I use the complex structure operator in this discussion. In case it is unfamiliar to you, think of it as doing the same thing to a vector in $\mathbb{R}^{2}$ as multiplication of a complex number by $\mathrm{i}=\sqrt{-1}.$ That is, it rotates by $\pi/2.$ For example, it relates to the wedge product as follows:
\begin{align*} \mathfrak{p}_{i}= & \begin{bmatrix}x_{i}\\ y_{i} \end{bmatrix}\\ \mathscr{J}\left[\mathfrak{p}_{1}\right]= & \begin{bmatrix}-y_{1}\\ x_{1} \end{bmatrix}=\begin{bmatrix}0 & -1\\ 1 & 0 \end{bmatrix}\begin{bmatrix}x_{1}\\ y_{1} \end{bmatrix}\\ \mathscr{J}\left[\mathfrak{p}_{1}\right]\cdot\mathfrak{p}_{2}= & \begin{bmatrix}-y_{1} & x_{1}\end{bmatrix}\begin{bmatrix}x_{2}\\ y_{2} \end{bmatrix}\\ = & \mathfrak{p}_{1}\wedge\mathfrak{p}_{2}\\ = & x_{1}y_{2}-x_{2}y_{1}. \end{align*}
In his 5th lecture on Universal Hyperbolic Geometry, Norman Wildberger provides the following theorem (using my notation):
https://youtu.be/YDGUnGGkaTs?t=450
Line through two points theorem: For any two points $\mathfrak{p}_{1}$ and $\mathfrak{p}_{2},$there is a unique line $\mathcal{L}$ that passes through them:
\begin{align*} \mathcal{L}= & \left\{ \left(y_{1}-y_{2}\right):\left(x_{2}-x_{1}\right):\left(x_{2}y_{1}-x_{1}y_{2}\right)\right\} . \end{align*}
The proof is left as an exercise.
Well, I got a different result. I certainly understand how he might get it backwards. I have flipped it around a few times while typing this question. Here's my attempted proof.
The line we seek will be parallel to $\Delta\mathfrak{p}=\mathfrak{p}_{2}-\mathfrak{p}_{1}.$ So we may write its homogeneous linear form using the complex structure operator $\tilde{\mathcal{L}}=\mathscr{J}\left[\Delta\mathfrak{p}\right].$ This gives us the general and rational forms of $\mathcal{L}$ as
\begin{align*} 0= & L_{x}x+L_{y}y+L_{o}\\ = & \tilde{\mathcal{L}}\cdot\mathfrak{p}+L_{o}\\ = & \Delta\mathfrak{p}\wedge\mathfrak{p}+L_{o}\\ = & -\Delta yx+\Delta xy+L_{o},\\ \mathcal{L}= & \left\{ -\Delta y:\Delta x:L_{o}\right\} \\ = & \left\{ \left(y_{1}-y_{2}\right):\left(x_{2}-x_{1}\right):L_{o}\right\} . \end{align*}
Which agrees with Wildberger so far. To find $L_{o}$ we put one of our given points $\mathfrak{p}_{1}$ into the general equation and leverage the anti-symmetry of the wedge product:
\begin{align*} 0= & \Delta\mathfrak{p}\wedge\mathfrak{p}_{1}+L_{o},\\ L_{o}= & \mathfrak{p}_{1}\wedge\left(\mathfrak{p}_{2}-\mathfrak{p}_{1}\right),\\ \mathcal{L}= & \left\{ -\Delta y:\Delta x:\left(\mathfrak{p}_{1}\wedge\mathfrak{p}_{2}\right)\right\} \\ = & \left\{ \left(y_{1}-y_{2}\right):\left(x_{2}-x_{1}\right):\left(x_{1}y_{2}-x_{2}y_{1}\right)\right\} . \end{align*}
Notice that my third component is $L_{o}=x_{1}y_{2}-x_{2}y_{1}$ which is the negative of Wildberger's. As a check we can put the other given point into our equation:
\begin{align*} 0= & \Delta\mathfrak{p}\wedge\mathfrak{p}_{2}+\mathfrak{p}_{1}\wedge\mathfrak{p}_{2}\\ = & -\mathfrak{p}_{1}\wedge\mathfrak{p}_{2}+\mathfrak{p}_{1}\wedge\mathfrak{p}_{2}\\ = & x_{2}y_{1}-x_{1}y_{2}+\left(x_{1}y_{2}-x_{2}y_{1}\right). \end{align*}
If my eyes are not lying to me, my result is correct.
So, who is correct; I or Wildberger?
My answer is correct. I reworded his theorem(s) (not to textbook standards) and summarized my demonstration. I used to be shy about using the wedge product in this way in public. Not anymore. This means of demonstration came to me while typing up my question; one of the hidden benefits of asking questions on MSE.
As for the proof of the second part (Wildberger's Collinear Points Theorem), the proof, as they say, is in the putting.