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

Consider the equation of perpendicular bisector is $y=m x+c$
The intercept is $1,$ thus the equation becomes $y=m x+1$The mid-point of the points $A(\alpha ,3)$ and $B(2,-1)$ is $\left(\frac{\alpha +2}{2},1\right)$
Since the equation passes through $\left(\frac{\alpha +2}{2},1\right)$
Thus,

$1=m\left(\frac{\alpha +2}{2}\right)+1$

$2=m(\alpha +2)+2$

$m(\alpha +2)=0$

$m\alpha +2m=0 \dots \left(\mathrm{I}\right)$

Here, $m$ is the gradient of the perpendicular line.
$m=-\left(\frac{\alpha -2}{4}\right)$

Substitute the value in equation $\left(\mathrm{I}\right)$.  

$\alpha \left(\frac{2-\alpha }{4}\right)+2\left(\frac{2-\alpha }{4}\right)=0$

                                      $\alpha =\pm 2$