Let \(\mathrm{PQR}\) be a right angled isosceles triangle, right angled at \(\mathrm{P}(2,1)\). If the equation of the line \(\mathrm{QR}\) is \(2 x+y=3\), then the equation representing the pair of lines \(P Q\) and \(P R\) is

\(\text { Slope of } Q R=-2 . \)
\( \text { Slope of } P Q=\mathrm{m}_1 \)
\( \therefore \tan 45^{\circ}=\left|\frac{\mathrm{m}_1+2}{1+\mathrm{m}_1(-2)}\right| \)
\( \Rightarrow 1=\left|\frac{\mathrm{m}_1+2}{1-2 \mathrm{~m}_1}\right| \)
\( \Rightarrow \mathrm{m}_1=-\frac{1}{3}\)

\(\therefore \) Equation of \(\mathrm{PQ}\) passing through point \(\mathrm{P}(2,1)\) and having slope \(\frac{-1}{3}\) is
\(
\begin{aligned}
& y-1=-\frac{1}{3}(x-2) \\
& \Rightarrow 3(y-1)+(x-2)=0
\end{aligned}
\)
Slope of \(\mathrm{PR}=\mathrm{m}_2=3\)
\(\therefore \) equation of \(\mathrm{PR}\) is
\(
\begin{aligned}
& y-1=3(x-2) \\
& \Rightarrow(y-1)-3(x-2)=0
\end{aligned}
\)
\(\therefore \) The joint equation of the lines is
\(
\begin{aligned}
& {[3(y-1)+(x-2)][(y-1)-3(x-2)]=0} \\
& \Rightarrow 3(y-1)^2-8(y-1)(x-2)-3(x-2)^2=0 \\
& \Rightarrow 3\left(x^2-4 x+4\right)+8(x y-x-2 y+2) \\
& -3\left(y^2-2 y+1\right)=0 \\
& \Rightarrow 3 x^2-3 y^2+8 x y-20 x-10 y+25=0
\end{aligned}
\)