<s>[INST] Equation of a line in homogenous coordinates given 2 points in affine coordinates So if I have 2 points $A$ and $B$ such that $F(A) = (1; a, a^3)$, and $F(B) = (1; b, b^3)$. how do I find the equation of this line in homogeneous coordinates? 
So I know how to get a line the "normal" way. If I take the points $a$ and $b$ and represent them as regular Cartesian coordinates $A = (a, a^3), B = (b, b^3)$ and then say a line equation is $y = mx + b$. I could find the slope to be $(b^3-a^3)/(b-a)$ and plug that in for $m$ and then find '$b$' by plugging in the coordinates for $A$ and $B$. 
But then I wouldn't know how to take that $y = mx+b$ and turn it into homogeneous coordinates. I don't think this is necessarily how I'm supposed to go about it. 
 [/INST] In planar geometry the line passing through two points A and B in homogeneous coordinates is found from their vector cross product:
$$ \begin{pmatrix} C_0 \\ C_1 \\ C_2 \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \\ A_2 \end{pmatrix} \times \begin{pmatrix} B_0 \\ B_1 \\ B_2 \end{pmatrix} = \begin{pmatrix} A_1 B_2 - A_2 B_1 \\ A_2 B_0 - A_0 B_2 \\ A_0 B_1 - A_1 B_0 \end{pmatrix} $$
The equation for the line is
$$ C_0 + C_1 x + C_2 y = 0 $$
Example
Two points A and B have coordinates $A=(1,7,2)$ and $B=(1,3,-2)$. The line that joints them is
$$ C = \begin{pmatrix} 1 \\ 7 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -20 \\ 4 \\ -4 \end{pmatrix}$$
with the equation $$ (-20)+(4)x+(-4)y=0 \\ x-y-5 = 0 $$

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