The perpendicular bisector of the line segmen | vedclass

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(C) The slope of $PQ$ is $m_{PQ} = \frac{3-4}{k-1} = \frac{-1}{k-1}$.The slope of the perpendicular bisector is $m = -(k-1) = 1-k$.The midpoint of $PQ

Vedclass · Accepted Answer

$-4$

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(C) The slope of $PQ$ is $m_{PQ} = \frac{3-4}{k-1} = \frac{-1}{k-1}$.The slope of the perpendicular bisector is $m = -(k-1) = 1-k$.The midpoint of $PQ$ is $M = \left(\frac{k+1}{2}, \frac{7}{2}\right)$.The equation of the perpendicular bisector is $y - \frac{7}{2} = (1-k)(x - \frac{k+1}{2})$.To find the $y-$intercept,set $x = 0$:$y - \frac{7}{2} = (1-k)(0 - \frac{k+1}{2})$$y = \frac{7}{2} - \frac{(1-k)(k+1)}{2}$$y = \frac{7 - (1-k^2)}{2} = \frac{7 - 1 + k^2}{2} = \frac{6 + k^2}{2}$.Given the $y-$intercept is $-4$:$\frac{6 + k^2}{2} = -4$$6 + k^2 = -8$$k^2 = -14$ (This implies no real solution for $k$ based on the provided options).Re-evaluating the slope calculation:Slope of $PQ = \frac{3-4}{k-1} = \frac{-1}{k-1}$.Slope of perpendicular bisector $= k-1$.Equation: $y - \frac{7}{2} = (k-1)(x - \frac{k+1}{2})$.At $x=0$,$y = \frac{7}{2} - (k-1)(\frac{k+1}{2}) = \frac{7 - (k^2-1)}{2} = \frac{8-k^2}{2}$.Given $y-$intercept is $-4$:$\frac{8-k^2}{2} = -4$ $\Rightarrow 8-k^2 = -8$ $\Rightarrow k^2 = 16$ $\Rightarrow k = \pm 4$.Thus,a possible value of $k$ is $-4$.

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

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