20. Find the condition under which points $A\left(x_{1} ; y_{1}\right), B\left(x_{2} ; y_{2}\right)$ and $C\left(x_{3} ; y_{3}\right)$ lie on the same line.
Solution. If these points lie on the same straight line, then triangle $A B C$ degenerates into a segment. Then, by theorem $1 S_{3}=0$, hence $F_{3}=0$, i.e.

$$
\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(y_{1} x_{2}+y_{2} x_{3}+y_{3} x_{1}\right)=0
$$

from which

$$
x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}=y_{1} x_{2}+y_{2} x_{3}+y_{3} x_{1}
$$