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 the value of $k$ is

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

Answer 1

The correct option is D

$-4$

Explanation for the correct answer:\

Step-1 Find the equation of perpendicular bisector:

Let $P(1,4)=\left(x_1,y_1\right)$ and $Q(k,3)=\left(x_2,y_2\right)$

Let $m_1$ be the slope of $PQ$ and let $m_2$ be the slope of the perpendicular bisector of $PQ$

The product of slopes of perpendicular lines is $-1$

$\Rightarrow$$m_1{m}_2=-1$

The slope of line joining the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$$=\frac{y_2-y_1}{x_2-x_1}$

$\Rightarrow$ $m_1$$=\frac{3-4}{k-1}=\frac{-1}{k-1}$

Hence, the slope of perpendicular bisector is $m_2=\frac{-1}{m_1}=\frac{-1}{{\displaystyle \frac{-1}{k-1}}}=k-1$

The equation of the perpendicular bisector can be written in slope intercept form as

$y=m_{2}x+c$ where $c$ is the $y-$intercept

Substituting the values we get

$y=\left(k-1\right)x-4$$...\left(i\right)$

Step-2: Find the valie $k$:

Let $M\left(x_m,y_m\right)$ be the midpoint of $PQ$

By midpoint formula

$M\left(x_m,y_m\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$\Rightarrow$$M\left(x_m,y_m\right)=\left(\frac{1+k}{2},\frac{7}{2}\right)$

The midpoint of a segment lies on the perpendicular bisector of the segment. Hence, its co-ordinates satisfy the equation.

Substituting the co-ordinates in equation of perpendicular bisector we get,

$\Rightarrow$ $\frac{7}{2}=\left(k-1\right)\frac{k+1}{2}-4$

$\Rightarrow$ $7=\left(k^2-1\right)-8$

$\Rightarrow$ $\left(k^2-16\right)=0$

$\Rightarrow$$\left(k-4\right)\left(k+4\right)=0$

$\Rightarrow$ $k=4$ or $k=-4$

So, option (D) is the correct answer.