CAMBRIDGE O LEVEL · 2025–2027

Perpendicular lines.

Use the negative reciprocal, then locate the line using a point or midpoint.

Syllabus 3.7Visual notesWorked examples

BY THE END

Find perpendicular gradients and equations · construct perpendicular bisectors

Perpendicular means a right angle

Perpendicular lines meet at a right angle. Their rise-and-run directions are related by a quarter-turn, which swaps the two changes and reverses one sign.

For finite non-zero gradients, this gives negative reciprocals: a gradient of 2 becomes −12. The product is −1. Horizontal and vertical lines form a special perpendicular pair, because the vertical gradient is undefined rather than an ordinary reciprocal.

For two non-vertical perpendicular lines, the product of their gradients is −1. Take the negative reciprocal to get the new gradient.

m₁m₂ = −1 → m₂ = −1m₁

Lines of gradients 2 and −12 meet at a right angle. Both axes use equal unit lengths.

Gradient 2Gradient −12

WORKED EXAMPLE

Find the gradient perpendicular to 2y = 3x + 1.

Rearrange: y = 32 x + 12, so the given gradient is 32.

Perpendicular gradient = −23

Check: (32) × (−23) = −1.

Pause and explainWhat is the gradient perpendicular to −3?

13. Reverse the sign and take the reciprocal; (−3) × 13 = −1.

Find the new line through a point

A perpendicular gradient gives the direction of the required line. As with parallel-line problems, a point is needed to locate it.

Find the original gradient, replace it with its negative reciprocal, then use the supplied point to find the new intercept. Keep the reciprocal as an exact fraction to avoid introducing rounding into the equation.

WORKED EXAMPLE

Find the line perpendicular to y = −2x + 3 through (4, 1).

New gradient = 12. Substitute into y = 12 x + c:

1 = 2 + c, so c = −1
y = 12 x − 1

WORKED EXAMPLE

Find the line perpendicular to y = 5 through (−2, 3).

The given line is horizontal. The required line is vertical, so x = −2.

Pause and explainIf the original gradient is 25, what gradient should the new perpendicular line have?

−52. Interchange numerator and denominator and reverse the sign. Multiplying the two gradients then gives −1.

A perpendicular bisector passes through the midpoint

A perpendicular bisector does two jobs at once: it crosses a segment at 90° and passes through the segment’s midpoint. Both conditions are needed; either one alone describes many possible lines.

Calculate the segment’s gradient and midpoint separately. Use the negative reciprocal for the bisector’s direction, then the midpoint to determine its position. Every point on the resulting line is equally distant from the original endpoints, which provides a geometric check.

“Bisector” adds a second requirement: the line must pass through the midpoint of the segment as well as meeting it at a right angle.

WORKED EXAMPLE

Find the perpendicular bisector of A (−3, 8) and B (9, −2).

Midpoint M = ((−3 + 9) ÷ 2, (8 − 2) ÷ 2) = (3, 3).
Gradient AB = (−2 − 8) ÷ (9 − (−3)) = −1012 = −56.
Perpendicular gradient = 65.
Use M: 3 = (65) × 3 + c, so c = −35.

y = 65 x − 35
Fully simplified integer form: 6x − 5y = 3

The perpendicular bisector meets AB at its midpoint M (3, 3). Its equation is 6x − 5y = 3.

Line through A and BPerpendicular bisector

Pause and explainWhy use the midpoint rather than either endpoint to find the bisector equation?

The bisector must pass halfway along the segment. An endpoint would usually locate a different perpendicular line.

Your turn. Check direction and position.

List the two conditions your answer must satisfy: perpendicular direction and passage through the required point or midpoint.

Treat horizontal or vertical segments separately. For all other cases, use an exact negative reciprocal, form the equation and substitute the required point as a final check.

Find the gradient perpendicular to 4x + 2y = 7.

Reveal worked answer

The given gradient is −2. Its negative reciprocal is 12.

Find the line perpendicular to y = 3x − 2 through (3, 4).

Reveal worked answer

Gradient = −13. 4 = (−13) × 3 + c gives c = 5. Equation: y = −13 x + 5, or x + 3y = 15.

Find the perpendicular bisector of (−2, 1) and (4, 5).

Reveal worked answer

Midpoint (1, 3), segment gradient 46 = 23, perpendicular gradient −32. Through the midpoint: y − 3 = −32 (x − 1), so 3x + 2y = 9.

Find the perpendicular bisector of (−4, 2) and (6, 2).

Reveal worked answer

The segment is horizontal, with midpoint (1, 2). Its perpendicular bisector is x = 1.