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QJEEM21Mabdc2: JEE Main 2021 Mathematics - Class 11 JEEM21Mabdc2 Solution: A man starts walking from the point P( 3, 4), touches the…

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((PR)2+(RQ)2)50\left( {{{(PR)}^2} + {{(RQ)}^2}} \right) is equal to ____________.

Step-by-Step Verified Solution

JEE Main 2021 (Online) 1st September Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 65 English Explanation
For minimum (PR+RQ)(P R+R Q)

RR lies on PQ′P Q^{\prime} (where Q′Q^{\prime} is image of QQ in XX -axis)

⇒\Rightarrow Equation on PQ′P Q^{\prime} is

2x+y+2=0⇒R(−1,0)2 x+y+2=0 \Rightarrow R(-1,0)

∴\therefore 50(PR2 + RQ2)

= 50(20 + 5)

= 50(25)

= 1250

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