Hi could someone show me the steps involved in solving for $U_a$ and $U_b$. Also what is this method called. $ 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) $
Cheers
Edit : Attept to solve as Karolis suggested
Ok so first I will try and solve for $ U_a $
$ x_1 + U_a(x_2 - x_1) = x_3 + U_b(x_4 - x_3) $ $ U_a(x_2 - x_1) = x_3 - x_1 + U_b(x_4 - x_3) $ $ U_a = \frac{x_3 - x_1 + U_b(x_4 - x_3)}{(x_2 - x_1)} $
So now I have $ U_a $ I can substitute that in and solve for $ U_b $
$ y_1 + U_a(y_2 - y_1) = y_3 + U_b(y_4 - y_3) $ $ y_1 + \left(\frac{x_3 - x_1 + U_b(x_4 - x_3)}{(x_2 - x_1)}\right)(y_2 - y_1) = y_3 + U_b(y_4 - y_3) $
Alight so that is substituted. I should be able to solve for $ U_b $
$ y_1 + \left(\frac{x_3 - x_1 + U_b(x_4 - x_3)}{(x_2 - x_1)}\right)(y_2 - y_1) = y_3 + U_b(y_4 - y_3) $
$ \left(\frac{x_3 - x_1 + U_b(x_4 - x_3)}{(x_2 - x_1)}\right)(y_2 - y_1) = y_3 - y_1 + U_b(y_4 - y_3) $
$ \frac{x_3 - x_1 + U_b(x_4 - x_3)}{(x_2 - x_1)} = \frac{ y_3 - y_1 + U_b(y_4 - y_3)}{ (y_2 - y_1) } $
$ x_3 - x_1 + U_b(x_4 - x_3) = \frac{ (y_3 - y_1 + U_b(y_4 - y_3))(x_2 - x_1)}{ (y_2 - y_1) } $
$ (x_3 - x_1 + U_b(x_4 - x_3))(y_2 - y_1) = (y_3 - y_1 + U_b(y_4 - y_3))(x_2 - x_1) $
$ (x_3 - x_1 + U_bx_4 - U_bx_3)(y_2 - y_1) = (y_3 - y_1 + U_by_4 - U_by_3)(x_2 - x_1) $
$ x_3(y_2 - y_1) - x_1(y_2 - y_1) + U_bx_4(y_2 - y_1) - U_bx_3(y_2 - y_1) = y_3(x_2 - x_1) - y_1(x_2 - x_1) + U_by_4(x_2 - x_1) - U_by_3(x_2 - x_1) $
$ U_bx_4(y_2 - y_1) - U_bx_3(y_2 - y_1) = y_3(x_2 - x_1) - y_1(x_2 - x_1) + U_by_4(x_2 - x_1) - U_by_3(x_2 - x_1) - x_3(y_2 - y_1) + x_1(y_2 - y_1) $
$ U_bx_4(y_2 - y_1) - U_bx_3(y_2 - y_1) + U_by_3(x_2 - x_1) - U_by_4(x_2 - x_1) = y_3(x_2 - x_1) - y_1(x_2 - x_1) - x_3(y_2 - y_1) + x_1(y_2 - y_1) $
$ U_b\left(x_4(y_2 - y_1) - x_3(y_2 - y_1) + y_3(x_2 - x_1) - y_4(x_2 - x_1)\right) = y_3(x_2 - x_1) - y_1(x_2 - x_1) - x_3(y_2 - y_1) + x_1(y_2 - y_1) $
$ U_b = \frac{y_3(x_2 - x_1) - y_1(x_2 - x_1) - x_3(y_2 - y_1) + x_1(y_2 - y_1)}{x_4(y_2 - y_1) - x_3(y_2 - y_1) + y_3(x_2 - x_1) - y_4(x_2 - x_1)} $
Now this solution seems to be correct for $U_b$ but I have seen this can be further simplified to
$ U_b = \frac{(x_2-x_1)(y_1-y_3)-(y_2-y_4)(x_1-x_3)}{(y_4-y_3)(x_2-x_1)-(x_4-x_3)(y_2-y_1)} $
I have checked my solution against this one and checked both return the same result. My next question is how do I simplify mine to something similar to the above?
Any guidance would be highly appreciated.