Question

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

• A. 1
• B. 2
• C. -2
• D. -4

Answer 1

The correct option is D -4

Slope of $PQ=\frac{3-4}{k-1}=\frac{-1}{k-1}$
$\therefore$ Slope of perpendicular bisector of $PQ=(k-1)$
Also mid point of $PQ(\frac{k+1}{2},\frac{7}{2})$ .
$\therefore$ Equation of perpendicular bisector is
$y-\frac{7}{2}=(k-1)(x-\frac{k+1}{2})\Rightarrow 2y-7=(k-1)x-(k^2-1)\Rightarrow 2(k-1)x-2y+(8-k^2)=0\therefore \text{y -intercept}=-\frac{8-k^2}{-2}=-4\Rightarrow 8-k^2=-8\text{or}{k}^2=16\Rightarrow k=\pm 4$