19. A.

From the conditions, we have
$$
\begin{array}{l}
A B^{2}=(5-1)^{2}+(3-1)^{2}=20, \\
A C^{2}=(4-1)^{2}+(4-3)^{2}=10, \\
B C^{2}=(5-4)^{2}+(4-1)^{2}=10 .
\end{array}
$$

Thus, $\triangle A B C$ is an isosceles right triangle.
$$
\text { Therefore, } \frac{S_{1}}{S}=\frac{\frac{1}{2} A C^{2}}{5 \times 6}=\frac{1}{6} \text {. }
$$
19. Divide the Cartesian coordinate system

into a $6 \times 5$ unit grid as shown in Figure 3. Let points $A(1,3)$, $B(5,1)$, and $C(4,4)$. Then the area $S_{1}$ of $\triangle ABC$ occupies what fraction of the area $S$ of this grid?
(A) $\frac{1}{6}$
(B) $\frac{1}{5}$
(C) $\frac{1}{4}$
(D) $\frac{1}{3}$
(E) $\frac{1}{2}$