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

Solution :
Let $$m$$ be the slope of $$PQ$$  then
$$\eqalign{ & \tan \,{45^ \circ } = \left| {\frac{{m - \left( { - 2} \right)}}{{1 + m\left( { - 2} \right)}}} \right| \cr & \Rightarrow 1 = \left| {\frac{{m + 2}}{{1 - 2m}}} \right| \cr & \Rightarrow \pm 1 = \frac{{m + 2}}{{1 - 2m}} \cr & \Rightarrow m + 2 = 1 - 2m\,\,\,{\text{or}}\,\, - 1 + 2m = m + 2 \cr & \Rightarrow m = - \frac{1}{3}\,\,\,\,\,\,{\text{or}}\,\,\,\,\,m = 3 \cr} $$

As $$PR$$  also makes $$\angle {45^ \circ }$$   with $$RQ.$$
$$\therefore $$ The above two values of $$m$$ are for $$PQ$$  and $$PR.$$
$$\therefore $$ Equation of $$PQ,$$
$$\eqalign{ & y - 1 = - \frac{1}{3}\left( {x - 2} \right) \cr & \Rightarrow 3y - 3 = - x + 2 \cr & \Rightarrow x + 3y - 5 = 0 \cr} $$
and equation of $$PR$$  is $$ \Rightarrow 3x - y - 5 = 0$$
$$\therefore $$ Combined equation of $$PQ$$  and $$PR$$  is
$$\eqalign{ & \left( {x - 3y - 5} \right)\left( {3x - y - 5} \right) = 0 \cr & \Rightarrow 3{x^2} - 3{y^2} + 8xy - 20x - 10y + 25 = 0 \cr} $$