Question

The perpendicular bisector of the line segment doing the points $P(1,4)$ & $Q(K,3)$ has $Y$-intercept $-4$, then a possible value of $k$ is

• A. $2$
• B. $-2$
• C. $-4$
• D. $1$

Answer 1

Answer: (C)

$-4$

Line segment $PQ$ has slope$,m$

$m=\frac{4-3}{1-k}=\frac{1}{1-k}$

Line segment $PQ$ has midpoint

$(\frac{1+k}{2},\frac{4+3}{2})=(\frac{1+k}{2},\frac{7}{2})$
Perpendicular bisector has slope $\frac{-(1-k)}{1}=k-1$ and passes through point $(\frac{1+k}{2},\frac{7}{2})$
Equation of perpendicular bisector:

$y-\frac{7}{2}=(k-1)(x-\frac{(1+k)}{2})$

$\Rightarrow y=(k-1)x-\frac{(k-1)(1+k)}{2}+\frac{7}{2}$

$\Rightarrow y=(k-1)x+\frac{(1-k)(1+k)+7}{2}$

$\Rightarrow y=(k-1)x+\frac{1-k²+7}{2}$

$\Rightarrow y=(k-1)x+\frac{8-k²}{2}$
Perpendicular bisector has y-intercept $=-4$

$\frac{8-k²}{2}=-4$

$\Rightarrow 8-k^2=-8$

$\Rightarrow k^2=16$ $\Rightarrow k=\pm 4$