Two equal sides of an isosceles triangle are 7x – y + 3 = 0 and x + y – 3 = 0 and its third side passes through the point (1, -10) determine the equation of the third side is

Given lines 7x – y + 3 =0 , Slope = 7

X + y – 3 = 0, Slope = -1

Let m be the slope of third side

$\begin{array}{l}\therefore \left|\frac{m-7}{1+7m}\right|=\left|\frac{m+1}{1+m\left(-1\right)}\right|\\ \Rightarrow \left|\left(m-7\right)\left(1-m\right)\right|=\left|\left(m+1\right)\left(1+7m\right)\right|\\ \Rightarrow \left|m-m^2-7+7m\right|=\left|m+7m^2+1+7m\right|\\ \Rightarrow \left|8m-m^2-7\right|=\left|7m^2+8m+1\right|\\ If8m-m^2-7=7m^2+8m+1\\ \Rightarrow 8m^2=-8\Rightarrow m^2\ne -1\\ If8m-m^2-7=-\left(7m^2+8m+1\right)\end{array}$

$\begin{array}{l}\Rightarrow 8m-m^2-7=-7m^2-8m-1\\ \Rightarrow 6m^2+16m-6=0\Rightarrow 3m^2+8m-3=0\\ \Rightarrow 3m^2+8m-3=0\Rightarrow 3m^2+9m-m-3=0\\ \Rightarrow 3m\left(m+3\right)-1\left(m+3\right)=0\\ \Rightarrow \left(m+3\right)\left(3m-1\right)=0\Rightarrow m=-3,m=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ Ifm=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.andP\left(1,-10\right)\end{array}$

Equation of third sides

 $\begin{array}{l}y+10=\frac{1}{3}\left(x-1\right)\\ \Rightarrow 3y+30=x-1\\ \Rightarrow x-3y-31=0\end{array}$