Perpendicular lines and perpendicular bisectors
Use a negative reciprocal gradient, including horizontal–vertical exceptions, and bisect a segment accurately.
Section E3.7 only · Perpendicular lines and bisectors
Extended only. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Find gradients, line equations and midpoints.
- Calculate a signed reciprocal.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the negative reciprocal of 4/3?
What you will learn
- Find a perpendicular line through a point.
- Find the equation of a perpendicular bisector.
- Handle horizontal and vertical lines without dividing by zero.
A right angle changes the gradient
For two non-vertical lines with non-zero gradients, perpendicularity means m₁m₂ = −1. Therefore a line of gradient 4/3 has perpendicular gradient −3/4. Both the sign and the reciprocal change. Merely negating 4/3 to −4/3 does not usually make a right angle.
Use the reciprocal of the fully simplified gradient and change its sign. A gradient −2 has perpendicular gradient 1/2. The product check (−2)(1/2) = −1 confirms the relationship. The rule assumes equal physical units on the two coordinate axes; a visually stretched graph can distort apparent angles.
Pass the perpendicular through the required point
Once the perpendicular gradient is known, write y = m₂x + c and substitute the specified point. For a line perpendicular to y = 2x + 3 through (4, 1), m₂ = −1/2. Then 1 = −2 + c gives c = 3, so the equation is y = −(1/2)x + 3.
A horizontal line has gradient zero, so its perpendicular is vertical. Through (4, 1) that vertical equation is x = 4. A line perpendicular to a vertical line is horizontal, here y = 1. These are geometric exceptions to using a finite negative-reciprocal calculation.
A bisector must also go through the midpoint
The perpendicular bisector of AB is a line at right angles to AB through its midpoint. For A = (−3, −2) and B = (3, 6), AB has gradient 8/6 = 4/3 and midpoint (0, 2). The bisector therefore has gradient −3/4 and equation y = −(3/4)x + 2.
Check both conditions: the gradient product is −1, and the midpoint satisfies the bisector equation. Passing a perpendicular through endpoint A would make a right angle but would not bisect the segment. Every point on the perpendicular bisector is equally distant from the two endpoints; this gives a useful map or construction interpretation.
Pause and explain
A perpendicular bisector must pass through which point?
Worked example
Find the perpendicular bisector of the segment joining (−3, −2) and (3, 6).
Show the worked solution
- The midpoint is ((−3 + 3)/2, (−2 + 6)/2) = (0, 2).
- The segment gradient is (6 + 2)/(3 + 3) = 4/3, so the perpendicular gradient is −3/4.
- Through (0, 2), the intercept is two: y = −(3/4)x + 2.
Answer y = −(3/4)x + 2
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the line perpendicular to y = 2x + 3 through (4, 1).
Give me a hint
Find the negative reciprocal before solving for the intercept.
Compare my reasoning
- The perpendicular gradient is −1/2.
- Use 1 = (−1/2) × 4 + c to get c = 3.
- The line is y = −(1/2)x + 3; its gradient product with the original is −1.
y = −(1/2)x + 3
Look for these in your work
- Used the negative reciprocal.
- Passed the line through the specified point.
Find the perpendicular bisector of the segment from (−2, 1) to (4, 1).
Give me a hint
First decide whether the segment is horizontal or vertical.
Compare my reasoning
- The segment is horizontal because both y-values are one.
- Its midpoint is (1, 1).
- The bisector is the vertical line x = 1; no finite gradient is needed.
x = 1
Look for these in your work
- Recognised the horizontal–vertical exception.
- Used the midpoint rather than an endpoint.
Two radio towers are at A = (−3, −2) and B = (3, 6). Verify that P = (4, −1) lies on their perpendicular bisector and is equally distant from both.
Give me a hint
Test the line equation, then compare squared distances.
Compare my reasoning
- The bisector is y = −(3/4)x + 2; at x = 4 it gives y = −1.
- PA² = (4 + 3)² + (−1 + 2)² = 49 + 1 = 50.
- PB² = (4 − 3)² + (−1 − 6)² = 1 + 49 = 50, so both distances are √50.
P is on the bisector; PA = PB = 5√2 units
Look for these in your work
- Checked both the equation and equal-distance interpretation.
- Compared distances without unnecessary rounding.
Common mistakes
- Changing the sign without taking the reciprocal.
- Passing a bisector through an endpoint.
- Dividing by zero for horizontal or vertical exceptions.
What can you explain now?
What line through (2, −3) is perpendicular to x = 5?
Compare with the explanation
y = −3
The original line is vertical, so its perpendicular is horizontal at the specified point's y-coordinate.
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Make it stick
Tomorrow, form a perpendicular bisector from new endpoints and test its gradient and midpoint conditions separately.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.