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The point $P$ is equidistant from $A(1,3)$, $B(-3,5)$ and $C(5,-1)$, then $P A$ is equal to :
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Verified Answer
The correct answer is:
$5 \sqrt{10}$
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, $P A^2=(-8-1)^2+(-10-3)^2$
$=81+169=250$
$P A=\sqrt{250}=5 \sqrt{10}$
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, $P A^2=(-8-1)^2+(-10-3)^2$
$=81+169=250$
$P A=\sqrt{250}=5 \sqrt{10}$
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