A point P moves so that the sum of squares of its distances from...

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

$⇒ {x}^{2} + {y}^{2} + x - 3y - 2 = 0$

Put $x = 0$

$⇒ {y}^{2} - 3y - 2 = 0$

$⇒ y = \dfrac{3 \pm \sqrt{17}}{2}$

Put $y = 0$

$⇒ {x}^{2} + x - 2 = 0$

\left(x + 2\right)\left(x - 1\right) = 0$</p><p>$∴ A\left(- 2, 0\right), B\left(1, 0\right), C\left(0, \dfrac{3 + \sqrt{17}}{2}\right), D\left(0, \dfrac{3 - \sqrt{17}}{2}\right)$</p><p>Area formed by$ACBD$will be$= \dfrac{1}{2}\left|$\begin{cases}- 2 & 0 \\ 0 & \dfrac{3 + \sqrt{17}}{2} \\ 1 & 0 \\ 0 & \dfrac{3 - \sqrt{17}}{2} \\ - 2 & 0\end{cases}$\right|$</p><p>$= \dfrac{1}{2}\left|- 3 - \sqrt{17} - \dfrac{3}{2} - \dfrac{\sqrt{17}}{2} + \dfrac{3}{2} - \dfrac{\sqrt{17}}{2} + 3 - \sqrt{17}\right|$</p><p>$= \dfrac{1}{2} \times 3 \times \sqrt{17} = \dfrac{3\sqrt{17}}{2} sq units.