Chapter 4 · Geometry and Equations · Concept 2
4-11 Perpendicular Bisector and Symmetric Points with Respect to a Line
Lesson 4-4 kept finding points equally far from A and B without ever naming the line they all sit on. Here is its name, its equation — and then the harder question of bouncing a point off a mirror.
Guiding question
What does it take to pin down a reflected point?
Learning outcome. Find the equation of a perpendicular bisector, and the point symmetric to another about a line.
Two jobs, and the second needs two conditions
The perpendicular bisector of AB
It is perpendicular to AB and passes through the midpoint of AB. So you need exactly two things, both from earlier lessons: the gradient of AB (turn it into its negative reciprocal) and the midpoint of AB. Then it is lesson 4-8.
The point B(s, t) symmetric to A about a line l
Write both of these as equations and solve them together:
1. AB is perpendicular to l
2. the midpoint of AB lies on l
Two unknowns, so two equations are needed. Neither condition on its own is enough — and trying to get by with one is the reason this lesson feels stuck.
The two halves are the same idea from opposite ends. A perpendicular bisector IS the mirror line that swaps A and B. Given A and B, the first half finds the mirror; given A and the mirror, the second half finds B.
Step through and watch the two conditions appear separately. The dashed line is the first; the red square is the second.
2. Find B, the point symmetric to A(−4, 3) about the line 3x − y + 5 = 0.
- 1
Name the unknown and get the mirror’s gradient.
B(s, t); 3x − y + 5 = 0 → y = 3x + 5, so the mirror’s gradient is 3Rearrange first, as always — the gradient cannot be read off ax + by + c = 0.
Work with a partner. When you have a mirror image, check it two ways: the midpoint must satisfy the line, and the two points must be the same distance from it. If both hold, it is right.
1. Perpendicular bisectors of AB.
2–3. Mirror images.
In an eye-surgery laser system, the targeting beam is the line y = 2x. A calibration sensor sits at A(5, 0), and the signal receiver must go at B — the exact mirror image of A across the beam. Find B.
Using only one condition.
Two unknowns need two equations. Perpendicular alone leaves a whole line of possible points, and so does the midpoint condition alone.
Using the mirror’s gradient for AB.
AB is PERPENDICULAR to the mirror, so its gradient is the negative reciprocal. Flip and change sign, both.
Multiplying only part of an equation by 2.
Clearing the halves in the midpoint condition doubles EVERY term, including the constant on the end.
Confusing the two halves of the lesson.
Given A and B, you are looking for the mirror line. Given A and the line, you are looking for B. Read which one the question is asking.
Not checking.
Put the midpoint into the line — it must give 0 — and multiply the two gradients — they must give −1. Both take seconds.
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