The point $P$ is equidistant from $A(1,3)$, $B(-3,5)$ and $C(5,-1)$, then $P A$ is equal to :

The point $P$ is equidistant from $A(1,3)$, $B(-3,5)$ and $C(5,-1)$, then $P A$ is equal to :
  1. 5
  2. $5 \sqrt{5}$
  3. 25
  4. $5 \sqrt{10}$

Solution

Let co-ordinates of $P$ are $(x, y)$. Since, $P$ is equidistant from $A, B, C$, then $P A^2=P B^2$...(i) and $P B^2=P C^2$...(ii) From Eq. (i), $(x-1)^2+(y-3)^2=(x+3)^2+(y-5)^2$ $\Rightarrow \quad x^2+1-2 x+y^2+9-6 y$ $=x^2+9+6 x+y^2+25-10 y$ $\Rightarrow \quad 8 x-4 y+24=0$ $\Rightarrow \quad 2 x-y+6=0$...(iii) From Eq. (ii), $(x+3)^2+(y-5)^2=(x-5)^2+(y+1)^2$ $\Rightarrow \quad x^2+9+6 x+y^2+25-10 y$ $=x^2+25-10 x+y^2+1+2 y$ $\Rightarrow \quad 16 x-12 y+8=0$ $\Rightarrow \quad 4 x-3 y+2=0$...(iv) On solving Eqs. (iii) and (iv), we get $x=-8, y=-10$ Now, $P A^2=(-8-1)^2+(-10-3)^2$ $=81+169=250$ $P A=\sqrt{250}=5 \sqrt{10}$

Asked in: AP EAMCET 2003

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