Question

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

Answer 1

Let $(x_1,y_1)$ and $(x_2,y_2)$ be the coordinates of B and C.

Perpendicular bisector of AB is $x-y+5=0$

It slope = 1

Coordinates of $F=(\frac{x_1+1}{2},\frac{y_1-2}{2})$

F lies on the $x-y+5=0$

$\Rightarrow \frac{x_1+1}{2}-\frac{y_1-2}{2}+5=0$

$\Rightarrow x_1+1-y_1+2+10=0$

$x_1-y_1+13=0...\left(1\right)$

AB is perpendicular to HF

(Slope of AB) (Slope of HF) = - 1

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

$x_1+y_1+1=0...\left(2\right)$

Solving equation (1) and (2),

$x_1=-7,y_1=6$

Thus, B is $(-7,6)$

Now, perpendicular bisector of AC is $x+2y=0$

Slope of this is $=-\frac{1}{2}$

Mid-point of ACE $=(\frac{x_2+1}{2},\frac{y_2-2}{2})$

E lies on perpendicular bisector of AC

$\Rightarrow \left(\frac{x_2+1}{2}\right)+2\left(\frac{y_2-2}{2}\right)=0$

$x_2+1+2y_2-4=0$

$x_2+2y_2-3=0...\left(3\right)$

AC is perpendicular to HE

(Slope of AC) (Slope of HE) = - 1

$\left(\frac{y_2+2}{x_2-1}\right)(-\frac{1}{2})=-1$

$y_2+2=2x_2-2$

$2x_2-y_2=4...\left(4\right)$

Solving equation (3) and (4), we get

$x_2=\frac{11}{5},y_2=\frac{2}{5}$

Thus, point C is $(\frac{11}{5},\frac{2}{5})$

Equation of BC is

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

$y-6=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}(x+7)$

$y-6=\frac{-\frac{28}{5}}{\frac{46}{5}}(x+7)$

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

$23y-138=-14x-98$

$14x+23y-40=0$