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

• A. $-2$
• B. $\sqrt{15}$
• C. $\sqrt{14}$
• D. $-4$

Answer 1

The correct option is D $-4$

Slope of the line $PQ$ :
$m_{PQ}=\frac{4-3}{1-k}$
Now, slope of line perpendicular to the line $PQ$ is $m=k-1$.
Midpoint of $PQ=(\frac{k+1}{2},\frac{7}{2})$
Equation of perpendicular bisector :
$y-\frac{7}{2}=(k-1)(x-\frac{k+1}{2})$
For $y$ intercept put $x=0$
$y=\frac{7}{2}-\left(\frac{k^2-1}{2}\right)=-4$ (given)
$\Rightarrow \frac{k^2-1}{2}=\frac{15}{2}$
$\therefore k=\pm 4$