2. Given that the vertices of $\triangle A B C$ are all lattice points (points with integer coordinates are called lattice points), and $A(0,0), B(36,15)$. Then the minimum value of the area of $\triangle A B C$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{3}{2}$
(C) $\frac{5}{2}$
(D) $\frac{7}{2}$
2. (B).

From the given information, the line $l_{AB}: y=\frac{15}{36} x=\frac{5}{12} x$, which can be rewritten as $5 x-12 y=0$.

Let $C(x, y), x \in \mathbb{Z}, y \in \mathbb{Z}$. The height $h$ from $C$ to the line $l_{AB}$ in $\triangle ABC$ is the distance from $C$ to $l_{AB}$, so $h=\frac{|5 x-12 y|}{13}$.
Also, $|AB|=\sqrt{36^2+15^2}=3 \times 13$, thus
$S_{\triangle ABC}=\frac{1}{2}|AB| \cdot h=\frac{3}{2}|5 x-12 y|$.
Since $|5 x-12 y| \in \mathbb{Z}, x \in \mathbb{Z}, y \in \mathbb{Z}$, when $x=5, y=2$, $|5 x-12 y|$ is minimized, with the minimum value being 1. Therefore, the minimum value of $S_{\triangle ABC}$ is $\frac{3}{2}$.