Solve problems involving midpoint and length — Paper 2 · IGCSE Mathematics - Additional 0606

8 Solutions to this question by accurate drawing will not be accepted. The points A and B are (−8, 8) and (4, 0) respectively. (i) Find the equation of the line AB. [2] (ii) Calculate the length of AB. [2] The point C is (0, 7) and D is the mid-point of AB. (iii) Show that angle ADC is a right angle. [3] J N 4 The point E is such that AE = KK OO. - 7 L P (iv) Write down the position vector of the point E. [1] (v) Show that ACBE is a parallelogram. [2]

10 marks

Mark scheme: 8(i) 8 B2 8 y − 8 = − ( x −−( 8 ) ) oe isw B1 for m AB = − oe 12 12 8 8 − 0 or y [ −0] = − ( x − 4) oe isw or M1 for oe −−8 4 12 or 3 y = −2 x + 8 oe isw 8(ii) ( −−8 4 ) 2 + ( 8[ −0] ) 2 oe M1 any valid method 208 isw or 4 13 isw or 14.4222051… rot to 3 or A1 implies M1 provided nfww more sf 8(iii) [coordinates of D =] (–2, 4) soi B1 If coordinates of D not stated then a calculation for mCD or a relevant length with the coordinates clearly embedded must be shown to imply B1 Gradient methods: M1 or Length of sides methods:  7 − their 4   3  2  mCD = =  their   finds or states AC = 65 or AC = 65  0 − their ( − 2)   2  2 2 2 or AC = ( −−8 0 ) + ( 8 − 7 ) oe y or AC = ( −−8 0 ) 2 + ( 8 − 7 ) 2 oe A 65 8 C and CD 2 = their13 or CD = their 13 6 13 2 2 2 or CD = ( 0 − their ( −2 ) ) + ( 7 − their 4 ) oe 2 13 D 4 or CD = ( 0 − their ( −2 ) ) 2 + ( 7 − their 4 ) 2 oe 2 B and AD 2 = their 52 or AD = their 2 13 -8 -6 -4 -2 0 2 4 x or AD 2 = ( −−8 their ( −2 ) ) 2 + ( 8 − their 4 ) 2 -2 2 2 or AD = ( −−8 their ( −2 ) ) + ( 8 − their 4 ) or uses a valid method with their coordinates of D to find the exact area of the triangle and equates to 1 ( AD )(CD )sin( ADC ) 2 3  8  3 A1 applies Pythagoras to confirm, using states ×  −  = − 1 oe or is the negative integer values, that 65 = 13 + 52 or finds 2  12  2 2 e.g. AC = 65 using (2 13) 2 + ( 13) 2 reciprocal of − oe 3 or finds the equation of the perpendicular bisector or 3 1 of AB as y = x + 7 independently of C and solves 2 13 13 sin ADC = 13 or 2 ( )( ) 2 states that C lies on this line. 2 65 = (2 13) 2 + ( 13) 2 ( ) − 2(2 13)( 13)cos ADC to show ADC is a right angle 8(iv)  −4  B1 condone coordinates   or −4i + j  1  8(v) Full valid method e.g. B2 B1 for incomplete method JJG 4 0  4  JJG  4  for showing that e.g. CB =  −  =   e.g. for stating that CB =   0 7  − 7   − 7  or showing that e.g. JJJG 0  − 8   8  JJJG  8  JJG AC =  −   =   oe or AC =   = EB 7  8   −1   − 1  JJG 4  −4   8  and EB =  −   =   oe or just showing that one pair of opposite 0  −1   −1  sides is parallel or has the same length or comparing gradients of both pairs of opposite or just showing that length DC = length sides and showing they are pairwise the same DE or just showing that C, D and E are collinear or comparing the lengths of both pairs of opposite sides and showing that they are 1 pairwise the same A(-8, 8) m AC = − 8 65 C(0, 7) or showing that length AC = length AE or that the length BC = length BE 65 7 m BC = − D(-2, 4) 4 or comparing the gradients and lengths of a mAE = − 7 65 pair of opposite sides 4 E(-4, 1) or showing that D is the midpoint of CE 65 1 B (4, 0) mEB =− 8 or showing that length DC = length DE and that C, D and E are collinear

This question in 0606/22 May/June 2017

11 A line with equation y =- 5x + k + 5 is a tangent to a curve with equation y = 7 - kx - x 2 . (i) Find the two possible values of k. [5] (ii) Find, for each of your values of k, • the equation of the tangent • the equation of the curve • the coordinates of the point of contact of the tangent and the curve. [5] (iii) Find the distance between the two points of contact. [2]

12 marks

Mark scheme: 11(i) −5 x + k + 5 = 7 − kx − x 2 M1 * 2 2 M1 Dep* b − 4 ac ( = 0 ) → ( k − 5 ) − 4 ( k − 2 ) ( = 0 ) k 2 − 14 k + 33 ( = 0 ) A1 ( k − 11)( k − 3 ) ( = 0 ) M1 Dep dep * solve quadratic in k k = 11 and k = 3 A1 11(ii) y = –5x + 16 and y = 7 – 11x – x2 B2 FT their k B1 for any two correct y = –5x + 8 and y = 7 – 3x – x2 solve one tangent/curve pair for one variable from M1 quadratic equation with repeated root (–3, 31) and (1, 3) A2 A1 for one correct point or two correct x values 11(iii) find distance between any two points found in (ii) M1 800 oe A1

This question in 0606/23 Oct/Nov 2018

3 The points A, B and C have coordinates (4, 7), (-3, 9) and (6, 4) respectively. (i) Find the equation of the line, L, that is parallel to the line AB and passes through C. Give your answer in the form ax + by = c, where a, b and c are integers. [3] (ii) The line L meets the x-axis at the point D and the y-axis at the point E. Find the length of DE. [2]

5 marks

Mark scheme: 3(i) 7 − 9 2 M1 oe or − seen 4 −−( 3) 7  2  M1 y − 4 = their  −  ( x − 6)  7   2  or y = their  −  x + c  7  40 and their c = oe 7 2 x + 7 y = 40 oe A1 3(ii) 2 2 M1 FT their equation from part (i)  40   40  their   + their    2   7  20.8[00…] A1

This question in 0606/23 May/June 2019

6 Do not use a calculator in this question. C 3– 3 A B 3 + 3 (i) Find tan ACB in the form r + s 3 , where r and s are integers. [3] (ii) Find AC in the form t u, where t and u are integers and t ! 1. [3]

6 marks

Mark scheme: 6(i) 3 + 3 B1 [ tan ACB = ] 3 − 3 rationalise with 3 + 3 M1 simplify showing at least 3 terms in A1 numerator to 2 + 3 3 + 3 + 3 − 3 oe6(ii) ( AC ) 2 = ( 2 2 M1 Pythagoras ) ( ) at least 4 terms 12 + 6 3 + 12 − 6 3 A1

This question in 0606/23 Oct/Nov 2019

6 (a) Find the equation of the tangent to the curve 2y = tan 2x + 7 at the point where x = r . 8 Give your answer in the form ax - y = r + c , where a, b and c are integers. [5] b (b) This tangent intersects the x-axis at P and the y-axis at Q. Find the length of PQ. [2]

7 marks

Mark scheme: 6(a) dy 2 B1 = sec 2 x dx dy B1 d y their = their 2 FT their dx x = π d x 8 π B1 x = , y = 4 8  π  M1 y − their 4 = ( their 2 )  x −  oe  8  π A1 2 x − y = − 4 4 6(b) 2 2 M1  π   π   − 2  +  4 −  oe  8   4  3.59 or 3.59[03…] rot to four or more A1 figs

This question in 0606/22 May/June 2020

3 (a) Find the equation of the perpendicular bisector of the line joining the points (12, 1) and (4, 3), giving your answer in the form y = mx + c . [5] (b) The perpendicular bisector cuts the axes at points A and B. Find the length of AB. [3]

8 marks

Mark scheme: 3(a) 3 − 1  1  B1 Gradient of line =  −  4 − 12  4  Gradient of perpendicular = 4 M1 − 1 their grad line Mid-point is (8, 2) B1 y − 2 M1 Using their perpendicular gradient and Equation: = 4 mid-point x − 8 y = 4x – 30 A1 3(b) x = 0 → (y) = –30 B1 FT equation must have 3 terms y = 0 → (x) = 7.5 B1 FT equation must have 3 terms B1 15 17 AB = 30 2 + 7.5 2 = 30.9 or better nfww Accept exact answer of 2

This question in 0606/22 Oct/Nov 2020

6 The points A(5, - 4 ) and C(11, 6) are such that AC is the diagonal of a square, ABCD. (a) Find the length of the line AC. [2] (b) (i) The coordinates of the centre, E, of the square are (8, y). Find the value of y. [1] (ii) Find the equation of the diagonal BD. [3] (iii) Given that the x‑coordinate of B is less than the x‑coordinate of D, write EB and ED as column vectors. [2]

8 marks

Mark scheme: 6(a) 2 2 M1 (11 − 5) + (6 −−4) oe 11.7 or 11.66[19...] rot to 4 or more A1 figs 6(b)(i) [y = ] 1 B1 6(b)(ii) 6 −−4 10 B1 m AC = or nfww oe 11 − 5 6 −1 M1 m BD = oe 10 their 6 A1 FT their 1 from (b)(i) and their perpendicular y – their 1 = − 3( x − 8) oe isw gradient 5 6(b)(iii)  − 5   5  B2 B1 for either   and    3   − 3  If 0 scored, SC1 for −5i + 3j and 5i −3j

This question in 0606/22 May/June 2021

11 The coordinates of points A and B are ( - 5, 6) and ( 4, - 6) respectively. The point C lies on the line AB, AC 1 between A and B, such that = . CB 2 (a) Find the coordinates of C. [2] (b) The line CD is perpendicular to AB. Find the equation of CD in the form y = mx + c . [4] (c) The length of BD is 125. Find the coordinates of the two possible positions of point D. [6]

12 marks

Mark scheme: 11(a) (−2, 2) B2 B1 for one correct coordinate nfww 1  9  M1 for AC =   3  −12  1  −9  or CA =   3  12  for x = −5 + 3 or x = 4 – 6 or for y = 6 – 4 or y = −6 + 8 4 − x y + 6 or for x + 5 = or 6 − y = 2 2   −5    4  or for 2  OC −    =   − OC oe   6    −6  11(b) −−6 6 12 4 B1 m AB = oe or − or − 4 −−( 5) 9 3 3 M1 −1 mCD = FT 4 their m AB M1 −1 y – 2 = 3( x + 2) oe FT their (−2, 2) and 4 their m AB 3  3  or y = x + c and 2 =   (−2)+c oe soi 4  4  3 7 A1 y = x + or equivalent in form 4 2 y = mx + c 11(c) ( x − 4 ) 2 + ( y + 6 ) 2 = 125 oe, soi B1 3 7 M1 if correct implies B1 Uses their y = x + to eliminate one 4 2 unknown Correct equation in one unknown A1 2  BD 2 = ( x − 4 ) 2 +  3 x + 7 + 6  = 125 oe  4 2  Writes in solvable form: A1 25x2 + 100x – 300 = 0 oe Factorises or solves a correct 3-term A1 quadratic (2, 5) and (−6, −1) A1 If B1, M0 award: SC2 for identifying one correct point by inspection from the length equation and testing it in the correct equation of CD and SC2 for identifying the second correct point by inspection from the length equation and testing it in the correct equation of CD 11(c) Alternative [ BC = ] (B1) FT their C (4 − (their − 2)) 2 + ( −−6 ( their 2 )) 2 CD = 125 – their100 (M1) FT their BC2 providing 125 – their 100 > 0 CD = 5 (A1) 4  −2   −4   −2  (A2) 4  −4   +   or   +   A1 for CD1 =  or CD2 =   soi 3  2   −3   2  3  −3  OR OR A1 for finds finds 25x2 + 100x – 300 = 0 oe 2 2  3 7  ( x + 2) + x + − 2 = 25    4 2  (2, 5) and (−6, −1) (A1)

This question in 0606/22 Oct/Nov 2022

1 (a) A straight line passes through the points (4, 23) and (-8, 29). Find the point of intersection, P, of this line with the line y = 2x + 5 . [5] (b) Find the distance of P from the origin. [2]

7 marks

Mark scheme: Question Answer Marks Guidance 1(a) 1 3 29 − 23 1 y = − x + 25 isw M1 for m = oe or − 2 −−8 4 2 and y − 23  1  M1 FT for = their  −  oe x − 4  2  or 1 1 y = their − x + c and 23 = −  4 + c oe ( 2 ) 2 OR M1 for solving 23 = 4m + c 29 = –8m + c 1 for m = − or c = 25 2 and M1 FT for correctly using their m or their c to find c or m Solves their linear equation simultaneously M1 1 FT their y = − x + 25 oe with y = 2x + 5 to find x or y 2 (8, 21) A1 1(b) 2 2 M1 FT their (8, 21) 8 + 21 oe 505 isw or 22.5 A1 or 22.47[22…] rot to 2 or more dp

This question in 0606/22 Oct/Nov 2023

7 (a) The curves 4x 2 - 3y 2 + xy = 24 and y = intersect at the points P and Q. Find the coordinates x of P and Q. [5] (b) Find the length of PQ. Give your answer in the form a b, where a is rational and b is the smallest possible integer. [2]

7 marks

Mark scheme: 7(a) Correctly eliminates x or y e.g. M1 2  2  2  2  4 x − 3   + x   = 24 oe or  x   x   2  2 2  2  4 − 3 y + y = 24 oe      y   y  Rearranges to a 3-term quadratic in x2 or y2 A1 soi e.g. 4 x 4 − 22 x 2 − 12  = 0  or 2 x 4 − 11x 2 − 6  = 0  or 3 y 4 + 22 y 2 − 16  = 0  oe Factorises or solves their 3-term quadratic in M1 x2 or y2 soi e.g. (2x2 + 1)(x2 − 6) or (3y2 – 2)(y2 + 8) 2 2 2 A1 x = 6 oe, nfww or y = nfww 3  2   2  A1 and no other values;  6,    or   6,   oe, nfww dep on at least the first M1 A1  6   3  7(b) 2 2 M1 FT providing their xP, xQ and their yP, ( xP − xQ ) + ( y P − yQ ) oe, soi yQ are non-zero 4 A1 15 3

This question in 0606/22 Feb/March 2024