Neat way to calculate intersection of 2 lines using 4 points. I'm a minecraft speedrunner. We throw pearls to locate a stronghold. It's hard to explain, but here's an explanation:
Image explaining it here
So we know the points (x1,y1), (x2,y2) which are on one line. and we know (x3,y3), (x4,y4), which are on another line. We want to know the intersection of these points, here called (x,y). We are not allowed to use calculators during runs, but we are doing a 10+ people speedrun so we can have some people crunching the numbers with pen and paper. We decided to put them in as vectors, and then take the x and y parts of those vectors, and solve for that. From the x equation we got:
$$a = \frac{x_3-x_1+b \cdot (x_3-x_4)}{x_1-x_2} $$
Filling that in into the y equation gives:
$$b=\frac{(y_1-y_2)\cdot(y_1(x_1-x_2) + x_3-x_1)-y_3(x_1-x_2)}{((x_3-x_4)(y_3-y_4)-(x_3-x_4))}$$
Filling those in in $x=x_3+b(x_3-x_4)$ and $y=y_3+b(y_3-y_4)$ gives:
$$x=x_3+\frac{(y_1-y_2)\cdot(y_1(x_1-x_2) + x_3-x_1)-y_3(x_1-x_2)}{((y_3-y_4)-1)}$$
$$y=y_3+\frac{
(y_1-y_2)\cdot(y_1(x_1-x_2) + x_3-x_1)-y_3(x_1-x_2)}{((x_3-x_4)(y_3-y_4)-(x_3-x_4))}(y_3-y_4)$$
However, this is still pretty complicated. We need to be able to do this during a run, by hand with just pen and paper. Are there any other, less convoluted ways to do this?
Thanks in advance!
EDIT: We can walk to 0,0 to set (x1,y1) = (0,0) which turns the x equation into $$x=x_3+\frac{
-y_2x_3+y_3x_2}{y_3-y_4-1}$$
and the y equation after setting (x3,y3) = (0,0) to $$y=\frac{
(y_1-y_2)\cdot(y_1(x_1-x_2) -x_1)}{(x_4y_4+x_4)}(-y_4)$$
Which is sliiightly nicer already I guess
EDIT 2:
To the person who suggested cramer's rule and deleted their comment: thank you so much! That really helps, we're gonna try to get a formula from that which is probably a LOT easier
 A: This is to document how Cramer's rule may be used to find the intersection point. First, we may express the equation of the first line in point-slope form as $$y-y_1=\frac{x-x_1}{x_2-x_1}(y_2-y_1).$$
After some rearrangement, we obtain
$$(y_2-y_1)x+(x_1-x_2)y=x_1y_2-x_2 y_1.$$
If we repeat this form the second line, we instead obtain
$$(y_4-y_3)x+(x_3-x_4)y=x_3y_4-x_4 y_3.$$
Cramer's rule then gives the solution in terms of determinants as
$$x=\begin{vmatrix} x_1 y_2-x_2 y_1 & x_1-x_2 \\ x_3y_4-x_4 y_3 & x_3-x_4\end{vmatrix}\Bigg/\begin{vmatrix} y_2-y_1 & x_1-x_2 \\ y_4-y_3 & x_3-x_4\end{vmatrix},$$
$$y=\begin{vmatrix} y_2-y_1 & x_1 y_2-x_2 y_1 \\  y_4-y_3 & x_3y_4-x_4 y_3\end{vmatrix}\Bigg/\begin{vmatrix} y_2-y_1 & x_1-x_2 \\ y_4-y_3 & x_3-x_4\end{vmatrix},$$
where $\begin{vmatrix} a & b \\ c & d\end{vmatrix}=ad-bc$. For that matter we may also write $$x_1 y_2 -x_2 y_1=\begin{vmatrix} x_1 & x_2 \\ y_1 & y_2 \end{vmatrix},\quad x_3 y_4 -x_4 y_3=\begin{vmatrix} x_3 & x_4 \\ y_3 & y_4 \end{vmatrix}$$
though this may be more tedious than useful. It should be noted that the above is mostly useful as a way of organizing the computation: Ultimately, the results are the same as would be obtained by the seemingly-uglier equations in the OP.
