The locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points.
Given,
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.
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 = \(\frac{180°}{2}\) = 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)
`● Loci
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