Question

In a triangle, $ABC$, the equation of the perpendicular bisector of $AC$ is $3x-2y+8=0$. If the coordinates of the points $A$ and $B$ are $(1,-1)$ and $(3,1)$ respectively, find the equation of the line $BC$.

Answer 1

Take $C$ to be $(h,k)$. Mid point
$(\frac{h+1}{2},\frac{k+1}{2})$ of $AC$ lies on the bisector
$3x-2y+8=0$ $\therefore$ $3h-2k+21=\mathrm{0.....}\left(1\right)$
$AC$ is $\perp$ to $3x-2y+8=0$
$\therefore m_1{m}_2=-1\Rightarrow 2h+3k+1=\mathrm{0...}\left(2\right)$
Solving (1) and (2) the point $C$ is $(-5,3)$
By two point formula, equation of $BC$ is
$x+4y-7=0$