The Equation y = mx + c

Read the gradient m and y-intercept c from y = mx + c, and write the equation of a line from a grid, a gradient, or two points.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024, Edexcel Further Pure Maths 4PM1 · 56 real past-paper questions below

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What m and c Mean

Every straight line (except an upright one) has an equation of this form:

Formula

y = m x + c

m is the gradient: how far the line goes up for each 1 across. Minus means it goes down. c is where the line crosses the y-axis, at the point $(0, c)$ .

Equation	m and c
$y = 2 x + 1$	$m = 2, c = 1$
$y = 4 - 3 x$	$m = - 3, c = 4$

Key idea

If y is not on its own, get it alone first: $2 y = 6 x + 4$ becomes $y = 3 x + 2$ , so $m = 3$ .

When y Is Not Alone

Exams often give a line as $3 y + 2 x = 12$ . You can only read m and c when y is on its own, so get y alone first.

Worked example

1
$3 y + 2 x = 12$ — The question. y is not alone: it has a 3 with it and a 2x beside it.
2
$3 y = 12 - 2 x$ — Take 2x from both sides, so only the y term is left on the left.
3
$y = \frac{12}{3} - \frac{2 x}{3}$ — Divide both sides by 3. Every term on the right is divided, not just the first.
4
$y = 4 - \frac{2}{3} x$ — 12 ÷ 3 = 4. Now y is alone. Writing this line earns a mark for working.
5
$m = - \frac{2}{3}, c = 4$ — m is the number with x, minus included. c = 4, so the line crosses the y-axis at (0, 4).

Common mistake

Dividing only the 12 by 3. The 2x must be divided by 3 too, or the gradient comes out as −2.

Key idea

A bracket hides c. Multiply it out first: $y = 3 (x - 4) = 3 x - 12$ , so c = −12, not −4.

The Equation of a Line on a Grid

The 3-mark question: “Find an equation for the line L.” Read c from the y-axis, then work out the gradient.

Worked example

1
$c = 1$ — L crosses the y-axis at (0, 1). That crossing is c.
2
$(0, 1) to (2, 4)$ — Pick two points where L goes exactly through grid corners, so you read them without guessing.
3
$up 3, across 2$ — From (0, 1) to (2, 4), y goes up 3 (1 to 4) while x goes across 2 (0 to 2).
4
$m = \frac{3}{2} = 1.5$ — Gradient = up ÷ across, so 3 ÷ 2 = 1.5.
5
$y = 1.5 x + 1$ — Put m = 1.5 in front of x and c = 1 on its own.

A line going down. Read it the same way, but down means a minus.

Worked example

1
$c = - 1$ — The line crosses the y-axis at (0, −1), so c = −1.
2
$(0, - 1) to (3, - 3)$ — Two grid corners on the line.
3
$down 2, across 3$ — y goes from −1 to −3, so it drops 2 while x goes across 3.
4
$m = \frac{- 2}{3} = - \frac{2}{3}$ — Down means the gradient is negative. Leave it as a fraction, because it is exact.
5
$y = - \frac{2}{3} x - 1$ — Put m and c in. c is −1, so the line ends “− 1”, not “+ 1”.

Common mistake

Doing across ÷ up. The gradient is up ÷ across. If the line goes down from left to right, the gradient is negative.

Key idea

Show your working: draw the triangle on the grid and write up ÷ across. That earns a mark even if your final answer slips. Start the answer with “y =”, not “L =”.

Gradient and y-Intercept Given

The 2-mark question: “Write down an equation of the line with gradient −4 through $(0, 7)$ .” A point with x = 0 is on the y-axis, so it gives c.

Worked example

1
$m = - 4$ — The gradient is given, so m = −4. Keep the minus.
2
$(0, 7)$ — Its x is 0, so this point is on the y-axis.
3
$c = 7$ — Where the line crosses the y-axis is c, so c = 7.
4
$y = - 4 x + 7$ — Put m in front of x and c on its own.

Common mistake

Giving the y-axis crossing as (7, 0). It is on the y-axis, so x is 0: the point is (0, 7).

Flat and Upright Lines

The 1-mark question: “Write down the equation of the line PQ”, where PQ is flat.

Worked example

1
$P (- 2, 2), Q (3, 2)$ — Look at the points. The y value is 2 both times.
2
$m = 0$ — y never changes, so the line goes up 0 for every 1 across.
3
$y = 0 x + 2$ — Put m = 0 and c = 2 into y = mx + c.
4
$y = 2$ — 0x is 0, so drop it. Every point on PQ has y = 2.

Upright lines. An upright (vertical) line is the one line that y = mx + c cannot describe. Here x stays fixed instead.

Worked example

1
$R (4, - 1), S (4, 3)$ — Look at the points. The x value is 4 both times.
2
$x = 4$ — Every point on RS has x = 4, whatever its y. So its equation is x = 4.

Key idea

The axes are lines too: the y-axis is x = 0 and the x-axis is y = 0.

Common mistake

Mixing them up. A flat line keeps y fixed, so it is y = a number. An upright line keeps x fixed, so it is x = a number.

A Line Through Two Points

The 3-mark question: “Find an equation of the line through A and B.” Find m, then put one point in to find c.

Worked example

1
$A (- 2, 7), B (4, - 5)$ — The two points. Keep each x with its own y.
2
$change in y = - 5 - 7 = - 12$ — y of B take away y of A. y falls by 12.
3
$change in x = 4 - (- 2) = 4 + 2 = 6$ — Same order: x of B take away x of A. Taking away −2 is the same as adding 2.
4
$m = \frac{- 12}{6} = - 2$ — Gradient = change in y ÷ change in x. It is negative, so the line goes down.
5
$7 = - 2 \times (- 2) + c$ — A is on the line, so put x = −2, y = 7 and m = −2 into y = mx + c.
6
$7 = 4 + c$ — (−2) × (−2) = +4, because minus times minus is plus.
7
$c = 3$ — Take 4 from both sides: 7 − 4 = 3.
8
$y = - 2 x + 3$ — Put m and c in. Check with B: −2 × 4 + 3 = −5, which matches.

Key idea

The gradient line and the line where you put a point in each earn a mark, so write both down.

Key idea

For more on this, see Equation of a Line Through Two Points.

Where a Line Crosses the x-Axis

Exams also ask where a line crosses the x-axis. Every point on the x-axis has y = 0, so put y = 0 and solve.

Worked example

1
$y = 2 x + 9$ — The line. We want the point where it meets the x-axis.
2
$0 = 2 x + 9$ — On the x-axis y is 0, so put 0 in place of y. This line earns a mark for working.
3
$- 9 = 2 x$ — Take 9 from both sides, so the x term is alone.
4
$x = - 4.5$ — Divide both sides by 2: −9 ÷ 2 = −4.5.
5
$(- 4.5, 0)$ — Give the point, not just x. Its y is 0 because it is on the x-axis.

Common mistake

Writing (0, −4.5). On the x-axis y is 0, and x always comes first, so the point is (−4.5, 0).

Key idea

For the y-axis put x = 0; for the x-axis put y = 0. This line crosses the y-axis at (0, 9).

A Point With a Missing Number

“The point (k, 11) is on the line $y = 3 - 4 x$ . Find k.” A point is on a line when its x and y fit the equation, so put in the number you know.

Worked example

1
$(k, 11)$ — x always comes first, so x = k is missing and y = 11 is known.
2
$11 = 3 - 4 k$ — Put y = 11 and x = k into y = 3 − 4x.
3
$8 = - 4 k$ — Take 3 from both sides: 11 − 3 = 8.
4
$k = \frac{8}{- 4} = - 2$ — Divide both sides by −4. Positive ÷ negative is negative.
5
$3 - 4 \times (- 2) = 3 + 8 = 11$ — Check: put k = −2 back in. We get 11, the y we were given.

Key idea

If the missing number is y, just work it out: for (5, p) on this line, $p = 3 - 4 \times 5 = - 17$ .

Common mistake

Putting 11 in place of x. In (k, 11), k is x and 11 is y.

Which Method? Read the Question

Find what you are given, then read off or work out m and c.

You see	You do
y not alone, e.g. $3 y + 2 x = 12$	get y alone first, then read m and c
A line drawn on a grid	c from the y-axis, m = up ÷ across
A gradient and (0, a number)	write y = mx + c straight away
A flat line	y = a number
An upright line	x = a number
Two points	m first, then put a point in for c
Where it crosses the x-axis	put y = 0 and solve
A point with a missing number	put the known number in and solve
Draw the line from its equation	Drawing Straight-Line Graphs

Key idea

Check your equation: put a point you know into it. Both sides must match.

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