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(C) Let the points be $A(\alpha, 3)$ and $B(2, -1)$. The midpoint $M$ of $AB$ is $\left(\frac{\alpha+2}{2}, \frac{3-1}{2}\right) = \left(\frac{\alpha+

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$\pm 2$

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(C) Let the points be $A(\alpha, 3)$ and $B(2, -1)$. The midpoint $M$ of $AB$ is $\left(\frac{\alpha+2}{2}, \frac{3-1}{2}ight) = \left(\frac{\alpha+2}{2}, 1ight)$.Since the perpendicular bisector passes through $M$ and has a $y$-intercept of $1$,the point $M$ must lie on the $y$-axis or the line must be horizontal. However,the $y$-intercept is $1$,which is the same as the $y$-coordinate of $M$. This implies the $x$-coordinate of $M$ must be $0$.$\frac{\alpha+2}{2} = 0 \implies \alpha = -2$.Alternatively,the slope of $AB$ is $m_{AB} = \frac{-1-3}{2-\alpha} = \frac{-4}{2-\alpha} = \frac{4}{\alpha-2}$.The slope of the perpendicular bisector is $m = -\frac{1}{m_{AB}} = \frac{2-\alpha}{4}$.The equation of the line is $y - 1 = m(x - \frac{\alpha+2}{2})$.Given the $y$-intercept is $1$,we set $x=0$ and $y=1$:$1 - 1 = m(0 - \frac{\alpha+2}{2}) \implies 0 = m(-\frac{\alpha+2}{2})$.Since $m 
eq 0$ (unless $\alpha=2$,which makes $AB$ vertical),we must have $\alpha+2 = 0$,so $\alpha = -2$.Wait,checking the case where the line is horizontal: if $m=0$,then $\alpha=2$. If $\alpha=2$,$A=(2,3)$ and $B=(2,-1)$. The perpendicular bisector is $y=1$,which has $y$-intercept $1$. Thus $\alpha=2$ is also a solution.Therefore,$\alpha = \pm 2$.

List-$I$	List-$II$
$A$. The equation of line passing through $(4,3)$ whose $X$-intercept is twice its $Y$-intercept	$I$. $x+y-2\sqrt{2}=0$
$B$. The equation of the line passing through the centroid and circumcentre of $\triangle ABC$ with vertices $A(1,1), B(3,3), C(6,-6)$	$II$. $7x+23y-8=0$
$C$. The equation of the line whose $X$-intercept is $(-3/5)$ and is perpendicular to $x-y+2=0$	$III$. $x+2y+\sqrt{2}=0$
$D$. The equation of the line whose distance from the origin is $2$ and the normal drawn from the origin makes an angle $45^{\circ}$ with the positive direction of $X$-axis	$IV$. $x+2y-10=0$
	$V$. $5x+5y+3=0$

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

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