We define as balanced line the line that connect one red point and one blue point that divides the plane with the same amount of $m$ red points and $m$ blue points in one side and $n$ red points and $n$ blue points in the other side. The image below shows that $r_1$ and $r_2$ are balanced lines, but $r_3$ is not. Note that $m$ is not necessarily equals to $n$.
Exercise: Given $2n$ points in the plane, $n$ blue points and $n$ red points, no $3$ are collinear. Show that we have at least $2$ balanced lines.
I know that the objective of this exercise is to use combinatorics associated with geometry.
I tried two different approachs:
The first one: Make a $s$ line that is not perpendicular to any line that connect one red point to one blue point. We do de projection of these points to the line and try to make some combinatoric, but I couldn't proceed.
The second one: I tried by absurd that doesn't exist any balanced line, but I couldn't reach a contradiction.