If the perpendicular bisector of the line segment joining the points P ( 1 , 4 ) and Q ( k , 3 ) has y…

Mid point of $\mathrm{PQ}$ is $\left(\frac{k+1}{2},\frac{7}{2}\right)$
And slope of $\mathrm{PQ}=\frac{1}{1-\mathrm{k}}$
So equation of perpendicular bisector of $\mathrm{PQ}$

$y-\frac{7}{2}=(k-1)\left(x-\frac{k+1}{2}\right)$
It's $y$ intercept $=-4$. So, this equation must satisfy $\left(0,-4\right)$.

$\Rightarrow -\frac{15}{2}=-\left(\frac{k^2-1}{2}\right)$
$k^2=16\Rightarrow k=\pm 4$