The equations of the perpendicular bisectors of the sides AB and AC of $ riangle \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

Question

The equations of the perpendicular bisectors of the sides AB and AC of $	riangle \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

Aicharya · Accepted Answer

$\because A B$ 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 $B C$ : $y-6=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}(x+7) \Rightarrow 14 x+23 y-40=0$

The equations of the perpendicular bisectors of the sides AB and AC of $\triangle \mathrm{ABC}$ are…

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