Question

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

Answer 1

Equation of perpendicular bisector of $AB$ is $x-y+5=0$ ----- $\left(1\right)$

Also equation of $AB$ can be written as

$y+x+\lambda =0$ (perpendicular to $x-y+5=0$)

Since, passes through $A=(1,-2)$

Therefore,

$-2+1+\lambda =0$

$\lambda =1$

And,

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

From $\left(1\right)$ and $\left(2\right)$

The coordinates of $D$, which is middle point of $AB$ is given by

$D=(3,2)$

Now,

Let the coordinates of $B$ be $(x_1,y_1)$

Then,

$\frac{x_1+1}{2}=-3$ and $\frac{y_1-2}{2}=2$ (since, $D$ is mid-point of $AB$.)

$x_1=-7$ and $y_1=6$

Hence, point of $B=(-7,6)$ --- $\left(3\right)$

Now equation $AC$

$2x-y-4=0$ ---- $\left(4\right)$

Solve this equation bey the given equation

$x+2y=0$

We get, the point $E$ which is the mid-point of $AC$

hence,

$E=(\frac{8}{5},\frac{-4}{5})$

Let the coordinates of $C$ be $(l,m)$

Then,

$\frac{l+1}{2}=\frac{8}{5}$ and $\frac{q-2}{2}=\frac{4}{5}$

Hence,

$C(l,m)=(\frac{11}{5},\frac{2}{5})$

Therefore equation of line $BC$ which is passing through $B$ and $C$ is given by,

$y-m=\frac{y_1-m}{x_1-l}(x-l)$

$14x+23y=40$

Which is required equation of line.