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)

x - y - 3 = 0

Answer 2

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

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