## What you need to know

Two straight lines are parallel if they are always the same distance away from each other, no matter how long the lines are extended. In other words, they’re going the exact same direction and will never meet. Additionally, two lines are perpendicular if, when they meet, they form a right-angle. Both of these words appear all over in maths so are worth getting used to.

Before going into this topic, you should be familiar with the equation of a straight line (https://mathsmadeeasy.co.uk/gcse-maths-revision/ymxc-gcse-maths-revision-worksheets/) as well as drawing straight-line graphs (https://mathsmadeeasy.co.uk/gcse-maths-revision/drawing-straight-line-graphs-gcse-maths-revision-worksheets/).

If two lines are parallel, then they have the same gradient. This means two things, 1) you can tell if two straight lines are parallel by looking at their equations, and 2) if you are told that a line is parallel to a different line whose gradient you know, then you also know the gradient of the first line.

Example: Is the line $y=3x-4$ parallel to the line $3y-9x=21$?

So, we need to know the gradient of both lines. The first line equation is given to us in the desired form of $y=mx+c$, so we know its gradient (the coefficient of $x$) is 3.

The second line equation requires some rearranging before we can know its gradient. Firstly, add $9x$ to both sides of the equation $3y-9x=21$ to get

$3y=9x+21$.

Then, if we divide both sides by 3 we get

$y=3x+7$.

This is now in the right form, and we can see that its gradient is 3 so must be parallel to the 1st line.

Example: Find the equation of the straight line that passes through $(1, 3)$ and is parallel to the line $y=5x-1$. Plot this straight line.

The gradient of the given line is 5, which means the gradient of the perpendicular line must also be 5. We’re given that it passes through $(1, 3)$, and we now know the gradient to be 5, so we can substitute these values into $y=mx+c$ in order to find $c$. Doing so, we get

$3 = 1\times 5 +c = 5+c, \text{ therefore }c =3-5=-2$.

So, the equation of this line is

$y=5x-2$.

We now have plenty of information to plot the graph. The result looks like the figure on the right.

The remainder of this topic is only relevant for the higher course.

If two lines are perpendicular, then the product of their two gradients (i.e., the result of multiplying them together) is -1. Another way of putting this is: if you have a straight line with gradient $m$, then a line which is perpendicular to it will have gradient $-\frac{1}{m}$. This is often referred to as the negative reciprocal of $m$.

Note: to find the reciprocal of a fraction, simply flip the fraction over.

Example: Is the line $x+4y=8$ perpendicular to the line $y=4x-13$?

The second line equation is in the desired form, but the first is not. So, subtracting $x$ from both sides of the first equation, we get

$4y = -x+8$.

Then, dividing both sides by 4, we get

$y=-\dfrac{x}{4}+2$

So, the gradient of this line is $-\frac{1}{4}$, and the gradient of the other line is 4. Multiplying these two values together, we get

$4 \times \left(-\dfrac{1}{4}\right) = -1$

Their product is -1, so the lines are perpendicular.

Example: Find the equation of the straight line that passes through $(-9, -2)$ and is perpendicular to $y=-3x+10$. Plot the straight line.

The gradient of the given line is -3, so the gradient of the perpendicular line must

$-\left(\dfrac{1}{-3}\right) = \dfrac{1}{3}$

The two minus signs cancel, so the result is a positive number. The gradient of one line and another line perpendicular to it will always have opposite signs – if one is negative the other will always be positive and vice versa.

We’re given that the line passes through $(-9, -2)$, and we now know the gradient is $\frac{1}{3}$, so we can substitute these values into $y=mx+c$ in order to find $c$. Doing so, we get

$-2=(-9)\times\dfrac{1}{3}+c=-3+c,\text{ therefore }c=-2+3=1$.

So, the equation of this line is

$y=\dfrac{1}{3}x+1$.

We now have plenty of information to plot the graph. The result looks like the figure below.

## Example Questions

#### 1) State which, if any, of the following 3 lines are parallel.a) $4y - 1 = 2x$b) $2y + 4x = 5$c) $y - \dfrac{1}{2}x = 45$

We need to write all 3 equations in the form $y=mx+c$ and see which ones have the same gradient.

1. a) Add 1 to both sides to get

$4y = 2x + 1$

Then, divide both sides by 4 to get

$y=\dfrac{1}{2}x + \dfrac{1}{4}$

1. b) Subtract $4x$ from both sides to get

$2y = -4x+5$

Then, divide both sides by 2 to get

$y = -2x + \dfrac{5}{2}$

1. c) Add $\frac{1}{2}x$ to both sides to get

$y=\dfrac{1}{2}x+45$

With all 3 equations written in the desired form, we can see that whilst b) has gradient -2, both a) and c) have gradient $\frac{1}{2}$, therefore a) and c) are parallel.

#### 2) Plot the graph, from $x=0$ to $x=4$, of the straight line that is parallel to the line $5y=-10x-3$ and passes through the point $(1, 6)$.

We need to find the gradient of the line given in the question by writing it in the form $y=mx+c$. Dividing both sides by 5, we get

$y = 2x - \dfrac{3}{5}$

Now, the line we need to draw is parallel to this one so must have the same gradient: -2. We’re not asked to work out the equation (although you’re welcome to do that if it helps) and knowing that the line has gradient -2 and passes through $(1, 6)$ is enough to draw it. The correct plotting is shown below.

#### 3) (HIGHER ONLY) Find the equation of the line that is perpendicular to $y=\dfrac{3}{7}x+9$ and passes through the point $(5, -4)$.

The line given in the question has gradient $\frac{3}{7}$. The negative reciprocal of this (and therefore the gradient of the perpendicular line) is

$-\dfrac{7}{3}$.

Now we have the gradient, and we know the lines passes through $(5, -4)$, we can substitute these values into $y=mx+c$ in order to find $c$. So, we get

$-4 = \left(-\dfrac{7}{3}\right)\times 5+c = -\dfrac{35}{3}+c$

The most important thing here is to be careful with your fraction operations. Adding $\frac{35}{3}$ to both sides, we get

\begin{aligned}c=-4+\dfrac{35}{3}&=-\dfrac{4}{1}+\dfrac{35}{3}\\&=-\dfrac{12}{3}+\dfrac{35}{3}=\dfrac{23}{3}\end{aligned}

Therefore, the equation of the line is

$y=-\dfrac{7}{3}x+\dfrac{23}{3}$

If you are looking for perpendicular lines revision notes and practice questions then you have arrived on the right page. The perpendicular and parallel line resources on this page can be used by Maths teachers and tutors everywhere to support your pupils learning. If you are interested in the other GCSE Maths resources we also offer then visit our GCSE Maths revision page.