STD 11 - rectangular cartensian co-ordinates — Maths STD 11 Science — Question

$ \Rightarrow \,\,2x(1 + 3) + 2y(3 - 5) = (1 + 9) - (9 + 25)$

$ \Rightarrow \,\,2x - y + 6 = 0$ .....$(i)$

Perpendicular bisector of $A\,(1,\,\,3)$ and $C\,(5,\,\, - 1)$ is

$2x\,(1 - 5) + 2y(3 + 1) = (1 + 9) - (25 + 1)$

$ \Rightarrow \,\,x - y - 2 = 0$ .....$(ii)$

Point of intersection of $(i)$ and $(ii)$ is $P = ( - 8,\,\, - 10)$

Then $PA = \sqrt {{{(1 + 8)}^2} + {{(3 + 10)}^2}} = \sqrt {81 + 169} $

$ = \sqrt {250} = 5\sqrt {10} $.