$$
\begin{array}{l}
\text { Given: Point } P\left(x_{0}, y_{0}\right) \text {, line } l: A x+B y \\
+C=0 \quad\left(A^{2}+B^{2} \neq 0\right) .
\end{array}
$$

Find: The distance from point $P$ to line $l$.
Solution: Let $H(x, y)$ be any point on the line $l$.
$\because$ In the line equation $A x+B y+C=0$,
$\therefore A, B$ at least one is not zero. Without loss of generality, let $B \neq 0$,
$$
y=-\frac{A x+C}{B}.
$$
$\therefore$ The coordinates of point $H$ are $\left(x,-\frac{A x+C}{B}\right)$.
The distance from point $P$ to point $M$ is $|P, M|$
$$
\begin{array}{l}
=\sqrt{\left(x-x_{0}\right)^{2}+\left(-\frac{A x+C}{B}-y_{0}\right)^{2}} \\
=\sqrt{\left(x-x_{0}\right)^{2}+\left(y_{0}+\frac{A x+C}{B}\right)^{2}}.
\end{array}
$$

According to the definition of the distance from a point to a line and the properties of the line, the problem of finding the distance from $P$ to $l$ can be converted into finding the minimum value of $|P M|$.
Set $\left|P M\right|^{2}=\left(x-x_{0}\right)^{2}$
$$
+\left(y_{0}+\frac{A x+C}{B}\right)^{2}=Y.
$$

Then when $\bar{Y}$ has the minimum value, $|P M|$ also has the minimum value.
$$
\begin{array}{l}
\bar{Y}=x^{2}-2 x_{0} x+x_{0}^{2}+y_{0}^{2}+2 y_{0} \frac{A x+C}{B} \\
+\left(\frac{A x+C}{B}\right)^{2} \\
B^{2}\left(x_{0}^{2}+\frac{A^{2} y_{0}^{2}}{B^{2}}+\frac{A^{2} C^{2}}{B^{4}}-\frac{2 A x_{0} y_{0}}{B}-\frac{2 A C x_{0}}{B^{2}+B^{2}}+\frac{2 A^{2} C y_{0}}{B^{3}}\right) \\
= \frac{\left(A^{2}+B^{2}\right)\left(B^{2} x_{0}^{2}+B^{2} y_{0}^{2}+2 B C y_{0}+C^{2}\right)}{B^{2}\left(A^{2}+B^{2}\right)} \\
+\frac{2 C y_{0}^{2}-2 x_{0} x+x_{0}^{2}+y_{0}^{2}+\frac{2 y_{0} A x}{B}}{B}+\frac{A^{2} x^{2}+2 A C x+C^{2}}{B^{2}} \\
=\left(\frac{A^{2}+B^{2}}{B^{2}}\right) x^{2}-2\left(x_{0}-\frac{A y_{0}}{B}-\frac{A C}{B^{2}}\right) x+\left(x_{0}^{2}+y_{0}^{2}+\frac{2 y_{0} C}{B}+\frac{C^{2}}{B^{2}}\right) \\
= a x^{2}+b x+c_{0}\left[\text { where, } a=\frac{A^{2}+B^{2}}{B^{2}},\right. \\
b= 2\left(x_{0}-\frac{A y_{0}}{B}-\frac{A C}{B^{2}}\right), c=\left(x_{0}^{2}+y_{0}^{2}\right. \\
\left.\left.+\frac{2 y_{0} C}{B}+\frac{C^{2}}{B^{2}}\right)\right) \\
\because a=\frac{A^{2}+B^{2}}{B^{2}}>0,
\end{array}
$$
$\therefore \bar{Y}$ has a minimum value.
$$
\begin{array}{l}
Y_{\text {min }}=\frac{4 a c-b^{2}}{4 a}=c-\frac{b^{2}}{4 a} \\
=\left(x_{0}^{2}+y_{0}^{2}+\frac{2 y_{0} C}{B}+\frac{C^{2}}{B^{2}}\right) \\
-\frac{4\left(x_{0}-\frac{A y_{0}}{B}-\frac{A C}{B^{2}}\right)^{2}}{4 \frac{A^{2}+B^{2}}{B^{2}}} \\
=\frac{B x_{0}^{2}+B^{2} y_{0}^{2}+2 B C y_{0}+C^{2}}{B^{2}} \\
-\frac{B^{2}\left(B^{2} x_{0}^{2}+A^{2} y_{0}^{2}+\frac{A^{2}}{B^{2}}-B \cdot 2 A x_{0} y_{0}-2 A C x_{0}+\frac{2 A^{2} C y_{0}}{B}\right)}{B^{2}\left(A^{2}+B^{2}\right)} \\
=\frac{A^{2} B^{2} x_{0}^{2}+A^{2} B^{2} y_{0}^{2}+2 A^{2} B C y_{0}+A^{4} C^{2}+B^{4} x_{0}^{2}+B^{4} y_{0}^{2}+2 B^{3} C y_{0}+B^{2} C^{2}}{B^{2}\left(A^{2}+B^{4}\right)} \\
-\frac{B^{4} x_{0}^{2}+A^{2} B^{2} y_{0}^{2}+A^{2} C^{2}-2 B^{3} A x_{0} y_{0}-2 B^{2} A C x_{0}+2 A^{2} B C y_{0}}{B^{2}\left(A^{2}+B^{2}\right)} \\
=\frac{A^{2} B^{2} x_{0}^{2}+B^{4} y_{0}^{2}+2 B^{3} C y_{0}+B^{2} C^{2}+2 B^{3} A x_{0} y_{0}+2 A B^{2} C x_{0}}{B^{2}\left(A^{2}+B^{2}\right)} \\
=\frac{B^{2}\left(A^{2} x_{0}^{2}+B^{2} y_{0}^{2}+C^{2}+2 A B x_{0} y_{0}+2 A C x_{0}+2 B C y_{0}\right)}{B^{2}\left(A^{2}+B^{2}\right)} \\
=\frac{\left(A x_{0}+B y_{0}+C\right)^{2}}{A^{2}+B^{2}}.
\end{array}
$$
$$
\begin{array}{l}
\text { That is, }|P M|^{2} \text { minimum }=\frac{\left(A x_{0}+B y_{0}+C\right)^{2}}{A^{2}+B^{2}}. \\
\text { Therefore, }|P M|_{\text {min }}=\frac{\left|A x_{0}+B y_{0}+C\right|}{\sqrt{A^{2}+B^{2}}} \\
=d.
\end{array}
$$

Using this method to derive the "distance from a point to a line" formula demonstrates the characteristics of analytic geometry, with a clear and natural thought process. Moreover, it only requires that one of $A, B$ is not zero, which avoids the confusion students might have due to inconsistent conditions in the derivation and application of the formula.
(Note: Author Ping Zhu, Baoding No.1 High School, Hebei)