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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.

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typo in the title – Pierre-Yves Gaillard Aug 20 '11 at 12:52
@Pierre, took it upon myself to fix it. – Gerry Myerson Aug 20 '11 at 12:56
A useful keyword in this context is Cramer's rule and a detailed answer to your question is written here:… – Did Aug 20 '11 at 13:10
@Gerry: Thanks! (I don't have edit privileges.) – Pierre-Yves Gaillard Aug 20 '11 at 13:17

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 $u_b$.

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See also Cramer's rule. – Hans Lundmark Aug 20 '11 at 13:08

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.

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Teeny-tiny note: you can use the vertical bars, or you can use $\det$. Using both at the same time, though, looks sorta kinda redundant... :) – J. M. Aug 21 '11 at 15:12
@J. M.: Thanks. I was thinking $\det$, so I subconsciously used vertical bars instead of square brackets. – robjohn Aug 21 '11 at 15:50

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