10、Person A and Person B start from locations $A$ and $B$ respectively at the same time and walk towards each other. When A reaches the halfway point, B doubles his speed. As a result, they meet 1200 meters away from location $B$ and arrive at their destinations simultaneously. The distance between the two locations is $\qquad$ meters.
[Answer] 2800
[Solution] Let the distance between the two places be $S$. The original speeds of A and B are $x, y$, respectively. When A reaches the halfway point, B has traveled $\frac{S y}{2 x}$. After that, B doubles his speed, and then B travels $1200-\frac{S y}{2 x}$, while A travels $\frac{S}{2}-1200$ until they meet. Therefore, we get the equation $\frac{1200-\frac{S y}{2 x}}{2 y}=\frac{\frac{S}{2}-1200}{x}$. Since both finally arrive at the same time, we have $\frac{S}{x}=\frac{S}{2 x}+\frac{S-\frac{S y}{2 x}}{2 y}$. Next, we solve these two equations:
$$
\begin{array}{l}
\frac{1200-\frac{S y}{2 x}}{2 y}=\frac{\frac{S}{2}-1200}{x} \Rightarrow x\left(1200-\frac{S y}{2 x}\right)=2 y\left(\frac{S}{2}-1200\right) \Rightarrow 1200 x-\frac{S y}{2}=y S-2400 y \\
\frac{S}{x}=\frac{S}{2 x}+\frac{S-\frac{S y}{2 x}}{2 y} \Rightarrow \frac{S}{2 x}=\frac{S-\frac{S y}{2 x}}{2 y} \Rightarrow 2 y S=2 x S-S y \Rightarrow 3 y S=2 x S \Rightarrow y=\frac{2}{3} x. \text{ Substituting } \\
y=\frac{2}{3} x \text{ into } 1200 x-\frac{S y}{2}=y S-2400 y \text{ gives } \\
1200 x-\frac{S}{3} x=\frac{2}{3} x S-2400 \cdot \frac{2}{3} x \Rightarrow 2800 x=x S \Rightarrow S=2800
\end{array}
$$