A rectangle is formed by the lines $x = 0,y = 0$, $x = 3$ and $y = 4$. Let the line L be perpendicul

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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 :

Accepted Answer

Line is $y = \frac{X}{3} + C$Line passes thru $\left( \frac{3}{2},2 \right)$$${2 = \frac{1}{2} + C \Rightarrow C = \frac{3}{2} }{y = \frac{x}{3} + \frac{3}{2} }{\Rightarrow 6y = 2x + 9 }$$Line is $2x - 6y + 9 = 0$ &Dist $= \left| \frac{1 + 30 + 9}{\sqrt{40}} \right| = \sqrt{40} = 2\sqrt{10}$

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