Question

Let $Q$ and $R$ be the images of a point $P(–3,1)$ in the mirror $x^2-y^2-x+y=0$. If $S$ is the area of triangle $PQR,$ then the value of$\left[\sqrt{S}\right]$ is
$($ $\left[k\right]$ denotes the greatest integer less than or equal to $k.)$

Answer 1

$x^2-y^2-(x-y)=0$
$(x-y)(x+y-1)=0$
$d_1=\left|\frac{-3-1}{\sqrt{2}}\right|=2\sqrt{2}$
$d_2=\left|\frac{-3+1-1}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}$

$\therefore$ Area of $\Delta PQR,$
$S=\frac{1}{2}\times 4\sqrt{2}\times 3\sqrt{2}$
$\Rightarrow S=12$
$\therefore \left[\sqrt{S}\right]=3$


Alternative :
$P\equiv (-3,1);Q\equiv (1,-3);R\equiv (0,4)$
$S=\frac{1}{2}\left|\begin{array}{ccc}-3& 1& 1\\ 1& -3& 1\\ 0& 4& 1\end{array}\right|=12$