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(C) Let $LM$ be the perpendicular bisector of $PQ$ at $R$.The midpoint $R$ of $PQ$ is given by:$R = \left(\frac{1+k}{2}, \frac{4+3}{2}\right) = \left(

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$-4$

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(C) Let $LM$ be the perpendicular bisector of $PQ$ at $R$.The midpoint $R$ of $PQ$ is given by:$R = \left(\frac{1+k}{2}, \frac{4+3}{2}\right) = \left(\frac{k+1}{2}, \frac{7}{2}\right)$The slope of $PQ$ is:$m_{PQ} = \frac{3-4}{k-1} = \frac{-1}{k-1}$Since $LM$ is perpendicular to $PQ$,the slope of $LM$ $(m_{LM})$ is:$m_{LM} = -\frac{1}{m_{PQ}} = -\frac{1}{-1/(k-1)} = k-1$The equation of the line $LM$ with slope $(k-1)$ and $y$-intercept $-4$ is:$y = (k-1)x - 4$Since the perpendicular bisector passes through the midpoint $R\left(\frac{k+1}{2}, \frac{7}{2}\right)$,it must satisfy the equation:$\frac{7}{2} = (k-1)\left(\frac{k+1}{2}\right) - 4$Multiply by $2$ on both sides:$7 = (k-1)(k+1) - 8$$7 = k^2 - 1 - 8$$7 = k^2 - 9$$k^2 = 16$$k = \pm 4$Thus,a possible value of $k$ is $-4$.

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

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