The equal sides of an isosceles triangle are given by equations $7 x-y+3=0$ and $x+y-3=0$. If the slope $m$ of the third side is an integer, then $\mathrm{m}=$

Let $\triangle \mathrm{ABC}$ be the isosceles triangle where $\mathrm{AB}=\mathrm{AC}$ Hence Bisector of Angle $\angle B A C$ will also be the perpendicular on line $\mathrm{BC}$.
Let $m_1$ and $m$ be the slopes of bisector lines AP and line $B C$ respectively. Equation of Bisector lines (AP) are $$ \begin{aligned} & \frac{7 x-y+3}{\sqrt{49+1}}= \pm \frac{(x+y-3)}{\sqrt{(1)+(1)}} \\ & \Rightarrow x-3 y+6=0 \text { and } 3 x+y-12=0 \end{aligned} $$ Here $m_1=\frac{1}{3}$ Since AP $\perp$ BC $$ \therefore m_1 \cdot m=-1 $$ $$ m=-3 $$ Here $m_1=-3$ Since AP $\perp$ BC $$ \therefore m_1 \cdot m=-1 $$ $$ \Rightarrow m=\left(\frac{1}{3}\right) $$ Since $m$ is integer (given) Hence $m=-3$