A ray of light coming from the point [P(1,2)] gets reflected from the point [Q] on the x-a…

Question

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

Accepted Answer

$$\begin{aligned}
& P^{\prime} R: y+2=\frac{5}{3}(x-1) \
& \text { For Point } Q \Rightarrow y=0 \
& \frac{6}{5}=a-1 \Rightarrow a=\frac{11}{5}
\end{aligned}$$

Now, $P Q R S$ is parallelogram

$$\begin{aligned}
& \therefore \frac{h+a}{2}=\frac{4+1}{2} \Rightarrow h=5-\frac{11}{5}=\frac{14}{5} \
& \text { and } \frac{2+3}{2}=\frac{k}{2} \Rightarrow K=5
\end{aligned}$$

Now $$h k^2=25 \times \frac{14}{5}=14 \times 5=70$$

Solutions · 1