We know that the area of the triangle formed by points $A\left(x_1, y_1\right), B\left(x_2, y_2\right) \mathrm{and} C\left(x_3, y_3\right)$

is, $\mathrm{Ar}\left(∆ABC\right)=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ x_1& x_2& x_3\\ y_1& y_2& y_3\end{array}\right|$

Here, we have points $A\left(1, 0\right)$, $B\left(0, 2\right)$ , $C\left(1, 2\right)$ and $P\left(x, y\right)$

Given, 

${\left|2\times \mathrm{Area}\left(∆\mathrm{ABC}\right)\right|}^2={\left|\mathrm{Area}\left(∆\mathrm{PAB}\right)\right|}^2$

${\left(2\times \frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 1& 0& 1\\ 0& 2& 2\end{array}\right|\right)}^2={\left(\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 1& 0& x\\ 0& 2& y\end{array}\right|\right)}^2\Rightarrow 4\times {\left|\begin{array}{ccc}1& 1& 1\\ 1& 0& 1\\ 0& 2& 2\end{array}\right|}^2={\left|\begin{array}{ccc}1& 1& 1\\ 1& 0& x\\ 0& 2& y\end{array}\right|}^2$

$\Rightarrow 4{\left[1\left(0-2\right)-1\left(2-0\right)+1\left(2-0\right)\right]}^2={\left[1\left(0-2x\right)-1\left(y-0\right)+1\left(2\right)\right]}^2$

$\Rightarrow 16={\left(-2x-y+2\right)}^2\Rightarrow 4x^2+y^2+4xy-4y-8x-12=0$