A man starts walking from the point P(-3, 4), | vedclass

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(D) To minimize the time taken at a constant speed,the total distance $PR + RQ$ must be minimized.Let $Q'(0, -2)$ be the reflection of $Q(0, 2)$ acros

Vedclass · Accepted Answer

$1250$

Vedclass · Answer

(D) To minimize the time taken at a constant speed,the total distance $PR + RQ$ must be minimized.Let $Q'(0, -2)$ be the reflection of $Q(0, 2)$ across the $x$-axis.The distance $RQ = RQ'$. Thus,$PR + RQ = PR + RQ'$.This sum is minimized when $P, R,$ and $Q'$ are collinear.The line passing through $P(-3, 4)$ and $Q'(0, -2)$ has the equation:$y - (-2) = \frac{4 - (-2)}{-3 - 0}(x - 0)$$y + 2 = \frac{6}{-3}x$$y + 2 = -2x \implies 2x + y + 2 = 0$.The point $R$ is the intersection of this line with the $x$-axis $(y=0)$:$2x + 0 + 2 = 0 \implies x = -1$.So,$R = (-1, 0)$.Now,calculate the squared distances:$PR^{2} = (-1 - (-3))^{2} + (0 - 4)^{2} = (2)^{2} + (-4)^{2} = 4 + 16 = 20$.$RQ^{2} = (0 - (-1))^{2} + (2 - 0)^{2} = (1)^{2} + (2)^{2} = 1 + 4 = 5$.Finally,calculate $50(PR^{2} + RQ^{2})$:$50(20 + 5) = 50(25) = 1250$.

$A$ man starts walking from the point $P(-3, 4)$,touches the $x$-axis at $R$,and then turns to reach 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 $50((PR)^{2} + (RQ)^{2})$ is equal to ..... .

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