Question

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
• D. $5\sqrt{10}$

Answer 1

Answer: (D)

$5\sqrt{10}$

Answer 2

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

$2 x \left( x _ { 1 } - x _ { 2 } \right) + 2 y \left( y _ { 1 } - y _ { 2 } \right) = \left( x _ { 1 } ^ { 2 } + y _ { 1 } ^ { 2 } \right) - \left( x _ { 2 } ^ { 2 } + y _ { 2 } ^ { 2 } \right)$

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

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

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

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

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

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

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

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