Parallel and Perpendicular Lines

Understanding the gradient relationships for parallel and perpendicular lines is a key Higher topic. If two lines have the same gradient they are parallel; if the product of their gradients is −1 they are perpendicular. These rules let you find equations of lines in many exam questions. This page covers the theory, the method for finding equations, and fully worked examples.

What Are Parallel and Perpendicular Lines?

Two lines are parallel if they run in exactly the same direction and never meet. In coordinate geometry, this means they have the same gradient.

Two lines are perpendicular if they meet at a right angle (90°). Their gradients are negative reciprocals of each other, meaning the product of their gradients is −1.

Key Formulas

Step-by-Step Method

1. Find the gradient of the given line. If the equation is y = mx + c, the gradient is m. If it is in the form ax + by = c, rearrange to y = mx + c.
2. For a parallel line, use the same gradient m.
3. For a perpendicular line, use the negative reciprocal: −1/m.
4. Substitute the gradient and the given point into y − y₁ = m(x − x₁).
5. Simplify to get y = mx + c.

Worked Example 1 — Foundation Level

Question: Line A has equation y = 3x + 2. Line B has equation y = 3x − 5. Are they parallel?

Working:

Gradient of Line A = 3. Gradient of Line B = 3.

Since the gradients are equal, the lines are parallel.

Answer: Yes, they are parallel (both have gradient 3).

Worked Example 2 — Higher Level

Question: Find the equation of the line perpendicular to y = 2x + 1 that passes through (4, 3).

Working:

• Gradient of the given line: m = 2.
• Perpendicular gradient: −1/2.
• Using y − y₁ = m(x − x₁): y − 3 = −1/2 × (x − 4).
• Expand: y − 3 = −x/2 + 2.
• Add 3: y = −x/2 + 5.

Check: at x = 4, y = −4/2 + 5 = −2 + 5 = 3 ✓

Answer: y = −x/2 + 5 (or equivalently y = −0.5x + 5)

Worked Example 3 — Exam Style

Question: Line L passes through (1, 4) and (3, 10). Find the equation of the line perpendicular to L that passes through the midpoint of (1, 4) and (3, 10).

Working:

• Gradient of L: (10 − 4)/(3 − 1) = 6/2 = 3.
• Perpendicular gradient: −1/3.
• Midpoint: ((1 + 3)/2, (4 + 10)/2) = (2, 7).
• Using y − y₁ = m(x − x₁): y − 7 = −1/3 × (x − 2).
• Expand: y − 7 = −x/3 + 2/3.
• Add 7: y = −x/3 + 2/3 + 21/3 = −x/3 + 23/3.

Answer: y = −x/3 + 23/3

Common Mistakes

• Using m instead of −1/m for perpendicular lines. The perpendicular gradient is the negative reciprocal, not just the negative. For example, if m = 2, the perpendicular gradient is −1/2, not −2.
• Forgetting to flip the fraction. If the gradient is 3/4, the perpendicular gradient is −4/3, not −3/4.
• Not rearranging to find the gradient. If the line is given as 2x + 3y = 12, you must rearrange to y = −2x/3 + 4 to see the gradient is −2/3.

Exam Tips

• State clearly whether lines are parallel or perpendicular and justify with the gradient rule.
• If you are finding the equation of a line, the point-gradient form y − y₁ = m(x − x₁) is the most efficient starting point.
• Questions often combine this topic with midpoints — find the midpoint first, then use it as the point for the perpendicular bisector.
• Double-check by substituting the given point into your final equation.

Practice Questions

Q1 (Foundation): Are the lines y = 4x + 1 and y = 4x − 7 parallel, perpendicular, or neither?

Q2 (Higher): Find the equation of the line parallel to y = −2x + 3 passing through (5, 1).

Q3 (Higher): Find the equation of the line perpendicular to y = (3/4)x − 2 passing through (6, 1).

Practise parallel and perpendicular lines with instant feedback free on GCSEMathsAI.

Related Topics

• Linear Graphs and Equation of a Line
• Finding the Equation of a Line
• Midpoint and Distance Between Points

Summary

• Parallel lines have equal gradients.
• Perpendicular lines have gradients whose product is −1 (negative reciprocals).
• Use y − y₁ = m(x − x₁) to find the equation once you know the gradient and a point.
• Always rearrange the given equation to y = mx + c to read off the gradient.
• This topic often pairs with midpoints and perpendicular bisectors in exam questions.