Parallel & Perpendicular Lines

Exam code: 4037

Written by: Ashika|Reviewed by: Caroline Carroll|Updated 2 July 2026

Parallel Linesvideo

Parallel Lines

Parallel lines

What are parallel lines?

Parallel lines are always equidistant meaning they never intersect
Parallel lines have the same gradient
- If the gradient of line l₁is m₁ and gradient of line l₂is m₂then...
  - $m_{1} = m_{2} \Rightarrow l_{1} & l_{2} are parallel$
  - $l_{1} & l_{2} are parallel \Rightarrow m_{1} = m_{2}$
To determine if two lines are parallel:
- Rearrange into the gradient-intercept form $y = m x + c$
- Compare the coefficients of $x$
- If they are equal then the lines are parallel

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:

$y = 3 x - 5$

Perpendicular Lines

Perpendicular lines

What are perpendicular lines?

Perpendicular lines intersect at right angles
The gradients of two perpendicular lines are negative reciprocals
- If the gradient of line l₁is m₁ and gradient of line l₂is m₂then...
  - $m_{1} \times m_{2} = - 1 \Rightarrow l_{1} & l_{2} are perpendicular$
  - $l_{1} & l_{2} are perpendicular \Rightarrow m_{1} \times m_{2} = - 1$
To determine if two lines are perpendicular:
- Rearrange into the gradient-intercept form $y = m x + c$
- Compare the coefficients of $x$
- If their product is -1 then they are perpendicular
Be careful with horizontal and vertical lines
- $x = p$ and $y = q$ are perpendicular where p and q are constants

Worked example

Find the equation of the line that is perpendicular to $y = 2 x - 2$ and passes through (2, -3).

Leave your answer in the form $a x + b y + c = 0$ where $a, b, c$ 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

$x + 2 y + 4 = 0$