A point P moves so that the sum of squares of its distances from the points 1 , 2 and - 2 , 1 is 14 . Let f…

Given,

A point $P$ moves so that the sum of squares of its distances from the points $\left(1,2\right)$ and $\left(-2,1\right)$ is $14$. 

So, ${\left(x-1\right)}^2+{\left(y-2\right)}^2+{\left(x+2\right)}^2+{\left(y-1\right)}^2=14$

$\Rightarrow x^2+y^2+x-3y-2=0$

Put $x=0$

$\Rightarrow y^2-3y-2=0$

$\Rightarrow \mathrm{y}=\frac{3\pm \sqrt{17}}{2}$

Put $y=0$

$\Rightarrow x^2+x-2=0$

$\left(x+2\right)\left(x-1\right)=0$

$\therefore A\left(-2,0\right),B\left(1,0\right),C\left(0,\frac{3+\sqrt{17}}{2}\right),D\left(0,\frac{3-\sqrt{17}}{2}\right)$

Area formed by $ACBD$ will be $=\frac{1}{2}\left|\begin{array}{cc}-2& 0\\ 0& \frac{3+\sqrt{17}}{2}\\ 1& 0\\ 0& \frac{3-\sqrt{17}}{2}\\ -2& 0\end{array}\right|$

$=\frac{1}{2}\left|-3-\sqrt{17}-\frac{3}{2}-\frac{\sqrt{17}}{2}+\frac{3}{2}-\frac{\sqrt{17}}{2}+3-\sqrt{17}\right|$

$=\frac{1}{2}\times 3\times \sqrt{17}=\frac{3\sqrt{17}}{2}$ sq units.