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 [tex](x_{1},y_{1}), and ,(x_{2},y_{2})[/tex] is given by [tex]=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2[/tex]
[tex]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}[/tex]
Hence, AP = BP
Where ,Point P, can be located anywhere on y axis.
