Question

Prove that the locus of a point equidistant from the extremities a line segment is the perpendicular bisector of it

Answer 1

Let X and Y be two given fixed points. PQ is the path traced out by the moving point P such that each point on it is equidistant from X and Y. Therefore, PX = PY.

To prove: PQ is the perpendicular bisector of the line segment XY.

Construction: Join X to Y. Let PQ cut XY at O.

Theorem on Locus

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Proof:

From △PXO and △PYO,

PX and PY (Given)

XO = YO (Since, every point of PQ is equidistant from X and Y, and O is a point on PQ.)

PO = PO (Common side.)

Therefore, by the SSS criterion of congruency△PXO ≅ △PYO.

Now ∠POX = ∠POY (since, corresponding parts of congruent triangles are congruent.)

Again ∠POX + ∠POY = 180° (Since, XOY is a straight line.

Therefore, ∠POX = ∠POY = 180°2 = 90°

Also, PQ bisects XY (Since, XO = YO)

Therefore, PQ ⊥ XY and PQ bisects XY, i.e., PQ is the perpendicular bisector of XY (Proved)