Coordinates, real-life graphs & straight-line graphs (y = mx + c) – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator3 marks

Points $P$ and $Q$ are marked on the grid.

(a) Write down the coordinates of the point $P$. [1]

Point $A$ has coordinates $(-3,\ 7)$ and point $B$ has coordinates $(5,\ -1)$.

(b) Find the coordinates of the midpoint of $AB$. [2]

(a) $(-2,\ -3)$ (do not accept $(-3,\ -2)$)B1
(b) for a correct method for either coordinate, e.g. $\dfrac{-3+5}{2}$ or $\dfrac{7+(-1)}{2}$, or one coordinate correctM1
(b) $(1,\ 3)$A1
Answer: (a) $(-2,\ -3)$ (b) $(1,\ 3)$
Examiner tip: Write the $x$-coordinate first (across, then up or down), and add the two coordinates before halving, taking care with the negative signs.

2Grade 1–3FoundationCalculator3 marks

$A$, $B$ and $C$ are three corners of a parallelogram $ABCD$.

$A$ is the point $(-2,\ 1)$, $B$ is the point $(3,\ 1)$ and $C$ is the point $(5,\ 4)$.

(a) Find the coordinates of $D$. [1]

(b) Write down the equation of the straight line that passes through $A$ and $B$. [1]

(a) $(0,\ 4)$B1
(b) $y = 1$B1
(c) $x = 3$B1
Answer: (a) $(0,\ 4)$ (b) $y = 1$ (c) $x = 3$
Examiner tip: A horizontal line has equation $y = \ldots$ and a vertical line has equation $x = \ldots$; candidates often swap these.

3Grade 4–5FoundationCalculator4 marks

The conversion graph can be used to change between pounds (£) and euros (€).

(a) Use the graph to change £30 to euros. [1]

(b) Use the graph to change €46 to pounds. [1]

Lena changes £200 into euros.

(a) answer in the range €34 to €36B1
(b) answer in the range £39 to £41B1
(c) for a correct method using a reading from the graph scaled up to £200, e.g. $4 \times 58$ (reading for £50) or $5 \times 46.4$ (reading for £40), or for finding the rate $58 \div 50 = 1.16$ and multiplying by 200M1
(c) answer in the range €228 to €236 (exact value €232)A1
Answer: (a) €35 (accept 34–36) (b) £40 (accept 39–41) (c) €232 (accept 228–236)
Examiner tip: The graph stops at £50, so split £200 into a multiple of a value you can read (e.g. 4 × £50), rather than extending the line by eye.

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4Grade 4–5FoundationCalculator4 marks

(a) Complete the table of values for $y = 3 - 2x$ [2]

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$7$			$1$		$-3$

(b) On the grid, draw the graph of $y = 3 - 2x$ for all values of $x$ from $-2$ to $3$ [2]

(a) for any two of $5,\ 3,\ -1$ correctB1
(a) all three correct: $5,\ 3,\ -1$B1
(b) for at least 4 points from their table plotted correctly, or a straight line of gradient $-2$, or a straight line through $(0,\ 3)$B1
(b) correct straight line from $(-2,\ 7)$ to $(3,\ -3)$ drawn with a ruler (ft their table if their points lie on a straight line)B1
Answer: (a) $5,\ 3,\ -1$ (b) straight line through $(-2,\ 7)$, $(0,\ 3)$ and $(3,\ -3)$
Examiner tip: Substituting a negative $x$ into $-2x$ gives a positive number: $3 - 2(-1) = 5$, not $1$.

5Grade 4–5Calculator3 marks

(a) Write down the gradient of the line with equation $y = 5 - 3x$ [1]

A straight line crosses the $y$-axis at $(0,\ -4)$ and has gradient $2$.

(b) Write down the equation of this line. [1]

(a) $-3$ (do not accept $-3x$)B1
(b) $y = 2x - 4$ (accept equivalent equations, e.g. $y - 2x = -4$)B1
(c) No, with a correct reason, e.g. when $x = 7$, $y = 2 \times 7 - 4 = 10$ (not $11$); ft their equation from (b)C1
Answer: (a) $-3$ (b) $y = 2x - 4$ (c) No, since $2(7) - 4 = 10 \neq 11$
Examiner tip: The gradient is the number multiplying $x$, wherever it appears: in $y = 5 - 3x$ it is $-3$, not $5$.

6Grade 4–5Calculator3 marks

The straight line $L$ is drawn on the grid.

Find an equation for $L$. [3]

for a correct method to find the gradient, e.g. $\dfrac{2-5}{4-(-2)}$ or $\dfrac{-3}{6}$ using two points on the lineM1
for $y = mx + 4$ with any non-zero $m$, or $y = -\tfrac{1}{2}x + c$ (ft their gradient), or gradient $-\tfrac{1}{2}$ and $y$-intercept $4$ both statedM1
$y = -\tfrac{1}{2}x + 4$ (accept equivalents, e.g. $x + 2y = 8$)A1
Answer: $y = -\tfrac{1}{2}x + 4$
Examiner tip: The line goes down from left to right, so the gradient must be negative; many candidates lose a mark by giving $\tfrac{1}{2}$.

7Grade 6–7HigherCalculator5 marks

The point $P$ has coordinates $(-3,\ 11)$ and the point $Q$ has coordinates $(5,\ -1)$.

(a) Work out the gradient of the line $PQ$. [2]

The line $M$ is parallel to $PQ$ and passes through the point $(4,\ 1)$.

(b) Find an equation of $M$. Give your answer in the form $ax + by = c$ where $a$, $b$ and $c$ are integers. [3]

(a) $\dfrac{-1-11}{5-(-3)}$ or $\dfrac{-12}{8}$ oeM1
(a) $-\tfrac{3}{2}$ oe (accept $-1.5$)A1
(b) for substituting $(4,\ 1)$ into $y = mx + c$ with their gradient from (a), e.g. $1 = -\tfrac{3}{2}(4) + c$, or $y - 1 = -\tfrac{3}{2}(x - 4)$M1
(b) $y = -\tfrac{3}{2}x + 7$ oe (ft their gradient)A1
(b) $3x + 2y = 14$ (accept any integer multiple, e.g. $-3x - 2y = -14$)A1
Answer: (a) $-\tfrac{3}{2}$ (b) $3x + 2y = 14$
Examiner tip: Parallel lines share the same gradient, so reuse $-\tfrac{3}{2}$, then clear the fraction by multiplying every term by 2.

8Grade 6–7HigherCalculator5 marks

The line $L_1$ has equation $2y = 5x - 6$

The line $L_2$ is perpendicular to $L_1$ and passes through the point $(10,\ -1)$.

(a) Find an equation of $L_2$. [4]

$L_2$ crosses the $x$-axis at the point $R$.

(b) Find the coordinates of $R$. [1]

(a) gradient of $L_1 = \tfrac{5}{2}$ (may be implied by $y = \tfrac{5}{2}x - 3$)B1
(a) for using $m_1 m_2 = -1$ to get gradient of $L_2 = -\tfrac{2}{5}$ (ft their gradient of $L_1$)M1
(a) for substituting $(10,\ -1)$ with their perpendicular gradient, e.g. $-1 = -\tfrac{2}{5}(10) + c$M1
(a) $y = -\tfrac{2}{5}x + 3$ oe (e.g. $2x + 5y = 15$)A1
(b) $(7.5,\ 0)$ (ft their equation in (a))B1
Answer: (a) $y = -\tfrac{2}{5}x + 3$ (b) $(7.5,\ 0)$
Examiner tip: Rearrange to $y = \ldots$ first: the gradient of $2y = 5x - 6$ is $\tfrac{5}{2}$, not $5$, so the perpendicular gradient is $-\tfrac{2}{5}$.

9Grade 8–9HigherCalculator6 marks

$A$ is the point $(-1,\ 2)$ and $B$ is the point $(5,\ 4)$.

The perpendicular bisector of $AB$ crosses the $y$-axis at the point $C$.

Work out the area of triangle $ABC$.

You must show all your working. [6]

for finding the midpoint of $AB$: $(2,\ 3)$M1
for finding the gradient of $AB$: $\dfrac{4-2}{5-(-1)} = \tfrac{1}{3}$M1
for using the perpendicular gradient $-3$ with their midpoint, e.g. $3 = -3(2) + c$ or $y - 3 = -3(x - 2)$M1
$C = (0,\ 9)$ (or equation $y = -3x + 9$ seen)A1
for a complete method for the area of triangle $ABC$, e.g. enclosing rectangle $6 \times 7 = 42$ minus three right-angled triangles $6 + 12.5 + 3.5$, or $\tfrac{1}{2} \times AB \times MC = \tfrac{1}{2}\sqrt{40}\sqrt{40}$M1
$20$A1
Answer: $20$ square units
Examiner tip: Use the midpoint (not $A$ or $B$) in the equation of a perpendicular bisector; then the neatest area method is $\tfrac{1}{2} \times AB \times MC$ because $MC$ is perpendicular to $AB$.

10Grade 4–5Calculator5 marks

The graph shows the cost, $C$ pounds, of a taxi journey of $d$ kilometres with Swift Cabs.

(a) What does the point where the graph crosses the cost axis represent? [1]

(b) Work out the gradient of the graph. [2]

Metro Taxis charge £2 per kilometre and have no fixed charge.

(a) the fixed (starting) charge of £4, paid before any distance is travelled (accept 'cost for 0 km is £4')C1
(b) for a correct method using two points on the line, e.g. $\dfrac{19-4}{10-0}$ or $\dfrac{22-4}{12}$M1
(b) $1.5$ (accept $\tfrac{3}{2}$; this is the cost in £ per km)A1
(c) for a correct method to compare, e.g. $1.5d + 4 = 2d$, or drawing $C = 2d$ on the graph, or comparing costs for two distances such as $d = 10$: £19 v £20M1
(c) journeys of more than 8 km (accept $d > 8$)A1
Answer: (a) the fixed charge of £4 (b) $1.5$ (c) more than 8 km
Examiner tip: In a real-life graph the $y$-intercept is the starting value and the gradient is a rate, so say what each means in context (e.g. £1.50 per km).

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