How was this equation calculated? I found this page on the intersection of 2 lines. And I'm really surprised about going from:
$$\begin{align*}
x_1 + u_a (x_2 - x_1) &= x_3 + u_b (x_4 - x_3)  \\\
y_1 + u_a (y_2 - y_1) &= y_3 + u_b (y_4 - y_3)   
\end{align*}$$
to this
$$\begin{align*}
u_a &= \frac{(x_4 - x_3)(y_1 - y_3) - (y_4-y_3)(x_1-x_3)}{(y_4-y_3)(x_2-x_1)-(x_4-x_3)(y_2-y_1)} \\\
u_b &= \frac{(x_2-x_1)(y_1-y_3)-(y_2-y_1)(x_1-x_3)}{(y_4-y_3)(x_2-x_1)-(x_4-x_3)(y_2-y_1)}
\end{align*}$$
Could somebody carry it for me cause I always fail and get other final equation.

 A: Rewrite as $Au_a+Bu_b=C,Du_a+Eu_b=F$ where $A=x_2-x_1,B=-(x_4-x_3),C=x_3-x_1,D=y_2-y-1,E=-(y_4-y_3),F=y_3-y_1$. Multiply first equation by $E$, second by $B$, subtract to get $(EA-BD)u_a=EC-BF$, so $$u_a={EC-BF\over EA-BD}={-(y_4-y_3)(x_3-x_1)+(x_4-x_3)(y_3-y_1)\over-(y_4-y_3)(x_2-x_1)+(x_4-x_3)(y_2-y-1)}$$ Then do a little fiddling to see if this is the given answer. Then do a similar thing to get $u_b$.  
A: Rewrite the equations as
$$
\begin{align}
u_a(x_2-x_1)+u_b(x_3-x_4)&=x_3-x_1\\
u_a(y_2-y_1)+u_b(y_3-y_4)&=y_3-y_1
\end{align}
$$
Then solve using Cramer's Rule:
$$
\begin{align*}
u_a&=\frac{\begin{vmatrix}x_3-x_1&&x_3-x_4\\y_3-y_1&&y_3-y_4\end{vmatrix}}{\begin{vmatrix}x_2-x_1&&x_3-x_4\\y_2-y_1&&y_3-y_4\end{vmatrix}}\\
&=\frac{(x_3-x_1)(y_3-y_4)-(y_3-y_1)(x_3-x_4)}{(x_2-x_1)(y_3-y_4)-(y_2-y_1)(x_3-x_4)}
\end{align*}
$$
and
$$
\begin{align*}
u_b&=\frac{\begin{vmatrix}x_2-x_1&&x_3-x_1\\y_2-y_1&&y_3-y_1\end{vmatrix}}{\begin{vmatrix}x_2-x_1&&x_3-x_4\\y_2-y_1&&y_3-y_4\end{vmatrix}}\\
&=\frac{(x_2-x_1)(y_3-y_1)-(y_2-y_1)(x_3-x_1)}{(x_2-x_1)(y_3-y_4)-(y_2-y_1)(x_3-x_4)}
\end{align*}
$$
which after some negations in the numerator and denominator that cancel, gives the answer you cite.
