A rectangle is formed by the lines x=0, x=3, | vedclass

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(D) The rectangle is bounded by $x=0, x=3, y=0, y=4$. The center of the rectangle is $(\frac{3}{2}, 2)$.Any line that divides the rectangle into two e

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$2\sqrt{10}$

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(D) The rectangle is bounded by $x=0, x=3, y=0, y=4$. The center of the rectangle is $(\frac{3}{2}, 2)$.Any line that divides the rectangle into two equal areas must pass through its center $(\frac{3}{2}, 2)$.The line $L$ is perpendicular to $3x+y+6=0$. The slope of $3x+y+6=0$ is $-3$, so the slope of line $L$ is $\frac{1}{3}$.The equation of line $L$ passing through $(\frac{3}{2}, 2)$ with slope $\frac{1}{3}$ is:$y - 2 = \frac{1}{3}(x - \frac{3}{2})$$3y - 6 = x - \frac{3}{2}$$6y - 12 = 2x - 3$$2x - 6y + 9 = 0$.The distance of the point $(\frac{1}{2}, -5)$ from the line $2x - 6y + 9 = 0$ is:$d = \frac{|2(\frac{1}{2}) - 6(-5) + 9|}{\sqrt{2^2 + (-6)^2}}$$d = \frac{|1 + 30 + 9|}{\sqrt{4 + 36}}$$d = \frac{40}{\sqrt{40}} = \sqrt{40} = 2\sqrt{10}$.

$A$ rectangle is formed by the lines $x=0, x=3, y=0$ and $y=4$. Let the line $L$ be perpendicular to $3x+y+6=0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $(\frac{1}{2}, -5)$ from the line $L$ is equal to:

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