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Perpendicular bisector theorem (equidistance)

20 practice questions 1 video lesson Theory + worked examples

The Perpendicular Bisector Theorem

California Geometry • Standard G-CO.9 • Lines & Angles

The Perpendicular Bisector Theorem is a topic in Lines & Angles in the California Common Core State Standards. It is aligned to Standard G-CO.9, which requires students to prove theorems about lines and angles, including that points on a perpendicular bisector are equidistant from the segment's endpoints.

The perpendicular bisector theorem states a point lies on the perpendicular bisector of a segment if and only if it is equidistant from the endpoints.

California Geometry › Lines & Angles › The Perpendicular Bisector Theorem  —  Standard G-CO.9

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Theory

The perpendicular bisector of a segment is the line that is perpendicular to it and passes through its midpoint.

Perpendicular bisector theorem: every point on the perpendicular bisector of \(\overline{AB}\) is equidistant from \(A\) and \(B\) — that is, \(PA=PB\). Converse: if a point is equidistant from \(A\) and \(B\), it lies on the perpendicular bisector of \(\overline{AB}\).
Equidistance is the key idea: the perpendicular bisector is exactly the set of points the same distance from both endpoints.
Perpendicular bisector theorem Any point P on the perpendicular bisector of a segment is equidistant from the two endpoints. A B P PA = PB
Point \(P\) on the perpendicular bisector: \(PA=PB\).
Converse of the theorem A point equidistant from the two endpoints of a segment lies on its perpendicular bisector. converse: equidistant ⇒ on the bisector
Converse: equidistant \(\Rightarrow\) on the perpendicular bisector.

The theorem and its converse:

\[P\text{ on perp. bisector of }\overline{AB}\iff PA=PB\]
a point is on the perpendicular bisector of AB exactly when its distances to A and B are equal
Two directions: on the bisector \(\Rightarrow\) equidistant (theorem); equidistant \(\Rightarrow\) on the bisector (converse).

How to use the theorem

  1. Identify the perpendicular bisector and the segment's endpoints.
  2. Set the distances equal: \(PA=PB\).
  3. Solve for the unknown, or conclude a point lies on the bisector (converse).
Example 1 — Use equidistance
Point \(P\) is on the perpendicular bisector of \(\overline{AB}\) and \(PA=7\). Find \(PB\).
Solution

Any point on the perpendicular bisector is equidistant from the endpoints.

\(PB\)\(=\)\(PA=7\)
PB equals 7
Example 2 — Solve for x
\(P\) is on the perpendicular bisector of \(\overline{AB}\). \(PA=3x-1\) and \(PB=2x+5\). Find \(x\).
Solution

Set the equal distances equal.

\(3x-1\)\(=\)\(2x+5\)
\(x\)\(=\)\(6\)
x equals 6
Example 3 — Apply the converse
Point \(Q\) satisfies \(QA=QB\). What can you conclude?
Solution

By the converse of the perpendicular bisector theorem, \(Q\) lies on the perpendicular bisector of \(\overline{AB}\).

Q lies on the perpendicular bisector of segment AB
Example 4 — Find a midpoint length
The perpendicular bisector of \(\overline{AB}\) meets it at \(M\) with \(AB=18\). Find \(AM\).
Solution

A bisector cuts the segment in half.

\(AM\)\(=\)\(\dfrac{18}{2}=9\)
AM equals 9

Common pitfalls

Both perpendicular AND through the midpoint. A line with only one of these is not the perpendicular bisector.
Equidistant means equal distances to the endpoints, not to the midpoint.
Use the converse to prove a point is on the bisector, and the theorem to get equal distances.

Frequently asked questions

What is a perpendicular bisector?

The line perpendicular to a segment that passes through its midpoint.

What does the perpendicular bisector theorem say?

Every point on the perpendicular bisector of a segment is equidistant from the segment's two endpoints.

What is the converse of the theorem?

If a point is equidistant from the two endpoints of a segment, it lies on the perpendicular bisector of that segment.

How is the theorem used to solve problems?

Set the two distances equal (\(PA=PB\)) and solve, since any point on the bisector satisfies this.

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