Equations of linear graphs — Paper 4 · IGCSE Mathematics 0580

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

(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

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

2 (a) p = q = 5 7 (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

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