IGCSE · Practice questions · The equation of a straight line

Find m, find c.

Six original Cambridge-style questions on the equation of a straight line, from reading gradients to modelling a charge with $y = mx + c$.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Gradient first, then substitute a point.

Read: in $y = mx + c$, $m$ is the gradient and $c$ the $y$-intercept.
Rearrange (Extended): make $y$ the subject, dividing every term.
Gradient: from the graph, or $\dfrac{y_2 - y_1}{x_2 - x_1}$ from two points.
Intercept: substitute one point into $y = mx + c$ and solve for $c$.
Check: the other point must fit; give the equation fully simplified.

Recall

[2 marks]

Write down the gradient and the $y$-intercept of

(a) $y = 6x + 3$ (b) $y = 4 - \dfrac{1}{2}x$.

(a) gradient 6, $y$-intercept 3 ✓
(b) gradient $-\dfrac{1}{2}$, $y$-intercept 4 ✓

Calculation

[2 marks]

Find the equation of the line with gradient $-4$ that passes through $(3, -5)$.

$-5 = -4 \times 3 + c = -12 + c$, so $c = 7$. ✓
$y = -4x + 7$ ✓

Calculation

[3 marks]

(Extended) The equation of a line is $4x + 3y = 12$.

(a) Find the gradient of the line.

(b) Find the coordinates of the point where the line crosses the $x$-axis.

(a) $3y = -4x + 12$, so $y = -\dfrac{4}{3}x + 4$. Gradient $-\dfrac{4}{3}$. ✓
(b) Put $y = 0$: $4x = 12$, so $x = 3$. The point is $(3, 0)$. ✓

Analysis

[3 marks]

(Extended) Find the equation of the line through $(-3, 9)$ and $(3, -3)$. Give your answer in the form $y = mx + c$.

$m = \dfrac{-3 - 9}{3 - (-3)} = \dfrac{-12}{6} = -2$. ✓
$9 = -2 \times (-3) + c = 6 + c$, so $c = 3$. ✓
$y = -2x + 3$ (check: $-2 \times 3 + 3 = -3$). ✓

Analysis

[3 marks]

A straight line is drawn on a grid. It passes through $(0, -1)$ and $(3, 5)$.

(a) Find the equation of the line.

(b) Does the point $(10, 19)$ lie on the line? Show your working.

(a) $c = -1$ and $m = \dfrac{5 - (-1)}{3} = 2$, so $y = 2x - 1$. ✓
(b) When $x = 10$, $y = 2 \times 10 - 1 = 19$. Yes, $(10, 19)$ lies on the line. ✓

Analysis

[4 marks]

A plumber's charge $C$ dollars for a job lasting $h$ hours is shown by a straight-line graph. A 1-hour job costs $70 and a 3-hour job costs $130.

(a) Find the equation of the line in the form $C = mh + c$.

(b) Explain what the values of $m$ and $c$ mean.

(a) $m = \dfrac{130 - 70}{3 - 1} = 30$. $70 = 30 \times 1 + c$, so $c = 40$. $C = 30h + 40$. ✓
(b) $m = 30$: the plumber charges $30 for each hour. $c = 40$: a fixed $40 call-out charge, even for a job of 0 hours. ✓
(c) $C = 30 \times 5.5 + 40 = 165 + 40 = 205$, so the charge is $205. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.