Equations of Lines: Parallel & Perpendicular

Mathematics· 3.5, 3.6, E3.7· 18 min read

1. Core: Equations of Parallel Lines★★☆☆☆⏱ 5 min

📘 Definition

Parallel Lines

Two or more straight lines that never intersect, no matter how far they are extended. All parallel lines have exactly equal gradient (m) values.

Example:

Lines $y = 2 x + 3$ and $y = 2 x - 7$ are parallel, as both have gradient $m = 2$ .

For Core tier, you will work with three line forms: $y = m x + c$ (sloped lines), $y = k$ (horizontal lines, gradient 0), and $x = k$ (vertical lines, undefined gradient). To find the equation of a parallel line through a given point, first copy the gradient of the original line, then substitute the gradient and point coordinates into $y = m x + c$ to solve for the y-intercept $c$ .

📐 Worked Example

Find the equation of the line parallel to $y = 3 x + 2$ that passes through the point $(2, 5)$ . Give your answer in $y = m x + c$ form.

1
1. Identify the gradient of the given line: $m = 3$ (parallel lines share the same gradient)
2
1. Substitute $m = 3$ , $x = 2$ , $y = 5$ into $y = m x + c$ :
3
$5 = 3(2) + c$
4
1. Solve for $c$ :
5
$5 = 6 + c \implies c = -1$
6
1. Write the final equation: $y = 3 x - 1$

Exam tip:

Always rearrange given line equations to y = mx + c first to extract the correct gradient before solving for parallel lines.

2. Extended Only: Perpendicular Lines & ax + by = c Form★★★☆☆Extended only⏱ 7 min

✓ Calculator OK

📘 Definition

Perpendicular Lines

Two lines that intersect at a 90° right angle. For non-vertical/non-horizontal lines, the product of their gradients equals -1: $m_{1} m_{2} = - 1$ . Horizontal lines ( $y = k$ ) and vertical lines ( $x = k$ ) are also perpendicular.

Example:

A line with gradient $2$ is perpendicular to a line with gradient $- \frac{1}{2}$ , as $2 \times - \frac{1}{2} = - 1$ .

Extended candidates must also be able to write line equations in $a x + b y = c$ form, where $a$ , $b$ , and $c$ are integers and $a$ is non-negative. To find a perpendicular line equation, first calculate the negative reciprocal of the original gradient, then substitute the given point coordinates to solve for the intercept.

📐 Worked Example

Find the equation of the line perpendicular to $y = 2 x + 1$ that passes through $(4, 3)$ . Give your answer in $a x + b y = c$ form with positive integer coefficients.

1
1. Gradient of given line: $m_{1} = 2$
2
1. Gradient of perpendicular line: $m_{2} = - \frac{1}{2}$ (since $2 \times - \frac{1}{2} = - 1$ )
3
1. Substitute $m = - \frac{1}{2}$ , $x = 4$ , $y = 3$ into $y = m x + c$ :
4
$3 = -\frac{1}{2}(4) + c \implies 3 = -2 + c \implies c = 5$
5
1. Write in $y = m x + c$ form: $y = - \frac{1}{2} x + 5$
6
1. Multiply all terms by 2 to eliminate fractions: $2 y = - x + 10$
7
1. Rearrange to $a x + b y = c$ form: $x + 2 y = 10$

📐 Worked Example

Find the perpendicular bisector of the line segment joining $A (1, 2)$ and $B (5, 6)$ . Give your answer in $y = m x + c$ form.

1
1. Calculate the midpoint of segment $A B$ :
2
$\text{Midpoint} = \left(\frac{1+5}{2}, \frac{2+6}{2}\right) = (3, 4)$
3
1. Calculate the gradient of $A B$ :
4
$m_1 = \frac{6-2}{5-1} = 1$
5
1. Gradient of perpendicular bisector: $m_{2} = - 1$
6
1. Substitute midpoint $(3, 4)$ and $m = - 1$ into $y = m x + c$ :
7
$4 = -1(3) + c \implies c = 7$
8
1. Final equation: $y = - x + 7$

✓ Quick check

What is the gradient of a line perpendicular to $y = 4 x - 3$ ?
Reveal answer
-1/4 —
The product of perpendicular gradients is -1, so $m = - 1/4$ .

3. Exam Cues & Quick Practice★★☆☆☆⏱ 4 min

✓ Quick check

What is the gradient of a line parallel to $y = - 5 x + 9$ ?
Reveal answer
-5 —
Parallel lines have identical gradients, so the gradient is exactly -5.

4. Common Pitfalls

Wrong move:

(Extended) Using the negative reciprocal gradient for parallel lines

Why:

Students mix up parallel and perpendicular gradient rules

Correct move:

Parallel lines have equal gradients; only use the negative reciprocal for perpendicular lines (Extended only)

Wrong move:

(Extended) Extracting the gradient directly from $a x + b y = c$ without rearranging to $y = m x + c$ first

Why:

For $2 x + y = 5$ , the gradient is not 2, it is -2

Correct move:

Always rearrange given line equations to $y = m x + c$ form before reading the gradient value

Wrong move:

(Extended) Using an endpoint of a segment instead of the midpoint when calculating the perpendicular bisector

Why:

The perpendicular bisector cuts the segment exactly in half, so it must pass through the midpoint

Correct move:

Calculate the midpoint of the segment first before finding the perpendicular bisector equation

Wrong move:

(Extended) Leaving a negative $a$ value when writing equations in $a x + b y = c$ form

Why:

Extended tier requires $a$ to be a non-negative integer

Correct move:

If you get $- 2 x + 3 y = 6$ , multiply all terms by -1 to get $2 x - 3 y = - 6$

Wrong move:

(Extended) Trying to apply the $m_{1} m_{2} = - 1$ rule to horizontal and vertical lines

Why:

Horizontal lines have gradient 0, vertical lines have undefined gradient, so their product does not exist

Correct move:

Recognize that all $y = k$ (horizontal) and $x = k$ (vertical) lines are automatically perpendicular to each other

5. Quick Reference Cheatsheet

Concept	Core Rule	Extended Rule
Parallel Lines	Equal gradients ( $m_{1} = m_{2}$ )	Same as Core, can write in $a x + b y = c$ form
Perpendicular Lines	Not assessed	Product of gradients = -1 ( $m_{1} m_{2} = - 1$ )
Perpendicular Bisector	Not assessed	Passes through segment midpoint, has perpendicular gradient
Allowed Line Forms	$y = m x + c$ , $y = k$ , $x = k$	All Core forms + $a x + b y = c$ (integer $a \geq 0$ )

6. Frequently Asked

How do I check if two lines are parallel?

Rearrange both lines to y = mx + c form. If their m (gradient) values are exactly equal, the lines are parallel.

What is the product of gradients of perpendicular lines?

For Extended tier, the product of gradients of two non-vertical/non-horizontal perpendicular lines is always -1. Horizontal (y=k) and vertical (x=k) lines are also perpendicular, even though their gradient product is undefined.

Do Core students need to use ax + by = c form?

No, Core students only need to work with y = mx + c, y = k, and x = k line forms. The ax + by = c form is assessed only for Extended tier candidates.

Going deeper

study_guideGradient of Straight Lines
study_guideBasics of Straight Line Equations

What's Next

Now that you have mastered parallel and perpendicular line equations, you can apply this knowledge to more complex coordinate geometry problems for your CIE IGCSE Math 0580 exam. Core students can move on to practice problems involving line intersections and graph sketching, while Extended students can progress to harder topics including 3D coordinates and shape problems that combine line equations with area or perimeter calculations. Be sure to practice past paper questions to build speed and accuracy under exam conditions.