Equations of linear graphs — Paper 4 · IGCSE Mathematics 0580

2 For 4 f(x) = O 3x, x ≠ 0 Examiner's 2 x Use (a) Complete the table. x O3 O2.5 O2 O1.5 O1 O0.5 0.5 1 1.5 2 2.5 3 f(x) 9.2 7.8 6.5 5.4 9.5 6.5 O3.6 O5.5 O7.2 O8.8 [2] (b) On the grid, draw the graph of y = f(x), for O3 Y x Y O0.5 and 0.5 Y x Y 3 . y 10 9 8 7 6 5 4 3 2 1 x –3 –2 –1 0 1 2 3 –1 –2 –3 –4 –5 –6 –7 –8 –9 [5] (c) Use your graph to solve the equations. For Examiner's (i) f(x) = 4 Use Answer(c)(i) x = [1] (ii) f(x) = 3x Answer(c)(ii) x = [2] (d) The equation f(x) = 3x can be written as x3 = k. Find the value of k. Answer(d) k = [2] (e) (i) Draw the straight line through the points (–1, 5) and (3, –9). [1] (ii) Find the equation of this line. Answer(e)(ii) [3] (iii) Complete the statement. The straight line in part (e)(ii) is a to the graph of y = f(x). [1]

17 marks

Mark scheme: 4 (a) 5, – 1 2 B1 B1 (b) 12 points plotted ft P3ft P2ft for 10 or 11, P1ft for 8 or 9 Smooth curve through at least 12 C1 In absence of plot[s], allow curve to imply plot[s]. points No ruled sections Two separate branches B1 Not touching y-axis (c) (i) 0.55 to 0.65 1 (ii) 0.65 to 0.75 2 M1 for y = 3x drawn (ruled) to cross curve 1 0.3& (d) 2 Accept 0.333[3….] or 3 2 M1 for 2 − 3 x = 3 x or better x IGCSE – October/November 2012 0580 43 (e) (i) Ruled line through (– 1, 5) 1 and (3, – 9) (ii) y = −5.3 x + 5.1 oe final 3 B2 for y = kx + 5.1 [k ≠ 0] oe or y = −5.3 x + d oe answer B1 for gradient = – 3.5 oe accept integer/integer or y = kx + 4.1[ to 6.1] oe SC2 for answer − 5.3 x + 5.1 [no ‘y =’ ] (iii) Tangent 1 5 (a) 0.57 B4 Condone use of other variables M1 for 2 w + 3l = 6.3 oe and M1 for l = w + .0 25 oe A1 for correct aw = b or cl = d or M2 for 2 w + (3 w + .025) = 6.3 oe or 2(l − .025) + 3l = 6.3 oe or M1 for w + 0.25 or l – 0.25 seen A1 for 2 w + 3w = 6.3 − .075 or better or 2l + 3l = 6.3 + 5.0 or better l = 0.82 implies M2A1 trial & error scores B4 or zero accept answer 57 if written 57 cents after M0, SC3 if answer 57

This question in 0580/43 Oct/Nov 2012

9 For Examiner′s y Use L2 5 4 3 R 2 L1 1 x 0 1 2 3 4 5 6 7 8 9 10 L3 (a) Find the equations of the lines L1, L2 and L3. Answer(a) L1 … L2 … L3 … [5] (b) Write down the three inequalities that defi ne the shaded region, R. Answer(b) … … … [3] (c) A gardener buys x bushes and y trees. For Examiner′s The cost of a bush is $30 and the cost of a tree is $200. Use The shaded region R shows the only possible numbers of bushes and trees the gardener can buy. (i) Find the number of bushes and the number of trees when the total cost is $720. Answer(c)(i) … bushes … trees [2] (ii) Find the number of bushes and the number of trees which give the greatest possible total cost. Write down this greatest possible total cost. Answer(c)(ii) … bushes … trees Greatest possible total cost = $ … [3] _____________________________________________________________________________________

13 marks

Mark scheme: 9 (a) y = 2 oe 1 y = 2x oe 2 M1 for y = kx, k ≠ 0 or gradient 2 soi 1 M1 for gradient – ½ soi or y = kx + 5oe y = – x + 5 oe 2 2 or x + 2y = k k ≠ 0 oe If L2 and L3 both correct but interchanged then SC3 (b) y ≥ 2 oe y ≤ 2x oe 1 y ≤ – x + 5 oe 3 B1 for each correct inequality, allow in any 2 order After 0 scored, SC1 for all inequalities reversed (c) (i) 4 [bushes], 3 [trees] 2 M1 for any correct trial using integer coordinates in region or 30x + 200y = 720 seen (ii) 2 [bushes], 4 [trees] 2 M1 for any correct trial using integer coordinates in region 860 1 IGCSE – October/November 2013 0580 41

This question in 0580/41 Oct/Nov 2013

7 (a) The co-ordinates of P are (–4, –4) and the co-ordinates of Q are (8, 14). For Examiner′s Use (i) Find the gradient of the line PQ. Answer(a)(i) … [2] (ii) Find the equation of the line PQ. Answer(a)(ii) … [2] (iii) Write as a column vector. Answer(a)(iii) = [1] f p (iv) Find the magnitude of . Answer(a)(iv) … [2] (b) For Examiner′s T Use A NOT TO SCALE R 4a O 3b B In the diagram, = 4a and = 3b. 1 R lies on AB such that = 5 (12a + 6b). 3 T is the point such that = 2 . (i) Find the following in terms of a and b, giving each answer in its simplest form. (a) Answer(b)(i)(a) = … [1] (b) Answer(b)(i)(b) = … [2] (c) Answer(b)(i)(c) = … [1] (ii) Complete the following statement. The points O, R and T are in a straight line because … … [1] (iii) Triangle OAR and triangle TBR are similar. area of triangle TBR Find the value of . area of triangle OAR Answer(b)(iii) … [2] _____________________________________________________________________________________

14 marks

Mark scheme: 3 14 ( 4 ) 7 (a) (i) or 1.5 2 M1 for oe 2 8 − ( −4 ) 3 3 (ii) y = x + 2 oe 2 B1 for y = their 2 x + c o.e. 2 or y = mx + 2, m ≠ 0 3 SC1 for x + 2 2 (iii)  12  1   18   (iv) 21.6 or 21.63[…] 2 M1 FT for their 122 + their 182 oe (b) (i) (a) 3b – 4a 1 1 1 (b) (6b – 8a) oe simplified 2 M1 for (12a + 6b) – 4a or AR = AO + OR 5 5 (c) 6a + 3b oe simplified 1 (ii) OR is parallel to OT 1 Dep on OT correct 9 2  3  (iii) or 2.25 2 M1 for   4  2  IGCSE – October/November 2013 0580 43 2 (s – ut )

This question in 0580/43 Oct/Nov 2013

8 (a) A straight line joins the points (–1, –4) and (3, 8). (i) Find the midpoint of this line. Answer(a)(i) ( … , … ) [2] (ii) Find the equation of this line. Give your answer in the form y = mx + c. Answer(a)(ii) y = … [3] (b) (i) Factorise x2 + 3x – 10. Answer(b)(i) … [2] (ii) The graph of y = x2 + 3x – 10 is sketched below. y NOT TO SCALE x (a, 0) 0 (b, 0) (0, c) Write down the values of a, b and c. Answer(b)(ii) a = … b = … c = … [3] (iii) Write down the equation of the line of symmetry of the graph of y = x2 + 3x – 10. Answer(b)(iii) … [1] (c) Sketch the graph of y = 18 + 7x – x2 on the axes below. Indicate clearly the values where the graph crosses the x and y axes. y NOT TO SCALE x 0 [4] (d) (i) x2 + 12x – 7 = (x + p)2 – q Find the value of p and the value of q. Answer(d)(i) p = … q = … [3] (ii) Write down the minimum value of y for the graph of y = x2 + 12x – 7. Answer(d)(ii) … [1] __________________________________________________________________________________________

19 marks

Mark scheme: 8 (a) (i) (1, 2) 1+1 8 − −4 (ii) y = 3x – 1 cao final answer 3 M1 for gradient = oe 3 − −1 and M1 for substituting (3, 8) or (–1, –4) into their y = 3x + c or for finding y-intercept is –1 (b) (i) (x + 5)(x – 2) isw solutions 2 SC1 for (x + a)(x + b) where ab = –10 or a + b = 3 (ii) [a =] –5 3FT B1FT for each of their 5 and their –2 from (b)(i) [b =] 2 and B1 for c = –10 [c =] –10 (iii) x = –1.5 1FT FT x = (their (a + b))/2 (c) Inverted parabola B1 x-axis intercepts at –2 and 9 B2 B1 for each After B0 allow SC1 for (9 – x)(2 + x) oe y-axis intercept at 18 B1 (d) (i) p = 6 3 B2 for (x + 6)2 – 43 or p = 6 or q = 43 q = 43 or M1 for (x + 6)2 or x2 + px + px + p2 and M1 for –7 – (their 6)2 or p2 – q = –7 or 2p = 12 (ii) –43 1FT FT – their q 16 × 11 + 17 × 10 + 18 p + 19 × 4 + 20 × 8

This question in 0580/43 Oct/Nov 2014

8 A line AB joins the points A (3, 4) and B (5, 8). (a) Write down the co-ordinates of the midpoint of the line AB. Answer(a) ( … , … ) [2] (b) Calculate the distance AB. Answer(b) AB = … [3] (c) Find the equation of the line AB. Answer(c) … [3] (d) A line perpendicular to AB passes through the origin and through the point (6, r). Find the value of r. Answer(d) r = … [3]

11 marks

Mark scheme: 8 (a) ( 4 , 6 ) 1, 1 (b) 4.47 or 4.472 3 M2 for (8 − 4 )2 + (5 − 3 )2 or better or M1 for (8 − 4 )2 + (5 − 3)2 or better (c) y = 2 x − 2 oe 3 B2 for 2 x − 2 or y = 2 x + c oe 8 − 4 or M1 for [m =] oe soi by 2x 5 − 3 and M1 for (3, 4) or (5, 8) or their midpoint substituted into their y = mx + c with m numerical (d) − 3 3 M1 for use of gradient × their m = − 1 soi by 1 – 2 M1 for r = their gradient × 6 [+0]

This question in 0580/42 Oct/Nov 2015

9 y 2 B A NOT TO x –4 0 4 SCALE –2 2 y x 2 The diagram shows a curve with equation = 1. 2 + 2 a b (a) A is the point (4, 0) and B is the point (0, 2). (i) Find the equation of the straight line that passes through A and B. Give your answer in the form y = mx + c. y = … [3] (ii) Show that a2 = 16 and b2 = 4. [2] (b) y 2 P NOT TO x –4 O 4 SCALE Q –2 2 y x 2 + = 1. P (2, k) and Q (2, -k) are points on the curve 16 4 (i) Find the value of k. k = … [3] (ii) Calculate angle POQ. Angle POQ = … [3] 2 y x 2 = 1 is r ab .(c) The area enclosed by a curve with equation 2 + 2 a b 2 y x 2 + = 1. (i) Find the area enclosed by the curve 16 4 Give your answer as a multiple of r. … [1] (ii) A curve, mathematically similar to the one in the diagrams, intersects the x-axis at (12, 0) and (-12, 0). Work out the area enclosed by this curve, giving your answer as a multiple of r. … [2]

14 marks

Mark scheme: 1 1 9 (a) (i) − x + 2 oe 3 SC2 for y = − x + c oe 2 2 or SC1 for y = kx + 2 oe, k ≠ 0 or − 2 M1 for [gradient =] 4 and M1 for substituting (4, 0) or (0, 2) into y = (their m)x + c (ii) 16  0[ 2 ]  4 2  0[ 2 ]  2  + 2  = 1 or 2  + 2  = 1 1 a  b  a  b  and a[2] = 4[2] 1  0[ 2 ]  4  0[ 2 ]  2 2  2  + 2 = 1 or  2  + 2 = 1  a  b  a  b and b[2] = 2[2] 3 2 k 3 (b) (i) = or better 1.73 or 1.732.. or 3 M2 for 4 4 2 2 k 2 or M1 for + = 1 oe 16 4 their 3  −1 (ii) 81.8 or 81.78 to 81.79 3 M2 for 2 × tan   oe   2   their 3 or M1 for tan = oe 2 (c) (i) 8π final answer 1 (ii) 72π final answer 2FT FT their (c)(i) × 9 in terms of π M1 for area factor of 32 or 9 or [new a] = 12, [new b] = 6

This question in 0580/41 May/June 2016

9 A line joins the points A (–2, –5) and B (4, 13). (a) Calculate the length AB. AB = … [3] (b) Find the equation of the line through A and B. Give your answer in the form y = mx + c. y = … [3] (c) Another line is parallel to AB and passes through the point (0, –5). Write down the equation of this line. … [2] (d) Find the equation of the perpendicular bisector of AB. … [5]

13 marks

Mark scheme: 2 + (13 −−5) 2 oe9 (a) 19[.0] or 18.97.. nfww 3 M2 for (4 −−2) or M1 for (4– – 2)² + (13 – – 5)² oe (b) [y =] 3x + 1 3 B2 for answer [y =]3x + c oe or answer kx + 1 (k ≠ 0 ) 13 −− 5 or M1 for oe or 3 4 −− 2 and M1 for correct substitution of (–2, –5) or (4, 13) into y = (their m)x + c oe (c) y = 3x – 5 oe 2FT FT their gradient from (b) M1 for y = mx – 5 with other m, m ≠ 0 or y = {their gradient from (b)}x + c If 0 scored, SC1 for answer 3x – 5 1 13 1 (d) y = − x + oe isw 5 B2FT for − x + c (c can be numeric or 3 3 3 algebraic) FT –1/ their gradient from (b) or M1 for –1/ their gradient from (b) soi and B1 for [midpoint of AB =] (1, 4) and M1 for substitution of (1, k) or (k, 4) into a linear equation

This question in 0580/42 May/June 2016

7 A line joins the points A (- 3, 8) and B (2, - 2) . (a) Find the co-ordinates of the midpoint of AB. ( … , … ) [2] (b) Find the equation of the line through A and B. Give your answer in the form y = mx + c . y = … [3] (c) Another line is parallel to AB and passes through the point (0, 7). Write down the equation of this line. … [2] (d) Find the equation of the line perpendicular to AB which passes through the point (1, 5). Give your answer in the form ax + by + c = 0 where a, b and c are integers. … [4]

11 marks

Mark scheme: 7(a) (–0.5, 3) 2 B1 for one correct value 7(b) [y = ] –2x + 2 final answer 3 −−2 8 M1 for better 2 −−or3 M1 for substitution of (–3, 8) or (2, –2) or their midpoint into y = mx + c with their m 7(c) y = –2x + 7 oe 2FT FT their (b) M1 for y = (their–2)x + k ( k ≠ 2) or y = kx + 7 (k ≠ 0) If zero scored, SC1 for ( their − 2 ) x + 7 7(d) x – 2y + 9 = 0 or 2y – x – 9 = 0 oe 4 B3 for any correct equivalent in wrong form Or M2 for y = ½ x + k oe (FT negative reciprocal of their gradient in (b)) or M1 for grad = ½ (FT negative reciprocal of their gradient in (b)) M1 for substitution of (1, 5) into y = mx + c oe with their m

This question in 0580/41 May/June 2017

8 Line A has equation y = 5x - 4 . Line B has equation 3x + 2y = 18 . (a) Find the gradient of (i) line A, … [1] (ii) line B. … [1] (b) Write down the co-ordinates of the point where line A crosses the x-axis. ( … , … ) [2] (c) Find the equation of the line perpendicular to line A which passes through the point (10, 9). Give your answer in the form y = mx + c . y = … [4] (d) Work out the co-ordinates of the point of intersection of line A and line B. ( … , … ) [3] (e) Work out the area enclosed by line A, line B and the y-axis. … [3]

14 marks

Mark scheme: 8(a)(i) 5 1 8(a)(ii) 3 1 − oe 2 8(b)  4  2 M1 for 5x – 4 = 0 soi  , 0  oe  5  8(c) y = –0.2x + 11 final answer 4 M2 for y = –0.2x + c oe (any form) FT their (a) or −1 B1FT for grad = soi their (a)(i) and M1 for substitution of (10, 9) into their equation 8(d) (2, 6) 3 M1 for elimination of one variable A1 for x = 2 or y = 6 8(e) 13 3 M2 for (4 + 9) × their 2 ÷ 2 oe or B1 for 9 oe or 4 or –4 seen

This question in 0580/43 Oct/Nov 2017

x 3 16 (a) Complete the table of values for y = - 2 , x ! 0. 3 2x x - 3 - 2 - 1 - 0.5 - 0.3 0.3 0.5 1 2 3 y - 9.1 - 2.8 - 0.8 - 5.6 - 5.5 - 2.0 8.9 [3] x 3 1 (b) On the grid, draw the graph of y = - 2 for - 3 G x G - 0.3 and 0.3 G x G 3. 3 2x y 10 9 8 7 6 5 4 3 2 1 x –3 –2 –1 0 1 2 3 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 [5] (c) (i) By drawing a suitable tangent, find an estimate of the gradient of the curve at x = - 2. … [3] (ii) Write down the equation of the tangent to the curve at x = - 2. Give your answer in the form y = mx + c. y = … [2] (d) Use your graph to solve the equations. x 3 1 (i) - 2 = 0 3 2x x = … [1] x 3 1 (ii) - 2 + 4 = 0 3 2x x = … or x = … or x = … [3] x 3 1 -3 + bxn - 3 = 0.(e) The equation - 2 + 4 = 0 can be written in the form axn 3 2x Find the value of a, the value of b and the value of n. a = … b = … n = … [3]

20 marks

Mark scheme: 6(a) – 2[.0], – 0.2, 2.5 3 B1 for each 6(b) Fully correct curve 5 B4 for correct curve, but branches joined or B3FT for 9 or 10 correct plots or B2FT for 7 or 8 correct plots or B1FT for 5 or 6 correct plots and B1 indep two separate branches not touching or cutting y-axis 6(c)(i) Correct tangent and 3 B2 for close attempt at tangent to curve 3 ⩽ grad ⩽ 5 at x = – 2 and answer in range OR B1 for ruled tangent at x = – 2, no daylight at x = –2 and M1dep (dep on B1 or close attempt rise at tangent) [at x = –2] for run 6(c)(ii) [y =] their(c)(i) x + their y-intercept final 2 Strict FT their y-intercept for their line answer M1 for y = their(c)(i) x + any value or ‘c’ oe seen or for y = any value(non-zero) x or ‘mx’ + their y-intercept seen oe 6(d)(i) 1.05 to 1.25 1 6(d)(ii) – 2.3 to – 2.2 3 B1 for each – 0.4 to – 0.3 After 0 scored B1 for y = –4 ruled 0.3 to 0.4 6(e) [a =] 2 3 B2 for 2 correct or for [b =] 24 2x5 + 24x2 [–3 = 0] [n =] 5 or B1 for 1 correct or for 2 x 5 − 3 + 4(6 x 2 ) [ = 0] oe 6 x 2 If 0 scored SC1 for 2x5 seen in final line of algebra

This question in 0580/42 May/June 2018

2 (a) (i) y = 2x Complete the table. x 0 1 2 3 4 y 2 4 8 [2] (ii) y = 14 - x 2 Complete the table. x 0 1 2 3 4 y 13 10 5 [2] (b) On the grid, draw the graphs of y = 2x and y = 14 - x 2 for 0 G x G 4 . y 16 14 12 10 8 6 4 2 0 x 1 2 3 4 −2 [6] (c) Use your graphs to solve the equations. (i) 2 x = 12 x = … [1] (ii) 2 x = 14 - x 2 x = … [1] (d) (i) On the grid, draw the line from the point (4, 2) that has a gradient of - 4 . [1] (ii) Complete the statement. This straight line is a … to the graph of y = 14 - x 2 at the point ( … , … ). [2]

15 marks

Mark scheme: 2(a)(i) 1, ….., ….., …., 16 2 B1 for each 2(a)(ii) 14, ….., ….., …., – 2 2 B1 for each 2(b) Fully correct smooth curves 6 B3 for correct curve of y = 2 x or B2FT for 4 or 5 correct points or B1FT for 2 or 3 correct points B3 for correct curve of y = 14 − x 2 or B2FT for 4 or 5 correct points or B1FT for 2 or 3 correct points 2(c)(i) 3.5 to 3.7 1 2(c)(ii) 2.65 to 2.8 1 2(d)(i) Correct line 1 Ruled, through (4, 2) and gradient −4 2(d)(ii) Tangent 2 B1 for each (2, 10)

This question in 0580/43 May/June 2018

8 y 8 l 7 A 6 5 4 3 2 1 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x –1 B –2 –3 (a) Write down the co-ordinates of A. ( … , … ) [1] (b) Find the equation of line l in the form y = mx + c . y = … [3] (c) Write down the equation of the line parallel to line l that passes through the point B. … [2] (d) C is the point (8, 14). (i) Write down the equation of the line perpendicular to line l that passes through the point C. … [3] (ii) Calculate the length of AC. … [3] (iii) Find the co-ordinates of the mid-point of BC. ( … , … ) [2]

14 marks

Mark scheme: 8(a) (5, 6) 1 8(b) 4 3 4 [ y = ] − x + 3 nfww B2 for [ y = ] − x + c nfww 5 5 rise or M1 for using any two of (–5, 7) run (0, 3) and (5, –1) and B1 for [ y = ]mx + 3 (m ≠ 0 ) 8(c) 4 2 FT their gradient from 8(b) y = − x − 2 oe 5 B1 for y = (their gradient)x + c (c not 0) or for y = mx − 2 (m ≠ 0 ) 4 or for − x − 2 alone 5 8(d)(i) 5 3 1 y = x + 4 oe M1 for −their gradient from 8(b) 4 M1 for (8, 14) substituted into y − 14 their y = mx + c or = m or better x − 8 8(d)(ii) 8.54 or 8.544... 3 M2 for (14 − their 6) 2 + (8 − their 5) 2 or better or M1 for 14 − their 6 and 8 − their 5 seen 8(d)(iii) (4, 6) 2 B1 for each

This question in 0580/41 Oct/Nov 2018

5 The table shows some values of y = x3 – 3x – 1. x –3 –2.5 –2 –1.5 –1 0 1 1.5 2 2.5 3 y –19 –9.1 0.1 1 –1 –3 –2.1 1 7.1 (a) Complete the table of values. [2] (b) Draw the graph of y = x 3 - 3x - 1 for - 3 G x G 3 . y 20 15 10 5 0 x –3 –2 –1 1 2 3 –5 –10 –15 –20 [4] (c) A straight line through (0, –17) is a tangent to the graph of y = x 3 - 3x - 1. (i) On the grid, draw this tangent. [1] (ii) Find the co-ordinates of the point where the tangent meets your graph. ( … , … ) [1] (iii) Find the equation of the tangent. Give your answer in the form y = mx + c. y = … [3] (d) By drawing a suitable straight line on the grid, solve the equation x 3 - 6x - 3 = 0 . x = … or x = … or x = … [4]

15 marks

Mark scheme: 5(a) –3, 17 2 B1 for each 5(b) Fully correct curve 4 B3 FT for 10 or 11 points or B2 FT for 8 or 9 points or B1 FT for 6 or 7 points 5(c)(i) Correct ruled tangent for their curve 1 through (0, −17) 5(c)(ii) (1.7 to 2.2, –1 to 2.5) 1 5(c)(iii) [y =] 9x – 17 final answer 3 M2dep for answer [y =] 9x[+] – c OR rise M1dep for gradient = for their tangent run at any point B1 for answer [y =] kx[+] – 17 (k ≠ 0) 5(d) y = 3x + 2 ruled correctly and 4 B2 for y = 3x + 2 ruled –2.2 … to –2.1 or B1 for [y =] 3x + 2 soi –0.6 to –0.4 or y = 3x + k ruled 2.6 to 2.8 or y = kx + 2 but not y = 2 B2 for all 3 values or B1 for 2 values

This question in 0580/42 Oct/Nov 2018

4 (a) The equation of a straight line is 2y = 3x + 4 . (i) Find the gradient of this line. … [1] (ii) Find the co-ordinates of the point where the line crosses the y-axis. ( … , … ) [1] (b) The diagram shows a straight line L. y 6 4 2 –2–2 0 22 4 6 x L –2 (i) Find the equation of line L. … [3] (ii) Find the equation of the line perpendicular to line L that passes through (9, 3). … [3] (c) A is the point (8, 5) and B is the point (- 4, 1). (i) Calculate the length of AB. … [3] (ii) Find the co-ordinates of the midpoint of AB. ( … , … ) [2]

13 marks

Mark scheme: 4(a)(i) 1.5 oe 1 4(a)(ii) (0, 2) 1 4(b)(i) y = −2 x + 6 oe final answer 3 B2 for y = − 2 x + c oe or y = mx + 6 oe m ≠ 0 or for answer −x2 + 6 6 or B1 for [gradient =] − oe or c = + 6 soi 3 4(b)(ii) y = 5.0 x − 5.1 oe final answer 3 B1 for [gradient = ] – 1 divided by their gradient from (b)(i) evaluated soi M1 for substitution of (9, 3) into y = (their m)x+ c seen in working 4(c)(i) 12.6 or 12.64 to 12.65 3 2 2 M2 for (8 − − 4) + ( 5 − )1 oe or M1 for (8 − −4 ) 2 + (5 − 1)2 oe 4(c)(ii) (2, 3) 2 B1 for each

This question in 0580/42 May/June 2019

1 x5 The table shows some values of y = - for 0.15 G x G 3.5 . 2x 4 x 0.15 0.2 0.5 1 1.5 2 2.5 3 3.5 y 3.30 0.88 - 0.04 - 0.43 - 0.58 - 0.73 (a) Complete the table. [3] 1 x (b) On the grid, draw the graph of y = - for 0.15 G x G 3.5 . 2x 4 The last two points have been plotted for you. y 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 x – 0.5 – 1.0 [4] 1 x 1(c) Use your graph to solve the equation - = for 0.15 G x G 3.5 . 2x 4 2 x = … [1] (d) (i) On the grid, draw the line y = 2 - x . [2] (ii) Write down the x co-ordinates of the points where the line y = 2 - x crosses the graph of 1 x y = - for 0.15 G x G 3.5 . 2x 4 x = … and x = … [2] 1 x(e) Show that the graph of y = - can be used to find the value of 2 for 0.15 G x G 3.5 . 2x 4 [2]

14 marks

Mark scheme: 5(a) 2.45, 0.25, − 0.25 3 B1 for each 5(b) Fully correct smooth curve 4 B3FT for 6 or 7 points or B2 FT for 4 or 5 points or B1 FT for 2 or 3 points 5(c) 0.7 to 0.8 1 FT their curve 5(d)(i) Correct ruled line 2 M1 for good freehand, or ruled line with gradient −1.05 to −0.95 or ruled line through (0, 2) but not line y = 2 5(d)(ii) Both intersections of their (b) and 2 Strict FT intersection of their (b) and their (d)(i) their (d)(i) B1FT for one correct OR B2 for 0.27 to 0.28 and 2.38 to 2.39 5(e) 1 x M1 Substitutes x = 2 into − 2 x 4 OR Identifies y = 0 oe OR Correctly manipulates to a single fraction 2 − x 2 e.g. oe seen 4 x Concludes ‘read the graph at y = 0’ A1 oe OR 1 x Manipulates 0 = − oe 2 x 4 leading to x 2 = 2 OR 2 − x 2 States oe = 0 leading to 4 x x 2 = 2

This question in 0580/42 May/June 2019

7 A straight line joins the points A (-2, -3) and C (1, 9). (a) Find the equation of the line AC in the form y = mx + c. y = … [3] (b) Calculate the acute angle between AC and the x-axis. … [2] (c) ABCD is a kite, where AC is the longer diagonal of the kite. B is the point (3.5, 2). (i) Find the equation of the line BD in the form y = mx + c. y = … [3] (ii) The diagonals AC and BD intersect at (-0.5, 3). Work out the co-ordinates of D. ( … , … ) [2]

10 marks

Mark scheme: 7(a) [y = ] 4x + 5 3 B2 for answer [y =] 4x + c oe (c can be numeric or algebraic) OR y − 9 9 −−( 3) M2 for = oe x − 1 1 −−( 2) OR 9 −−3 M1 for oe or for 1 −−2 M1 for correct substitution of (–2, –3) or (1, 9) into y = (their m)x + c oe 7(b) 76[.0] or 75.96... 2 M1 for tan[ ] = 4 oe 7(c)(i) 1 23 3 1 [y =] − x + oe B2FT for [y =] − x + c 4 8 their gradient from (a) oe (c can be numeric or algebraic) OR y − 2 1 M2 for = − oe x − 3.5 their gradient from (a) OR 1 M1 for −their gradient from (a) soi M1 for correct substitution of (3.5, 2) into y = (their m)x + c oe 7(c)(ii) (–4.5, 4) 2  − 8  B1 for each value or for   seen  2 

This question in 0580/43 May/June 2019

3 A line joins A (1, 3) to B (5, 8). (a) (i) Find the midpoint of AB. ( … , … ) [2] (ii) Find the equation of the line AB. Give your answer in the form y = mx + c . y = … [3] (b) The line AB is transformed to the line PQ. Find the co-ordinates of P and the co-ordinates of Q after AB is transformed by 5 (i) a translation by the vector e- 2o, P ( … , … ) Q ( … , … ) [2] (ii) a rotation through 90° anticlockwise about the origin, P ( … , … ) Q ( … , … ) [2] (iii) a reflection in the line x = 2 , P ( … , … ) Q ( … , … ) [2] - 1 2 (iv) a transformation by the matrix e 0 - 1o. P ( … , … ) Q ( … , … ) [2] (c) Describe fully the single transformation that maps the line AB onto the line PQ where P is the point (-2, -6) and Q is the point (-10, -16). … … [3]

16 marks

Mark scheme: 3(a)(i) (3, 5.5) 2 B1 for either value correct 3(a)(ii) 5 7 3 5 x + final answer B2 for answer x + c oe or for correct 4 4 4 equation in different form 8 − 3 or M1 for oe 5 − 1 and M1 for correct substitution shown of (1, 3) or (5, 8) or their (a)(i) into y = (their m)x + c oe 3(b)(i) (6, 1) 2 B1 for 2 or 3 values correct (10, 6) 3(b)(ii) (–3, 1) 2 B1 for 2 or 3 values correct (–8, 5) If 0 scored, SC1 for (3, –1) and (8, –5) 3(b)(iii) (3, 3) 2 B1 for 2 or 3 values correct but not for (–1, 8) (1, 3) and (5, 8) 3(b)(iv) (5, –3) 2 B1 for either (11, –8)  − 1 2  1   − 1 2  5  or M1 for    or     0 − 1  3   0 − 1  8  3(c) Enlargement 3 B1 for each –2 Origin oe

This question in 0580/42 Oct/Nov 2019

4 - 2 2 (a) p = q = e 5 o e 7o (i) Find 2 p + q . [2] f p (ii) Find p . … [2] - 3 (b) A is the point (4, 1) and AB = e 1o. Find the coordinates of B. ( … , … ) [1] (c) The line y = 3 x - 2 crosses the y-axis at G. Write down the coordinates of G. ( … , … ) [1] (d) D NOT TO T SCALE M O C In the diagram, O is the origin, OT = 2TD and M is the midpoint of TC. OC = c and OD = d . Find the position vector of M. Give your answer in terms of c and d in its simplest form. … [3]

9 marks

Mark scheme: 2(a)(i)  6  2 B1 for each    17  2(a)(ii) 6.4[0] or 6.403... 2 M1 for 42 + 52 2(b) (1, 2) 1 2(c) (0, –2) 1 2(d) 1 1 3 B2 for correct unsimplified answer c + d  2 2 3 or M1 for CT = – c + d oe 3  2 or TC = c – d oe 3 or for correct route

This question in 0580/42 May/June 2020

9 (a) The equation of line L is 3x - 8y + 20 = 0 . (i) Find the gradient of line L. … [2] (ii) Find the coordinates of the point where line L cuts the y-axis. ( … , … ) [1] (b) The coordinates of P are (-3, 8) and the coordinates of Q are (9, -2). (i) Calculate the length PQ. … [3] (ii) Find the equation of the line parallel to PQ that passes through the point (6, -1). … [3] (iii) Find the equation of the perpendicular bisector of PQ. … [4]

13 marks

Mark scheme: 9(a)(i) 3 2 M1 for 8 y = 3 x + 20 or better 8 9(a)(ii) (0, 2.5) oe 1 (b)(i) 15.6 or 15.62… 3 2 2 M2 for ( 9 −−3 ) + ( −−2 8 ) oe seen 2 2 or M1 for ( 9 −−3) or ( −−2 8 ) oe seen 9(b)(ii) 5 3 −−2 8 y = − x + 4 oe M1 for gradient oe 6 9 −−3 M1 for substituting (6, −1) into a linear equation oe 9(b)(iii) 6 3 4 5  y = x − oe M1 for gradient −1 / their −   5 5  6  B1 for midpoint at (3, 3) M1 for their midpoint substituted into y = their m × x + c oe

This question in 0580/43 May/June 2020

10 (a) The diagrams show the graphs of two functions. Write down each function. (i) f(x) 5 – 5 0 x f(x) = … [2] (ii) f(x) 2 0 x 180° 360° – 2 f(x) = … [2] (b) f(x) 4 3 P 2 1 – 0.5 0 0.5 1 1.5 2 2.5 x – 1 The diagram shows the graph of another function. By drawing a suitable tangent, find an estimate for the gradient of the function at the point P. … [3]

7 marks

Mark scheme: 10(a)(i) x + 5 2 B1 for linear equation with positive gradient or intercept 5 10(a)(ii) 2 sin x oe 2 B1 for recognition of sin or cos(x – 90) 10(b) tangent ruled at P B1 1.3 to 1.4 B2 dep on tangent drawn M1 for rise/run

This question in 0580/43 May/June 2020

8 AB is a line with midpoint M. A is the point (2, 3) and M is the point (12, 7). (a) Find the coordinates of B. ( … , … ) [2] (b) Show that the equation of the perpendicular bisector of AB is 2y + 5x = 74 . [4] (c) The perpendicular bisector of AB passes through the point N. The point N has coordinates (2, n). Find the value of n. n = … [1] (d) Points A, M and N form a triangle. Find the area of the triangle. … [2]

9 marks

Mark scheme: 8(a) (22, 11) 2 B1 for each value 8(b) their11 − 3 M1 oe or better their 22 − 2 1 M1 −their m Substitution of (12, 7) into M1 Accept y – 7 = their m(x – 12) oe y = (their m)x + c leading to 2y + 5x = 74 final answer A1 Without error or omission 8(c) 32 1 8(d) 145 2 1 M1 for × (their 32 – 3) × 10 oe 2 or 1 2 2 2 2  (7 − 3) + (12 − 2)  (their 32 − 7) + (2 − 12) oe 2

This question in 0580/42 Oct/Nov 2022

6 y A 4 L1 NOT TO SCALE B 0 8 x L2 A is the point (0, 4) and B is the point (8, 0). The line L1 is parallel to the x-axis. The line L2 passes through A and B. (a) Write down the equation of L1. … [1] (b) Find the equation of L2. Give your answer in the form y = mx + c . y = … [2] (c) C is the point (2, 3). The line L3 passes through C and is perpendicular to L2. (i) Show that the equation of L3 is y = 2x - 1. [3] (ii) L3 crosses the x-axis at D. Find the length of CD. … [5]

11 marks

Mark scheme: 6(a) y = 4 oe 1 6(b) 1 2 4 [ y = ] − x + 4 final answer B1 for grad = − oe soi 2 8 or  y =  kx + 4 6(c)(i) −1 M1 1 Gradient = Accept e.g. 2 × − = –1 oe their gradient in ( b ) 2 1 or states negative reciprocal of − = 2 2 Substituting (2, 3) in their equation. M1 3 = 2 × their m + c leading to y = 2x – 1 A1 No errors or omissions 6(c)(ii) 3.35 or 3.354... 5   1 B2 for 1,0 soi or x-coordinate of D =    2  2 or M1 for 2x – 1 = 0 M2 for (2 − their 12 ) 2 + (3 − their 0) 2 oe or M1 for (2 − their 12 ) and (3 − their 0) oe

This question in 0580/42 Feb/March 2023

12 y B NOT TO SCALE A x O O is the origin (0, 0), A is the point (8, 1) and B is the point (2, 5). (a) Write as column vectors. (i) OB OB = [1] f p (ii) AB AB = [1] f p (b) Find the equation of the line AB. Give your answer in the form y = mx + c . y = … [3] (c) Find the equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c . y = … [4] (d) The line AB meets the y-axis at P. The perpendicular bisector of AB meets the y-axis at Q. Find the length of PQ. … [2]

11 marks

Mark scheme: 12(a)(i) 2 1  5 12(a)(ii)  −6  1    4  12(b) 2 19 3 1 − 5 [ y =] − x + oe M1 for gradient = oe 3 3 8 − 2 M1 for substituting (8, 1) or (2, 5) into y = their mx + c 12(c) 3 9 4 B1 for (5, 3) oe [ y = x − oe 1 ]2 2 M1 for gradient = −their gradient of AB M1 substituting their midpoint into y = their mx + c 12(d) 65 2 19 9 oe M1 for their – their − oe 6 3 2

This question in 0580/42 Oct/Nov 2023

9 A is the point (0, 2), B is the point (3, 3) and C is the point (4, 0). (a) Determine if triangle ABC is scalene, isosceles or equilateral. You must show all your working. [4] (b) (i) Find the equation of the line AC. Give your answer in the form y = mx + c . y = … [3] (ii) Find the equation of the perpendicular bisector of AC. Give your answer in the form y = mx + c . y = … [4] (iii) ABCD is a kite. The point D has coordinates (w, 4w + 1). Find the coordinates of D. ( … , … ) [3]

14 marks

Mark scheme: 9(a) [AB2 =] (3 – 0)2 + (3 – 2)2 oe or M1 3 or  oe better 1 [AC2 =] (0 – 2)2 + (4 – 0)2 oe or M1  4  or   oe better  −2  [BC2 =] (0 – 3)2 + (4 – 3)2 oe or M1  1  or   oe better  −3  Triangle is isosceles [with 10, 20 A1 or Triangle is isosceles and only vector AB and BC and 10 or better shown] have the same magnitude [because they have the same components] 9(b)(i) 1 3 0 − 2 [y =] − x + 2 oe M1 for oe 2 4 − 0 M1 for substituting (0, 2) or (4, 0) into y = their mx + c oe or B1 for answer y = kx + 2 9(b)(ii) [y =] 2x – 3 4 −1 M1 for their grad (b)(i) B1 for (2, 1) M1 for substituting their (2, 1) into y = their px + d oe 9(b)(iii) (–2, –7) 3 B2 for w = – 2 or M1 for 4w + 1 = 2w – 3 FT their (b)(ii) 4 w + 1 − 3 or for 2 = w − 3

This question in 0580/43 Oct/Nov 2023

13 Line L has the equation y = 2 ( x - 5) . (a) Find the coordinates of the point where line L intersects the y-axis. ( … , … ) [1] (b) Find the x-coordinate of the point where line L intersects the line y = 19 . x = … [2] (c) Line P is perpendicular to line L. Line P passes through the point (2, 3). Find the equation of line P. Give your answer in the form y = mx + c . y = … [3]

6 marks

Mark scheme: 13(a) (0, –10) 1 13(b) 1 2 19 14.5 or 14 M1 for = x – 5 2 2 or for 19 = 2x – 10 or better 13(c) 1 3 B2 for a correct answer in the wrong form y = − x + 4 or y = –0.5x + 4 final 2 OR answer 1 1 M1 for y = − x + c or gradient = − soi 2 2 M1 for correct substitution of (2, 3) 1 into y = (their − )x + c or 2 1 into y – k = (their − )(x – h) 2

This question in 0580/43 May/June 2025

17 Find the equation of the straight line that passes through the points (2, 0) and (0, 4). Give your answer in the form y = mx + c . y = … [3]

3 marks

Mark scheme: 17 [y =] –2x + 4 3 4 − 0 M1 for gradient = oe 0 − 2 M1 for answer mx + 4

This question in 0580/41 Oct/Nov 2025