The equation of perpendicular bisectors of AB and AC of a triangle ABC are x-y+5 =0 and x+2y=0 respectively. If A=(1, -2), then the equation of $\overline{BC}is$

B (h  k) is Image of A (1,  -2) = (x₁,  y₁)

W.r.to x – y + 5 = 0

$\begin{array}{l}\frac{h-1}{1}=\frac{k+2}{-1}=\frac{-2\left(1+2+5\right)}{1+1}\\ \Rightarrow \frac{h-1}{1}=\frac{k+2}{-1}=-8\\ \Rightarrow h=-8+1\Rightarrow -7,k=8-2\Rightarrow 6\\ \therefore B\left(-7,6\right)\end{array}$

B (h  k) is Image of A (1,  -2) = (x₁,  y₁)

W.r.to x  + 2y  = 0

$\begin{array}{l}\frac{h-1}{1}=\frac{k+2}{2}=\frac{-2\left(1-4\right)}{1+4}\\ \frac{h-1}{1}=\frac{k+2}{2}=\frac{6}{5}\\ h=\frac{6}{5}+1\Rightarrow \frac{11}{5},k=\frac{12}{5}-2\Rightarrow \frac{2}{5}\\ \therefore C\left(\frac{11}{5},\frac{2}{3}\right)\\ slopeof\overline{BC}=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}=\frac{-28}{46}=\frac{-14}{23}\end{array}$

$\begin{array}{l}Equationof\overline{BC}y-6=\frac{-14}{23}\left(x+7\right)\\ \Rightarrow 23y-138=-14x-98\\ \Rightarrow 14x+23y-40=0\end{array}$