If the y-intercept of the perpendicular bisec | vedclass

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

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$-4$

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(C) The midpoint $M$ of $PQ$ is given by $M = \left(\frac{1+k}{2}, \frac{4+3}{2}ight) = \left(\frac{1+k}{2}, \frac{7}{2}ight)$.The slope of line $PQ$ is $m = \frac{3-4}{k-1} = \frac{-1}{k-1}$.The slope of the perpendicular bisector $m'$ is given by $m 	imes m' = -1$,so $m' = \frac{1}{m} = -(k-1) = 1-k$.The equation of the perpendicular bisector is $y - \frac{7}{2} = (1-k)(x - \frac{1+k}{2})$.For the $y$-intercept,we set $x = 0$ and $y = -4$:$-4 - \frac{7}{2} = (1-k)(0 - \frac{1+k}{2})$$-\frac{15}{2} = (1-k)(-\frac{1+k}{2})$$15 = (1-k)(1+k)$$15 = 1 - k^2$$k^2 = 1 - 15 = -14$.Wait,re-evaluating the slope calculation: $m = \frac{3-4}{k-1} = \frac{-1}{k-1}$. Thus $m' = k-1$.The equation is $y - \frac{7}{2} = (k-1)(x - \frac{1+k}{2})$.Setting $x=0, y=-4$:$-4 - \frac{7}{2} = (k-1)(-\frac{1+k}{2})$$-\frac{15}{2} = -\frac{k^2-1}{2}$$15 = k^2 - 1$ $\Rightarrow k^2 = 16$ $\Rightarrow k = \pm 4$.Thus,a possible value of $k$ is $-4$.

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

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