3 Variables x and y are such that when y is plotted against x3 a straight line graph passing through the points (2, 5) and (10, 21) is obtained. Find y in terms of x. [4]

3 3 = + y mx c soi M1 m = 21 5 10 2 − − oe or 2 M1 c = 5 – 2(2) oe or 1 M1 FT their m y = (2x3 + 1)2 oe, isw A1 Alternative method 3 = + y mx csoi (M1) 21 10 = + m c and 5 2 = + m c and solves to find m = 2 or c = 1 (M1) Finds the other unknown (M1) FT their m or c y = (2x3 + 1)2 oe, isw (A1)

2 Points A and C have coordinates ( , ) 4 6 - and ( , ) 2 18 respectively. The point B lies on the line AC such that . AB AC 3 2 = (a) Find the coordinates of B. [2] (b) Find the equation of the line l, which is perpendicular to AC and passes through B. [2] (c) Find the area enclosed by the line l and the coordinate axes. [2]

2(a) ( ) 0, 14 2 B1 for x-coordinate B1 for y-coordinate 2(b) 1 14 2 − = − y x 2 M1 for finding the gradient of a perpendicular line and attempt at the straight line equation using their B A1 Allow unsimplified 2(c) Area = 1 14 28 2 × × M1 Must be a complete method making use of their answer to (b) 196 A1

3 The line AB is such that the points A and B have coordinates ( , ) 4 6 - and (2, 14) respectively. (a) The point C, with coordinates (7, a) lies on the perpendicular bisector of AB. Find the value of a. [4] (b) Given that the point D also lies on the perpendicular bisector of AB, find the coordinates of D such that the line AB bisects the line CD. [2]

3(a) Gradient of line perp to AB 3 4 = − B1 Mid-point of AB ( ) 1, 10 − soi B1 ( ) 3 10 1 4 − = − + y x soi M1 For attempt at straight line using their perp gradient and their mid-point ( ) 3 10 7 1 4 − = − + a 4 = a A1 Allow 4 = y 3(b) ( ) 9, 16 − 2 B1 for 9 = − x B1 FT on their a, dep on M1 from (a) for 16 = y or 20 −their a B1 for 9,16 −

5 When ey is plotted against x2 a straight line graph passing through the points (2.24, 5) and (4.74, 10) is obtained. Find y in terms of x. [5]

5 2 e = + y mx c B1 May be implied by later work 10 4.74 = + m c 5 2.24 = + m c M1 For at least one correct equation 5 2.5 = m M1 Dep for attempt to solve for m 2, 0.52 = = m c A1 For both ( ) 2 ln 2 0.52 = + y x A1 Alternative 2 e = + y mx c (B1) May be implied by later work Gradient = 10 5 4.74 2.24 − = − m (M1) 10 4.74( ) = + their m c or 5 2.24( ) = + their m c (M1) 2, 0.52 = = m c (A1) For both ( ) 2 ln 2 0.52 = + y x (A1) To earn marks, you must first recognize the linear relationship between e^y and x^2. Then, you need to set up and solve simultaneous equations using the given points to find the gradient (m) and y-intercept (c), and finally express y in terms of x by taking the natural logarithm.

8 Variables x and y are such that y Axb = , where A and b are constants. When lgy is plotted against lgx, a straight line graph passing through the points (0.61, 0.57) and (5.36, 4.37) is obtained. (a) Find the value of A and of b. [5] Using your values of A and b, find (b) the value of y when x 3 = , [2] (c) the value of x when y 3 = . [2]

8(a) lg lg lg = + y b x A B1 May be implied by subsequent work 4.37 5.36 lg = + b A 0.57 0.61 lg = + b A M1 For at least one correct equation 0.8 = b A1 ( ) lg 0.082 = A k 10 = k A M1 Dep for substitution to obtain lg = A k and hence A 1.21 = A A1 Alternative 1 lg lg lg = + y b x A (B1) May be implied by subsequent work Gradient = 4.37 0.57 5.36 0.61 − − (M1) 0.8 = b (A1) ( ) lg 0.082 = A k 10 = k A (M1) Dep for substitution into a correct equation to obtain lg = A k and hence A 1.21 = A (A1) Alternative 2 4.37 5.36 10 10 = × b A or 0.57 0.61 10 10 = × b A (B1) 3.8 4.75 = b (M1) For eliminating A correctly Must have B1. 0.8 = b (A1) ( ) 4.37 5.36 ( ) 10 − × = theirb A oe (M1) For a correct attempt to find A. Must have B1 A = 1.21 (A1) 8(b) ( ) 0.8 1.21 3 or lg 0.8lg3 0.082 = = + y y B1 FT for substitution into their equation y = awrt 2.9 B1 8(c) 0.8 3 1.21 or lg3 0.8lg 0.082 = = + x x B1 FT for substitution into their equation x = awrt 3.1 B1 (a) To earn marks, take the logarithm of the given equation to linearize it (lg y = b lg x + lg A). Use the two given points to form a system of two linear equations. Solve these equations simultaneously to find the values of b and lg A, then convert lg A to A. (b) Substitute x=3 into the derived equation y = Ax^b and calculate y. (c) Substitute y=3 into the derived equation y = Ax^b and solve for x.

Cambridge IGCSE Additional Mathematics · 0606

IGCSE Additional Mathematics (0606) Straight-line graphs past paper questions

Straight-line questions convert a non-linear relationship to y = mx + c form to read constants off a graph — a favourite structured-question setup.

80 free questions from 2016–2025, each with its marking scheme. Open the full topical bank to filter by subtopic, paper, and year.

Straight-line graphs in IGCSE Additional Mathematics 0606 exams, 2020–2025

Across 2020–2025, Straight-line graphs accounted for 82 questions worth 608 marks in Cambridge IGCSE Additional Mathematics (0606) papers.

By year
Year	Questions	Marks
2020	11	76
2021	14	110
2022	13	89
2023	14	103
2024	16	139
2025	14	91

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Gradient and equation of intercept form287 marks
Midpoint240 marks
Distance between two points209 marks

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2025 May/Jun · Paper 2 · Q3 · 4 marks
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2021 Oct/Nov · Paper 1 · Q2 · 6 marks
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2021 Feb/Mar · Paper 1 · Q3 · 6 marks
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2021 May/Jun · Paper 1 · Q5 · 5 marks
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2021 May/Jun · Paper 1 · Q8 · 9 marks
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Straight-line graphs in IGCSE Additional Mathematics 0606 exams, 2020–2025

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