MCQ
The points $P$ is equidistant from $A(1,3), \,B (-3,5)$ and $C(5,-1)$. Then $PA =$
- A$5$
- B$5\sqrt 5 $
- C$25$
- ✓$5\sqrt {10} $
$ \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} $.
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