The Sylvester–Gallai theorem states that every finite set of points in the Euclidean plane has a line that passes through exactly two of the points or a line that passes through all of them.
I am trying to solve a related question to the theorem but I am not able to do it.
Question: Let $n$ distinct points are drawn in the plane such that all of them do not lie on a line. (1) Draw all possible lines through at least two points. And (2) Prove that there are at least $n$ lines.
Any help and references would be appreciated
We can do it by induction, as suggested by saulspatz (though Silvester-Gallai is in fact useful here).
The base case $n=3$ (with not all points on the same line) is clear.
Now if $n\geq 3$ and we are given $n+1$ non-collinear points, by Silvester-Gallai we know that there is a line $L$ with only two points on it. Remove one of those two points, so that the remaining $n$ points are not collinear (*). By induction hypothesis the remaining points determine at least $n$ lines (but $L$ is not one of them!). So the original $n+1$ points determine at least $n+1$ lines.
(*) this is always possible: if taking away one of the points (say, $A$) makes the remaining $n$ points lie on the same line $G$, then $A$ can't also lie on $G$ (since the $n+1$ points are non-collinear) and we may remove the other point instead.