Use the equation of a straight line — 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

9 y y = 2√x A (4, 4) O B x The diagram shows part of the curve y = 2 x . The normal to the curve at the point A (4, 4) meets the x-axis at the point B. (i) Find the equation of the line AB. [4] (ii) Find the coordinates of B. [1]

9 marks

Mark scheme: 9(i) dy − 12 B1 = x dx dy 1 B1 x = 4 → = dx 2 grad of normal = −2 M1 y − 4 A1 = −→2 [ y = −2 x + 12 ] x − 4 9(ii) (6, 0) B1 FT 9(iii) 1 B1 FT Area of triangle = × 2 × 4 = 4 2 1 M1 Area under curve 2 x 2 d x =∫ 3 A1 4 2 = x 3 2 A1 FT Total area = 14 [14.7 ] 3 OR Area of trapezium OBAP B1 FT 1 = ( 6 + 4 ) × 4 = 20 2 Area between curve and y- axis M1 y 2 = dy ∫ 4 y 3 A1 = 12 2 A1 FT Total area = 14 [14.7 ] 3

This question in 0606/22 Oct/Nov 2018

1 Solve the equation 5x - 3 =- 3x + 13 . [3]

3 marks

Mark scheme: Question Answer Marks Partial Marks 1 x = 2 B1 3 − 5 x = −3 x + 13 oe M1 x = –5 A1

This question in 0606/23 Oct/Nov 2018

8 Variables x and y are such that when y2 is plotted against e2x a straight line is obtained which passes through the points (1.5, 5.5) and (3.7, 12.1). Find (i) y in terms of e2x, [3] (ii) the value of y when x = 3 , [1] (iii) the value of x when y = 50 . [3]

7 marks

Mark scheme: 8(i) 12.1 − 5.5 B1 correct expression for gradient [= 3] 3.7 − 1.5 y 2 − 5.5 M1 = their grad e 2 x − 1.5 or correctly use y2 = (their m) e2x + c with one point to find c 2 x A1 y = [ ± ] 3e + 1 8(ii) [±]34.8 1 8(iii) 2 x B1 * 50 = ( their 3 ) e + their1 or 2500 = ( their 3 ) e 2 x + their1  2499  M1 Dep* 2 x = ln   obtain 2x explicitly  3  3.36 cao A1

This question in 0606/23 Oct/Nov 2018

6 The relationship between experimental values of two variables, x and y, is given by y = Abx , where A and b are constants. (i) Transform the relationship y = Abx into straight line form. [2] The diagram shows ln y plotted against x for ten different pairs of values of x and y. The line of best fit has been drawn. ln y 8 7 6 5 4 3 2 1 0 1 2 3 4 x (ii) Find the equation of the line of best fit and the value, correct to 1 significant figure, of A and of b. [4] (iii) Find the value, correct to 1 significant figure, of y when x = 2.7. [2]

8 marks

Mark scheme: 6(i) Takes logs, to any base, of both sides and applies M1 the addition/multiplication law for logs ln y = ln( Ab x ) ⇒ ln y = ln A + ln b x ⇒ ln y = ln A + x ln b A1 6(ii) ln y = 1.4x + 2.2 oe B2 B1 for either m = 1.4 or ln b = 1.4 or c = or ln y = xln 4 + ln 9 oe 2.2 or ln A = 2.2 [ A = e their 2.2 = ]9 and B2 FT their 2.2 and their 1.4 [b = e their 1.4 = ] 4 their 2.2 their 1.4 B1 FT for A = e or b = e or correct FT decimal rounded to more than 1 sf 6(iii) ln y = 6 M1 or y = their 9(their 4 2.7 ) or y = e their 2.2 (e their 1.4 × 2.7 ) or ln y = their1.4(2.7) + their 2.2 or ln y = (2.7)ln(their 4) + ln( their 9) awrt 400 correct to 1 sf A1

This question in 0606/22 Feb/March 2019

3 (i) Sketch the graph of y = 5x - 3 on the axes below, showing the coordinates of the points where the graph meets the coordinate axes. y O x [3] (ii) Solve the equation 5x - 3 = 2 - x . [3] 2

6 marks

Mark scheme: 3(i) Correct shape 3 B1 correct shape must have cusp on x- 0.6 oe indicated on x-axis axis 3 indicated on y-axis B1 for each correct point There must be a sketch to award the marks for the intercepts and sketch should be continuous with one intersection only on each axis 3(ii) Solves 5 x − 3 = x − 2 oe M1 or (5 x − 3) 2 = (2 − x ) 2 1 A1 [ x = ] oe 4 5 B1 [ x = ] oe 6

This question in 0606/21 May/June 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

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

1 (i) On the axes below, draw the graph of y = 2x - 3 . y 8 6 4 2 -2 0 2 4 x -2 -4 [2] (ii) Solve the equation 7 - 2x - 3 = 0 . [3]

5 marks

Mark scheme: Question Answer Marks Partial Marks 1(i) 6 y B2 B1 shape 4 B1 Correct intersection with axes. 2 x −2 2 4 −2 −4 1(ii) 7 = 2x – 3 → x = 5 B1 Uses 7 = 3 – 2x oe M1 x = –2 A1

This question in 0606/21 Oct/Nov 2019

5 (a) On the axes below, sketch the graph of y = 5x - 7 , showing the coordinates of the points where the graph meets the coordinate axes. [3] y O x (b) Solve 5 5x - 7 - 1 = 14 . [3]

6 marks

Mark scheme: 5(a) Correct V shape with vertex on positive x- B1 axis (0, 7) B1  7  B1  , 0   5  5(b) x = 2 B1 5 x − 7 = their ( − 3) oe, soi M1 or 25 x − 35 = their ( − 15) oe, soi 4 A1 x = oe 5 Alternative method 25 x 2 − 70 x + 40 = 0 oe (B1 factorising e.g. ( 5 x − 4 )( x − 2 ) M1 4 A1) x = 2, 5

This question in 0606/22 Feb/March 2020

1 Solve the equation 4x + 9 = 6 - 5 x . [3]

3 marks

Mark scheme: Question Answer Marks Partial Marks 1 4x + 9 = 6 − 5x oe M1 or 4x + 9 = 5x – 6 oe 1 A2 not from wrong working; no extras x = − , x = 15 3 A1 for x = 15 ignoring extras implies M1 if no extras seen mark final answer If M0 then SC1 for any correct value with at most one extra value Alternative method: M1 for (4x + 9)2 = (6 – 5x)2 oe soi A1 for 9 x 2 − 132 x − 45 = 0 oe 1 A1 for x = − , x = 15 only; mark 3 final answer

This question in 0606/22 Feb/March 2021

12 DO NOT USE A CALCULATOR IN THIS QUESTION. y A 1 y = 2x + 1 5y = x - 1 C O B x 1 The diagram shows part of the curve y = and part of the line 5y = x - 1. 2x + 1 The curve meets the y‑axis at point A. The line meets the x‑axis at point B. The line and curve intersect at point C. (a) (i) Find the coordinates of A and B. [1] (ii) Verify that the x‑coordinate of C is 2. [2] (b) Find the exact area of the shaded region. [5] Question 13 is printed on the next page.

8 marks

Mark scheme: 12(a)(i) A(0,1) and B(1, 0) B1 12(a)(ii) 1 2 − 1 B2 1 [ y = ] and [ y = ] B1 for [ y = ] and 5 y = 2 − 1 oe 2(2) + 1 5 2(2) + 1 and 1 evaluates both expressions as 5 Alternative 1 (B2) 1 1 2 − 1 1 1 1 1 [ y = ] = or[ y = ] = B1 for = and 5 × = x − 1 oe 2(2) + 1 5 5 5 2(2) + 1 5 5 and 2 1 1 1 1 or −= and = oe 1 5 5 5 2 x + 1 solves 5 × = x − 1 oe to get x = 2 5 1 1 or = oe to get x = 2 5 2 x + 1 Alternative 2 (B2) 2x2− x – 6 = 0 B1 for (2x + 1)(x – 1) = 5 or 2x2− x – 6 = 0 and solves or factorises to get (2x + 3)(x – 2) and states x = 2 OR shows 2(22) – 2 – 6 = 0 oe Alternative 3 (B2) (2x + 1)(x – 1) = 5 oe B1 for (2x + 1)(x – 1) = 5 and shows (2 × 2 + 1)(2 – 1) = 5 12(b) 1 B1 × 1 × 0.2 oe 2 2 2 2  12 1  or − −  −  oe 5 × 2 5  5 × 2 5  1 B2 [ F( x ) = ] ln(2 x + 1) [+c] oe 1 1 2 B1 for ln 2 x + 1 or ln x + 0.5 2 2 1 or ln( x + 0.5) [+c] oe or k ln(2 x + 1) or k ln( x + 0.5) , k ≠ 0.5 or 0 2 F(2) – F(0) – their 0.1 M1 FT their F(x) providing at least B1 for integration of curve awarded 0.5ln5 − 0.1 or exact equivalent A1

This question in 0606/22 May/June 2021

8 Variables x and y are such that when y is plotted against log 2 ( x + 1), where x 2- 1, a straight line is obtained which passes through ( 2, 10.4) and ( 4 , 15.4) . (a) Find y in terms of log 2 ( x + 1) . [4] (b) Find the value of y when x = 15 . [1]

8 marks

Mark scheme: 8(a) 15.4 − 10.4 M1 [Gradient =] oe soi 4 − 2 10.4 = their2.5 × 2 + c or 15.4 = their2.5 × 4 + c M1 FT their gradient or y − 10.4 y − 15.4 = their 2.5 or = their 2.5 x − 2 x − 4 [Gradient = ] 2.5 soi and [intercept =] 5.4 soi A1 y = 2.5log 2 ( x + 1) + 5.4 oe isw A1 Alternative method 10.4 = 2m + c and 15.4 = 4m + c (M1) and solving to find m or c Use their m or c to find their c or m (M1) m = 2.5 and c = 5.4 (A1) y = 2.5log 2 ( x + 1) + 5.4 oe isw (A1) 8(b) 5929 B1 or 237.16 25 8(c) 5 = their 2.5log 2 ( x + 1) + their 5.4 M1 FT their equation from (a) of correct form with m ≠ 1 or 0, and and rearrange to make log 2 ( x + 1) the subject c ≠ 0 Condone any base 4 A1 Condone any base − = log 2 ( x + 1) oe 25 x = − 0.105 or −0.1049[74…] rot to 4 or more sf A1

This question in 0606/22 Oct/Nov 2021

1 A line, L, has equation 4x + 5y = 9 . Points A and B have coordinates ( - 6 , 7) and (1, 9) respectively. Find the equation of the line parallel to L which passes through the mid‑point of AB. [3]

3 marks

Mark scheme: Question Answer Marks Partial Marks 1  5  B1 Mid-point:  − , 8  soi  2  4 B1 Gradient: − soi or substitution of mid- 5 point into e.g. 4 x + 5 y = k 4  5  B1 y − 8 = −  x +  5  2  or 4 x + 5 y = 30 oe, isw

This question in 0606/22 Feb/March 2022

5 Variables P and T are known to be connected by the relationship P = AbT , where A and b are constants. Values of P are found for certain values of time, T. (a) Show that a graph of lgP against T will be a straight line. [2] (b) lg P 12 10 8 6 4 2 T 0 2 4 6 8 10 12 14 The diagram shows the graph of lgP against T. The graph passes through (0, 6) and (14, 12). Find the values of A and b. [4] (c) Using the graph or otherwise, find the length of time for which P is between 100 million and 1000 million. [3]

9 marks

Mark scheme: 5(a) lg P  lg A  T lg b oe nfww B2 Must be seen and not from wrong working and correct comparison with y = mx + c soi B1 for lg P  lg A  T lg b isw, nfww 5(b) A = 106 oe isw and 4 B2 for A = 106 oe isw 3 or B1correct method which could be used b = 10 7 oe isw to find A e.g. 3 lgA = 6 or 12 =  14  lg A 7 3 B2 for b = 10 7 oe isw or B1correct method which could be used to find b e.g. 12  6 lgb = oe or 12 = 14lg b  6 14  0 5(c) lg P1 = 8 and lg P2 = 9 soi M1 If graph not used then allow M1 for substitution of their A and their b in the leading to exponential equation as far as 108 10 9  (theirb )T and  (theirb )T T1 = 4.6 to 4.8 theirA theirA or T2 = 6.8 to 7.2 OR substitution of their A and their b or their lgA and their lg b in the log equation lg108  their lg A  T (their lg b ) or better and lg109  their lg A  T (their lg b ) or better Difference of correct times: M1 T2  T1 where T2 = 6.8 to 7.2 T1 = 4.6 to 4.8 Answer in range 2.2 to 2.4 nfww A1 Alternative method 9  8 (M2) M1 for lg108 = 8 and lg109 = 9 and Change in T = 3 Change in lg P 3  7 Change in T 7 Answer in range 2.2 to 2.4 nfww (A1)

This question in 0606/22 May/June 2023

1 (a) On the axes, sketch the graph of y = 4x - 6 , showing the points where the graph meets the axes. [2] y O x (b) Solve the equation 4x - 6 = 2 x . [3]

5 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) Fully correct graph with intercepts marked B2 B1 for a graph of correct shape with vertex on x-axis y (0, 6) O (1.5, 0) x 1(b) 4x – 6 = 2x and 4x – 6 = –2x oe M1 x = 3 x = 1 A2 A1 for either correct Alternative method 12x2 – 48x + 36 = 0 oe (B1) Factorises or solves (M1) x = 1, x = 3 (A1)

This question in 0606/21 May/June 2024

1 The point A has coordinates (1, 4) and the point B has coordinates (5, 6). The perpendicular bisector of AB intersects the x-axis at the point C and the y-axis at the point D. Given that O is the origin, find the area of triangle OCD. [5]

5 marks

Mark scheme: Question Answer Marks Partial Marks 1 y  5 2  x  3  oe 3  5  1 6  4  M1 for midpoint  ,  or  3, 5   2 2  1 M1for m  oe or  2 1 2 121 2 B1 FT for x-intercept (5.5, 0) and or 30.25 oe cao y-intercept (0, 11) soi; 4 FT their perpendicular bisector providing M1 M1 awarded

This question in 0606/23 May/June 2024