A rectangle is formed by the lines x=0, y=0, x=3 and y=4.

Question

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 $\left(\dfrac12,-5\right)$ from the line $L$ is equal to:

Accepted Answer

For $L$ to bisect the rectangle's area (perpendicular to $3x+y+6=0$, i.e. slope $\frac13$), $L$ must pass through the centre of the rectangle, $\left(\frac32,2\right)$.
$$L: y-2=\frac13\left(x-\frac32\right) \implies 6y-12 = 2x-3 \implies 2x-6y+9=0$$

Distance from $\left(\frac12,-5\right)$:
$$\frac{\left|2\left(\frac12\right)-6(-5)+9\right|}{\sqrt{4+36}} = \frac{|1+30+9|}{\sqrt{40}} = \frac{40}{\sqrt{40}} = \sqrt{40}=2\sqrt{10}$$

Solutions · 1