Question:
If the perpendicular bisector of the line segment joining the points $P(1,4)$ and $Q(k, 3)$ has $y$-intercept equal to $-4$, then a value of $k$ is :
Correct Option: , 2
Solution:
Mid point of line segment $P Q$ be $\left(\frac{k+1}{2}, \frac{7}{2}\right)$.
$\therefore$ Slope of perpendicular line passing through
$(0,-4)$ and $\left(\frac{k+1}{2}, \frac{7}{2}\right)=\frac{\frac{7}{2}+4}{\frac{k+1}{2}-0}=\frac{15}{k+1}$
Slope of $P Q=\frac{4-3}{1-k}=\frac{1}{1-k}$
$\therefore \frac{15}{1+k} \times \frac{1}{1-k}=-1$
$1-k^{2}=-15 \Rightarrow k=\pm 4$
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