Every point on the perpendicular bisector of a segment is equally far from its two endpoints. To find a point equally far from A, B and C, find where two perpendicular bisectors meet.
You can construct the bisectors on a scale drawing or calculate them from coordinates. Two bisectors are enough when the three points do not lie on one straight line.
- Pick two segments that share an endpoint.
- Find each midpoint and draw or calculate the perpendicular line.
- Check the common point against the three distances.
Worked example | Find the common point on a coordinate grid
Let A = (0, 0), B = (8, 0) and C = (2, 6). Segment AB is horizontal with midpoint (4, 0), so its perpendicular bisector is the vertical line x = 4.
Segment BC has gradient (6 − 0)/(2 − 8) = −1 and midpoint (5, 3). A perpendicular line has gradient 1. Through (5, 3), its equation is y − 3 = x − 5, or y = x − 2.
Where x = 4, y = 2. The point (4, 2) is equally far from A and B, and equally far from B and C, so all three distances are equal.
Check by comparing squared distances
From (4, 2), the squared distance to A is 4² + 2² = 20. The squared distance to B is also 4² + 2² = 20. The squared distance to C is 2² + 4² = 20. Since distances are nonnegative, equal squared distances give equal distances.
Try a right-angled layout
Find the point equidistant from (0, 0), (6, 0) and (0, 8).
Answer. The horizontal segment has perpendicular bisector x = 3. The vertical segment has perpendicular bisector y = 4. The common point is (3, 4).
Frequently asked questions
Do I have to construct the third bisector?
You can use it as a check. Equality of the first and second distances and of the second and third already gives equality of all three.
What if the three points are distinct and collinear?
The two perpendicular bisectors are parallel and distinct, so there is no point equidistant from all three.
Official information
SEAB, 2027 SEC G3 Mathematics syllabus (K310), Geometry and Measurement, perpendicular bisectors. Syllabus checked 11 November 2026. The examples are original.