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# The perpendicular bisector of the line segment joining $P(1,4)$ and $Q(K,3)$ has y-intercept $-4$, Then a possible value of $k$ is

$\begin{array}{1 1}(A)\;1\\(B)\;2\\(C)\;-2 \\(D)\;-4 \end{array}$

Slope of $PQ=\large\frac{3-4}{k-1}=\frac{-1}{k-1}$
slope of perpendicular bisector of $PQ=(k-1)$
Also mid point of $PQ\bigg( \large\frac{k+1}{2},\frac{7}{2}\bigg)$
equation of perpendicular bisector is $y-\large\frac{7}{2}$$=(k-1) \bigg( x- \large\frac{k+1}{2} \bigg) => 2x-7=2(k-1)-(k^2-1) => 2 (k-1)x-2y+(8-k^2)=0 y- intercept =\large\frac{-8-k^2}{-2}$$=-4$
=> $8-k^2=-8$
$k^2=16$
$k=\pm4$
Hence D is the correct answer.