Length and midpoint — Paper 4 · IGCSE Mathematics - International 0607

10 y C (14, 8) NOT TO SCALE B (11, 4) A (2, 2) x 0 A is the point (2, 2), B is the point (11, 4) and C is the point (14, 8). (a) Find the equation, in the form y = mx + c, of (i) the line AC, y = … [3] (ii) the line through B that is perpendicular to AC. y = … [3] (b) Show that the point (10, 6) is on both the lines you found in part (a). [2] (c) AC is the perpendicular bisector of BD. Find the co-ordinates of D. ( … , … ) [1] (d) Find the exact area of the quadrilateral ABCD. … [4]

13 marks

Mark scheme: 10(a)(i) 1 3 8 − 2 [y =] x + 1 M1 for gradient = oe 2 14 − 2 M1 for correct substitution of (2, 2) or (14, 8) into y = (their m)x + c oe soi 10(a)(ii) [y =] –2x + 26 3 −1 M1 for gradient = their 12 M1for substituting (11, 4) into y = (their – 2 )x + c oe soi 10(b) Correct substitution and completion 2 B1 for either of (10, 6) for both lines oe OR M1 for correct elimination of x or y from equations A1 for completion to solution (10, 6) 10(c) (9, 8) 1 10(d) 30 cao 4 1 2 2 2 2 M3 for × 12 + 6 × 2 + 4 oe  2 or B2 for two of 12 2 + 6 2 oe (AC), 2 2 + 4 2 oe (BD or MC), 8 2 + 4 2 oe (AM), 2 2 + 12 oe (MD or MB) or B1 for one of these. (M is the intersection of AC and BD) OR M3 for full area e.g. [0.5 × 12 × 6 – 0.5 × 6 × 7] × 2 or B2 for 2 correct areas evaluated or B1 for 1 correct area evaluated

This question in 0607/42 May/June 2017

14 A is the point (1, 9) and B is the point (7, 1). (a) Find the length of AB. … [3] (b) Find the co-ordinates of the midpoint of AB. ( … , … ) [2] (c) B is the reflection of A in the line L. Find the equation of the line L. … [4]

9 marks

Mark scheme: 14(a) 10 3 2 2 M2 for 6 + 8 or B1 for 6 and 8 seen nfww 14(b) (4, 5) 2 B1 for each co-ordinate 14(c) 3 4 Must be 3 term equation y = x + 2 oe 4 3 B2 for gradient = 4 4 or B1 for gradient of AB = – 3 M1 for substituting their (b) into y = (their m) x + c oe

This question in 0607/42 Oct/Nov 2018

10 The points A (1, 2) and B (7, 5) are shown on the diagram below. y 12 NOT TO SCALE B A 0 16 x (a) Write AB as a column vector. [1] f p (b) Calculate the length of the line AB. … [2] (c) The point C has co-ordinates (10, k). AB = BC and k 2 0. Show that k = 11. [3] (d) Find the equation of the line that is perpendicular to AC that passes through the midpoint of AC. Give your answer in the form y = mx + c. y = … [4] (e) The points A, B, C and D form a rhombus. Find the co-ordinates of D. ( … , … ) [3]

13 marks

Mark scheme: 10(a) 6 1  3 10(b) 6.71 or 6.708… or 45 oe 2 2 2 M1 for (7 −1) + (5 − 2) oe 10(c) 2 2 2 2 2 k −=5 (their (b)) − 3 M2 M1 for (k − 5) + (10 − 7) = (their(b)) oe Reverse process scores 0. k −=5 6 A1 10(d) [ y = ] − x + 12 oe 4 11 − 2 M1 for grad AC = oe 10 − 1 1 M1 for grad perp = − their grad B1 for midpoint (5.5, 6.5) 10(e) (4, 8) 3 10  6 7 −3  M2 for   −  or −  oe 11  3 5 3 6  −3  or M1 for CD = or BD =  oe 3  3

This question in 0607/43 May/June 2019

6 (a) P is the point (3, 5) and Q is the point (7, - 2). Q is the midpoint of PR. Find the co-ordinates of the point R. ( … , … ) [2] (b) A NOT TO SCALE a C O b B OA = a and OB = b . C divides AB in the ratio 4 : 3. Find these vectors, in terms of a and b, in their simplest form. (i) AB AB = … [1] (ii) OC OC = … [3]

6 marks

Mark scheme: 6(a) (11, –9) 2 B1 for each co-ordinate 6(b)(i) –a + b 1 6(b)(ii) 3 4 1 3 B2 for unsimplified a + b or ( 3a + 4b) JJJG JJJG 7 7 7 4 or B1 for OA + AB oe or a correct 7 route

This question in 0607/43 Oct/Nov 2019

3 y C NOT TO SCALE D B A O x ABCD is a parallelogram. A is the point (3, 1), B is the point (10, 2) and D is the point (2, 3). (a) Find the coordinates of C. ( … , … ) [2] (b) Calculate the length of AB. Give your answer as a surd in its simplest form. AB = … [3] (c) The diagonals of the parallelogram meet at X. Find the coordinates of X. ( … , … ) [2] (d) The straight line BA is extended to meet the y-axis at P and the x-axis at Q. Find the coordinates of P and the coordinates of Q. P ( … , … ) Q ( … , … ) [5]

12 marks

This question in 0607/41 May/June 2020

9 y A NOT TO SCALE B O x C A is the point (-2, 6), B is the point (3, 2) and C is the point (3, -4). (a) Write down the equation of BC. … [1] (b) Find the coordinates of the point M, the mid-point of AC. ( … , … ) [1] (c) The quadrilateral ABCD has rotational symmetry of order 2 about the point M. Find the coordinates of the point D. ( … , … ) [2] (d) Find the equation of the perpendicular bisector of AC. … [4]

8 marks

Mark scheme: 9(a) x = 3 oe 1 9(b)  1  1  , 1  oe  2  9(c) (–2, 0) 2 B1 for each coordinate 9(d) 1 3 4 3 term equivalent y = x + oe −−4 6 2 4 M1 for gradient of AC = 3 −−( 2) −1 M1 for m = theirgradient M1 for substituting their (b) into their y = mx + c

This question in 0607/43 May/June 2020

9 p = q = 3 - 1 A is the point (3, 4). (a) Find p - q . [1] f p (b) A is translated onto H by the vector p. Find the coordinates of H. ( … , … ) [1] (c) J is translated onto A by the vector q. Find the coordinates of J. ( … , … ) [1] (d) Find the coordinates of the mid-point of HJ. ( … , … ) [1] (e) Find the length of HJ. HJ = … [3] 1 (f) A line L, parallel to the vector q, has gradient - . 2 Find the equation of the line perpendicular to the line L that passes through the point A. … [3]

10 marks

Mark scheme: 9(a)  − 3  1    4  9(b) (2, 7) 1 9(c) (1, 5) 1 9(d) (1.5, 6) 1 FT their (b) and (c). 9(e) 2.24 or 2.236... 3 FT their (b) and (c). M2 for (their 2 – their 1)2 + (their 7 – their 5)2 oe or M1 for (their 2 – their 1) and (their 7 – their 5) seen 9(f) y = 2x – 2 oe 3 − 1 M1 for gradient = oe soi 2 1 − 2 M1 for substituting (3, 4) in y = their m x + c Answer 2x – 2 implies M1 M1

This question in 0607/43 Oct/Nov 2020

11 y A (– 2, 4) NOT TO SCALE P O x B (8, – 1) A is the point (-2, 4) and B is the point (8, -1). P divides AB in the ratio 3 : 2. (a) Show that the coordinates of P are (4, 1). ( … , … ) [2] (b) The line L is perpendicular to AB and passes through P. Find the equation of line L. … [4] (c) The point C has coordinates (6, 5). Show that point C lies on line L. [1] (d) (i) Find the distance AB. Give your answer in surd form. … [2] (ii) Calculate the area of triangle ABC. … [3]

12 marks

Mark scheme: 11(a) 8 – –2 = 10, 3 : 2 = 6 : 4, M2 M1 for each coordinate x = –2 + 6 = 4 oe 4 to –1 = 5, y = 4 – 3 = 1 oe 11(b) y = 2x – 7 oe final answer 4 B3 for 2x – 7 as final answer OR −−1 4 M1 for gradient of AB = 8 −−( 2 ) −1 M1 for m =  1  their  −   2  M1 for 1 = (their2) × 4 + c or y – 1 = their2(x – 4)) 11(c) 2 × 6 – 7 = 5 oe 1 11(d)(i) 5 5 or 125 final answer 2 M1 for (8 – (–2))2 + ((–1) – 4)2 oe 11(d)(ii) 25 [.0] cao nfww 3 M1 for (6 – 4)2 + (5 – 1)2 M1 dep on first M1 for 1 × their ( d )( i ) × their 20 2

This question in 0607/42 May/June 2021

8 (a) A is the point ( - 11, 7) and B is the point ( 8 , - 13) . Find the length of AB. … [3] (b) P is the point ( 2, - 5) and Q is the point ( 6, 11) . Line L is perpendicular to PQ and crosses PQ at point R. The ratio PR : RQ = 3 : 1. Find the equation of line L. … [6]

9 marks

Mark scheme: 8(a) 27.6 or 27.58 to 27.59 3 M2 for (( 11)  8) 2  (7 ( 13)) 2 oe or M1 for (( 11)  8) or (7 ( 13)) oe 8(b) y  14 x  8 14 oe 6 B5 for answer  14 x  8 14 OR B2 for (5, 7) or B1 for (5, k) or (k, 7) 11 5 M1 for oe (=m1) 6  2 1 M1 for grad = their m1 M1 for substituting their (5, 7) into y = (their m)x + c

This question in 0607/41 May/June 2022

4 A is the point ( - 2 , - 3) and B is the point (4, 9). (a) Find the length of AB. … [3] (b) Find the equation of the perpendicular bisector of AB. … [5] (c) C is a point on AB. C divides AB in the ratio 2 : 1. Find the coordinates of C. ( … , … ) [2]

10 marks

Mark scheme: 4(a) 13.4 or 13.41 to 13.42 3 M2 for (4 – (–2))2 + (9 – (–3))2 oe or M1 for (4 – (–2)) oe and (9 – (–3)) oe soi by 6 and 12 4(b) 1 7 5 1 7 y = – x + oe B4 for – x + 2 2 2 2 OR 9 −−( 3) M1 for oe 4 −−( 2) M1 for –1 ÷ (their 2) B1 for mid-point = (1, 3) M1 for substituting their (1, 3) into 1 y = (their(– )x) + c 2 4(c) (2, 5) 2 B1 for each coordinate

This question in 0607/42 Oct/Nov 2022

9 y A NOT TO SCALE B O x A is the point (-4, 6) and B is the point (8, 2). (a) Find the coordinates of the mid-point of AB. ( … , … ) [2] (b) Find the equation of AB. … [3] (c) Show that the equation of the perpendicular bisector of AB is y = 3x - 2 . [3] (d) The point C has coordinates (3, 7). Show that C lies on the perpendicular bisector of AB. [1] (e) Find the area of triangle ABC. … [4]

13 marks

Mark scheme: 9(a) (2, 4) 2 B1 for each coordinate 9(b) 1 2 3 2 − 6 y = − x + 4 oe cao M1 for 3 3 8 −−( 4) final answer M1 for substituting (2, 8) or (–4, 6) into  1  y = their  −  x + c oe  3  9(c)  1  M1 Gradient = for –1 ÷  their −  oe  3  substituting their (2, 4) into M1 y = their 3 x + c oe Completion to y = 3x – 2 with no errors A1 Dep on M1, M1 or omissions 9(d) 3 × 3 – 2 = 7 1 9(e) 20 4 2 2 M1 for [AB =] ( 8 + 4 ) + ( 2 − 6 ) M1 for [h =] ( 7 − their 4 ) 2 + ( 3 − their 2 ) 2 1 M1 for  their 160  their 10 2

This question in 0607/42 Feb/March 2024

3 (a) AC is a straight line. B is the mid-point of AC. A is the point (-1, 11) and B is the point (3, 8). (i) Find the length of AB. … [3] (ii) Find the coordinates of C. ( … , … ) [2] (iii) Find the equation of the perpendicular bisector of AC. … [4] (b) PQR is a straight line. P is the point (-6, -1) and Q is the point (-3, 1). Q divides the line PR in the ratio PQ : QR = 1 : 2 . Find the coordinates of R. ( … , … ) [2]

11 marks

Mark scheme: 3(a)(i) 5 3 M2 for (3 −−( 1)) 2 + (8 − 11) 2 soi or M1 for (3 −−( 1)) and (8 − 11) soi 3(a)(ii) (7, 5) 2 B1 for each 3(a)(iii) y = 43 x + 4 oe 4 M1 for gradient AC = 8 − 11 or better 3 −−( 1) M1 for perpendicular gradient = –1÷ their – 34 M1 for substituting (3, 8) into y = their mx + c 3(b) (3, 5) 2 B1 for one correct value or 3 M1 for  soi 2 or for suitable diagram seen e.g. a correct triangle with 2 sides marked

This question in 0607/43 Oct/Nov 2024