JEE Main 2021 (Online) 1st September Evening ShiftQuestion 1
A man starts walking from the point $P(-3,4)$, touches the $x$-axis at $R$, and then turns to reach at the point $Q(0,2)$. The man is walking at a constant speed. If the man reaches the point $Q$ in the minimum time, then $50\left((PR)^2+(RQ)^2\right)$ is equal to ______.
Points and Straight Lines
Answer$1250$
Solution
For the least time the path $P\to R\to Q$ must be shortest. Reflect $Q$ in the $x$-axis: $Q'=(0,-2)$. For every $R$ on the axis $RQ=RQ'$, so $PR+RQ=PR+RQ'\ge PQ'$, with equality when $R$ lies on $PQ'$. <svg fill="none" stroke="#000" font-family="serif" font-size="10" viewBox="0 0 210 276"><path stroke-width=".8" d="M12 171.4h186m-53.1 93V12"/><path fill="#000" stroke="none" d="m198 171.4-5-2.2v4.4zM144.9 12l-2.2 5h4.4z"/><text x="203.6" y="181.2" fill="#000" stroke="none" font-style="italic">x</text><text x="150.9" y="15.5" fill="#000" stroke="none" font-style="italic">y</text><text x="139.3" y="181.2" fill="#000" stroke="none" font-style="italic" text-anchor="end">O</text><path stroke-width="1.2" d="m45.2 38.6 66.4 132.8"/><path fill="#000" stroke="none" d="m83.3 114.8-5.4-5.1 4.6-2.4z"/><path stroke-width="1.2" d="m111.6 171.4 33.3-66.4"/><path fill="#000" stroke="none" d="m131.5 131.8-.8 7.4-4.7-2.3z"/><path stroke-dasharray="4 3" d="m111.6 171.4 33.3 66.5"/><path stroke-dasharray="4 3" stroke-width=".7" d="M144.9 105v132.9"/><circle cx="45.2" cy="38.6" r="1.9" fill="#000" stroke-width=".7"/><text x="52.2" y="42.1" fill="#000" stroke="none">P(-3,4)</text><circle cx="144.9" cy="105" r="1.9" fill="#000" stroke-width=".7"/><text x="151.9" y="108.5" fill="#000" stroke="none">Q(0,2)</text><circle cx="144.9" cy="237.9" r="1.9" fill="#000" stroke-width=".7"/><text x="151.9" y="241.4" fill="#000" stroke="none">Q'(0,-2)</text><circle cx="111.6" cy="171.4" r="1.9" fill="#000" stroke-width=".7"/><text x="106" y="181.2" fill="#000" stroke="none" text-anchor="end">R</text></svg> $PQ'$: slope $\frac{-2-4}{0+3}=-2$, $y=-2x-2$. At $y=0$: $x=-1$, so $R=(-1,0)$. $PR^2=2^2+4^2=20$, $RQ^2=1^2+2^2=5$. $50\left(PR^2+RQ^2\right)=50\cdot25=1250$.