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 $3 x+y+6=0$ 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 li...

Accepted Answer

For $L$ to bisect area, line must pass through the centre of rectangle $(\frac{3}{2},2)$

$\begin{gathered} \Rightarrow \, L\equiv (y-2)=\frac{1}{3}(x-\frac{3}{2}) \ \Rightarrow \, 3y-6=x-\frac{3}{2}\Rightarrow 6y-12=2x-3 \ 6y-2x-9=0 \end{gathered}$

The distance from $L$ of $(\frac{1}{2},-5)$ is

$\begin{gathered} |\frac{6(-5)-2(\frac{1}{2})-9}{\sqrt{40}}|=\sqrt{40} \ =2\sqrt{10} \end{gathered}$

Solutions · 1