If the equations of the perpendicular bisectors of the sides AB and AC of a $\Delta ABCarex-y+5=0andx+2y=0$respectively and if A is (1, -2) then the equation of the perpendicular bisector of the side BC is

detailed solution

Correct option is D

$\begin{array}{l}Ifx-y+5=0is\perp rbi\sec torofABthen\\ Bis\mathrm{Im}ageofA(1,-2)=(x_1,y_1)\\ \frac{h-1}{1}=\frac{k+2}{-1}=\frac{-2(1+2+5)}{1+1}\\ \frac{h-1}{1}=\frac{k+2}{-1}=-8\\ B\left(hk\right)=(-7,6)\\ Ifx+2y=0is\perp rbi\sec torofACthen\\ cis\mathrm{Im}ageofA(1,-2)\\ \frac{h-1}{1}=\frac{k+2}{2}=\frac{-2(1-4)}{1+4}\\ \frac{h-1}{1}=\frac{k+2}{2}=\frac{6}{5}\end{array}$

$\begin{array}{l}h=\frac{6}{5}+1\Rightarrow \frac{11}{5},k=\frac{12}{5}-2,\Rightarrow \frac{2}{5}\\ \therefore C\left(hk\right)=(\frac{11}{5},\frac{2}{5})\\ slopeof\overline{BC}=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}=\frac{-28}{46}=\frac{-14}{23}\\ Midpo\mathrm{int}ofBC=(\frac{\frac{11}{5}-7}{2},\frac{\frac{2}{5}+6}{2})\Rightarrow (\frac{-12}{5},\frac{16}{5})\\ Equationof\perp rbi\sec torof\overline{BC}is\\ y-\frac{16}{5}=\frac{23}{14}(x+\frac{12}{5})\\ \Rightarrow 5y-16=\frac{23}{14}(5x+12)\\ \Rightarrow 70y-224=115x+276\\ \Rightarrow 115x-70y+500=0\end{array}$