JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 18)
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:
60
70
80
90
Explanation
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$$\begin{aligned} & P^{\prime} R: y+2=\frac{5}{3}(x-1) \\ & \text { For Point } Q \Rightarrow y=0 \\ & \frac{6}{5}=a-1 \Rightarrow a=\frac{11}{5} \end{aligned}$$
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 $$h k^2=25 \times \frac{14}{5}=14 \times 5=70$$
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