1. Given that $AB$ is a line segment of length 8, and point $P$ is at a distance of 3 from the line containing $AB$. Then the minimum value of $AP \cdot PB$ is $\qquad$.
$-, 1.24$
Draw a perpendicular from point $P$ to the line $AB$, with the foot of the perpendicular being $H$. Then
$$
\begin{array}{l}
\frac{1}{2} AB \cdot PH = S_{\triangle PAB} = \frac{1}{2} AP \cdot PB \sin \angle APB \\
\Rightarrow AP \cdot PB = \frac{24}{\sin \angle APB} \geqslant 24 .
\end{array}
$$

When $\angle APB = 90^{\circ}$, the equality holds.
In fact, as shown in the figure
2, with $AB = 8$ as the
diameter, draw $\odot O$, and on one side of $AB$,
draw a parallel line at a distance
of 3 from $AB$ intersecting
$\odot O$ at points $P$ and $P_1$,
connect $AP$ and $BP$. Then in
$\triangle APB$,
$$
\angle APB = 90^{\circ}, AP \cdot PB = AB \cdot PH = 24 \text{. }
$$