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

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Aicharya · Accepted Answer

Let $L R M$ is the perpendicular bisector of $P Q$ at R

$\left[\because\right.$ Slope of line passing through $\left(x_1, y_1\right)$ and$\left.\left(x_2, y_2\right) \text { is } \frac{y_2-y_1}{x_2-x_1}\right]$ $\begin{aligned} R & =\left(\frac{1+k}{2}, \frac{4+3}{2}\right) \\ R & =\left(\frac{k+1}{2}, \frac{7}{2}\right)\end{aligned}$ $\left\{\begin{array}{c}\because \text { mid-point of line with end points }\left(x_1, y_1\right),\left(x_2, y_2\right) \\ =\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\end{array}\right\}$ Also slope of $P Q=\frac{3-4}{k-1}$ $m_{P Q}=\frac{-1}{k-1}$ $\because P Q$ and $L M$ are perpendicular to each other. $\therefore \quad m_{L M}=\frac{-1}{m_{P Q}}=\frac{-1}{-1 / k-1}$ $m_{L M}=k-1$ $\therefore$ Equation of $L M$ which has slope $(k-1)$ has $Y$-intercept -4 $\Rightarrow$ Equation of $L M: y=(k-1) x-4$ ...(i) Since, Eq. (i) passes through $R$. Hence, point $R$ will satisfy Eq. (i) $y=(k-1) x-4$ $\Rightarrow \quad \frac{7}{2}=(k-1)\left(\frac{k+1}{2}\right)-4$ $\begin{aligned} & \Rightarrow & 7 & =k^2-1-8 \\ & \Rightarrow & k^2-16 & =0 \\ & \Rightarrow & k & = \pm 4\end{aligned}$

The perpendicular bisector of line segment joining the points $P(1,4)$ and $Q(k, 3)$ has $y$-intercept -4 .…

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