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} $

Answer

Correct option: D.
$5\sqrt {10} $
d
(d) Perpendicular bisector of $A\,(1,\,\,3)$ and $B\,( - 3,\,\,5)$ is $2x({x_1} - {x_2}) + 2y\,({y_1} - {y_2}) = (x_1^2 + y_1^2) - (x_2^2 + y_2^2)$

$ \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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