A ray of light coming from the point $P(1,2)$ gets reflected from the point $Q$ on the $x$-axis and then passes through the point $R(4,3)$. If the point $S(h, k)$ is such that PQRS is a parallelogram, then $h k^2$ is equal to :

Question

Aicharya · Accepted Answer

Image of $P$ wrt $x$-axis will be $P^{\prime}(1,-2)$ equation of line joining $P^{\prime} R$ will be $y-3=\frac{5}{3}(x-4)$
Above line will meet $\mathrm{x}$-axis at $\mathrm{Q}$ where $\begin{aligned} & \mathrm{y}=0 \Rightarrow \mathrm{x}=\frac{11}{5} \\ & \therefore \mathrm{Q}\left(\frac{11}{5}, 0\right) \end{aligned}$ $\because \mathrm{PQRS}$ is parallelogram so their diagonals will bisects each other $\begin{aligned} & \Rightarrow \frac{4+1}{2}=\frac{\frac{11}{5}+\mathrm{h}}{2} \& \frac{2+3}{2}=\frac{\mathrm{k}+0}{2} \\ & \Rightarrow \mathrm{h}=\frac{14}{5} \& \mathrm{k}=5 \\ & \therefore \mathrm{hk}^2=\frac{14}{5} \times 5^2=70\end{aligned}$

A ray of light coming from the point $P(1,2)$ gets reflected from the point $Q$ on the $x$-axis and then…

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