JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 23)
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 ____________.
Answer
1250
Explanation
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For minimum $(P R+R Q)$
$R$ lies on $P Q^{\prime}$ (where $Q^{\prime}$ is image of $Q$ in $X$-axis)
$\Rightarrow$ Equation on $P Q^{\prime}$ is
$$ 2 x+y+2=0 \Rightarrow R(-1,0) $$
$$ \therefore $$ 50(PR2 + RQ2)
= 50(20 + 5)
= 50(25)
= 1250
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