Question

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

Answer 1

Answer: 11

Slope of line $AC=\frac{-1}{\frac{-3}{-2}}=\frac{-2}{3}$

Equation of line $AC$ is $(y+1)=\frac{-2}{3}(x-1)$
$\Rightarrow 3y+3=-2x+2$
$\Rightarrow 6x+9y=-3\dots (1)$
$\Rightarrow 2x+3y=-1$
Equation of line perpendicular is $6x-4y=-16$ ......(2)
$\therefore$ On solving (1) and (2), we get $M(-2,1)$
For $C\left(x_1,y_1\right)$, we have
$\frac{x_1+1}{2}=-2\Rightarrow x_1=-4-1=-5$
$\frac{y_1-1}{2}=1\Rightarrow y_1=2+1=3$
$\therefore C(-5,3)\text{ and }B(3,1)$
$\therefore C(-5,3)$ and $B(3,1)$
Equation of $BC$ is
$(y-1)=\frac{3-1}{-5-3}(x-3)\Rightarrow -8(y-1)=2(x-3)\Rightarrow -4(y-1)=x-3$
$\Rightarrow -4y+4=x-3\Rightarrow x+4y=7\Rightarrow a=4$ and $b=7\Rightarrow a+b=11$