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Cambridge IGCSE Mathematics 0580 · Coordinate geometry

Parallel lines

Why parallel lines have equal gradients, how to spot parallel lines from their equations, and how to find the equation of a line parallel to a given line through a given point. Extended: given lines written as $a x + b y = c$ , and parallel sides in shapes found from coordinates.

C3.6E3.6Core and ExtendedFree lesson

What the syllabus says

C3.6 (Core)

Find the gradient and equation of a straight line parallel to a given line.

Notes and examples

e.g. find the equation of the line parallel to $y = 4 x - 1$ that passes through $(1, - 3)$ .

E3.6 (Extended)

Find the gradient and equation of a straight line parallel to a given line.

Notes and examples

e.g. find the equation of the line parallel to $y = 4 x - 1$ that passes through $(1, - 3)$ .

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

Parallel lines have equal gradients

Parallel lines are lines that run in the same direction and never meet, however far they are extended. On a graph, that means they have exactly the same steepness, so:

$parallel lines have the same gradient$

Why. The gradient measures how far a line rises for each step across. If two lines rise by different amounts for each step, one is gaining on the other and they must meet somewhere. Only when they rise by exactly the same amount does the gap between them stay the same everywhere. Another way to see it: both lines make the same angle with the $x$ -axis (corresponding angles), and the same angle means the same gradient.

So $y = 4 x - 1$ and $y = 4 x + 6$ are parallel: both have gradient 4. They are different lines because they cross the $y$ -axis at different places. Two lines with the same gradient and the same intercept are not two lines at all; they are the same line.

Spotting parallel lines

To decide whether lines are parallel, write each equation as $y = m x + c$ and compare only the gradients.

$y = 2 x - 5$ and $y = 3 + 2 x$ are parallel: both have gradient 2. The order of the terms does not matter.
$y = 2 x + 1$ and $y = \frac{1}{2} x + 1$ are not parallel: the gradients are 2 and $\frac{1}{2}$ .
$y = 5 - 3 x$ and $y = 3 x + 5$ are not parallel: the gradients are $- 3$ and 3. The sign is part of the gradient.

The intercepts are irrelevant to being parallel. They only tell you where each line sits.

A family of parallel lines

All three lines have gradient 2, so they are parallel. Each one is the line $y = 2 x$ moved up or down: the intercept changes, the gradient does not.

Finding the equation of a parallel line

A typical question gives a line and a point, and asks for the equation of the line parallel to the given line that passes through the point.

Gradient: read the gradient $m$ of the given line. The new line has the same $m$ .
Intercept: substitute the coordinates of the point into $y = m x + c$ and solve for $c$ .
Equation: write $y = m x + c$ in full, fully simplified.
Check: the point should fit your equation, and the gradient should match the given line.

The most common error is to keep the old intercept as well as the old gradient. That gives back the original line, which does not pass through the new point. Only the gradient carries over.

Worked example 1

Find the equation of the line parallel to $y = 4 x - 1$ that passes through $(1, - 3)$ .

The given line has gradient 4, so the new line is $y = 4 x + c$ .
Substitute $x = 1$ , $y = - 3$ : $- 3 = 4 \times 1 + c$ .
$- 3 = 4 + c$ , so $c = - 7$ .
The equation is $y = 4 x - 7$ .
Check: $4 \times 1 - 7 = - 3$ , and the gradient is 4.

Answer: $y = 4 x - 7$

Worked example 2

Line $L$ is parallel to $y = 5 - 2 x$ and passes through the point $(- 3, 8)$ . Find the equation of $L$ .

Rewrite the given line as $y = - 2 x + 5$ : its gradient is $- 2$ .
So $L$ is $y = - 2 x + c$ .
Substitute $(- 3, 8)$ : $8 = - 2 \times (- 3) + c = 6 + c$ .
$c = 2$ .
$L$ is $y = - 2 x + 2$ . Check: $- 2 \times (- 3) + 2 = 8$ .

Answer: $y = - 2 x + 2$

Check yourself

Write down the gradient of a line parallel to $y = 7 - 3 x$ .

Answer it in the portal, where your answer is checked as you type →

Check yourself

Find the equation of the line parallel to $y = 3 x + 2$ that passes through $(2, 1)$ .

Answer it in the portal, where your answer is checked as you type →

When the given line is not in the form y = mx + c Extended

On the Extended papers the given line may be written as $a x + b y = c$ . Make $y$ the subject first to find its gradient. For $4 x + 2 y = 9$ : $2 y = - 4 x + 9$ , so $y = - 2 x + 4.5$ , and the gradient is $- 2$ .

A useful shortcut. Every line parallel to $a x + b y = c$ can be written $a x + b y = d$ , with the same $a$ and $b$ and a different number on the right. Changing only the right-hand side slides the line without turning it. To make it pass through a point, substitute the point to find $d$ .

For example, the line parallel to $3 x - 4 y = 12$ through $(2, 5)$ is $3 x - 4 y = d$ with $d = 3 \times 2 - 4 \times 5 = - 14$ : so $3 x - 4 y = - 14$ . Both methods give the same line; use the one that fits the form the question asks for.

Worked example 3 Extended

Find the equation of the line parallel to $3 x + 2 y = 6$ that passes through $(4, - 1)$ . Give your answer in the form $y = m x + c$ .

Make $y$ the subject of the given line: $2 y = - 3 x + 6$ , so $y = - \frac{3}{2} x + 3$ . Gradient $- \frac{3}{2}$ .
New line: $y = - \frac{3}{2} x + c$ .
Substitute $(4, - 1)$ : $- 1 = - \frac{3}{2} \times 4 + c = - 6 + c$ , so $c = 5$ .
$y = - \frac{3}{2} x + 5$ .
Check with the shortcut: $3 \times 4 + 2 \times (- 1) = 10$ , so the line is $3 x + 2 y = 10$ , which rearranges to $y = - \frac{3}{2} x + 5$ .

Answer: $y = - \frac{3}{2} x + 5$

Parallel sides in shapes Extended

Because Extended candidates can calculate a gradient from two points, they can test whether two sides of a shape are parallel without a grid.

A parallelogram has two pairs of parallel sides: show that opposite sides have equal gradients.
A trapezium has exactly one pair of parallel sides.

Label the vertices in order round the shape, so that you compare the right pairs ( $A B$ with $D C$ , and $B C$ with $A D$ ). In a "show that" answer, calculate every gradient you use, then state the conclusion in words.

Check yourself Extended

Write down the gradient of a line parallel to $2 x + 5 y = 7$ .

Answer it in the portal, where your answer is checked as you type →

Key points

Parallel lines have equal gradients. The intercepts do not matter.
To compare, write each line as $y = m x + c$ and look only at $m$ , including its sign.
Parallel line through a point: keep $m$ , substitute the point to find the new $c$ .
Extended: lines parallel to $a x + b y = c$ are $a x + b y = d$ .
Extended: equal gradients of opposite sides show a shape has parallel sides.

Where marks are lost

Keeping the old intercept as well as the gradient, which gives back the original line.
Using the $y$ -coordinate of the point as the new intercept, which is only right if the point is on the $y$ -axis.
Sign slips when solving for $c$ : from $8 = 6 + c$ , $c = 2$ , not 14.
Reading the gradient of $y = 5 - 2 x$ as 5.
Confusing parallel with perpendicular and taking the negative reciprocal.
Extended: reading the gradient of $3 x + 2 y = 6$ as 3 instead of $- \frac{3}{2}$ .

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E3.7

All IGCSE Mathematics 0580 lessons →

An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.

Parallel lines

Parallel lines have equal gradients

Spotting parallel lines

A family of parallel lines

Finding the equation of a parallel line

Worked example 1

Worked example 2

Check yourself

Check yourself

When the given line is not in the form y = mx + c Extended

Worked example 3 Extended

Parallel sides in shapes Extended

Check yourself Extended

Key points

Where marks are lost

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Perpendicular lines

Equations of linear graphs

Length and midpoint

Gradient of linear graphs