Let $A(2,3), B(3,-6), C(5,-7)$ be three points. If $P$ is a point satisfying the condition $P A^2+P B^2=2 P…
Let $A(2,3), B(3,-6), C(5,-7)$ be three points. If $P$ is a point satisfying the condition $P A^2+P B^2=2 P C^2$, then a point that lies on the locus of $P$ is
$(2,-5)$
$(-2,5)$
$(13,10)$
$(-13,-10)$
Solution
Given, points are $A(2,3), B(3,-6), C(5,-7)$.
Let point $P$ be $(x, y)$, then according to condition
$\begin{aligned}
& P A^2+P B^2=2 P C^2 \\
& \begin{aligned}
\Rightarrow(x-2)^2+(y-3)^2+(x-3)^2+(y+6)^2 \\
=2\left[(x-5)^2+(y+7)^2\right]
\end{aligned} \\
& \begin{aligned}
\Rightarrow x^2+4-4 x+y^2+9-6 y+x^2+9-6 x \\
\quad+y^2+36+12 y
\end{aligned} \\
& =2\left[x^2+25-10 x+y^2+14 y+49\right] \\
& \Rightarrow 2 x^2+2 y^2-10 x+6 y+58 \\
& =2 x^2+2 y^2-20 x+28 y+148 \\
& \Rightarrow 10 x-22 y=90
\end{aligned}$
By checking options,
1. 10(2) − 22(− 5) = 20 + 110 =130
2. 10(− 2) − 22(5) = − 20 −110 = −130
3. 10(13) − 22(10) =130 − 220 = − 90
4. 10(−13) − 22(−10) = −130 + 220 = 90
So, point (− 13, − 10) lies on the locus of P