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 $-4$

Given,

co-ordinates $P(1,4)$ and $Q(k,3)$

perpendicular bisector has a y-intercept of $-4$ units

$\therefore theco-ordinateatthatpointwillbe(0,-4)$

let, M be the point where the perp. bisects the line

So, M is the midpoint
$\therefore$ co-ordinate of M is $(\frac{k+1}{2},\frac{7}{2})$

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

$=\left(\frac{1}{1-k}\right)=m_1\left(let\right)$

$\because m_1=\frac{-1}{m_2}$

$m_2=\frac{-1}{m_1}=\frac{-1}{\frac{1}{1-k}}=(k-1)$

$Slopeof\perp line=m_2=\left(k-1\right)$

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

$\Rightarrow \frac{15}{2}=\frac{\left(k+1\right)\left(k-1\right)}{2}$.....$(puttingx=0andy=-4andsimplifying)$

$\Rightarrow k^2-1=15$

$\Rightarrow k^2=16$

$\Rightarrow k=\pm 4$
So, one of the possible value is $-4$
hence, Option $D$ is the answer.