It is well known that a finite set of $n$ points cannot form more than
$$\bigg\lfloor \frac{n(n-3)}{6} \bigg\rfloor+1 $$
lines that include $3$ points. Would this result still hold if we assume that the set $P$ of $n$ points has some sort of central symmetry i.e. $k$-fold symmetry for some $k\in \mathbb{N}$? Or would the upper bound be smaller?