322. The Equidistant Locus · always the perpendicular bisector

Try to break it

Drag P along the perpendicular bisector. PA and PB always read the same distance, no matter where P sits on the line. Imagine pulling P off the line: PA and PB instantly drift apart. The perpendicular bisector is exactly the locus of points equidistant from A and B.

How you build it

Construct the perpendicular bisector of AB, and see that PA = PB everywhere on it.

• Point tool: mark A on the left — the first fixed point.
• Point tool: mark B to the right of A — the second fixed point.
• Segment tool: join A to B.
• Bisector tool: click A then B — it draws the line through the midpoint of AB, exactly square to AB. This is the locus.
• Point tool: mark P anywhere on the perpendicular bisector you just drew.
• Segment tool: join P to A.
• Segment tool: join P to B. Because P sits on the perpendicular bisector, PA = PB — and this stays true wherever P is on the line. That line is the locus of all points equidistant from A and B.

The proof, step by step

Prove that the perpendicular bisector of AB is the locus of points equidistant from A and B.

• In ΔPMA and ΔPMB, AM = MB (by construction, M is the midpoint of AB).
• ∠PMA = ∠PMB = 90° (PM is perpendicular to AB).
• PM = PM (common side).
• By SAS Congruence Rule, ΔPMA ≅ ΔPMB.
• Therefore, PA = PB (c.p.c.t.). Hence, P lies on the perpendicular bisector of AB.

Worked example

In the figure, AB is a line segment of length 10 cm. P is a point such that PA = PB = 13 cm. What is the distance of P from the midpoint M of AB?

Since PA = PB, P lies on the perpendicular bisector of AB. In right ΔPMA, AM = 5 cm, PA = 13 cm. By Pythagoras theorem, PM = √(13² - 5²) = √(169 - 25) = √144 = 12 cm.

• 10 cm
• 12 cm — correct
• 13 cm
• 5 cm