The equation of perpendicular bisectors of sides A B and A C of a ∆   A B C are x - y + 5 = 0 and…

Let $B(x_1, x_1)$ and $C(x_2, y_2)$ be two vertices and

$P\left(\frac{x_1+1}{2},\frac{y_1-2}{2}\right)$ being mid point of AB lies on its perpendicular bisector PN having equation $x-y+5=0$

$\therefore \frac{x_1+1}{2}-\left(\frac{y_1-2}{2}\right)=-5$

$\Rightarrow x_1-y_1=-13$ ...(i)




Also, $PN$ is perpendicular to $AB$.

$\therefore \frac{y_1+2}{x_1-1}\times 1=-1$ (Slopes of AB and PN satistfy the condition for slopes of perpendicular lines , Product of slopes= -1 )

$\Rightarrow x_1+y_1=-1$ ...(ii)

On solving Eqs. (i) and (ii), we get

$x_1=-7, y_1=6$

$\therefore$ The coordinates of $B$ are $\left(-\mathrm{7,6}\right)$ Similarly, the coordinates of $C$ are $\left(\frac{11}{5},\frac{2}{5}\right)$

Hence, the equation of $BC$ is

$$\begin{gathered} y-6=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}\left(x+7\right) \\ \Rightarrow y-6=\frac{-28}{46}\left(x+7\right) \end{gathered}$$

$\Rightarrow y-6=\frac{-14}{23}\left(x+7\right)$

$\Rightarrow 14x+23y-40=0$