Straight-line graphs: gradient, y = mx + c, parallel & perpendicular – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator4 marks

The diagram shows a straight line drawn on a grid. Two points on the line are marked with dots.

(a) Write down the coordinates of the point where the line crosses the $y$-axis. [1]

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

(a) $(0, 1)$B1
(b) for $\dfrac{\text{change in } y}{\text{change in } x}$ using two points read correctly from the line, e.g. $\dfrac{7-1}{3-0}$ or $\dfrac{6}{3}$M1
(b) $2$A1
(c) $y = 2x + 1$ (ft their gradient from (b) with intercept 1; must be an equation, not just $2x + 1$)B1
Answer: (a) $(0, 1)$ (b) $2$ (c) $y = 2x + 1$
Examiner tip: Gradient is rise ÷ run – read the 'change in $y$' using the $y$-axis scale (labelled every 2 units), not by counting labels.

2Grade 1–3FoundationNon-calculator3 marks

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

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

(b) Does the point $(5, 12)$ lie on the line $y = 3x - 2$? You must show how you get your answer. [1]

(a) for 2 correct values from $-5$, $1$, $4$B1
(a) (total B2) all three correct: $-5$, $1$, $4$B1
(b) No, with supporting working, e.g. when $x = 5$, $y = 3 \times 5 - 2 = 13$ and $13 \ne 12$; accept when $y = 12$, $x = \tfrac{14}{3} \ne 5$, or $3 \times 5 - 2 \ne 12$ shownC1
Answer: (a) $-5$, $1$, $4$ (b) No, since $3 \times 5 - 2 = 13 \ne 12$
Examiner tip: In (b) a bare 'No' scores nothing – you must show the substitution that gives 13.

3Grade 4–5Non-calculator4 marks

A straight line passes through the points $A(2, 5)$ and $B(6, -3)$.

Find the equation of the line. Give your answer in the form $y = mx + c$ [4]

for method to find the gradient, e.g. $\dfrac{-3-5}{6-2}$M1
gradient $= -2$A1
for substituting a point into $y = mx + c$ with their gradient, e.g. $5 = -2 \times 2 + c$, or $y - 5 = -2(x - 2)$M1
$y = -2x + 9$A1
Answer: $y = -2x + 9$
Examiner tip: Keep the signs consistent: subtract the coordinates in the same order top and bottom, and expect a negative gradient for a line going downhill.

4Grade 4–5Non-calculator4 marks

Here are the equations of four straight lines.

$$\textbf{P: } 2y = 6x - 1 \qquad \textbf{Q: } y = 4 - 3x \qquad \textbf{R: } 3x - y = 7 \qquad \textbf{S: } 6x + 2y = 5$$

(a) Which of these lines are parallel to $y = 3x + 4$? You must show your working. [2]

(b) Find the equation of the line that is parallel to $y = 3x + 4$ and passes through the point $(1, -2)$. [2]

(a) for rearranging at least one of P, R or S into the form $y = mx + c$, e.g. $y = 3x - \tfrac12$ or $y = 3x - 7$M1
(a) P and R only, with no incorrect lines includedA1
(b) for $y = 3x + c$ with $(1, -2)$ substituted, e.g. $-2 = 3 \times 1 + c$M1
(b) $y = 3x - 5$A1
Answer: (a) P and R (b) $y = 3x - 5$
Examiner tip: Rearrange to $y = \ldots$ before reading the gradient – $3x - y = 7$ has gradient $+3$, not $-1$, and $6x + 2y = 5$ has gradient $-3$.

5Grade 6–7HigherNon-calculator4 marks

The line $L$ has equation $2x + 3y = 12$

Find an equation of the straight line that is perpendicular to $L$ and passes through the point $(4, -1)$. Give your answer in the form $y = mx + c$ [4]

for rearranging to find the gradient of $L$: $y = -\tfrac{2}{3}x + 4$, so gradient $-\tfrac23$M1
for using $m_1 \times m_2 = -1$ to get perpendicular gradient $\tfrac{3}{2}$ (ft their gradient of $L$)M1
for substituting $(4, -1)$ with their perpendicular gradient, e.g. $-1 = \tfrac32 \times 4 + c$M1
$y = \tfrac{3}{2}x - 7$ (oe, e.g. $y = 1.5x - 7$)A1
Answer: $y = \tfrac{3}{2}x - 7$
Examiner tip: The gradient of $2x + 3y = 12$ is $-\tfrac23$, not $2$ – rearrange first, then take the negative reciprocal.

6Grade 8–9HigherNon-calculator6 marks

$A$ is the point $(-2, 1)$ and $B$ is the point $(6, 5)$.

The line $L$ is the perpendicular bisector of $AB$.

$L$ crosses the $x$-axis at $P$ and the $y$-axis at $Q$. $O$ is the origin.

Work out the area of triangle $OPQ$. [6]

for finding the midpoint of $AB$: $(2, 3)$M1
for finding the gradient of $AB$: $\dfrac{5-1}{6-(-2)} = \tfrac12$M1
for perpendicular gradient $-2$ used with their midpoint, e.g. $3 = -2 \times 2 + c$M1
$y = -2x + 7$A1
for finding both intercepts from their line: $P(3.5, 0)$ and $Q(0, 7)$M1
$12.25$ (oe, e.g. $\tfrac{49}{4}$) square unitsA1
Answer: $12.25$ square units
Examiner tip: A perpendicular bisector must pass through the MIDPOINT of $AB$ – using $A$ or $B$ instead gives the wrong line and loses most of the marks.

7Grade 6–7HigherNon-calculator3 marks

The points $A(1, 2)$, $B(4, 8)$ and $C(8, 6)$ are the vertices of a triangle.

Show that angle $ABC$ is a right angle. [3]

for gradient of $AB$: $\dfrac{8-2}{4-1} = 2$
(Alternative: M1 for two of $AB^2 = 45$, $BC^2 = 20$, $AC^2 = 65$)M1
for gradient of $BC$: $\dfrac{6-8}{8-4} = -\tfrac12$
(Alternative: M1 for all three of $AB^2 = 45$, $BC^2 = 20$, $AC^2 = 65$)M1
for a complete argument: $2 \times -\tfrac12 = -1$, so $AB$ is perpendicular to $BC$ and angle $ABC = 90°$
(Alternative: $45 + 20 = 65$, so $AB^2 + BC^2 = AC^2$ and angle $ABC = 90°$ by the converse of Pythagoras)C1
Answer: Gradients $2$ and $-\tfrac12$ multiply to $-1$, so $AB \perp BC$
Examiner tip: In a 'show that' you must state the conclusion – write that the product is $-1$ AND that this means the lines are perpendicular.

8Grade 1–3FoundationNon-calculator4 marks

The points $A$, $B$ and $C$ are plotted on the grid.

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

(b) $ABCD$ is a parallelogram. Write down the coordinates of point $D$. [1]

(a) $(4, -1)$ (do not accept $(-1, 4)$)B1
(b) $(-2, -2)$B1
(c) for a correct method for one coordinate, e.g. $\dfrac{-4 + 4}{2}$ or $\dfrac{3 + (-1)}{2}$, or one coordinate of the midpoint correctM1
(c) $(0, 1)$ (ft their $C$ from (a))A1
Answer: (a) $(4, -1)$ (b) $(-2, -2)$ (c) $(0, 1)$
Examiner tip: Write the $x$-coordinate first (across, then up/down) – swapping them to $(-1, 4)$ loses the mark straight away.

9Grade 1–3FoundationNon-calculator3 marks

Two straight lines, $L_1$ and $L_2$, are drawn on the grid.

(a) Write down the equation of line $L_1$ [1]

(b) Write down the equation of line $L_2$ [1]

$$(3, -3) \qquad (-3, 5) \qquad (5, -3) \qquad (2, -3)$$

Which one of these points lies on line $L_1$? [1]

(a) $x = -3$ (do not accept $y = -3$ or just $-3$)B1
(b) $y = 2$ (do not accept $x = 2$)B1
(c) $(-3, 5)$B1
Answer: (a) $x = -3$ (b) $y = 2$ (c) $(-3, 5)$
Examiner tip: A vertical line has every $x$-coordinate the same, so its equation is $x = \ldots$ – most students who lose this mark write $y = -3$.

10Grade 4–5Calculator5 marks

A phone repair shop charges a fixed call-out fee plus an amount for each hour of work.

The graph shows the total cost, $C$ pounds, of a repair that takes $t$ hours.

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

(b) Explain what the gradient represents in this context. [1]

A second shop has no call-out fee and charges £40 for each hour of work.

(a) for $\dfrac{\text{change in } C}{\text{change in } t}$ using two points read from the line, e.g. $\dfrac{130 - 80}{4 - 2}$ or $\dfrac{180 - 30}{6}$M1
(a) $25$A1
(b) the cost per hour of work, e.g. "it costs £25 for each extra hour" (ft their gradient; do not accept just "the cost" or "the fixed fee")C1
(c) for a correct method to compare, e.g. $40t = 25t + 30$, or finding that both shops charge £80 at $t = 2$, or drawing $C = 40t$ to cross the line at $t = 2$M1
(c) less than 2 hours (accept $t < 2$ oe; do not accept just "2")A1
Answer: (a) $25$ (b) the cost per hour, £25 (c) repairs shorter than 2 hours
Examiner tip: The answer to (c) is a range, not a value – write "less than 2 hours", because at exactly 2 hours the two shops cost the same.

11Grade 4–5Non-calculator4 marks

The straight line $L$ has gradient $-\tfrac{3}{4}$ and passes through the point $(-4, 6)$.

(a) Find the equation of $L$. Give your answer in the form $y = mx + c$ [2]

(b) Find the coordinates of the point where $L$ crosses the $x$-axis. [2]

(a) for substituting $(-4, 6)$ into $y = -\tfrac34 x + c$, e.g. $6 = -\tfrac34 \times (-4) + c$, or using $y - 6 = -\tfrac34(x + 4)$M1
(a) $y = -\tfrac{3}{4}x + 3$ (oe, e.g. $y = -0.75x + 3$)A1
(b) for setting $y = 0$ in their equation, e.g. $0 = -\tfrac34 x + 3$M1
(b) $(4, 0)$ (ft their equation of the form $y = mx + c$)A1
Answer: (a) $y = -\tfrac{3}{4}x + 3$ (b) $(4, 0)$
Examiner tip: Take care with $-\tfrac34 \times (-4) = +3$: a sign slip here gives $c = 9$ and loses the accuracy marks in both parts.

12Grade 6–7Non-calculator4 marks

$P$ is the point $(k, 3)$ and $Q$ is the point $(4, 11)$.

The line $PQ$ is parallel to the line with equation $2y = 8x - 1$

(a) Find the value of $k$. [3]

(b) Ellie says, "The line $y = 2 - 4x$ is also parallel to $PQ$."

Is Ellie correct? Give a reason for your answer. [1]

(a) for finding the gradient of $2y = 8x - 1$ as $4$, e.g. $y = 4x - \tfrac12$M1
(a) for forming an equation using the gradient of $PQ$, e.g. $\dfrac{11 - 3}{4 - k} = 4$ (ft their gradient)M1
(a) $k = 2$A1
(b) No, with a reason, e.g. "its gradient is $-4$, not $4$" (accept "it slopes down but $PQ$ slopes up")C1
Answer: (a) $k = 2$ (b) No – $y = 2 - 4x$ has gradient $-4$, but $PQ$ has gradient $4$
Examiner tip: In $2y = 8x - 1$ the gradient is not $8$: divide everything by $2$ first so the equation starts $y = \ldots$

13Grade 6–7HigherNon-calculator5 marks

Here are the equations of four straight lines.

$$\textbf{A: } 4y = x + 8 \qquad \textbf{B: } x + 4y = 12 \qquad \textbf{C: } y = 4 - 4x \qquad \textbf{D: } 2y = 3 - \tfrac{1}{2}x$$

(a) Which two of these lines are perpendicular to the line $y = 4x - 3$? You must show your working. [2]

(b) The line $M$ is perpendicular to $y = 4x - 3$ and passes through the point $(-8, 7)$.

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

(a) for rearranging at least two of the lines into the form $y = mx + c$ or stating the perpendicular gradient $-\tfrac14$M1
(a) B and D (with gradients shown; both needed and no extras)A1
(b) for substituting $(-8, 7)$ into $y = -\tfrac14 x + c$, e.g. $7 = -\tfrac14 \times (-8) + c$M1
(b) $y = -\tfrac14 x + 5$ (oe)A1
(b) $x + 4y = 20$ (accept any integer multiple, e.g. $2x + 8y = 40$, or $-x - 4y = -20$)A1
Answer: (a) B and D (b) $x + 4y = 20$
Examiner tip: Perpendicular means negative reciprocal: $-\tfrac14$, not $\tfrac14$ (that is line A) and not $-4$ (that is line C).

14Grade 8–9HigherNon-calculator6 marks

$ABCD$ is a rectangle.

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

The point $D$ lies on the $y$-axis.

(a) Find the coordinates of $D$. [3]

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

Diagram not drawn accurately.

(a) for gradient of $AB = \dfrac{4 - 2}{5 - 1} = \tfrac12$ and so gradient of $AD = -2$M1
(a) for equation of line $AD$, e.g. $y = -2x + 4$ or $y - 2 = -2(x - 1)$ (ft their perpendicular gradient)M1
(a) $(0, 4)$A1
(b) $(4, 6)$ (ft their $D$, using $C = B + (D - A)$)B1
(c) for finding the length of $AB$ or $AD$ using Pythagoras, e.g. $AB = \sqrt{4^2 + 2^2} = \sqrt{20}$ or $AD = \sqrt{1^2 + 2^2} = \sqrt{5}$ (ft their $D$), or another complete method, e.g. enclosing rectangle $5 \times 4 - 4 - 1 - 4 - 1$M1
(c) $10$ (square units) from $\sqrt{20} \times \sqrt{5}$A1
Answer: (a) $D(0, 4)$ (b) $C(4, 6)$ (c) $10$ square units
Examiner tip: Angle $DAB$ is $90°$, so $AD$ is perpendicular to $AB$ – find the gradient $-2$ first; guessing $D$ from the sketch scores nothing.

15Grade 8–9HigherNon-calculator6 marks

The line $L$ has equation $y = 2x - 1$

$P$ is the point $(6, 1)$.

The point $F$ is the point on $L$ that is closest to $P$.

Work out the shortest distance from $P$ to the line $L$.

Give your answer as a surd in its simplest form. [6]

for recognising that $PF$ is perpendicular to $L$ and so has gradient $-\tfrac12$M1
for equation of line $PF$, e.g. $y = -\tfrac12 x + 4$ or $y - 1 = -\tfrac12(x - 6)$M1
for solving their equation simultaneously with $y = 2x - 1$, e.g. $2x - 1 = -\tfrac12 x + 4$M1
$F = (2, 3)$A1
for using Pythagoras with their $F$, e.g. $\sqrt{(6 - 2)^2 + (1 - 3)^2} = \sqrt{20}$M1
$2\sqrt{5}$ (do not accept $\sqrt{20}$ or $4.47$ for this mark)A1
Answer: $2\sqrt{5}$
Examiner tip: The shortest distance to a line is always along the perpendicular – a quick distance to the $y$-intercept $(0, -1)$ is not the shortest and scores no marks.

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