Example 2 (2000 National High School League Question) The minimum value of the distance from a lattice point (a point with integer coordinates) to the line $y=$ $\frac{5}{3} x+\frac{4}{5}$ is ( ).
A. $\frac{\sqrt{34}}{170}$
B. $\frac{\sqrt{34}}{85}$
C. $\frac{1}{20}$
D. $\frac{1}{30}$
Solution: Choose B. Reason: Let the integer point be $\left(x_{0}, y_{0}\right)$, then the distance from it to the line $y=\frac{5}{3} x+\frac{4}{5}$ is
$$
d=\frac{\left|25 x_{0}-15 y_{0}+12\right|}{\sqrt{25^{2}+15^{2}}}=\frac{\left|25 x_{0}-15 y_{0}+12\right|}{5 \sqrt{34}} .
$$

Since $x_{0}, y_{0} \in \mathbf{Z}$, $25 x_{0}+15 y_{0}$ is a multiple of 5, so $\left|25 x_{0}-15 y_{0}+12\right| \geqslant 2$. And when $x_{0}=-1$, $y_{0}=-1$, $\left|25 x_{0}-15 y_{0}+12\right|=2$, hence the minimum value of the required $d$ is $\frac{2}{5 \sqrt{34}}=\frac{\sqrt{34}}{85}$.