Example 1 Find the distance from point $P\left(x_{0}, y_{0}\right)$ to the line $l: A x+B y+C=$ $0\left(A^{2}+B^{2} \neq 0\right)$.
Let point $Q(x, y)$ be any point on the line, then the problem is transformed into finding the minimum value of
$$t=\sqrt{\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}}$$
under the condition $A x+B y+C=0$.
(1) can be transformed into
$$\frac{\left(A x-A x_{0}\right)^{2}}{(A t)^{2}}+\frac{\left(B y-B y_{0}\right)^{2}}{(B t)^{2}}=1$$

By Theorem 1, we get $\left(A^{2}+B^{2}\right) t^{2} \geqslant\left(A x-A x_{0}+B y-\right.$ $\left.B y_{0}\right)^{2}=\left(A x_{0}+B y_{0}+C\right)^{2}$.

Thus, we have $t \geqslant \frac{\left|A x_{0}+B y_{0}+C\right|}{\sqrt{A^{2}+B^{2}}}$.
Therefore, $d=t_{\min }=\frac{\left|A x_{0}+B y_{0}+C\right|}{\sqrt{A^{2}+B^{2}}}$.