I have 3 lines:
$$ A_1x + B_1y + C_1 = 0 $$ $$ A_2x + B_2y + C_2 = 0 $$ $$ A_3x + B_3y + C_3 = 0 $$
That I believe are dependent. That is, all of the intersections of any pair of two of these lines are exactly the same. Normally I could just show that there is some $\alpha$ such that
$$\alpha(A_1 x + B_1y + C_1) + (A_2x + B_2y + C_2) = (A_3x + B_3y + C_3)$$
but I only have symbolic expressions for the A's, B's, and C's, and they are quite complicated, so finding that $\alpha$ is infeasible. I thought I could instead show that
$$\alpha_A A_1 + A_2 = A_3$$ $$\alpha_B B_1 + B_2 = B_3$$ $$\alpha_C C_1 + C_2 = C_3$$
and show that $\alpha_A = \alpha_B = \alpha_C$, but the $\alpha$'s I find do not seem to be equal (even though plotting the lines shows that they do in fact share a common intersection.
Can anyone comment on what is wrong with this approach? Or suggest a better approach?