Know and use the condition for two lines to — Paper 2 · IGCSE Mathematics - Additional 0606

8 The points A(3, 7) and B(8, 4) lie on the line L. The line through the point C(6, −4) with gradient 6 meets the line L at the point D. Calculate (i) the coordinates of D, [6] (ii) the equation of the line through D perpendicular to the line 3y - 2x = 10 . [2]

8 marks

This question in 0606/22 Feb/March 2017

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

10 The line y = 12 - 2x is a tangent to two curves. Each curve has an equation of the form y = k + 6 + kx - x 2 , where k is a constant. (i) Find the two values of k. [5] The line y = 12 - 2x is a tangent to one curve at the point A and the other curve at the point B. (ii) Find the coordinates of A and of B. [3] (iii) Find the equation of the perpendicular bisector of AB. [3]

11 marks

Mark scheme: 10(i) 12 − 2 x = k + 6 + kx − x 2 M1 * Equate and collect terms → x 2 − ( 2 + k ) x + 6 − k = 0 b 2 − 4ac = 0 M1 Dep* → ( 2 + k ) 2 = 4 ( 6 − k ) k 2 + 8k − 20 = 0 A1 ( k + 10 )( k − 2 ) = 0 M1 k = −10 or 2 A1 M1 Insert values of k in equations10(ii) ( − 4, 20 ) and ( 2, 8 ) 3 and solve for x A1 x 2 + 8 x + 16 = 0 → x = −4 → y = 20 A1 x 2 − 4 x + 4 = 0 → x = 2 → y = 8 10(iii) 1 B1 Grad of perpendicular = 2 Midpoint ( − 1 ,1 4 ) B1 FT y − 14 1 1 B1 FT Eqn = → y = x + 14.5 x + 1 2 2

This question in 0606/21 Oct/Nov 2018

5 Solutions to this question by accurate drawing will not be accepted. The points A(3, 2), B(7, -4), C(2, -3) and D(k, 3) are such that CD is perpendicular to AB. Find the equation of the perpendicular bisector of CD. [6]

6 marks

Mark scheme: 5 2 + 4 3 M1 [ m AB = ] oe or − soi 3 − 7 2 2 M1 [ mCD = ] their oe, soi 3 2 3 + 3 M1 their = oe or 3 k − 2 2 3 + 3 = their ( x − 2) oe 3 k = 11 nfww A1  (their 11) + 2 3 + −3  M1  ,  oe  2 2  3 A1 FT their mAB and (their 6.5, 0) y = − ( x − 6.5 ) oe isw 2

This question in 0606/22 Feb/March 2019

10 Solutions to this question by accurate drawing will not be accepted. The points A and B have coordinates ( p, 3) and (1, 4) respectively and the line L has equation 3x + y = 2 . 1 (i) Given that the gradient of AB is , find the value of p. [2] 3 (ii) Show that L is the perpendicular bisector of AB. [3] (iii) Given that C q, - 10 lies on L, find the value of q. [1] ` j (iv) Find the area of triangle ABC. [2]

8 marks

Mark scheme: 10(i) 4 − 3 1 M1 ALT uses y = mx + c with A and B as = oe 1 −p 3 far as an equation in p only −2 A1 10(ii) Either: Finds midpoint AB B1 FT their p  their p + 1 3 + 4   ,   2 2  Verifies ( −0.5, 3.5 ) is on L B1 y = −3 x + 2 therefore m = −3 oe B1 1 and ×−=3 −1 oe 3 Or: finds midpoint AB B1 FT their p  their p + 1 3 + 4   ,   2 2  1 B1 ×−=3 −1 oe 3 y − 3.5 = − 3( x + 0.5) and completion to B1 y = −3 x + 2 10(iii) q = 4 B1 10(iv) 22.5 nfww B2 B1 for correct method to find area using correct values 1 e.g. × AB × MC where M is the 2 midpoint of AB

This question in 0606/21 May/June 2019

5 Solutions to this question by accurate drawing will not be accepted. The points A and B are (4, 3) and (12, −7) respectively. (a) Find the equation of the line L, the perpendicular bisector of the line AB. [4] (b) The line parallel to AB which passes through the point (5, 12) intersects L at the point C. Find the coordinates of C. [4]

8 marks

Mark scheme: 5(a) Finds coordinates of mid-point B1 (8, –2) 3 + 7  5  B1 m AB =  = −  oe soi 4 − 12  4  −1 M1 mL = oe − 54 4 A1 y + 2 = ( x − 8) oe isw 5 5(b) 5 B1 y − 12 = − ( x − 5) 4 Attempts to solve their equations M1 (13, 2) A2 A1 for x = 13 or y = 2

This question in 0606/22 May/June 2020

1 Solutions to this question by accurate drawing will not be accepted. Find the equation of the perpendicular bisector of the line joining the points (4, - 7 ) and ( - 8 , 9). [4]

4 marks

Mark scheme: Question Answer Marks Partial Marks 1 Coordinates of mid-point B1 ( −2,1) 9 −−7  16  B1 m AB =  = −  −−8 4  12  −1 M1 m⊥ = −1612 3 A1 y −=1 ( x + 2) oe 4

This question in 0606/23 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

3 The points A, B and C have coordinates (2, 6), (6, 1) and (p, q) respectively. Given that B is the mid‑point of AC, find the equation of the line that passes through C and is perpendicular to AB. Give your answer in the form ax + by = c , where a, b and c are integers. [5]

5 marks

Mark scheme: 3 C(10, 4) B1  1 5 M1  m AC  6 oe or oe 2  6 4 4 M1 1 m FT 5 5 their 4 4 A1 FT their coordinates of C providing y – (4) = ( x  10) oe that one coordinate is correct and their 5 perpendicular gradient 4x – 5y = 60 oe A1

This question in 0606/23 May/June 2022

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