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For six distinct points on an irreducible conic in the complex projective plane, the three intersections of opposite sides of the resulting hexagon are collinear. Form the two cubics given by the products of alternate side equations. They meet at the six conic points and the three opposite-side intersections. Applying residual intersection through a curve pencil with degrees three and two places the last three points on a line. This is the conic counterpart of Pappus's hexagon theorem for two lines.

Let $u=(1,1,1)$ and $v=(1,\alpha ,\beta )$. The vector representing $r_{23}$ satisfiesConsequently the three representative vectors are linearly dependent, so $r_{12},r_{13},r_{23}$ are collinear in the projective plane. Equivalently their three-by-three determinant vanishes. This is the two-line configuration of Pappus's hexagon theorem.