Question

In a $\Delta ABC,2x+3y+1=0,x+2y-2=0$ are the perpendicular bisectors of its sides $AB$ and $AC$ respectively and if $A=(3,2)$, then the equation of the side $BC$ is

• A. x + y - 3 = 0
• B. x - y - 3 = 0
• C. 2x - y - 2 = 0
• D. 2x + y - 2 = 0

Answer 1

Answer: (B)

Given that, equation of $AB$ is $3x-2y+c=0$
passes through $A(3,2)$
$\Rightarrow C=-5$

$\therefore$ Equation becomes $3x-2y-5=0$
Here, $D=(1,-1)$
Since, $D$ is mid-point of $AB$.
Let coordinates of $B$ are $(\alpha ,\beta )$.
$\therefore \frac{3+\alpha }{2}=1$ and $\frac{2+\beta }{2}=-1$
$\to 3+\alpha =2$ and $2+\beta =-2$
$\Rightarrow \alpha =-1$ and $\beta =-4$
$\therefore B$ is $(-1,-4)$.
Similarly, equation of $AC$ is $2x-y+c=0$ is passes through $(3,2)$
$\Rightarrow C=-4\Rightarrow 2x-y-4=0$
Here, $E$ is a point of intersection of $x+2y-2=0$
and $2x-y-4=0$.
$\therefore E=(2,0)$
Since, $E$ is mid-point of $AC$.
Let coordinates of $C$ are $\left({\alpha }_1,{\beta }_1\right)$
$\therefore \frac{3+{\alpha }_1}{2}=2$ and $\frac{2+{\beta }_1}{2}=0$
$\Rightarrow {\alpha }_1=4-3$ and ${\beta }_1=-2$
$\Rightarrow {\alpha }_1=1$ and ${\beta }_1=-2$
$\therefore C$ is $(1,-2)$
$\therefore$ Required equation of $BC$ is
$(y+2)=\frac{-2+4}{1+1}(x-1)$
$\Rightarrow y+2=\frac{2}{2}(x-1)$
$\Rightarrow y+2=(x-1)$
$\Rightarrow x-y-3=0$