Parallel & Perpendicular Lines (Cambridge O Level Maths)

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Amber

Last updated

24 October 2023

Parallel Lines

What are parallel lines?

Parallel lines are lines that have the same gradient, but are not the same line
Parallel lines do not intersect with each other
You can easily spot that two lines are parallel when they are written in the form $y = m x + c$ , as they will have the same value of $m$ (gradient)

$y = 3 x + 7$ and $y = 3 x - 4$ are parallel
$y = 2 x + 3$ and $y = 3 x + 3$ are not parallel
$y = 4 x + 9$ and $y = 4 x + 9$ are not parallel; they are the exact same line

How do I find the equation of a line parallel to another line?

As parallel lines have the same gradient, a line of the form $y = m x + c$ will be parallel to a line in the form $y = m x + d$ , where $m$ is the same for both lines

If $c = d$ then they would be the same line and therefore not parallel

If you are asked to find the equation of a line parallel to $y = m x + c$ , you will also be given some information about a point that the parallel line, $y = m x + d$ passes through; $(x_{1}, y_{1})$
You can then substitute this point into $y = m x + d$ and solve to find $d$

Worked example

Find the equation of the line that is parallel to $y = 3 x + 7$ and passes through (2,1)

As the gradient is the same, the line that is parallel will be in the form:

$y = 3 x + d$

Substitute in the coordinate that the line passes through:

$1 = 3 (2) + d$

Simplify:

$1 = 6 + d$

Subtract 6 from both sides:

$- 5 = d$

Final answer:

$y = 3 x - 5$

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Perpendicular Lines

What are perpendicular lines?

You should already know that parallel lines have equal gradients
Perpendicular lines meet each other at right angles – ie they meet at 90°

What’s the deal with perpendicular gradients (and lines)?

Before you start trying to work with perpendicular gradients and lines, make sure you understand how to find the equation of a straight line – that will help you do the sorts of questions you will meet
Gradients m₁ and m₂ are perpendicular if m₁ × m₂ = −1
- For example
  - 1 and −1
  - $\frac{1}{3}$ and 3
  - $- \frac{2}{3}$ and $\frac{3}{2}$
We can use m₂ = −1 ÷ m₁ to find a perpendicular gradient. This is called the negative reciprocal.
If in doubt, SKETCH IT!

Worked example

The line L has equation $y = 2 x - 2$ .
Find an equation of the line perpendicular to L which passes through the point $(2, - 3)$ .
Leave your answer in the form $a x + b y + c = 0$ where $a$ , $b$ and $c$ are integers.

L is in the form $y = m x + c$ so we can see that its gradient is 2

$m_{1} = 2$

Therefore the gradient of the line perpendicular to L will be the negative reciprocal of 2

$m_{2} = - \frac{1}{2}$

Now we need to find $c$ for the line we're after. Do this by substituting the point $(2, - 3)$ into the equation $y = - \frac{1}{2} x + c$ and solving for $c$

$\begin{array}{rcl} - 3 & = & - \frac{1}{2} \times 2 + c \\ - 3 & = & - 1 + c \\ c & = & - 2 \end{array}$

Now we know the line we want is

$\begin{array}{rcl} y & = & - \frac{1}{2} x - 2 \end{array}$

But this is not in the form asked for in the question. So rearrange into the form $a x + b y + c = 0$ where $a$ , $b$ and $c$ are integers

$\begin{array}{rcl} y + \frac{1}{2} x + 2 & = & 0 \\ 2 y + x + 4 & = & 0 \end{array}$

Write the final answer

$x + 2 y + 4 = 0$

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