JEE Main 2024 (Online) 9th April Morning ShiftQuestion 1
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 :
Points and Straight Lines
Solution
Reflect $P$ in the $x$-axis: $P'=(1,-2)$. The reflected ray $QR$, extended backwards, passes through $P'$. <svg fill="none" stroke="#000" font-family="serif" font-size="10" viewBox="0 0 210 300"><path stroke-width=".8" d="M12 201.3h186M31.9 287.7V12"/><path fill="#000" stroke="none" d="m198 201.3-5-2.2v4.4zM31.9 12l-2.2 5h4.4z"/><text x="203.6" y="211.1" fill="#000" stroke="none" font-style="italic">x</text><text x="37.9" y="15.5" fill="#000" stroke="none" font-style="italic">y</text><text x="26.3" y="211.1" fill="#000" stroke="none" font-style="italic" text-anchor="end">O</text><path stroke-width="1.2" d="m65.1 134.9 39.9 66.4"/><path fill="#000" stroke="none" d="m88.9 174.4-5.9-4.6 4.5-2.7z"/><path stroke-width="1.2" d="m105 201.3 59.8-99.6"/><path fill="#000" stroke="none" d="m139.7 143.5-1.4 7.4-4.4-2.7z"/><path stroke-dasharray="4 3" d="m105 201.3-39.9 66.5m0-132.9 59.8-99.6m0 0 39.9 66.4"/><circle cx="65.1" cy="134.9" r="1.9" fill="#000" stroke-width=".7"/><text x="58.1" y="138.4" fill="#000" stroke="none" text-anchor="end">P(1,2)</text><circle cx="105" cy="201.3" r="1.9" fill="#000" stroke-width=".7"/><text x="110.6" y="211.1" fill="#000" stroke="none">Q</text><circle cx="164.8" cy="101.7" r="1.9" fill="#000" stroke-width=".7"/><text x="171.8" y="105.2" fill="#000" stroke="none">R(4,3)</text><circle cx="65.1" cy="267.8" r="1.9" fill="#000" stroke-width=".7"/><text x="72.1" y="271.2" fill="#000" stroke="none">P'(1,-2)</text><circle cx="124.9" cy="35.3" r="1.9" fill="#000" stroke-width=".7"/><text x="131.9" y="38.8" fill="#000" stroke="none">S(h,k)</text></svg> $P'R$: slope $\frac{3+2}{4-1}=\frac53$; at $y=0$: $x=1+\frac65=\frac{11}{5}$, so $Q=\left(\frac{11}{5},0\right)$. In parallelogram $PQRS$ the diagonals $PR$ and $QS$ bisect each other: $S=P+R-Q=\left(\frac{14}{5},5\right)$. $hk^2=\frac{14}{5}\cdot25=70$. Answer (B).