Question:
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 ____________.
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 ____________.
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Minimum time means minimum path length PR + RQ; reflect Q in the x-axis and take R on the line joining P to the image.
Updated On: Nov 4, 2026
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Correct Answer: 1250
Solution and Explanation
Step 1: Constant speed, so the time is minimum when \(PR+RQ\) is minimum.
Step 2: P and Q lie on the same side of the x-axis. The reflection of Q in the x-axis is \(Q'(0,-2)\); \(PR+RQ=PR+RQ'\) is minimum when P, R, Q' are collinear.
Step 3: Line \(PQ'\): slope \(=\frac{-2-4}{0+3}=-2\), \(y-4=-2(x+3)\). At \(y=0\): \(x+3=2\Rightarrow x=-1\). So \(R(-1,0)\).
Step 4: \(PR^2=(-1+3)^2+4^2=20\), \(RQ^2=1^2+2^2=5\).
Step 5: \(50(20+5)=1250\).
Answer: 1250
Step 2: P and Q lie on the same side of the x-axis. The reflection of Q in the x-axis is \(Q'(0,-2)\); \(PR+RQ=PR+RQ'\) is minimum when P, R, Q' are collinear.
Step 3: Line \(PQ'\): slope \(=\frac{-2-4}{0+3}=-2\), \(y-4=-2(x+3)\). At \(y=0\): \(x+3=2\Rightarrow x=-1\). So \(R(-1,0)\).
Step 4: \(PR^2=(-1+3)^2+4^2=20\), \(RQ^2=1^2+2^2=5\).
Step 5: \(50(20+5)=1250\).
Answer: 1250
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