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(D) The midpoint $M$ of the line segment $PQ$ is given by $M = \left(\frac{1+k}{2}, \frac{4+3}{2}\right) = \left(\frac{k+1}{2}, \frac{7}{2}\right)$.Th

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$-4$

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(D) The midpoint $M$ of the line segment $PQ$ is given by $M = \left(\frac{1+k}{2}, \frac{4+3}{2}\right) = \left(\frac{k+1}{2}, \frac{7}{2}\right)$.The slope of the line segment $PQ$ is $m_{PQ} = \frac{3-4}{k-1} = \frac{-1}{k-1}$.The slope of the perpendicular bisector is $m = -\frac{1}{m_{PQ}} = k-1$.The equation of the perpendicular bisector passing through $M$ with slope $m$ is $y - \frac{7}{2} = (k-1)\left(x - \frac{k+1}{2}\right)$.The $y$-intercept is the value of $y$ when $x=0$. Given the $y$-intercept is $-4$,we substitute $x=0$ and $y=-4$ into the equation:$-4 - \frac{7}{2} = (k-1)\left(0 - \frac{k+1}{2}\right)$$-\frac{15}{2} = (k-1)\left(-\frac{k+1}{2}\right)$$\frac{15}{2} = \frac{(k-1)(k+1)}{2}$$15 = k^2 - 1$$k^2 = 16$$k = \pm 4$.Since $-4$ is an option,the correct value is $-4$.

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

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