Perpendicular lines — Paper 2 · IGCSE Mathematics - International 0607

13 A is the point (1, 8) and B is the point (5, 0). Find the equation of the perpendicular bisector of AB in the form y = mx + c . y = … [4]

4 marks

Mark scheme: 13 1 5 4 B1 for (3, 4) seen [ y = ] x + 2 2 8 B1 for − oe seen 4 −1 M1 for grad = their ( −2)

This question in 0607/23 Oct/Nov 2017

13 Find the equation of the straight line perpendicular to the line y = 2x + 1 that passes through the point (2, 5). Give your answer in the form y = mx + c. y = … [3]

3 marks

Mark scheme: 13 1 3 1 − x + 6 B1 for gradient = − 2 2 M1 for substitution of (2, 5) into y = ( their m ) x + c

This question in 0607/23 Oct/Nov 2018

11 The point A has co-ordinates (3, 8). The point B has co-ordinates (7, 0). (a) Find the co-ordinates of the midpoint of AB. ( … , … ) [1] (b) Find the equation of the perpendicular bisector of AB. Write your answer in the form y = mx + c. y = … [3]

4 marks

Mark scheme: 11(a) (5, 4) 1 11(b) 1 1 3 8 − 0 [y =] x + 1 oe M1 for oe or gradient = −2 2 2 3 − 7 M1 for gradient of perpendicular −1 = their gradient

This question in 0607/21 May/June 2019

16 A is the point (0, 8) and B is the point (6, 0). The line L passes through B and is perpendicular to AB. Find the equation of L. … [4]

4 marks

Mark scheme: 16 3 9 4 4 y = x − oe M1 for gradient of AB = − oe 4 2 3 1 M1 for gradient of L = – 4 their − 3 M1 for substitution of (6, 0) in their y = mx + c oe

This question in 0607/21 Oct/Nov 2019

10 The point A has co-ordinates (1, - 5) and the point B has co-ordinates (9, 1) . Find the equation of the perpendicular bisector of AB in the form y = mx + c . y = … [5]

5 marks

Mark scheme: 10 4 14 5 1 −−( 5) y = − x + M1 for oe 3 3 9 − 1 1 M1 for grad = − their gradient  9 + 1 −+5 1  M1 for midpoint  ,  oe  2 2  M1 for subst (their midpoint) and (their gradient) into y = mx + c

This question in 0607/22 Oct/Nov 2019

12 The equation of the line L is y = 3 x - 2 . (a) Find the co-ordinates of the point A, where the line L crosses the y-axis. ( … , … ) [1] (b) Find the co-ordinates of the point B, where the line L crosses the x-axis. ( … , … ) [1] (c) The line M passes through the point A and is perpendicular to the line L. Find the equation of the line M. … [2]

4 marks

Mark scheme: 12(a) (0, –2) 1 12(b)  2  1  , 0  oe  3  12(c) 1 2 FT their (a) y = – x – 2 oe 1 3 B1 for m = – 3

This question in 0607/23 Oct/Nov 2019

14 A is the point (1, 7) and B is the point (4, 13). Find the equation of the perpendicular bisector of AB in the form y = mx + c . y = … [5] Question 15 is printed on the next page.

5 marks

Mark scheme: 14 [ y = ] − 0.5 x + 11.25 5 B1 (2.5, 10) seen − 7 M1 for gradient = 13 oe 4 − 1 M1 grad of perp = –1/their grad M1 for subst their grad and their (2.5, 10) into y = mx + c

This question in 0607/22 May/June 2020

13 A is the point (1, 7) and B is the point (4, 1). Find the equation of the perpendicular bisector of AB in the form y = mx + c . y = … [5]

5 marks

Mark scheme: Question Answer Marks Partial Marks 12(a) and (b) are dependent on an increasing curve. 12(a) Correct graph passing through 4 B2 for correct vertical plots (0, 0) (10, 8) (15, 24) (20, 49) or B1 for 3 or 4 correct vertical plots or (30, 66) (50, 80) all cf’s seen correct B1 for plotting points at upper group limit B1 for smooth curve 12(b) 17 to 19 1 FT their curve 13 1 11 7 − 1 y = x + 5 M1 for gradient = oe 2 4 1 − 4 − 1 M1 for gradient perp = their gradient B1 for midpoint (2.5, 4) seen M1 for subst their gradient (perp) and their mid-point into y = mx + c

This question in 0607/21 Oct/Nov 2020

10 A is the point (-5, 7) and C is the point (1, -2). (a) B is the mid-point of AC. Find the coordinates of B. ( … , … ) [2] (b) The line CD is perpendicular to the line AC. Find the equation of line CD. … [4]

6 marks

Mark scheme: 10(a) (–2, 2.5) oe 2 B1 for each coordinate 10(b) 2 8 4 Equivalent 3 term equation. y  x  oe 3 3 2 7 M1 for gradient of BC = oe 1 ( 5) 3 M1 for gradient of CD = –1 ÷ (their  ) 2 M1 for substituting (1, – 2) and their m into y = mx + c oe

This question in 0607/23 May/June 2022

10 A is the point (1, 11) and B is the point (4, 5). Find the equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c . y = … [5]

5 marks

Mark scheme: 10 1 27 5 11 − 5 y = x + M1 for [ = −2] 2 4 1 − 4 1 M1 for Grad perp = − their ( −2) B1 for mid-point (2.5, 8) M1 for correct substitution of their mid-point and gradient into y = mx + c.

This question in 0607/21 Oct/Nov 2022

16 The point A has coordinates (2, 3) and the point B has coordinates (6, 5). The point C lies on the line AB. The point D has coordinates (2, 5.5). CD is perpendicular to AB. Find the coordinates of C. ( … , … ) [5]

5 marks

Mark scheme: 16 (3, 3.5) oe 5 5  3 M1 for grad AB = oe or better 6  2 M1 for equation of AB y  0.5 x  2 oe  1 M1 for grad CD = their grad AB M1 for 5.5  their (0.5 p  2)  their grad CD 2  p oe where ‘p’ is x-coordinate of C If 0 scored, SC1 for (3, k ) or ( k , 3.5)

This question in 0607/23 May/June 2023

14 The line L is perpendicular to the line 2y = 5 - x and passes through the point (2, 3). Find the equation of line L. Give your answer in the form y = mx + c . y = … [4] Questions 15 and 16 are printed on the next page.

4 marks

Mark scheme: 14 2x – 1 4 1 B1 for grad of given line = − 2 1 M1 for grad of L = −  1  their  −   2  M1 for 3 = their grad × 2 + c y − 3 or = their (grad) x − 2 1 (their grad not − ) 2

This question in 0607/21 Oct/Nov 2023

10 (a) P is the point (-5, 3) and Q is the point (2, -1). Find the coordinates of the mid-point of PQ. ( … , … ) [2] (b) Line L is perpendicular to the line y = 3x - 2. The point (6, 1) is on line L. Find the equation of line L. Give your answer in the form y = mx + c. y = … [3]

5 marks

Mark scheme: 10(a)  3  2 B1 for each  − , 1  oe  2  10(b) 1 3 1 y = − x + 3 B1 for gradient − 3 3 M1 for correct substitution of (6, 1) into y = (their m) x + c m  3

This question in 0607/22 Oct/Nov 2023

9 A is the point (1, 3) and B is the point (3, -7). The line l passes through A and is perpendicular to AB. Find the equation of line l. Give your answer in the form py + qx = r where p, q and r are integers. … [4]

4 marks

Mark scheme: 9 5 y  x  14 oe 4 Correct answer in the form py  qx  r p, q, r integers B3 for a correct equation in the wrong form 1 or B2 for y  x  k oe 5 7 3 M1 for [Grad  ] oe 3  1 1 M1 for [Grad perp  ] their gradient M1 for subst (their grad perp) and (1, 3) into y  mx  c oe

This question in 0607/21 May/June 2024

13 The point A has coordinates (4, -1) and the point B has coordinates (8, -3). Find the equation of the perpendicular bisector of the line AB. Give your answer in the form y = mx + c . y = … [5]

5 marks

Mark scheme: 13 y = 2 x − 14 5 −−−1 3 M1 for [Grad = ] oe 4 − 8  4 − 8  M1 for [Grad perp = ] − their    −−−1 3  B1 for (6, –2) as mid-point M1 for subst (their grad perp) and their (6, –2) into y = mx + c

This question in 0607/21 Oct/Nov 2024

9 These are the equations of two lines. 4y = x + 7 y + 4x = 6 (a) Find the coordinates of the point where these two lines intersect. ( … , … ) [3] (b) Are the two lines perpendicular? Give a reason for your answer. … because … … [2]

5 marks

Mark scheme: 9(a) (1, 2) final answer 3 M1 for correctly equating one set of coefficients A1 for x = 1 A1 for y = 2 Correct answers spoiled scores SC2 If 0 scored SC1 for their solutions satisfying one equation 9(b) 1 B1 both gradients seen –4 and 4 1 B1 FT their gradients and correct conclusion Yes and −4 = −1 4

This question in 0607/22 Feb/March 2025

16 A is the point (-3, 2) and B is the point (5, -8). Find the equation of the perpendicular bisector of AB. … [5]

5 marks

Mark scheme: 16 4 19 5 −−8 2 y = x – oe M1 for oe 5 5 5 −−( 3) final answer − 1 M1 for their m B1 for (1, –3) M1 for substituting their (1, –3) and 4 their into y = mx + c 5

This question in 0607/21 May/June 2025