Given $2n$ points in the plane, $n$ blue points and $n$ red points, no $3$ are collinear. Prove that we have at least two "balanced lines".

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.

At least $3$ of the points must be vertices of their convex hull. Let $p$ be one of these, and let $\ell$ be a line through $p$ that has all of the other $2n-1$ points on one side of it. Without loss of generality we may establish a coordinate system with $p$ at the origin, $\ell$ as the $x$-axis, and all of the other points above the $x$-axis, and we may suppose that $p$ is red. The open half space above $\ell$ has $n$ blue points and $n-1$ red points, so that half space has a surplus of $1$ blue point. Now rotate the coordinate system counterclockwise about the origin $p$. The $x$-axis passes through the other $2n-1$ points one at a time. For $0\le\theta\le\pi$ let $d(\theta)$ be the excess of blue points over red points in the open upper half plane, so that $d(0)=1$, and $d(\pi)=0$. Clearly $d(\theta)$ is piecewise constant, changing only when the $x$-axis hits one of the other $2n-1$ points. Specifically, $d(\theta)$ decreases by $1$ when the $x$-axis hits a blue point and increases by $1$ when it hits a red point.

Let $\theta_0=\min\{\theta\in[0,\pi]:d(\theta)=0\}$. Since $d(0)=1>0$, and $d(\theta)$ can change only in steps of $\pm1$, the $x$-axis at angle $\theta_0$ must pass through $p$ and a blue point $q$ and must therefore be a balanced line.

As noted at the beginning, there must be at least one vertex of the convex hull of the $2n$ points that is neither $p$ nor $q$, and the same argument shows that there is a balanced line through it. Thus, there are at least two balanced lines.