Example 2. Given a directed line segment $P Q$ with the coordinates of the starting point $P$ and the end point $Q$ being $(-1,1)$ and $(2,2)$ respectively. If the line $l_{1} x+m y+m=0$ intersects the extension of $P Q$, then the range of values for $m$ is $\qquad$
Let $k_{l}$ denote the solution for the line $l$, then
$$
k_{i}=-\frac{1}{m} \text {. }
$$

One of the extreme positions of this slope is the line $l_{1}$ passing through $Q$. Substituting the coordinates of $Q$ $(2,2)$ into the equation of $l_{1}$, we get
$$
2+2 m_{i}+m_{1}=
$$
0,
$$
\begin{aligned}
\therefore k_{l_{1}} & =-\frac{1}{m_{1}} \\
& =\frac{3}{2} .
\end{aligned}
$$

The other extreme position is
$l_{2} / / P Q$ (as shown in the figure).
$$
\begin{array}{c}
\because \quad k_{t_{2}}=k_{P Q}=\frac{2-1}{2+1} \\
=\frac{1}{3}, \\
\therefore \frac{1}{3}<-\frac{1}{m}<\frac{3}{2} .
\end{array}
$$

From this, we can deduce that the range of values for $m$ is
$$
-3<m<-\frac{2}{3} \text {. }
$$

Using extremes in conjunction with special cases can yield even greater effectiveness. Consider the following example.