Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$.  Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$.  This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$, where these fractions are in lowest terms.  What is $p+q+r+s$?

$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$
First, we shall find the area of quadrilateral $ABCD$. This can be done in any of three ways:
[Pick's Theorem](https://artofproblemsolving.com/wiki/index.php/Pick%27s_Theorem): $[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.$
Splitting: Drop perpendiculars from $B$ and $C$ to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is $1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.$
[Shoelace Theorem](https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem): The area is half of $|1 \cdot 3 - 2 \cdot 3 - 3 \cdot 4| = 15$, or $\dfrac{15}{2}$.
$[ABCD] = \frac{15}{2}$.  Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$.
Call the point where the line through $A$ intersects $\overline{CD}$ $E$.  We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$.  Furthermore, we know that $b = 4$, as $AD = 4$.  Thus, solving for $h$, we find that $2h = \frac{15}{4}$, so $h = \frac{15}{8}$.  This gives that the y coordinate of E is $\frac{15}{8}$.  
Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$.  Solving for $x$, we find that $x = \frac{27}{8}$.
From this, we know that $E = \left(\frac{27}{8}, \frac{15}{8}\right)$.  $27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}$