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

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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.

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