I was in need of a method to compute the intersection of two lines given two points along each line. While searching for such a method, I came across one on Wikipedia that listed the results from determinants. It however does not show how these determinants were derived. I would like to know how they were derived and not just blindly use the method without understanding it.
To clarify, I am not asking how the final version of $(P_{x}, P_{y})$ is derived; thats just the cross product of the determinants themselves. Rather, how did the original determinants come to be? What was the linear system when given 4 points, 2 for each line?
Suppose we have that $(x_1,y_1)$ and $(x_2,y_2)$ determine the line $a_1 x+b_1 y+c_1=0$.
Now write $u_1:=(x_1,y_1,1)$ and $u_2:=(x_2,y_2,1)$, and $w_1:=(a_1,b_1,c_1)$.
Then we have $w_1 \cdot u_1 =0$ and $w_1 \cdot u_2 =0$. Hence $w_1$ is parallel to the cross product $u_1\wedge u_2$.
For the second line $a_2 x+b_2 y+c_2=0$ we have $u_3:=(x_3,y_3,1)$ and $u_4:=(x_4,y_4,1)$, and $w_2:=(a_2,b_2,c_2)$. Hence $w_2$ is parallel to $u_3\wedge u_4$.
The common point $(x,y)$ of the two lines is got by considering $p:=(x,y,1)$. We have that $w_1 \cdot p=0$ and $w_2 \cdot p=0$, so that $p$ is parallel to $w_1 \wedge w_2$.
That is $p=\lambda[ (x_1 \wedge x_2) \wedge (x_3 \wedge x_4)]$, where $\lambda$ is chosen to ensure the third component of $p$ is $1$.
If you use the usual expression for a cross-product as a triple of $2\times 2$ determinants you'll easily see where all the terms in the expression you reference come from.