Question

Prove: Every point on the perpendicular bisector of a segment is equidistant from the ends of the segment.

Answer 1

AP=√(a² + b² )
BP=√(a² + b² )

Answer 2

Answer with explanation:

Here ,let line, x=0 intersect ,segment AB at point M.

So, PM ⊥ AB.

And, AM =BM →→→∵ y axis or segment PM , is Perpendicular bisector of segment AB.

In Δ AMP and ΔB MP

∠AMP = ∠ B MP=90°[→→ Each being 90°]

AM = BM →Line PM , is perpendicular Bisector.

Side MP , is Common.

Δ AMP ≅ ΔB MP→→→[S AS]

So, AP= BP →→[C P C T]

2. We will use distance formula to find AP and BP.

Distance between two points $(x_{1},y_{1}), and ,(x_{2},y_{2})$ is given by $=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2$

 $AP=\sqrt{(-a-0)^2+(0-b)^2}\\\\ AP=\sqrt{a^2+b^2}\\\\ BP=\sqrt{(a-0)^2+(0-b)^2}\\\\ BP=\sqrt{a^2+b^2}$

Hence, AP = BP

Where ,Point P, can be located anywhere on y axis.