Question

If the perpendicular bisector of the line segment joining $A(\alpha, 3)$ and $B (2, -1)$ has $y$-intercept $1$, then $\alpha$ =

• A. 0
• B. $\pm$ 1
• C. $\pm$ 2
• D. $\pm$ 3

Answer 1

Answer: (C)

$\pm$ 2

Answer 2

Let the equation of perpendicular bisector is
$y=m x+c$
Here, $c=1$
$\therefore y=m x+1 \dots$(i)
Mid-point of points $A(\alpha, 3)$ and $B(2,-1)$ is
$\left(\frac{\alpha+2}{2}, 1\right)$
Since, E (i) passes through $\left(\frac{\alpha+2}{2}, 1\right)$.
So, $1=m\left(\frac{\alpha+2}{2}\right)+1$
$\Rightarrow 1=\frac{m(\alpha+2)}{2}+1$
$\Rightarrow 2=m(\alpha+2)+2$
$\Rightarrow m(\alpha+2)=0$
$\Rightarrow m \alpha+2 m=0 \dots$(ii)
$m=$ gradient of the perpendicular line
$m=-($ gradient of A B)=-\left\\{\frac{\alpha-2}{4}\right\\}=\left(\frac{2-\alpha}{4}\right)
Put value of $m$ in E (ii), we get
$\alpha\left(\frac{2-\alpha}{4}\right)+2\left(\frac{2-\alpha}{4}\right) =0$
$\frac{2 \alpha-\alpha^{2}}{4}+\frac{4-2 \alpha}{4} =0$
$\Rightarrow -\alpha^{2}+4=0$
$\Rightarrow \alpha^{2}=4$
$\Rightarrow \alpha=\pm 2$