Question

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

Answer 1

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

Equation of line $AC$ is $( y +1)=\frac{-2}{3}( x -1)$
$\Rightarrow 3 y+3=-2 x+2 $
$\Rightarrow 6 x+9 y=-3 \ldots(1)$
$ \Rightarrow 2 x+3 y=-1$
Equation of line perpendicular is $6 x-4 y=-16$ ......(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 $B C$ is
$(y-1)=\frac{3-1}{-5-3}(x-3) \Rightarrow-8(y-1)=2(x-3) \Rightarrow-4(y-1)=x-3$
$\Rightarrow -4 y+4=x-3 \Rightarrow x+4 y=7 \Rightarrow a=4$ and $b=7 \Rightarrow a+b=11$