Find a line equation that divides a triangle in half Let $(0,0), (2,2), (4,0)$ be the vertices of a triangle. Then find the line $l$ that passes through $(1,-1)$ and divides the triangle in half (equal area).
I think can solve this problem by
Letting $P$ and $Q$ be the intersection of the line and the triangle and $m$ be the slope of $l$.
$$
 \frac{y-1}{x+1}=m\\
$$
Then the height of the $\triangle APQ $ is
$$
\frac{|mx-y+m+1|}{\sqrt{m^2+1}}=\frac{|5m+1|}{\sqrt{m^2+1}}
$$
and then finding the distance between $P$ and $Q$... Finding the half area in terms of $m$...But is there more clean and easy way to do this? Am I missing something obvious?
 A: First, it should be $$
 \frac{y+1}{x-1}=m\\
$$
Say $A=(0,0)$, $B=(4,0)$. So $$Area BPQ = {1\over 2}Area ABC = 2$$ 
If $P$ is intersection of this line with $AB$ then $P= (1+{1\over m},0)$ and if $Q$ is intersection of $l$ with line $BC: \;\;x+y=4$ then $Q= (...,{3m-1\over 3m+1})$. 
So $$BP\cdot y_Q= {3m-1\over 3m+1} \cdot (3-{1\over m}) = 4$$
But we get nothing nice...
A: Just an alternative solution. Not very nice.
Let $A=(0,0)$, $B=(4,0)$, $C=(2,2)$, $P=(1,-1)$ and $D=(2,0)$.
Suppose that $l$ meets $AB$ at $H$, $CD$ at $K$ and $BC$ at $L$. Let$ DK=a$.
Then $m=a+1$,  $K=(0,a)$, $\displaystyle H=\left(2-\frac{a}{a+1},0\right)$ and $L=(4-b,b)$, where $\displaystyle \frac{b-a}{2-b}=a+1$. Hence $\displaystyle L=\left(\frac{a+6}{a+2},\frac{3a+2}{a+2}\right)$. 
Note that $\triangle HKD$ and $\triangle KLC$ have the same area.
\begin{align*}
\frac{1}{2}\left(\frac{a}{a+1}\right)(a)&=\frac{1}{2}(2-a)\left(\frac{a+6}{a+2}-2\right)\\
a^3+2a^2&=a^3-3a^2+4\\
a&=\frac{2}{\sqrt{5}}
\end{align*}
So, $\displaystyle m=1+\frac{2}{\sqrt{5}}$.
