A rectangle is formed by the lines x= 0, y = 0, x=3 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 $$\left(\frac{1}{2},-5\right)$$ from the line L is equal to :

Rectangle vertices are bounded by $$x=0, \ y=0, \ x=3, \ y=4$$

Center of the rectangle: $$C = \left(\frac{3}{2}, 2\right)$$

Line $$L$$ is perpendicular to $$3x + y + 6 = 0$$: $$\text{Slope of } L \ (m_L) = \frac{-1}{-3} = \frac{1}{3}$$

Equation of line $$L$$ passing through $$C\left(\frac{3}{2}, 2\right)$$:

$$y - 2 = \frac{1}{3}\left(x - \frac{3}{2}\right) \implies 3y - 6 = x - \frac{3}{2} \implies 2x - 6y + 9 = 0$$

Perpendicular distance $$d$$ of point $$P\left(\frac{1}{2}, -5\right)$$ from line $$L$$:

$$d = \frac{\left\vert{}2\left(\frac{1}{2}\right) - 6(-5) + 9\right\vert{}}{\sqrt{2^2 + (-6)^2}}$$

$$d = \frac{\vert{}1 + 30 + 9\vert{}}{\sqrt{40}} = \frac{40}{\sqrt{40}} = \sqrt{40} = 2\sqrt{10}$$