smc

There are $8$ choices for the first endpoint of the line segment, and $7$ choices for the second endpoint, giving a total of $8\cdot 7=56$ segments. However, both $\overline{AB}$ and $\overline{BA}$ were counted, while they really are the same line segment. Every segment got double counted in a similar manner, so there are really $\frac{56}{2}=28$ line segments, and the answer is $\boxed{28}$. In shorthand notation, we're choosing $2$ endpoints from a set of $8$ endpoints, and the answer is $\left(\genfrac{}{}{0ex}{}{8}{2}\right)=\frac{8!}{6!2!}=28$.