Question

The equation of the perpendicular bisectors of the sides AB and AC of triangle ABC an $x-y+5=0$ and $x+2y=0$, respectively. If the point A is (1, -2) find the equation of the line BC.

Answer 1

We have the equation of perpendicular bisector of $AB$ is

$x-y+5=0$ $\Rightarrow y=x+5......\left(1\right)$

On comparing that with

$y=mx+c$ Then, $m=1$
Since product of slopes of two perpendicular lines is $-1$, slope of line $AB$ is $-1$

The equation of $AB$ passing through the point $(1,-2)$.

Then, equation of line

$y-y_1=m(x-x_1)$

$\Rightarrow y+2=-1(x-1)$

$\Rightarrow y+2=-x+1$

$\Rightarrow x+y+1=0......\left(2\right)$

Given equation

$x+2y=0......\left(3\right)$

According to question from equation (2)

$x+y+1=0$

$\Rightarrow y=-(x+1)$

Substituting $y$ in (3) ,

$x+2y=0$

$\Rightarrow x-2(x+1)=0$

$\Rightarrow x-2x-2=0$

$\Rightarrow -x-2=0$

$\Rightarrow x=-2$

Put in equation (3)

$y=1$

Then,$(x_1,y_1)=(-2,1)$

From equation (1) and (3) ,

Point of intersection

$(x_2,y_2)=(\frac{-10}{3},\frac{5}{3})$

Then equation of

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

$\Rightarrow y+2=\frac{\frac{5}{3}-1}{\frac{-10}{3}+2}(x+2)$

$\Rightarrow y+2=\frac{\frac{2}{3}}{\frac{-7}{3}}(x+2)$

$\Rightarrow y+2=\frac{-2}{7}(x+2)$

$\Rightarrow 7y+14=-2x-4$

$\Rightarrow 2x+7y=-4-14$

$\Rightarrow 2x+7y=-18$
Line equation of $BC$ is $2x+7y=-18$