Cambridge A Level Mathematics 9709 — 2024 May/June Paper 1 · Variant 2

1 The coefficient of x2 in the expansion of ( 1 - 4)x 6 is 12 times the coefficient of x2 in the expansion of ( 2 + ax) 5 . Find the value of the positive constant a. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 2 240    x or 2 80a [x2] B1 May be seen in an expansion. 2 240 12 80   a M1 Their 240 equated to 12 × their 2 80a which must contain a2. 0.5 A1 OE Condone ± 0.5 3

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2 The curve y = x2 is transformed to the curve y = 4 ( x - 3) 2 - 8 . Describe fully a sequence of transformations that have been combined, making clear the order in which the transformations have been applied. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Stretch factor 4 in y-direction/parallel to the y axis/vertically. B1 Allow use of SF in place of factor. Allow in/on/along the y axis or ‘the x axis is invariant.’ Translation 3 0    or 3 parallel to the x axis or in the x direction, allow horizontally. 0 8        or 8 parallel to the y axis or in the y direction, allow vertically. B2 Condone ‘Shift’. These translations can be combined as 3 8        , this counts as 2 elements. Give priority to a correct vector over any incorrect wording. B2 for all 3 B1 for 2 out of 3 Two translations, one in each direction, and a stretch only. M1 Condone inaccurate terminology, such as up, down, left and right, if the intention is clear. Correct order of operations. The stretch which must be in the in the y direction must come before the translation in the y direction. A1 Condone inaccurate terminology if the intention is clear but numerical values must be correct. Question Answer Marks Guidance 2 Alternative Method for Question 2 Translation 3 0    or 3 parallel to the x axis or in the x direction, allow horizontally. 0 2        or 2 parallel to the y axis or in the y direction, allow vertically. (B2) Condone ‘Shift’. These translations can be combined as 3 2        , this counts as 2 elements. Give priority to a correct vector over any incorrect wording. B2 for all 3. B1 for 2 out of 3. Stretch factor 4 in y-direction/parallel to the y axis/vertically. (B1) Allow use of SF in place of factor. Allow in/on/along the y axis or “the x axis is invariant.” Two translations, one in each direction, and a stretch only. (M1) Condone inaccurate terminology, such as transform, move, up, down, left and right, if the intention is clear. Correct order of operations. The stretch which must be in the in the y direction must come after the translation in the y direction. (A1) Condone inaccurate terminology if the intention is clear but numerical values must be correct. 5

More questions on Functions

7 tan i3 (a) Show that the equation + 12 = 0 can be expressed as cos i 12 sin 2i - 7 sin i - 12 = 0 . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 7 tan i (b) Hence solve the equation + 12 = 0 for 0c G i G 360c. [3] cos i ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a)   sin 7 cos 12 0 cos      sin leading to 7 12cos 0 cos            Use of sin tan cos     .    2 7sin 12 1 sin 0      DM1 Use of 2 2 1   s c . ⇒ 2 12sin 7sin 12 0      A1 AG, WWW Condone use of s, c and t and/or omission of θ throughout working but the A1 is for cao. 3 3(b)    2 12sin 7sin 12 0 leading to 4sin 3 3sin 4              M1 3 sin 4  4 or 3       B1 OE, WWW Can be implied by a correct value for 1 3 sin 4        e.g.  48.6°.  228.6 ,311.4    B1 AWRT, WWW No others in the range 0 360    . Ignore any answers outside this range. Condone 229°, 311°. 3

More questions on Trigonometry

4 The function f is defined as follows: f ( x) = x - 1 for x 2 1. (a) Find an expression for f - 1 ( x) . [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ y O x 1 The diagram shows the graph of y = g ( x) where g ( x) = 2 for x ! R . x + 2 (b) State the range of g and explain whether g -1 exists. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1The function h is defined by h ( x) = 2 for x H 0 . x + 2 25(c) Solve the equation hf ( x) = f Give your answer in the form a + b c , where a, b and c are e 16 o. integers. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 2 1 f 1        x x B1 ISW Condone ‘y =’. 1 4(b)  1 0 g 2  x  or   1 1 g 0 and g or 0, 2 2 x x         B1 Do not allow   1 g 0, g 2  x x  . Do not allow   1 g 0 or g 2 x x   . Condone g or y in place of  g . x g–1 does not exist because it is one to many or g–1 does not exist because it is not one to one. Or g–1 does not exist because g is not one to one or g–1 does not exist because g is many to one or g–1 does not exist because g fails the horizontal line test. B1 g–1 can be replaced by ‘It’ throughout. A correct statement followed by any further incorrect explanation can be awarded B1. 2 Question Answer Marks Guidance 4(c) 25 1 f 16 4       B1 SOI   2 1 1 4 1 2    x M1 Equating   2 1 1 2   x , or their ‘simplified’ version, to their 25 f . 16         2 1 2 4 leading to 1 2 x x             2 leading to 1 2 x   Or   2 2 1 2 4 leading to 2 1 0 leading to 1 2 x x x x x            Or 2 6 36 4 1 2 leading to 6 1 0 leading to 2 x x x x x           A1 Simplification as far as x =… Allow just + in the results because  can be disregarded at this stage. Can be implied by the final answer. Note: 1 2  x scores A0. 3 2 2  A1 Must discount the solution 3 2 2  . 4

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5 The first and second terms of an arithmetic progression are tan i and sin i respectively, where r . 0 1 i 1 12 r , find the exact sum of the first 40 terms of the progression. [4] (a) Given that i = 14 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The first and second terms of a geometric progression are tan i and sin i respectively, where r .0 1 i 1 12 (b) (i) Find the sum to infinity of the progression in terms of i. [2] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... r , find the sum of the first 10 terms of the progression. Give your answer (ii) Given that i = 13 correct to 3 significant figures. [3] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................

Mark scheme: 5(a) sin tan     d term. Condone incorrect evaluation before subtraction.   2 1 2   d A1 OE Sight of 0.29  AWRT can be awarded M1A1.       40 40 2 tan 39 sin tan 2        S M1 Use of a correct formula for S40. Condone use of their, clearly identified, incorrect values for a and d for this mark. 780 390 2 740 or 740 2   A1 ISW If A0 then sight of 188 AWRT, 188.5 or 189  should be awarded M1A1M1A0. 4 5(b)(i)   sin cos tan      r B1 Condone omission of . tan 1 cos     S or 2 sin cos    cos or 2 tan tan sin     B1 ISW Do not allow fractions within fractions nor omission of . 2 Question Answer Marks Guidance 5(b)(ii)   1 3 1.73.. and 2   a r B1 OE, SOI.   10 10 1 1 2 3 1 1 2                        S M1 This mark can be awarded for a correct formula with their values for a and r or sin tan and or cos . tan a r       Condone 10 1 2 . = 3.46 A1 AWRT Condone 1023 3 512 . 3 Note: S9 gives the same answer but scores B1M0A0.

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2 has a minimum point at A and intersects the positive x-axis at B.6 The curve with equation y = 2x - 8x 1 (a) Find the coordinates of A and B. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) y O B x 2x - 32 y = 3 1 2 y = 22x - 88x A 2 and the line AB. It is given that the The diagram shows the curve with equation y = 2x - 8x 1 2x - 32 equation of AB is y = . 3 Find the area of the shaded region between the curve and the line. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) 1 2 d 1 2 8 d 2     y x x 1 2 2 4 0    x M1 Equating their two term d d y x , with at least one term correct, to 0. [A is]   4, 8  or 4, 8 x y   A1 [B is]   16,0 or 16, 0 x y   B1 4 Note: Correct answers without use of d d y x can be awarded 4/4. Question Answer Marks Guidance 6(b)    3 2 2 2 8 3 2 2 x x C    B1 Seen correct in unsimplified form or better.     2 2 2 32 32 or 3 12     x x x C B1 Seen correct in unsimplified form or better. Attempt to integrate, defined by at least one correct power in each expression, and then subtract. M1 Multiplying by 3 before integration scores M0. 3 3 2 2 2 2 8 8 16 .16 4 .4 3 3 2 2                                        2 2 16 32 16 4 32 4 3 3                            M1 Use of their x values, > 0, from (a) as limits in their integrated expressions. Allow, for correct limits, sight of  256 80 256 112 3 3 3 3                                                 . If incorrect limits are used, then clear substitution must be seen. Question Answer Marks Guidance 6(b) Alternative Method 1 for first 4 marks of Question 6(b)    3 1 2 2 2 8 2 8 3 2 x x dx x x C              (B1) Seen correct in unsimplified form or better. [Area of triangle =] 48 (B1) Attempt to integrate, defined by at least one correct power, and then subtract their triangle area. (M1) 3 3 2 2 2 2 8 8 16 .16 4 .4 3 3 2 2                                      (M1) Use of their x values, > 0, from (a) as limits in their integrated expression. Allow sight of 256 80 3 3                            . If incorrect limits are used, then clear substitution must be seen. Question Answer Marks Guidance 6(b) Alternative Method 2 for first 4 marks of Question 6(b) Subtract and then integrate, defined by at least two correct powers. Condone functions being the wrong way round. (M1) If terms in x have not been combined use the first scheme.  3 2 2 4 8 32 3 3 2 3 2               x x x (B2,1,0) B2 for 3 correct terms, B1 for any 2 correct terms.  3 3 2 2 2 2 4 8 32 16 4 8 32 4 16 16 4 4 3 3 3 2 3 3 2 3 2 2                                               (M1) Use of their x values, >0, from (a) as limits in their integrated expression. Allow sight of 32 0 3         . If incorrect limits are used, then clear substitution must be seen. 32 3 , 10 2 3 or 10.7 (B1) AWRT Allow 32 3  or 32 3  changed to + 32 3 for this mark. (5) Condone the inclusion of π for the first 4 marks but use of 2 y  scores a maximum of B1 for the triangle.

More questions on Integration

7 The equation of a circle is ( x - 6) 2 + ( y + a) 2 = 18 . The line with equation y = 2a - x is a tangent to the circle. (a) Find the two possible values of the constant a. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) For the greater value of a, find the equation of the diameter which is perpendicular to the given tangent. 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Mark scheme: 7(a) 2 6   x 2 2 18    a x a M1* Replacing y with 2a – x in the circle equation, condone incorrect expansion before substitution.   2 2 2 12 6 9 36 18 0       x x ax a A1 All terms collected on one side of the equation. May be implied by the discriminant.      2 2 12 6 4 2 9 18 0      a a DM1 Correct use of “b2 — 4ac” from their 3 term quadratic equation in x , with an x term of the form    m na x with both m and n  0.   2 36 144 0 0    a a A1 0, 4 a a   A1 5 7(b) [Centre is] (6, −4) or [Point of intersection is] (9, −1) B1 [Gradient of diameter] 1  B1   4 6 or 1 9 leading to 10 y x y x y x        B1FT FT on their point of intersection or their centre with an x co-ordinate of ±6 and gradient = 1. 3

More questions on Coordinate geometry

8 B C A 2 cm D 1 1 r rad 3 r rad 3 F E The diagram shows a symmetrical plate ABCDEF. The line ABCD is straight and the length of BC is 2 cm. Each of the two sectors ABF and DCE is of radius r cm and each of the angles ABF and DCE is r radians. equal to 13 (a) It is given that r = 0.4 cm. (i) Show that the length EF = 2.4 cm. [2] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) Find the area of the plate. Give your answer correct to 3 significant figures. [4] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ........................................................................................................................................................... ........................................................................................................................................................... ........................................................................................................................................................... ........................................................................................................................................................... ........................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (b) It is given instead that the perimeter of the plate is 6 cm. Find the value of r. Give your answer correct to 3 significant figures. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 8(a)(i) C π ˆ 6 XCE  π ˆ 3 CEX  X E 0.4  XE π π sin or cos 6 0.4 3 XE  [XE = 0.2] M1 A correct trig expression involving XE. Do not condone a mixture of degrees and radians. Length EF = 2 2 0.2  = 2.4 A1 AG 2 Question Answer Marks Guidance 8(a)(ii)   π 0.4cos 6 CX  or π 0.4sin 3 or 2 2 0.4 0.2  B1 OE, SOI Expect 3 5 or 0.3464.     2 1 π Sector 0.4 2 3    B1 SOI Expect 0.0838 or 2π 75 . Allow use of   2 60 π 0.4 . 360 Either Area of their (rectangle + two triangles + two sectors) Or Area of their (trapezium + two sectors) M1 Either implied by a correct answer or areas clearly labelled. Expect 0.6928 + 0.06928 + 0.1676 or 2 3 3 4π 5 25 75   . Or 0.7621 + 0.1676 or 11 3 4π 25 75  . 0.930 A1 AWRT 11 3 4π Condone 25 75  . 4 Question Answer Marks Guidance 8(b) [Length AD =] 2 2 r B1 Must be seen alone or part of a list and not part of a product. [Arc length =] π 3 r  B1 May be implied by π 2 3 r   . Must be seen alone or part of a list.   π π EF 2 2 sin or 2 2 cos 6 3 r r    or 2 + r B1 Must be seen alone or part of a list and not part of a product. [4+ 3r + 2π 6 leading to 3 r  ] 0.393 B1 AWRT Condone 6 . 2π 9  NB: Using EF = 2.4 gives 0.391. 4

More questions on Circular measure

9 A function f is such that f l ( x) = 6 ( 2x - 3) 2 - 6x for x ! R . (a) Determine the set of values of x for which f ( x) is decreasing. 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(b) Given that f ( 1) = - 1, find f ( x) . 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Mark scheme: 9(a) 2 6 2 3 6 0 x x    or = 0 2 6 2 3  x 2 6 2 3 6   x is used, do not treat as a MR.    2 2 24 78 54 or 4 13 9 or 1 4 9 x x x x x x       OR   2 6 2 3 6   x x   leading to 2 3 leading to 2 3 x x x x     M1 Expanding brackets and collecting terms to arrive at a three term quadratic, only condone sign errors.   9 1 , 4  x B1 9 1 4 x   or 9 1 and 4 x x   or 9 1, 4       DB1FT OE Condone consistent use of ⩽ and ⩾ or [ ]. Do not allow 9 1 or 4   x x nor 9 1, 4   x x . FT on their values coming from a correct initial statement. 4 Question Answer Marks Guidance 9(b)      3 2 6 6 f 2 3 3 2 2 x x x C                  B1 B1 B1 for each   Correct integral .   3 2 1 1 3 1 C    M1  f 1  x equated to their integrated expression, defined by two terms with at least one correct power + C, with x = 1.    3 2 2 3 3 3 f x x x        A1 CAO Only condone C = 3 as final answer if coefficients have been simplified earlier. Do not ISW if the result is of the form   y mx c . Alternative method for Question 9(b)     2 3 2 24 78 54 leading to f 8 39 54 f x x x x x x x C                (B2,1,0) B2 completely correct, B1 any two correct terms. 1 8 39 54 C    (M1)  f 1  x equated to their integrated expression, defined by three terms with at least one correct power + C, with x = 1.  3 2 8 39 54 24 f x x x x        (A1) Only condone C = 24 as final answer if coefficients have been simplified earlier. Do not ISW if the result is of the form . y mx c   4

More questions on Differentiation

3 210 The equation of a curve is y = ( 5 - 2 x) + 5 for x 1 52 . (a) A point P is moving along the curve in such a way that the y-coordinate of point P is decreasing at 5 units per second. Find the rate at which the x-coordinate of point P is increasing when y = 32 . 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(b) Point A on the curve has y-coordinate 32. Point B on the curve is such that the gradient of the curve at B is - 3 . Find the equation of the perpendicular bisector of AB. Give your answer in the form ax + by + c = 0 , where a, b and c are integers. 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Mark scheme: 10(a) 2  x     1 1 2 2 d 3 5 2 2 5 2 d 2 y k x x x           M1* OE Differentiating to get   1 2 5 2  k x only. d d d leading to d d d y y t x t x         d 9 5 d t x   DM1 Correct statement linking their numerical expression for d d y x with d d t x and 5.  5 9 or 0.556 = A1 AWRT 4 Question Answer Marks Guidance 10(b)   1 2 5 2 3   k x M1 Equating their d d y x of the form   1 2 5 2  k x to 3 . [B is]   2, 6 A1 1 32 6 Gradient 2 2 AB m    1 1 4 , gradient of perpendicular 26   m M1* For A, y must be 32. Clear use of difference in y co-ordinates difference in x co-ordinates for points A and B, condone inconsistent order, and using m1m2 = 1 . If incorrect values or another complete method used, then working must be clear.   2 2 6 32 Mid point is , 0,19 2 2         M1* Finding the midpoint of AB using A and B. If incorrect values used then all working must be clear. For A, y must be 32.   2 19 0 13    y x DM1 Finding the equation of the perpendicular bisector using their midpoint and their perpendicular gradient. 2 13 247 0 x y    or integer multiples of this. A1 6

More questions on Coordinate geometry