$37 \cdot 2$ In the Cartesian coordinate system, a point whose both coordinates are integers is called an integer point. Let the line passing through points $A(-2 \sqrt{2},-1+\sqrt{2}), B(0, \sqrt{2})$ be $l$; and the line passing through points $C\left(0, \frac{3}{2}\right)$, $D\left(\frac{3}{2}, \frac{3}{4}\right)$ be $m$. Then
(A) There are no integer points on line $l$.
(B) There are a finite number of integer points on line $l$.
(C) There are no integer points on line $m$.
(D) There are a finite number of integer points on line $m$.
(China Guangzhou, Wuhan, Fuzhou, and other five cities Junior High School Mathematics League, 1990)
[Solution] Let $(x, y)$ be a point on $l$, then we have
$$
\frac{x-0}{y-\sqrt{2}}=\frac{0+2 \sqrt{2}}{\sqrt{2}-(\sqrt{2}-1)},
$$

which simplifies to
$$
\frac{x}{y-\sqrt{2}}=2 \sqrt{2} \text {, or } x=2 \sqrt{2} y-4 \text { . }
$$

Thus, its integer solutions are only $(-4,0)$, meaning $l$ passes through only one integer point. Similarly, it can be shown that $m$ passes through infinitely many integer points.
In conclusion, the answer is $(B)$.