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JEE Main20211 Sep 2021Evening ShiftMathematicsStraight LinesActual

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 ( P R ) 2 + ( R Q ) 2 is equal to ______ .

Correct answer

0

Step-by-step solution

For minimum time, value of P R + R Q must be minimum so R lies on P Q ' (where Q '   is image of Q w.r.t x -axis) From the diagram we can easily say Q ' ( 0 , - 2 ) As we know that y - y 1 = y 2 - y 1 x 2 - x 1 x - x 1 Equation of P Q '     y + 2 = 4 + 2 - 3 - 0 x - 0 ⇒ Equation of P Q ' is 2 x + y + 2 = 0 For the coordinate of point R put y = 0 ⇒ R - 1 ,   0 ⇒ 50 ( P R ) 2 + ( R Q ) 2 Distance formula = ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 ⇒ P R = ( - 3

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