• # question_answer The equation of perpendicular bisectors of sides AB and AC of a$\Delta ABC$are$x-y+5=0$and$x+2y=0,$respectively. If the coordinates of vertex A are$(1,-2),$the equation of BC is A)  $14x+23y-40=0$ B)  $14x-23y+40=0$ C)  $23x+14y-40=0$ D)  $23x-14y+40=0$

Let$B({{x}_{1}},{{y}_{1}})$and$C({{x}_{2}},{{y}_{2}})$be two vertices and$P\left( \frac{{{x}_{1}}+1}{2},\frac{{{y}_{1}}-2}{2} \right)$lies on perpendicular bisector$x-y+5=0$. $\therefore$ $\frac{{{x}_{1}}+1}{2}-\frac{{{y}_{1}}-2}{2}=-5$ $\Rightarrow$ ${{x}_{1}}-{{y}_{1}}=-13$         ...(i) Also, PAT is perpendicular to AB. $\therefore$ $\frac{{{y}_{1}}+2}{{{x}_{1}}-1}\times 1=-1$ $\Rightarrow$ ${{y}_{1}}+2=-{{x}_{1}}+1$ $\Rightarrow$ ${{x}_{1}}+{{y}_{1}}=-1$ On solving Eqs. (i) and (ii), we get ${{x}_{1}}=-7,{{y}_{1}}=6$ $\therefore$The coordinates of B are$(-7,\text{ }6)$. Similarly, the coordinates of C are$\left( \frac{11}{5},\frac{2}{5} \right)$ Hence, the equation of BC is $y-6=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}\times (x+7)$ $\Rightarrow$ $(y-6)=\frac{-14}{23}\times (x+7)$ $\Rightarrow$ $14x+23y-40=0$