Pascal's theorem has a short proof using the Cayley–Bacharach theorem that given any 8 points in general position, there is a unique ninth point such that all cubics through the first 8 also pass through the ninth point. In particular, if 2 general cubics intersect in 8 points then any other cubic through the same 8 points meets the ninth point of intersection of the first two cubics. Pascal's theorem follows by taking the 8 points as the 6 points on the hexagon and two of the points (say, and in the figure) on the would-be Pascal line, and the ninth point as the third point ( in the figure). The first two cubics are two sets of 3 lines through the 6 points on the hexagon (for instance, the set , and the set ), and the third cubic is the union of the conic and the line . Here the "ninth intersection" cannot lie on the conic by genericity, and hence it lies on .