Is there is any other method to produce a third set of collinear points rathar than the Pappus's hexagon method?

Question

Pappus's hexagon theorem: Given one set of collinear points $A,B,C$, and another set of collinear points $a,b,c$, then the intersection points $X,Y,Z$ of line pairs $Ab$ and $aB$, $Ac$ and $aC,Bc$ and $bC$ are collinear, lying on the Pappus line.

This is the known method used to produce a third set of collinear points. However, the expression for the coordinates of the points $X,Y,Z$ are very long and complicated.

My question is: Is there is any other simpler method to produce a third set of collinear points rathar than the Pappus's hexagon method?

Answer 1

Essentially, no. Pappus and Desargues, at a deep level, follow from the properties of linear systems of cubic curves in the projective plane. By a counting argument, any cubic that contains eight of the nine intersection points of two fixed cubics must necessarily also contain the ninth intersection point as well. This and the fact that a degenerate cubic (one that is a union of curves) must be a conic and a line, or three lines, is the proof of Desargues and Pappus respectively. In that sense, there is no simpler theorem in that category - in fact there is no other theorem in that category, period.

This algebraic argument for results that seem to be pure geometry is what attracted some people to the study of algebraic geometry, a difficult but really cool field in mathematics.

Answer 2

Let $\ell$ be the line containing $A,B,C$ and $m$ the line containing $a,b,c$. Take any line $n$ different from $\ell$ and $m$. Then the intersection points $Aa\cap n$, $Bb\cap n$, and $Cc\cap n$ are collinear.