smc

Let the vertices be $A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)$. It is easy to see that the line must pass through $CD$. Let the line intersect $CD$ at the point $G=(3,3-x)$ (i.e. the point $x$ units below $C$). Since the quadrilateral $ABCG$ and pentagon $GDEFA$ must have the same area, we have the equation $3\times \frac{1}{2}\times (x+3)=\frac{1}{2}\times 3\times (3-x)+2$. This simplifies into $3x=2$, or $x=\frac{2}{3}$, so $G=(3,\frac{7}{3})$. Therefore the slope of the line is $\boxed{\frac{7}{9}}$