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)
$y=(k-1) x-4$
$\Rightarrow \quad \frac{7}{2}=(k-1)\left(\frac{k+1}{2}\right)-4$
$\begin{aligned} & \Rightarrow & 7 & =k^2-1-8 \\ & \Rightarrow & k^2-16 & =0 \\ & \Rightarrow & k & = \pm 4\end{aligned}$