A rectangle is formed by the lines x= O, y = O, 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 :

$$3x + y + 6 = 0$$ $$m_2 = -3$$.

A line perpendicular to this must satisfy $$m_1\,m_2 = -1$$, hence $$m_1 = \frac{1}{3}\,. $$ Thus the required line L can be written as $$y = \frac{1}{3}x + k\,. $$

The rectangle has vertices $$(0,0),(3,0),(3,4),(0,4)$$ and area $$12$$.

We want L to cut off half this areaAt $$x=0$$, $$y=k$$, so we need $$0 \amp;le; k \amp;le; 4\,. $$ At $$x=3$$, $$y = 1 + k$$, so we need $$0 \amp;le; 1 + k \amp;le; 4 \implies -1 \amp;le; k \amp;le; 3\,. $$ Combining gives $$0 \amp;le; k \amp;le; 3\,. $$

For $$k$$ in this range, the portion of the rectangle below L is bounded by $$y=0$$ and $$y=\tfrac{1}{3}x + k$$ for $$0 \amp;le; x \amp;le; 3$$. Its area is $$\int_{0}^{3}\Bigl(\tfrac{1}{3}x + k\Bigr)\,dx = \Bigl[\tfrac{1}{6}x^2 + kx\Bigr]_{0}^{3} = \frac{9}{6} + 3k = 1.5 + 3k\,. \quad-(1)$$

Setting this equal to half the rectangle’s area gives $$1.5 + 3k = 6 \quad\implies\quad 3k = 4.5 \quad\implies\quad k = 1.5\,. $$

Therefore the equation of L is $$3y = x + \frac{9}{2} \quad\Longrightarrow\quad x - 3y + \frac{9}{2} = 0\,. $$

For a line $$ax+by+c=0$$ the distance from $$(x_0,y_0)$$ is $$\text{Distance} = \frac{\bigl|a x_0 + b y_0 + c\bigr|}{\sqrt{a^2 + b^2}}\,. $$ Here $$a=1\,,\; b=-3\,,\; c=\tfrac{9}{2}$$ and $$(x_0,y_0)=\bigl(\tfrac12,-5\bigr)$$, so:

$$ \text{Distance} = \frac{\Bigl|1\cdot\tfrac12 + (-3)\cdot(-5) + \tfrac{9}{2}\Bigr|}{\sqrt{1^2 + (-3)^2}} = \frac{\bigl|\tfrac12 + 15 + \tfrac{9}{2}\bigr|}{\sqrt{10}} = \frac{20}{\sqrt{10}} = 2\sqrt{10}\,. $$