19. (3 points) A male and a female track and field athletes are practicing running on a 110-meter slope (the top of the slope is $A$, the bottom is $B$). Both start from point $A$ at the same time, running back and forth between $A$ and $B$ continuously. If the male athlete's uphill speed is 3 meters per second and downhill speed is 5 meters per second; the female athlete's uphill speed is 2 meters per second and downhill speed is 3 meters per second, then the location of their second face-to-face meeting is $\qquad$ meters away from point $A$.
【Solution】Solution: Let the second meeting point of the two people be $X$ meters away from point $A$,
$$
\begin{array}{l}
\text { then } \frac{110}{5}+\frac{110}{3}+\frac{x}{5}=\frac{110}{3}+\frac{110-x}{2}, \\
\frac{110}{5}+\frac{x}{5}=\frac{110-x}{2}, \\
220+2 x=550-5 x, \\
7 x=330, \\
x=47 \frac{1}{7} ;
\end{array}
$$

Answer: The second meeting point of the two people is $47 \frac{1}{7}$ meters away from point $A$. Therefore, the answer to this problem is: $47 \frac{1}{7}$.