Perpendicular lines — 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

19 (a) Find the equation of the straight line that is perpendicular to the line y = x + 1 and passes through 2 the point (1, 3). … [3] (b) y 8 7 6 5 4 R 3 2 1 x 0 1 2 3 4 5 6 7 8 9 10 11 12 (i) Find the three inequalities that define the region R. … … … [4] (ii) Find the point (x, y), with integer co-ordinates, inside the region R such that 3x + 5y = 35 . ( … , … ) [2]

9 marks

Mark scheme: 9(a) y = −2 x + 5 oe 3 B2 for –2x + 5 or 1 M1 for gradient = −÷1 or better 2 M1 for substituting (1, 3) into y = (their m)x + c oe If 0 scored SC1 for (1, 3) satisfying their wrong 1 equation (c ≠ 0) with gradient ≠ 2 9(b)(i) x . 2 oe 1 SC3 for x > 2 and y < 5 and y > x 2 y - 5 oe OR B1 for x ⩾ 2 1 B1 for y ⩽ 5 y . x oe 4 1 2 B2 for y ⩾ x 2 or M1 for y ⩾ kx (k > 0) OR SC2 for all three boundary lines identified but with incorrect sign(s) If 0 scored SC1 for one or two correct boundary lines with incorrect sign(s) 9(b)(ii) (5, 4) 2 M1 for one trial of an integer point inside region or for 3 x + 5 y = 35 drawn

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, l, joins point F (3, 2) and point G (- 5, 4). (a) Calculate the length of line l. … [3] (b) Find the equation of the perpendicular bisector of line l in the form y = mx + c . y = … [5] (c) A point H lies on the y-axis such that the distance GH = 13 units. Find the coordinates of the two possible positions of H. ( … , … ) and ( … , … ) [4]

12 marks

Mark scheme: 3(a) 8.25 or 8.246… 3 2 2 M2 for  3 5    2  4  oe or better or M1 for  3  5  and  2  4  oe seen 3(b) [ y  ] 4 x  7 5 B1 for [midpoint] (− 1, 3) soi 4  2 M1 for [gradient of l =] oe 5 3  1  M1 for gradient 1 / their     4  M1dep on at least M1 for their (− 1, 3) substituted into y = their m  x + c oe 3(c) (0, − 8) and (0, 16) 4 B3 for (0, −8) or (0, 16) or for –8 and 16 OR B2 for distance = [±]12 soi or M1 for 132 – (5[–0])2 oe B1 for both answers (0, k), k ≠ 0 or 4 ALT METHOD B3 for (0, −8) or (0 , 16) or for – 8 and 16 OR M2 for y2 – 8y – 128 [= 0] or for (y – 4)2 = 144 or better or M1 for 132 = (–5 – 0)2 + (4 – y)2 oe B1 for both answers (0, k), k ≠ 0 or 4

This question in 0580/42 May/June 2022

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

11 M has coordinates (4, 1) and N has coordinates ( -2, -7). (a) Find the length of MN. … [3] (b) Find the gradient of MN. … [2] (c) Find the equation of the perpendicular bisector of MN. … [4] Question 12 is printed on the next page.

9 marks

Mark scheme: 11(a) 10 3 M2 for (1 – –7)2 + (4 – –2)2 oe or M1 for (1 – –7) or (4 – –2) oe 11(b) 4 8 2 1 7 or M1 for oe 3 6 4 2 11(c) 3 9 4 3 9 y  x  B3 for  x  4 4 4 4 or 4 y  3 x  9  0 oe OR final answers B1 for midpoint (1, − 3) 3 1 M1 for gradient  or  4 their (b) M1 for substituting their (1, −3) into y = (their m)x + c or for y 3 their m = oe x  1

This question in 0580/43 May/June 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

25 P is the point (8, 0) and Q is the point (20, 6). Find the equation of the perpendicular bisector of PQ. Give your answer in the form y = mx + c . y = … [5]

5 marks

Mark scheme: 25 [y =] –2x + 31 5 B1 for (14, 3) AND 6 − 0 M1 for (m1) oe 20 − 8 1 M1 for m = − their m1 AND M1dep for their 3 = their m × their 14 + c oe

This question in 0580/42 May/June 2025