2. A helicopter and an airplane flew towards each other at the same time. By the time they met, the helicopter had flown $100 \mathrm{~km}$ less than the airplane, and the helicopter reached the airplane's takeoff location 3 hours after the meeting. The airplane reached the helicopter's takeoff location 1 hour and 20 minutes after the meeting. Determine the speeds of the helicopter and the airplane, and the distance between the takeoff locations.
Solution. Let the helicopter fly $x \mathrm{~km}$ to the meeting. Then the airplane flew $(x+100) \mathrm{~km}$ to the meeting. The speed of the helicopter is $\frac{x+100}{3} \mathrm{~km} / \mathrm{h}$, and the speed of the airplane is $\frac{x}{1 \frac{1}{3}} \mathrm{~km} / \mathrm{h}$. Thus, the time the helicopter flew from its takeoff point to the meeting is $x: \frac{x+100}{3}=\frac{3 x}{x+100}$ hours, and the time the airplane flew from its takeoff point to the meeting is $\frac{1 \frac{1}{3}(x+100)}{x}$ hours. They flew the same amount of time to the meeting, so

$$
\frac{1 \frac{1}{3}(x+100)}{x}=\frac{3 x}{x+100}, \text { i.e. }\left(\frac{x+100}{x}\right)^{2}=\frac{9}{4}
$$

From this, since $\frac{9}{4}=\left(\frac{3}{2}\right)^{2}=\left(-\frac{3}{2}\right)^{2}$, we get

$$
\frac{x+100}{x}=\frac{3}{2} \text { or } \frac{x+100}{x}=-\frac{3}{2} \text {. }
$$

However, $\frac{x+100}{x}>0$, so $\frac{x+100}{x}=\frac{3}{2}$, from which we get $x=200 \mathrm{~km}$.

Thus, the helicopter flew $200 \mathrm{~km}$, the airplane flew $300 \mathrm{~km}$, the speed of the helicopter is $100 \mathrm{~km} / \mathrm{h}$, and the speed of the airplane is $150 \mathrm{~km} / \mathrm{h}$. The distance between the takeoff points is $500 \mathrm{~km}$.