A triangle has vertices $(0,0)$, $(1,1)$, and $(6m,0)$, and the line $y = mx$ divides the triangle into two triangles of equal area. What is the sum of all possible values of $m$?
$\textbf{A} - \!\frac {1}{3} \qquad \textbf{(B)} - \!\frac {1}{6} \qquad \textbf{(C)}\ \frac {1}{6} \qquad \textbf{(D)}\ \frac {1}{3} \qquad \textbf{(E)}\ \frac {1}{2}$
Let's label the three points as $A=(0,0)$, $B=(1,1)$, and $C=(6m,0)$.
Clearly, whenever the line $y=mx$ intersects the inside of the triangle, it will intersect the side $BC$. Let $D$ be the point of intersection. 
The triangles $ABD$ and $ACD$ have the same height, which is the distance between the point $A$ and the line $BC$.
Hence they have equal areas if and only if $D$ is the midpoint of $BC$.
The midpoint of the segment $BC$ has coordinates $\left( \frac{6m+1}2, \frac 12 \right)$. This point lies on the line $y=mx$ if and only if $\frac 12 = m \cdot \frac{6m+1}2$. This simplifies to $6m^2 + m - 1 = 0$. Using Vieta's formulas, we find that the sum of the roots is $\boxed{\textbf{(B)} - \!\frac {1}{6}}$.

For illustration, below are pictures of the situation for $m=1.5$, $m=0.5$, $m=1/3$, and $m=-1/2$.