig 3, for eample, the gradient of AB is 5 ( ) ( ) = + 3+ = 5 But notice that in Fig 4 the gradient is = = 5 5 ; the negative gradient tells ou that the line slopes downhill as ou move from left to right As with the other formulae, it doesn t matter which point has the suffi and which has the suffi In Fig, ou can calculate the gradient as either = 6 =, or = = 3 Two lines are parallel if the have the same gradient

8 CHAPTER : COORDINATES, POINTS AND LINES 5 Eample 3 The ends of a line segment are ( p q, p+ q) and ( p+ q, p q) Find the length of the line segment, its gradient and the coordinates of its mid-point For the length and gradient ou have to calculate = ( p+ q) ( p q)= p+ q p+ q = q and = ( p q) ( p+ q)= p q p q = q ( ) + ( ) = ( ) + The length is q ( q) = 4q + 4q = 8q The gradient is q = = q For the mid-point ou have to calculate + = ( p q)+ ( p+ q)= p q+ p+ q = p and + = ( p+ q)+ ( p q)= p+ q+ p q = p The mid-point is ( ( + ), ( + ) )= ( ( p), ( p) )= ( p, p) Tr drawing our own figure to illustrate the results in this eample Eample 3 Prove that the points A(, ), B( 53, ), C( 30, ) and D(, ) form a parallelogram You can approach this problem in a number of was, but whichever method ou use, it is worth drawing a sketch This is shown in Fig 8 Method (using distances) In this method, find the lengths of the opposite sides If both pairs of opposite sides are equal, then ABCD is a parallelogram A(,) ( ) B 5,3 AB = DC = CB = ( 5 ) + ( 3 ) = 0 ( 3 ( ) ) + ( 0 ( ) ) = 0 ( 5 3) + ( 3 0) = 3 D(, ) C( 3, 0) DA = ( ( ) ) + ( ( ) ) = 3 Fig 8 Therefore AB = DC and CB = DA, so ABCD is a parallelogram Method (using mid-points) In this method, begin b finding the mid-points of the diagonals AC and BD If these points are the same, then the diagonals bisect each other, so the quadrilateral is a parallelogram ( ) ( ) The mid-point of AC is ( + 3), ( + 0), which is, The mid-point of BD is ( 5+ 3 ( ( )), + ( ( ))), which is also (, ) So ABCD is a parallelogram

9 6 PURE MATHEMATICS Method 3 (using gradients) In this method, find the gradients of the opposite sides If both pairs of opposite sides are parallel, then ABCD is a parallelogram The gradients of AB and DC are 3 5 = = and 0 ( ) 4 3 ( ) = = respectivel, 4 so AB is parallel to DC The gradients of DA and CB are both 3, so DA is parallel to CB As the opposite sides are parallel, ABCD is a parallelogram Eercise A Do not use a calculator Where appropriate, leave square roots in our answers Find the lengths of the line segments joining these pairs of points In parts (e) and (h) assume that a > 0 ; in parts (i) and (j) assume that p> q > 0 (a) (, 5) and ( 7, 7) (b) ( 3, ) and (, ) (c) ( 4, 5) and (, 0) (d) ( 3, 3) and ( 7, 3) (e) ( aa, ) and ( 0a, 4a) (f) ( a+, a+ 3) and ( a, a ) (g) (, 9) and (, 4) (h) ( a, 5b) and ( 3a, 5b) (i) ( pq, ) and ( qp, ) (j) ( p+ 4q, p q) and ( p 3q, p) Show that the points (, ), ( 6, ), ( 9, 3) and ( 4, ) are vertices of a parallelogram 3 Show that the triangle formed b the points ( 3, ), (, 7) and ( 5, ) is isosceles 4 Show that the points ( 7, ), ( 3, ) and ( 4, 5) lie on a circle with centre (, 0) 5 Find the coordinates of the mid-points of the line segments joining these pairs of points (a) (, ),( 6, 5) (b) ( 5, 7), ( 3, 9) (c) (, 3),(, 6) (d) ( 34, ), ( 85, ) (e) ( p+ 3, p ), ( 3p+ 4, p 5) (f) ( p+ 3, q 7), ( p+ 5, 3 q) (g) ( p+ q, p+ 3q),( 5p q, p 7q) (h) ( a+ 3, b 5), ( a+ 3, b+ 7) 6 A (, ) and B ( 65, ) are the opposite ends of the diameter of a circle Find the coordinates of its centre 7 M( 57, ) is the mid-point of the line segment joining A ( 34, ) to B Find the coordinates of B 8 A(, ), B( 6, ), C( 9, 3) and D( 4, ) are the vertices of a parallelogram Verif that the mid-points of the diagonals AC and BD coincide 9 Which one of the points A( 5, ), B( 6, 3) and C( 47, ) is the mid-point of the other two? Check our answer b calculating two distances 0 Find the gradients of the lines joining the following pairs of points (a) ( 3, 8),( 5, ) (b) (, 3), (, 6) (c) ( 4, 3),( 0, ) (d) ( 5, 3), ( 3, 9) (e) ( p+ 3, p 3), ( p+ 4, p 5) (f) ( p+ 3, q 5), ( q 5, p+ 3) (g) ( p+ q, q+ p 3),( p q+, q p+ 3 ) (h) ( 7, p),(, p)

10 CHAPTER : COORDINATES, POINTS AND LINES 7 Find the gradients of the lines AB and BC where A is ( 3, 4), B is ( 7, 6) and C is ( 3, ) What can ou deduce about the points A, B and C? The point P (, ) lies on the straight line joining A( 30, ) and B( 56, ) Find epressions for the gradients of AP and PB Hence show that = A line joining a verte of a triangle to the mid-point of the opposite side is called a median Find the length of the median AM in the triangle A(, ), B( 0, 3), C( 4, 7) 4 A triangle has vertices A(, ), B( 3, 4) and C( 57, ) (a) Find the coordinates of M, the mid-point of AB, and N, the mid-point of AC (b) Show that MN is parallel to BC ( ) 5 The points A(, ), B( 7, ) and C 4, form a triangle M is the mid-point of AB and N is the mid-point of AC (a) Find the lengths of MN and BC (b) Show that BC = MN 6 The vertices of a quadrilateral ABCD are A(, ), B( 7, 3), C( 9, 7) and D( 3, 3) The points P, Q, R and S are the mid-points of AB, BC, CD and DA respectivel (a) Find the gradient of each side of PQRS (b) What tpe of quadrilateral is PQRS? 7 The origin O and the points P( 4, ), Q( 55, ) and R( 4, ) form a quadrilateral (a) Show that OR is parallel to PQ (b) Show that OP is parallel to RQ (c) Show that OP = OR (d) What shape is OPQR? 8 The origin O and the points L( 3, ), M( 47, ) and N( 64, ) form a quadrilateral (a) Show that ON = LM (b) Show that ON is parallel to LM (c) Show that OM = LN (d) What shape is OLMN? 9 The vertices of a quadrilateral PQRS are P(, ), Q( 70, ), R( 6, 4) and S( 3, ) (a) Find the gradient of each side of the quadrilateral (b) What tpe of quadrilateral is PQRS? 0 The vertices of a quadrilateral are T( 3, ), U( 5, ), V( 87, ) and W( 6, ) The mid-points of UV and VW are M and N respectivel Show that the triangle TMN is isosceles The vertices of a quadrilateral DEFG are D( 3, ), E( 0, 3), F(, 3) and G( 4, ) (a) Find the length of each side of the quadrilateral (b) What tpe of quadrilateral is DEFG? The points A(, ), B( 60, ) and C( 0, ) form an isosceles triangle with AB and BC of equal length The point G is ( 6, 4) (a) Write down the coordinates of M, the mid-point of AC (b) Show that BG = GM and that BGM is a straight line (c) Write down the coordinates of N, the mid-point of BC (d) Show that AGN is a straight line and that AG = GN

11 8 PURE MATHEMATICS 4 What is meant b the equation of a straight line or of a curve? How can ou tell whether or not the points ( 3, 7) and (, 5) lie on the curve = 3 +? The answer is to substitute the coordinates of the points into the equation and see whether the fit; that is, whether the equation is satisfied b the coordinates of the point For ( 3, 7): the right side is = 9 and the left side is 7, so the equation is not satisfied The point ( 3, 7) does not lie on the curve = 3 + For (, 5): the right side is 3 + = 5 and the left side is 5, so the equation is satisfied The point (, 5) lies on the curve = 3 + The equation of a line or curve is a rule for determining whether or not the point with coordinates (, ) lies on the line or curve This is an important wa of thinking about the equation of a line or curve 5 The equation of a line Eample 5 Find the equation of the line with gradient which passes through the point (, ) Fig 9 shows the line of gradient through A(, ), with another point P (, ) ling on it P lies on the line if (and onl if) the gradient of AP is The gradient of AP is Equating this to gives =, which is = 4, or = 3 In the general case, ou need to find the equation of the line with gradient m through the point A with coordinates (, ) Fig 0 shows this line and another point P with coordinates (, ) on it The gradient of AP is Equating to m gives = m, or = m( ) A(,) Fig 9 A(, ) Fig 0 gradient P(,) gradient m P(,) ( ) The equation of the line through, with gradient m is = m ( ) Notice that the coordinates of A, ( ) satisf this equation

12 CHAPTER : COORDINATES, POINTS AND LINES 9 Eample 5 Find the equation of the line through the point ( 3, ) with gradient ( ) ( ) Using the equation = m gives the equation 3= ( ), which is 3= or = + As a check, substitute the coordinates ( 3, ) into both sides of the equation, to make sure that the given point does actuall lie on the line Eample 53 Find the equation of the line joining the points ( 3, 4) and (, ) To find this equation, first find the gradient of the line joining ( 34, ) to (, ) Then ou can use the equation = m ( ) 4 The gradient of the line joining ( 3, 4) to (, ) is ( ) 3 = = 4 The equation of the line through ( 3, 4) with gradient is 4 = ( 3) After multipling out and simplifing ou get 8 = 3, or = + 5 Check this equation mentall b substituting the coordinates of the other point 6 Recognising the equation of a line The answers to Eamples can all be written in the form = m+ c, where m and c are numbers It is eas to show that an equation of this form is the equation of a straight line If = m+ c, then c = m( 0, ) or c m 0 = (ecept when = 0) This equation tells ou that, for all points (, ) whose coordinates satisf the equation, the line joining ( 0,c) to (, ) has gradient m That is, (, ) lies on the line through ( 0,c) with gradient m The point ( 0,c) lies on the -ais The number c is called the -intercept of the line To find the -intercept, put = 0 in the equation, which gives ou can t do this division if m = 0 In that case the line is parallel to the -ais, so there is no - intercept When m = 0, all the points on the line have coordinates of the form ( something,c) Thus the points (, ), (, ), ( 5, ), all lie on the straight line =, shown in Fig As a special case, the -ais has equation = 0 (,) c = But notice that m The straight line = (, ) ( 5, ) Fig

13 0 PURE MATHEMATICS Similarl, a straight line parallel to the -ais has an equation of the form = k All points on it have coordinates ( k,something) Thus the points ( 3, 0), ( 3, ), ( 3, 4), all lie on the line = 3, shown in Fig The -ais itself has equation = 0 ( 3, 4) ( 3, ) The straight line = 3 The line = k does not have a gradient; its gradient is undefined Its equation cannot be written in the form = m+ c ( 3, 0) Fig 7 The equation a + b + c =0 Suppose ou have the equation = + 4 It is natural to multipl b 3 to get = + 4, which can be rearranged to get 3+ 4 = 0 This equation is in the form a + b + c = 0 where a, b and c are constants Notice that the straight lines = m+ c and a + b + c = 0 both contain the letter c, but it doesn t have the same meaning For = m+ c, c is the -intercept, but there is no similar meaning for the c in a + b + c = 0 A simple wa to find the gradient of a + b + c = 0 is to rearrange it into the form = Here are some eamples Eample 7 Find the gradient of the line = 0 Write this equation in the form =, and then use the fact that the straight line m c = + has gradient m From = 0 ou find that 3 = + 4 and = Therefore, comparing this equation with = m+ c, the gradient is 3 Eample 7 One side of a parallelogram lies along the straight line with equation = 0 The point (, 3) is a verte of the parallelogram Find the equation of one other side The line = 0 is the same as = , so its gradient is 3 The line 4 through (, 3) with gradient 3 4 is 3= 3 4 ( ), or = 0 8 The point of intersection of two lines Suppose that ou have two lines with equations = 4 and 3+ = How do ou find the coordinates of the point of intersection of these lines? You want the point (, ) which lies on both lines, so the coordinates (, ) satisf both equations Therefore ou need to solve the equations simultaneousl

14 CHAPTER : COORDINATES, POINTS AND LINES From these two equations, ou find =, =, so the point of intersection is (, ) This argument applies to straight lines with an equations provided the are not parallel To find points of intersection, solve the equations simultaneousl The method can also be used to find the points of intersection of two curves Eercise B Test whether the given point lies on the straight line (or curve) with the given equation (a) (, ) on = 5 3 (b) ( 3, ) on = 3 7 (c) ( 3, 4) on + = 5 (d) (, ) on 3 + = 40 (e) (, ) on = (f) 5p, on = 3 p (g) p, ( p ) + on = + (h) t,t on 4 ( ) ( ) = Find the equations of the straight lines through the given points with the gradients shown Your final answers should not contain an fractions (a) (, 3), gradient 5 (b) (, ), gradient 3 (c) ( 0, 4), gradient (d) (, ), gradient 3 8 (e) ( 0, 0), gradient 3 (f) ( 3, 8), gradient 0 (g) ( 5, ), gradient 3 4 (h) ( 30, ), gradient (i) ( 3, ), gradient 3 8 (j) ( 3, 4), gradient (k) (, ), gradient (l) (, 5), gradient 3 (m) ( 0, 4), gradient 7 (n) ( 0, ), gradient (o) ( 3, ), gradient 5 8 (p) ( 3, 0), gradient 3 5 (q) ( d,0), gradient 7 (r) ( 0, 4), gradient m (s) ( 0,c), gradient 3 (t) ( c,0), gradient m 3 Find the equations of the lines joining the following pairs of points Leave our final answer without fractions and in one of the forms = m+ c or a + b + c = 0 (a) (, 4) and ( 3, 0) (b) ( 4, 5) and (, 7) (c) ( 3, ) and ( 04, ) (d) ( 3, 7) and ( 3, ) (e) ( 0, 3) and ( 5, ) (f) ( 3, ) and ( 4, 0) (g) (, 3) and (, 3) (h) (, 0) and ( 5, ) (i) ( 4, ) and (, 3) (j) (, ) and ( 5, 3) (k) ( 34, ) and ( 39, ) (l) ( 0, ) and ( 0, ) (m) (, 7) and ( 3, 0) (n) ( 54, ) and (, ) (o) ( 0, 0) and ( 5, 3) (p) ( 0, 0) and ( pq, ) (q) ( pq, ) and ( p+ 3, q ) (r) ( p, q) and ( p, q) (s) ( pq, ) and ( p+, q+ ) (t) ( p, 0) and ( 0, q)

15 PURE MATHEMATICS 4 Find the gradients of the following lines (a) + = 7 (b) 3 4 = 8 (c) 5+ = 3 (d) = 5 (e) 3 = 4 (f) 5 = 7 (g) + = 3 (h) = 3( + 4 ) (i) 7 = (j) 3( 4)= 7 (k) = m( d) (l) p + q = pq 5 Find the equation of the line through (, ) parallel to = 3 6 Find the equation of the line through ( 4, 3) parallel to + = 7 7 Find the equation of the line through (, )parallel to the line joining ( 3, ) and ( 5, ) 8 Find the equation of the line through ( 3, 9) parallel to the line joining ( 3, ) and (, 3) 9 Find the equation of the line through (, 7) parallel to the -ais 0 Find the equation of the line through ( d,0) parallel to = m+ c Find the points of intersection of the following pairs of straight lines (a) 3+ 4 = 33, = (b) = 3+, = 4 (c) = 7, 3 = (d) = 3+ 8, = 7 (e) + 5 =, 3+ = 4 (f) + 7 = 47, 5+ 4 = 50 (g) + 3 = 7, 6+ 9 = (h) 3+ = 5, + 3 = (i) = + 3, 4 = 6 (j) a + b = c, = a (k) = m+ c, = m+ d (l) a b =, = Let P, with coordinates ( pq, ), be a fied point on the curve with equation = m+ c and let Q, with coordinates ( rs, ), be an other point on = m+ c Use the fact that the coordinates of P and Q satisf the equation = m+ c to show that the gradient of PQ is m for all positions of Q 3 There are some values of a, b and c for which the equation a + b + c = 0 does not represent a straight line Give an eample of such values 9 The gradients of perpendicular lines In Section 3 it is stated that two lines are parallel if the have the same gradient But what can ou sa about the gradients of two lines which are perpendicular? Firstl, if a line has a positive gradient, then the perpendicular line has a negative gradient, and vice versa But ou can be more eact than this In Fig 3, if the gradient of PB is m, ou can draw a gradient triangle PAB in which PA is one unit and AB is m units Gradient m B m P A Fig 3

16 CHAPTER : COORDINATES, POINTS AND LINES 3 In Fig 4, the gradient triangle PAB has been Gradient m rotated through a right-angle to PA B B, so that PB is perpendicular to PB The -step for B m A m PA B is and the -step is m, so Gradient m -step gradient of PB = = = P A -step m m Fig 4 Therefore the gradient of the line perpendicular to PB is m Thus if the gradients of the two perpendicular lines are m and m, then mm = It is also true that if two lines have gradients m and m, and if mm =, then the lines are perpendicular To prove this, see Miscellaneous eercise Question Two lines with gradients m and m are perpendicular if mm =, or m =, or m = m m Notice that the condition does not work if the lines are parallel to the aes However, ou can see that a line = constant is perpendicular to one of the form = constant Eample 9 Show that the points ( 0, 5), (, ), ( 4, 7) and ( 5, 0) form a rhombus You could tackle this question in several was This solution shows that the points form a parallelogram, and then that its diagonals are perpendicular ( ) The mid-points of the diagonals are ( 0+ 4), ( 5+ 7), or (, ), and ( (( )+ 5), ( + 0) ), or (, ) As these are the same point, the quadrilateral is a parallelogram The gradients of the diagonals are 7 ( 5 ) 0 = = and 5 ( ) = = 6 3 As the product of the gradients is, the diagonals are perpendicular Therefore the parallelogram is a rhombus Eample 9 Find the coordinates of the foot of the perpendicular from A(, 4) to the line joining B( 0, ) and C( 4, ) Alwas draw a diagram, like Fig 5; it need not be to scale The foot of the perpendicular is the point of intersection, P, of BC and the line through A perpendicular to BC First find the gradient of BC and its equation

17 4 PURE MATHEMATICS 4 The gradient of BC is 0 = = The equation of BC is = ( 0 ), which simplifies to + = C(, 4) The gradient of the line through A perpendicular to BC is = The equation of this line is B( 0,) ( 4 )= ( ( ) ), or = 6 A(, 4) P These lines meet at the point P, whose coordinates satisf the simultaneous equations + = and = 6 This is the point (, ) Fig 5 Eercise C In each part write down the gradient of a line which is perpendicular to one with the given gradient (a) (b) 3 (c) 3 4 (d) 5 6 (e) (f) 3 (g) (h) m 4 m p a (i) (j) 0 (k) m (l) q b c In each part find the equation of the line through the given point which is perpendicular to the given line Write our final answer so that it doesn t contain fractions (a) (, 3), = 4+ 3 (b) ( 3, ), = + 3 (c) (, 5), = 5 (d) ( 7, 4), = (e) ( 4, ), + 3 = 8 (f) ( 4, 3), 3 5 = 8 (g) ( 5, 3), = 3 (h) ( 0, 3), = (i) ( 0, 0), = m+ c (j) ( ab, ), = m+ c (k) ( cd, ), n = p (l) (, ), a + b = c 3 Find the equation of the line through the point ( 5, ) which is perpendicular to the line = 3+ Find also the point of intersection of the two lines 4 Find the equation of the line through the point (, ) which is perpendicular to the line 3 = Find also the point of intersection of the two lines 5 A line through a verte of a triangle which is perpendicular to the opposite side is called an altitude Find the equation of the altitude through the verte A of the triangle ABC where A is the point (, 3), B is (, 7) and C is ( 4, )

18 CHAPTER : COORDINATES, POINTS AND LINES 5 6 P( 5, ), Q( 5, ) and R( 8, 7) form a triangle (a) Find the equations of the altitudes (see Question 5) through (i) R and (ii) Q (b) Find the point of intersection of these altitudes (c) Show that the altitude through P also passes through this point Miscellaneous eercise Show that the triangle formed b the points ( 5, ), ( 3, ) and ( 59, ) is right-angled Find the coordinates of the point where the lines + = 3 and 3+ 5 = 0 meet 3 A triangle is formed b the points A( 3, ), B( 57, ) and C( 08, ) (a) Show that the angle ACB is a right angle (b) Find the coordinates of the point where the line through B parallel to AC cuts the -ais 4 A( 7, ) and C( 4, ) are two vertices of a square ABCD (a) Find the equation of the diagonal BD (b) Find the coordinates of B and of D 5 A quadrilateral ABCD is formed b the points A( 3, ), B( 43, ), C( 9, ) and D(, 3) (a) Show that all four sides are equal in length (b) Show that ABCD is not a square 6 P is the point ( 7, 5) and l is the line with equation 3+ 4 = 6 (a) Find the equation of the line l which passes through P and is perpendicular to l (b) Find the point of intersection of the lines l and l (c) Find the perpendicular distance of P from the line l 7 Prove that the triangle with vertices ( 8, ), ( 30, ) and ( 8, ) is isosceles Find its area 8 The three straight lines =, 7 = and 4+ = 60 form a triangle Find the coordinates of its vertices 9 Find the equation of the line through (, 3) which is parallel to + 7 = 5 Give our answer in the form a + b = c 0 Find the equation of the perpendicular bisector of the line joining (, 5) and ( 43) The points A(, ), B( 35, ), C( 66, ) and D form a parallelogram Find the coordinates of the mid-point of AC Use our answer to find the coordinates of D The point P is the foot of the perpendicular from the point A( 03, ) to the line = 3 (a) Find the equation of the line AP (b) Find the coordinates of the point P (c) Find the perpendicular distance of A from the line = 3,

19 6 PURE MATHEMATICS 3 Points which lie on the same straight line are called collinear Show that the points ( 3, ), ( 47, ) and (, 5) are collinear 4 Find the equation of the straight line that passes through the points ( 3, ) and (, ), giving our answer in the form a + b + c = 0 Hence find the coordinates of the point of intersection of the line and the -ais (OCR) 5 The coordinates of the points A and B are ( 3, ) and ( 4, 5) respectivel Find the coordinates of the mid-point of AB, and the gradient of AB Hence find the equation of the perpendicular bisector of AB, giving our answer in the form a + b + c = 0, where a, b and c are integers (OCR) 6 The curve = + crosses the -ais at the point A and the -ais at the point B + (a) Calculate the coordinates of A and of B (b) Find the equation of the line AB (c) Calculate the coordinates of the point of intersection of the line AB and the line with equation 3 = 4 (OCR) 7 The straight line p passes through the point ( 0, ) and is perpendicular to the line r with equation + = Find the equation of p Find also the coordinates of the point of intersection of p and r, and deduce the perpendicular distance from the point ( 0, ) to the line r (OCR) 8 Show b calculation that the points P( 07, ), Q( 65, ), R( 5, ) and S( 4, ) are the vertices of a rectangle 9 The line 3 4 = 8 meets the -ais at A The point C has coordinates ( 9, ) The line through C perpendicular to 3 4 = 8 meets it at B Calculate the area of the triangle ABC 0 The points A( 3, 4) and C( 5, 4) are the ends of the diagonal of a rhombus ABCD (a) Find the equation of the diagonal BD (b) Given that the side BC has gradient 5, find the coordinates of B and hence of D 3 Find the equations of the medians (see Eercise A Question 3) of the triangle with vertices ( 0, ), ( 6, 0) and ( 4, 4) Show that the medians are concurrent (all pass through the same point) Two lines have equations = m+ c and = m + c lines are perpendicular, and mm = Prove that the