If these points are not collinear, then for any straight line passing any two two points, there exists at least one point which does not locate on this line. Let the distance from this point to that line be $h_i$, $i=1, 2, \cdots$. Assume the smallest of these $h_i$'s be $h$ and the corresponding point is $A$ and corresponding line is $\mathcal{l}$. Because $\mathcal{l}$ passes at least three points, let them be point $B$, $C$, and $D$ in that order, as shown. Let the distance from $C$ to $AB$ and $C$ to $AD$ be $h_1$ and $h_2$, respectively. Then we have $h_1 \ge h$ and $h_2\ge h$.It follows that $$S_{\triangle{ABD}}=\frac{1}{2}\cdot AB\cdot h_1 + \frac{1}{2}\cdot AD\cdot h_2 \ge \frac{1}{2}(AB+AD)\cdot h > \frac{1}{2}\cdot BC\cdot h = S_{\triangle{ABD}}$$ This is impossible. Therefore, the assumption cannot hold which means the original claim is true.