jmc

To find the vertices of the triangle, we find where the two lines $y=4x-6$ and $y=-2x+12$ intersect. Solving $4x-6=-2x+12$, we get $x=3$. Substituting $x=3$ back into $y=4x-6$, we find $y=6$. Therefore, $(3,6)$ is one of the vertices of the triangle. The other two vertices are the $y$-intercepts of the two lines, namely $(0,12)$ and $(0,-6)$. Taking the side joining $(0,12)$ and $(0,-6)$ as the base of the triangle, we find that the area of the triangle is $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(12-(-6))(3)=\boxed{27}$ square units.