This is the famous Sylvester theorem. Let $\mathcal {S}$ be the set of the given $n$ points on the plane. Let $\mathcal {L}$ be the set of all lines passing through at least two of the points in $\mathcal {S}$ Consider a pair $(l,v)$, such that $l \in \mathcal {L}$ and $ v \in \mathcal {S}$. And it satisifies $v$ isn't on $l$, and the distance from $v$ to $l$ is the smallest among all such pairs. Since the points in $\mathcal {S}$ isn't all aligned, the pair exists and have only finite many, which implies that we are able to find the smallest. If $l$ passes exactly $2$ points, then the problem is done. Otherwise we assume that points $A,B,C$ are on line $l$, and WLOG $B$ is between $A$ and $C$. Notice that one of the angles $ \angle vBA$ and $ \angle vBC$ is $\geq \frac{ \pi}{2}$, WLOG $ \angle vBA \geq \frac{ \pi}{2}$, it tells us that $|vA| > |AB|$ , so the distance frome $v$ to $AB$ is greater than the distance from $B$ to $vA$, a contradiction. We are done! $\blacksquare$