Example 8 When the algebraic expression
$$
\begin{array}{l}
\sqrt{9 x^{2}+4}+\sqrt{9 x^{2}-12 x y+4 y^{2}+1}+ \\
\sqrt{4 y^{2}-16 y+20}
\end{array}
$$

reaches its minimum value, find the values of $x$ and $y$.
Explanation: Although the ratio 7 is complex, the problem-solving approach is the same, the key is to transform the given expression. At this point, one can think of the distance formula between two points.
Transform the original algebraic expression into
$$
\begin{array}{c}
\sqrt{[0-(-2)]^{2}+(3 x-0)^{2}}+ \\
\sqrt{(1-0)^{2}+(2 y-3 x)^{2}}+ \\
\sqrt{(3-1)^{2}+(4-2 y)^{2}} . \\
=\sqrt{[0-(-2)]^{2}+(3 x-0)^{2}}, \\
|B C|=\sqrt{(1-0)^{2}+(2 y-3 x)^{2}},
\end{array}
$$
$$
\begin{array}{l}
|C D|=\sqrt{(3-1)^{2}+(4-2 y)^{2}}, \\
|A D|=\sqrt{5^{2}+4^{2}}=\sqrt{41} .
\end{array}
$$

By the shortest distance between two points, we know
$$
A B+B C+C D \geqslant A D=\sqrt{41},
$$

with equality holding if and only if points $B$ and $C$ lie on segment $A D$. When the original expression takes its minimum value $\sqrt{41}$, we have
$$
\frac{3 x}{2}=\frac{4}{5}, \frac{2 y}{3}=\frac{4}{5} \text {. }
$$

Solving these, we get $x=\frac{8}{15}, y=\frac{6}{5}$.