If the $y$-intercept of the perpendicular bisector of the line segment joining $\mathrm{P}(1,4)$ and…
If the $y$-intercept of the perpendicular bisector of the line segment joining $\mathrm{P}(1,4)$ and $\mathrm{Q}(\mathrm{k}, 3)$ is -4 , then a possible value of $k$ is
$2$
$-2$
$-4$
$1$
Solution
Mid point of $P Q$ is $\Rightarrow M=\left(\frac{1+k}{2} ; \frac{7}{2}\right)$ Slope of line $P Q \Rightarrow m=\frac{4-3}{1-k}=\frac{1}{1-k}$ Slope of perpendicular bisector is $m m^{\prime}=-1 \Rightarrow m^{\prime}=(k-1)$
The equation of perpendicular bisector is
$\begin{aligned}
& \left(y-\frac{7}{2}\right)=m^{\prime}\left(x-\frac{1+k}{2}\right) \\
& y=(k-1)\left(x-\frac{1+k}{2}\right)+\frac{7}{2}
\end{aligned}$
Now, for $y$-intercept of perpendicular bisector:
$\begin{aligned}
& x=0 \\
& \therefore y=(k-1)\left(0-\frac{1+k}{2}\right)+\frac{7}{2} \\
& \Rightarrow y=-\frac{(k-1)(1+k)}{2}+\frac{7}{2} \\
& \Rightarrow-4=\frac{-\left(k^2-1\right)}{2}+\frac{7}{2} \\
& \Rightarrow k^2=16 \Rightarrow k=4,-4 \\
& \therefore k=-4
\end{aligned}$