Equation of a Perpendicular Bisector (GCSE Maths)

What is a perpendicular bisector?

The perpendicular bisector of a line segment AB has two defining properties:

1. It passes through the midpoint of AB.
2. It is perpendicular (at right angles, 90°) to AB.

Every point on the perpendicular bisector is equidistant from A and B — this is its geometric meaning. In coordinate geometry, you find its equation using the coordinates of A and B.

What formulae do you need?

Two key formulae:

Midpoint formula: The midpoint of (x₁, y₁) and (x₂, y₂) is:

$$M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$$

Perpendicular gradient rule: If a line has gradient m, any line perpendicular to it has gradient −1/m (the negative reciprocal).

• Gradient 2 → perpendicular gradient −1/2.
• Gradient −3 → perpendicular gradient 1/3.
• Gradient 1/4 → perpendicular gradient −4.

How do you find the equation of a perpendicular bisector?

Four-step method:

1. Find the midpoint M of the given line segment.
2. Find the gradient of the given line segment.
3. Find the perpendicular gradient (negative reciprocal).
4. Use y − y₁ = m(x − x₁) with the midpoint and perpendicular gradient, then rearrange to y = mx + c if required.

Worked example 1: straightforward case

Find the equation of the perpendicular bisector of the line segment joining A(2, 1) and B(6, 5).

Step 1 — Midpoint: M = ((2 + 6)/2, (1 + 5)/2) = (8/2, 6/2) = (4, 3).

Step 2 — Gradient of AB: m = (5 − 1)/(6 − 2) = 4/4 = 1.

Step 3 — Perpendicular gradient: Perpendicular gradient = −1/1 = −1.

Step 4 — Equation: y − 3 = −1(x − 4) y − 3 = −x + 4 y = −x + 7 (or equivalently x + y = 7).

Check: Does the line pass through (4, 3)? −4 + 7 = 3. ✓ Is the gradient −1 (perpendicular to slope 1)? ✓

Worked example 2: fractional gradient

Find the equation of the perpendicular bisector of the segment joining P(1, 4) and Q(5, 2).

Step 1 — Midpoint: M = ((1 + 5)/2, (4 + 2)/2) = (3, 3).

Step 2 — Gradient of PQ: m = (2 − 4)/(5 − 1) = −2/4 = −1/2.

Step 3 — Perpendicular gradient: Negative reciprocal of −1/2 is 2.

Step 4 — Equation: y − 3 = 2(x − 3) y − 3 = 2x − 6 y = 2x − 3.

Summary table for both worked examples:

What if the segment is horizontal or vertical?

• Horizontal segment (gradient = 0): perpendicular gradient is undefined — the perpendicular bisector is a vertical line. Equation: x = midpoint's x-coordinate.
• Vertical segment (gradient undefined): the perpendicular bisector is a horizontal line. Equation: y = midpoint's y-coordinate.

Example: Segment from (3, 2) to (3, 8) is vertical (x stays at 3). Midpoint = (3, 5). Perpendicular bisector: y = 5.

Frequently asked questions

Why does the perpendicular gradient rule give −1/m and not just −m?

Two lines are perpendicular when the product of their gradients equals −1: m₁ × m₂ = −1. Solving for m₂ gives m₂ = −1/m₁. This is the negative reciprocal — both the sign change and the flip of the fraction are needed. A common error is to negate without flipping (−m instead of −1/m), which does not give a perpendicular line.

Does the perpendicular bisector always cross inside the segment?

The perpendicular bisector always passes through the midpoint, which lies on the segment. However, the bisector itself is an infinite line extending in both directions. For any triangle, the three perpendicular bisectors of the sides all meet at a single point called the circumcentre — a GCSE Higher topic.

How do I check my final equation?

Substitute the midpoint into your equation and verify it satisfies it. Also check: if you worked out the gradient of the perpendicular bisector, multiply it by the gradient of the original segment — the product must be −1.

What if the question asks for the equation in a specific form?

Common forms are y = mx + c (slope-intercept) and ax + by = c (where a, b, c are integers). To convert: y = 2x − 3 can be written as 2x − y = 3. Rearrange to match the form requested by multiplying through to clear fractions if needed.

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Professor Pi can guide you step by step through coordinate geometry problems — visit aitutors.me.