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(D) The slope of $PQ$ is $m_{PQ} = \frac{3-4}{k-1} = \frac{-1}{k-1}$.The slope of the perpendicular bisector is $m_{\perp} = -(k-1) = 1-k$.The midpoin

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$-4$

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(D) The slope of $PQ$ is $m_{PQ} = \frac{3-4}{k-1} = \frac{-1}{k-1}$.The slope of the perpendicular bisector is $m_{\perp} = -(k-1) = 1-k$.The midpoint of $PQ$ is $M = \left(\frac{k+1}{2}, \frac{7}{2}\right)$.The equation of the perpendicular bisector is $y - \frac{7}{2} = (1-k)\left(x - \frac{k+1}{2}\right)$.Setting $x=0$ for the $y$-intercept,we get $y - \frac{7}{2} = (1-k)\left(-\frac{k+1}{2}\right)$.Given the $y$-intercept is $-4$,we have $-4 - \frac{7}{2} = -\frac{(1-k)(k+1)}{2}$.$-\frac{15}{2} = -\frac{1-k^2}{2} \implies 15 = 1-k^2$.$k^2 = 1 - 15 = -14$. Since $k$ must be real,let us re-evaluate the slope calculation.Slope of perpendicular bisector is $m_{\perp} = k-1$. Midpoint is $(\frac{k+1}{2}, 3.5)$.Equation: $y - 3.5 = (k-1)(x - \frac{k+1}{2})$.At $x=0$,$y = -4$: $-4 - 3.5 = (k-1)(-\frac{k+1}{2})$.$-7.5 = -\frac{k^2-1}{2} \implies 15 = k^2-1 \implies k^2 = 16 \implies k = \pm 4$.Given the options,$k = -4$ is the correct choice.

If the perpendicular bisector of the line segment joining $P(1, 4)$ and $Q(k, 3)$ has a $y$-intercept of $-4$,then the possible value of $k$ is:

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