Parallel And Perpendicular Lines - GCSE Maths Revision - BBC

In this guide

1. Revise
2. Test

Pages

1. Coordinates
2. Straight line graphs
3. y = mx + c
4. Parallel and perpendicular lines
5. The equation of a line through two points

Parallel and perpendicular lines

Parallel lines

Parallel lines are a fixed distance apart and will never meet, no matter how long they are extended. Lines that are parallel have the same gradient.

The graphs above, \(y = 2x + 1\) and \(y = 2x - 2\) have the same gradient of 2.

The lines are parallel.

Key factTwo lines will be parallel if they have the same gradient.

Example

State the equation of a line that is parallel to \(y = 3x + 7\).

To be parallel, two lines must have the same gradient. The gradient of \(y = 3x + 7\) is 3.

Any line with a gradient of 3 will be parallel to \(y = 3x + 7\).

Two examples are \(y = 3x - 2\) and \(y = 3x + 11.6\).

Perpendicular graphs - Higher

Two lines are perpendicular if they meet at a right angle.

Two lines will be perpendicular if the product of their gradients is -1.

To find the equation of a perpendicular line, first find the gradient of the line and use this to find the equation.

Example

Find the equation of a straight line that is perpendicular to \(y = 2x + 1\).

The gradient of \(y = 2x + 1\) is 2.

To find the perpendicular gradient, find the number which will multiply by 2 to give -1. This is the negative reciprocal of the gradient.

The reciprocal of 2 is \(\frac{1}{2}\), so the negative reciprocal of 2 is \(-\frac{1}{2}\).

This gives \(y = - \frac{1}{2}x + c\).

Examples of equations of lines that are perpendicular to \(y = 2x + 1\) would include \(y = -\frac{1}{2}x + 5\) or \(y = -\frac{1}{2}x - 4\).

Question

Find the equation of the line that is perpendicular to \(y = 3x - 1\) and goes through point (2, 5).

Show answer

First, find the perpendicular gradient and substitute this into the equation for all straight lines, \(y = mx + c\).

If \(y = 3x - 1\) the gradient of the graph is 3. This means the gradient of the perpendicular graph is \(-\frac{1}{3}\) (by finding the negative reciprocal).

\(y = -\frac{1}{3}x + c\)

To find the value of \(c\), the question states that the graph goes through the point (2, 5). This means \(x = 2\) and \(y = 5\). Substitute these values into \(y = mx + c\).

\(y = -\frac{1}{3}x + c\)

\(x = 2\) and \(y = 5\).

\(5 = -\frac{1}{3} \times 2 + c\)

\(5 = -\frac{2}{3} + c\). Add \(\frac{2}{3}\) to each side to find \(c\)

\(5 \frac{2}{3} = c\)

The final equation is:

\(y = -\frac{1}{3}x + 5 \frac{2}{3}\)

Next pageThe equation of a line through two pointsPrevious pagey = mx + c

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