Parallel Lines and Their Equations

Parallel lines have the same gradient: use this to write the equation of a line parallel to another one.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, IGCSE Additional Maths 0606, Cambridge O Level Maths 4024 · 18 real past-paper questions below

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Same Gradient, Different c

Parallel lines point the same way and never meet. So they have the same gradient. Only c, where they cross the y-axis, is different.

Parallel

y = 2 x + 3

y = 2 x - 1

both gradients are 2

Not parallel

y = 2 x + 3

y = 3 x + 3

gradients 2 and 3

Rule

To write a parallel line: keep m, then find the new c.

Key idea

Parallel lines have so much in common: the same gradient, the same direction. It is a shame they will never meet!

Parallel, Through a Point on the y-Axis

The 2-mark question: a line is parallel to $y = 4 - 3 x$ and passes through $(0, - 2)$ . Find its equation.

Worked example

1
$y = 4 - 3 x$ — The given line. The gradient is the number in front of x, with its sign: −3. The 4 is c, even though it comes first.
2
$m = - 3$ — Parallel lines have the same gradient, so the new line also has gradient −3.
3
$y = - 3 x + c$ — Write the new line with c still unknown. Only c can change.
4
$(0, - 2) \Rightarrow c = - 2$ — The point has x = 0, so it sits on the y-axis. That is where the line crosses, so c = −2.
5
$y = - 3 x - 2$ — Put m and c into $y = m x + c$ . Adding −2 is written as − 2.

Common mistake

Taking 4 as the gradient because it comes first. The gradient is the number in front of $x$ .

Common mistake

Writing only $- 3 x - 2$ . Without “y =” it is not an equation, and it loses a mark.

Key idea

“Write down a line parallel to $y = 4 - 3 x$ ” (1 mark): keep $- 3 x$ and change c. $y = - 3 x + 10$ is fine.

Parallel, Through Any Point

If the point is not on the y-axis, put its x and y into $y = m x + c$ to find c.

Worked example

1
$parallel to y = 2 x + 5$ — The given line. Its gradient is 2, the number in front of x.
2
$y = 2 x + c$ — Parallel, so keep gradient 2. c is unknown, because the new line has moved.
3
$1 = 2 (3) + c$ — The line passes through (3, 1), so put x = 3 and y = 1 in. This line is true only for the right c.
4
$1 = 6 + c$ — Work out 2 × 3 = 6.
5
$c = 1 - 6 = - 5$ — Take 6 from both sides to get c on its own.
6
$y = 2 x - 5$ — Put c = −5 back in. Check: 2 × 3 − 5 = 1, so (3, 1) is on the line.

Common mistake

Keeping the old c and writing $y = 2 x + 5$ . That is the same line, not a new one through (3, 1).

Key idea

Show the line $1 = 2 (3) + c$ . It earns a mark for working, even if a slip spoils the final answer. Writing $y = 2 x + c$ with a new c can earn a mark too.

Watch the Minus Signs

Exams like a negative gradient with a negative point. Here is how to keep every sign right.

Find the line parallel to $y = 7 - 2 x$ through $(- 3, 4)$ .

Worked example

1
$y = 7 - 2 x$ — The given line. The number in front of x is −2, so that is the gradient.
2
$y = - 2 x + c$ — Parallel, so keep gradient −2. c is unknown.
3
$4 = - 2 \times (- 3) + c$ — Put in x = −3 and y = 4. Keep −3 in brackets so its minus is not lost.
4
$- 2 \times (- 3) = + 6$ — Minus times minus is plus.
5
$4 = 6 + c$ — So the line now reads 4 = 6 + c.
6
$c = 4 - 6 = - 2$ — Take 6 from both sides to get c on its own.
7
$y = - 2 x - 2$ — Put c = −2 back in. Check: −2 × (−3) − 2 = 6 − 2 = 4. It fits.

Common mistake

Writing −2 × −3 = −6. Then 4 = −6 + c gives c = 10, and the line misses the point.

Parallel to a Line in Another Form

Sometimes the given line is not in the form $y = m x + c$ . Get y on its own first, or you will read the wrong gradient.

Find the line parallel to $3 x + 2 y = 10$ through $(4, 1)$ .

Worked example

1
$3 x + 2 y = 10$ — The given line. y is not on its own, so you cannot read the gradient yet.
2
$2 y = 10 - 3 x$ — Take 3x from both sides.
3
$y = 5 - 1.5 x$ — Divide every term by 2: 10 ÷ 2 = 5 and 3 ÷ 2 = 1.5.
4
$m = - 1.5$ — Now y is alone. The number in front of x is −1.5.
5
$y = - 1.5 x + c$ — Parallel, so keep gradient −1.5. c is unknown.
6
$1 = - 1.5 \times 4 + c$ — Put in x = 4 and y = 1.
7
$1 = - 6 + c$ — Work out −1.5 × 4 = −6.
8
$c = 1 + 6 = 7$ — Add 6 to both sides to get c on its own.
9
$y = - 1.5 x + 7$ — Put c = 7 back in. Check: −1.5 × 4 + 7 = 1. It fits.

Common mistake

Reading the gradient as 3 or −3 straight from $3 x + 2 y = 10$ . You must divide by the 2 in front of y first.

Where the Parallel Line Crosses

Some questions give a point on the x-axis and ask where the parallel line crosses the y-axis. On the x-axis, y = 0.

Line M is parallel to $y = \frac{x}{3} + 4$ and crosses the x-axis at A(6, 0). Show that it crosses the y-axis at B(0, −2).

Worked example

1
$y = \frac{x}{3} + 4 \Rightarrow m = \frac{1}{3}$ — x ÷ 3 is the same as $\frac{1}{3} \times x$ , so the gradient is one third.
2
$y = \frac{1}{3} x + c$ — M is parallel, so keep gradient one third. c is unknown.
3
$0 = \frac{1}{3} \times 6 + c$ — A(6, 0) is on M, so put x = 6 and y = 0 in.
4
$0 = 2 + c$ — A third of 6 is 2.
5
$c = - 2$ — Take 2 from both sides.
6
$B = (0, - 2)$ — B is on the y-axis, where x = 0, so its y value is c. That is what we had to show.

Key idea

In a “show that” question the answer is given, so the marks are for the working. Write every line, ending with the answer you were asked to show.

Which Method? Read the Question

Find the gradient first, then read what else you are given.

You see	You do
Parallel, through (0, a number)	same m, that number is c
Parallel, through another point	same m, substitute to find c
Is $2 y = 8 x + 3$ parallel?	get y on its own, compare m
Parallel, through (a number, 0) on the x-axis	same m, put y = 0 in to find c
At right angles to the line	Perpendicular Lines
No equation, just two points	Equation of a Line Through Two Points

Key idea

Compare gradients only when y is on its own. $2 y = 8 x + 3$ has gradient 4, not 8.

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