Resources For Teachers For Tutors For Students & Parents Pricing
Create Free Account
USA - Geometry Lines and angles

Perpendicular bisector theorem (equidistance)

20 practice questions 1 video lesson Theory + worked examples

The Perpendicular Bisector Theorem

Texas Geometry (TEKS) • Standard G.6(A) • Lines & Angles

The Perpendicular Bisector Theorem is a topic in Lines & Angles in the Texas Essential Knowledge and Skills (Geometry, §111.41). It is aligned to Standard G.6(A), which requires students to verify and apply the perpendicular bisector theorem.

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.

Texas Geometry (TEKS) › Lines & Angles › The Perpendicular Bisector Theorem  —  Standard G.6(A)

Practice 20 questions
Practice questions

Every question with a fully worked solution.

Start practising
Watch 1 video(s)
  • What is the Perpendicular Bisector Theorem? (and Converse) Watch
Create a free accountTrack your progress and save your work as you go.
Create free account

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.

More in Lines and angles
Angle theorems (vertical, complementary, supplementary)
20 questions · Video
Parallel lines cut by a transversal (alternate interior, corresponding, co-interior)
20 questions · Video