A ray of light coming from the point P(1, 2) | vedclass

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(D) The image of $P(1, 2)$ with respect to the $x$-axis is $P'(1, -2)$.The equation of the line joining $P'(1, -2)$ and $R(4, 3)$ is given by:$y - 3 =

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(D) The image of $P(1, 2)$ with respect to the $x$-axis is $P'(1, -2)$.The equation of the line joining $P'(1, -2)$ and $R(4, 3)$ is given by:$y - 3 = \frac{3 - (-2)}{4 - 1}(x - 4)$$y - 3 = \frac{5}{3}(x - 4)$This line meets the $x$-axis at $Q$,where $y = 0$:$0 - 3 = \frac{5}{3}(x - 4)$$-9 = 5x - 20$$5x = 11 \Rightarrow x = \frac{11}{5}$So,$Q = \left(\frac{11}{5}, 0ight)$.Since $PQRS$ is a parallelogram,its diagonals $PR$ and $QS$ bisect each other at the same midpoint.The midpoint of diagonal $PR$ is $\left(\frac{1 + 4}{2}, \frac{2 + 3}{2}ight) = \left(\frac{5}{2}, \frac{5}{2}ight)$.The midpoint of diagonal $QS$ is $\left(\frac{\frac{11}{5} + h}{2}, \frac{0 + k}{2}ight)$.Equating the midpoints:$\frac{11/5 + h}{2} = \frac{5}{2} \Rightarrow h = 5 - \frac{11}{5} = \frac{14}{5}$$\frac{k}{2} = \frac{5}{2} \Rightarrow k = 5$Therefore,$hk^2 = \frac{14}{5} 	imes 5^2 = 14 	imes 5 = 70$.

$A$ 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 $hk^2$ is equal to :

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