This is really a comment but will be too long and, anyway, I'm not entitled. There is a standard approach to this problem but I haven't carried through the computations.  Let me indicate the method for euclidean case first. Suppose we have points $A_1,\dots,A_6$ in the plane, with the first five fixed, say $A_1=(a_{11},a_{12})$ etc. but $A_6=(x,y)$ variable. Now compute $X,Y,Z$,the intersections of $A_2 A_6$ and $A_5A_3$, etc.  Then apply the condition for these three points to be collinear. This turns out to be a quadratic in $x$ and $y$ which proves Pascal's theorem AND its converse.
These computations are rather tedious but can be carried in a few minutes with, say, Mathematica.  They can even be done by hand by assuming wlog that three of the original points are $(0,0),(1,0),(0,1)$.