The equations of the perpendicular bisectors of the sides AB and AC of $\triangle \mathrm{ABC}$ are $x-y+5=0$ and $x+2 y=0$ respectively. If the coordinates of $A$ are $(1,-2)$, then the equation of the line $B C$ is

• Correct answer
  A

  $14 x+23 y-40=0$
• B

  $13 x-9 y-14=0$
• C

  $9 x-14 y-25=0$
• D

  $8 x+15 y-30=0$

Solution

$\because A B$ line is perpendicular to $E O$ and passes through vertex $A$, equation of $A B$ is $x+y+1=0$
Co-ordinates of $E=\left(\frac{x_1+1}{2}, \frac{y_1-2}{2}\right)$ which is also intersection point of $A B$ with $\begin{aligned} & x-y+5=0 . \text { So, } \frac{x_1+1}{2}=-3 ; \frac{y_1-2}{2}=2 \\ & \Rightarrow\left(x_1, y_1\right)=(-7,6) \end{aligned}$
Similarly, equation of $A C$ is $y=2 x-4$ And co-ordinates of $F=\left(\frac{x_2+1}{2}, \frac{y_2-2}{2}\right)$ Which is also intersection point of $A C$ with $x+2 y=0$ So, $\frac{x_2+1}{2}=\frac{8}{5} ; \frac{y_2-2}{2}=\frac{4}{5} \Rightarrow\left(x_2, y_2\right)=\left(\frac{11}{5}, \frac{2}{5}\right)$ So, the equation of side $B C$ : $y-6=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}(x+7) \Rightarrow 14 x+23 y-40=0$

Hence, the correct answer is option: 'A'.