Cambridge O Level Additional Mathematics · Syllabus 4037 · Straight-Line Graphs
Perpendicular Bisector
What is Perpendicular Bisector?
The perpendicular bisector of a line segment is the unique straight line that passes through the midpoint of the segment and is perpendicular to it. Both conditions are required: a line through the midpoint with the wrong gradient is not a perpendicular bisector, and neither is a line of the correct gradient that misses the midpoint. To construct it, find the midpoint by averaging the coordinates of the endpoints, find the gradient of the segment, take the negative reciprocal of that gradient, and substitute both results into the point-gradient form. Its defining geometric property is that every point on it is equidistant from the two endpoints of the segment, which is why it is used to locate points at equal distance from two fixed points.
This definition is part of the Straight-Line Graphs chapter in Cambridge O Level Additional Mathematics.
Common mistakes with Perpendicular Bisector
- 9. A “perpendicular bisector” that does not pass through the midpoint The error Finding the perpendicular gradient correctly, then using endpoint \(A\) rather than the midpoint \(M\). For \(A(-3,8)\), \(B(9,-2)\) this produces \(6x-5y=-58\). Why it fails That line genuinely is perpendicular to \(AB\), but it crosses \(AB\) at \(A\), so it bisects nothing. Half of the definition has been satisfied and half ignored. Fix Substitute the midpoint into your final answer. Here \(6(3)-5(3)=3\ne-58\), which exposes the error in one line. The correct bisector is \(6x-5y=3\).
Questions students ask about Perpendicular Bisector
How do you find the equation of a perpendicular bisector?
Find the midpoint of the segment by averaging the coordinates, find the gradient of the segment, take its negative reciprocal, then use \(y-y_1=m(x-x_1)\) with the midpoint and the new gradient. Both conditions are needed: a line through an endpoint with the right gradient is perpendicular but not a bisector, and a line through the midpoint with the segment's own gradient is just the segment again. Substitute the midpoint back into your answer to prove both conditions hold. “Equidistant from two points” is always a perpendicular-bisector question.