I have always wanted to solve geometry problems more simply using algebraic methods (for example, point geometry is one way, which uses one variable instead of two in 2D plane geometry). Still, the problem down below is so frustrating (and common, too) that I cannot find a good way to solve it.
I have four points $A(x_A,y_A),B(x_B,y_B),C(x_C,y_C),D(x_D,y_D)$. I want to calculate $E$, the intersection of lines $AC$ and $BD$. I have calculated this simple question for a long time, but it's too hard. Is there any simple way to calculate the coordinates of $E$? For example, using polar coordinates or other coordinate systems?
Since we are working in the plane we can use tricks that are not available in higher dimensions.
(i) Suppose the line $ax+by+c=0$ passes through $(x_1,y_1)$ and $(x_2,y_2)$. Then we have that $(a,b,c)\cdot(x_1,y_1,1)=(a,b,c)\cdot(x_2,y_2,1)=0$. That is, the vector $(a,b,c)$ is perpendicular to the vectors $(x_1,y_1,1)$ and $(x_2,y_2,1)$; and so is parallel to their vector product $(x_1,y_1,1)\times(x_2,y_2,1)=(y_1-y_2,x_2-x_1,x_1y_2-x_2 y_1)$. As $a,b,c$ is only determined up to a scalar we may take $$ (a,b,c)= (y_1-y_2,x_2-x_1,x_1y_2-x_2 y_1). $$
(ii) This lets you write down easily the equations of the lines $AC$ and $BD$.
(iii) Now suppose that we have two lines whose equations are $a_1 x +b_1 y +c_1=0$ and $a_2 x +b_2 y +c_2=0$. Suppose that $(\xi, \eta)$ is the intersection of these lines. Then we have that $(\xi,\eta,1)\cdot(a_1,b_1,c_1)=(\xi,\eta,1)\cdot(a_2,b_2,c_2)=0$, and so $(\xi,\eta,1)$ is perpendicular to both the vectors $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ and so must be parallel to their vector product $(a_1,b_1,c_2)\times(a_2,b_2,c_2)=(b_1c_2-b_2c_1,c_1a_2-c_2a_1,a_1b_2-a_2b_1)$. That is (recall that we want $(\xi,\eta,1)$ with a $1$ in the third place) $$ (\xi,\eta)=(\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}). $$
(iv) This lets you write down easily the intersection of the two lines $AC$ and $BD$.
(v) To sum up: you must calculate the double vector product $$ \left((x_A,y_A,1)\times(x_C,y_C,1)\right)\times\left((x_B,y_B,1)\times(x_D,y_D,1)\right) $$ and then the coordinates of your intersection are got by dividing the first two coordinates of this vector by the third.
(vi) I have assumed that the points $A,B,C,D$ are distinct and that the lines $AC$ and $BD$ are not parallel.
(vii) In the comments the point is made that you can solve your problem by inverting $2\times 2$ matrices. If you do that by Cramer's Rule you are doing exactly the same calculations as I suggest here.