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. $-4$
• B. $1$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $-4$

The mid point of the line joining of the given two points is $(\frac{k+1}{2},\frac{7}{2})$ and the slope of this line is $\frac{1}{1-k}$.

$\therefore$ The slope of the perpendicular bisector line is $k-1$.
Let the equation of this line be $y=(k-1)x+c\to \left(1\right)$

$y$- intercept is $-4\to \left(2\right)$ (given)

From $\left(1\right)$ and $\left(2\right)$, equation of the perpendicular bisector is $y=(k-1)x-4$.

This line passes through the point $(\frac{k+1}{2},\frac{7}{2})$.

On substituting this point in above line equation, we get $k=\pm 4$.
So, the possible value of $k$ is $-4$.