Past exam of the geometry course of the University of Oxford / 2007 / B / 5 / ii

By the preceding frame result take the four points to be $[1:0:0],[0:1:0],[0:0:1],[1:1:1]$. The conics through them have equationsCoefficient vectors differing by a common nonzero scalar define the same conic. Their parameter space is the projectivization of a two-dimensional vector space, hence a projective line.

Each partition of the four points into two unordered pairs gives a reducible conic formed by the two joining lines. There are three pairings. Conversely a reducible conic is a union of two lines, possibly repeated. General position prevents any one line from carrying three of the points, so the lines must be distinct and carry exactly one of these pairings. Thus there are exactly three reducible members. In these coordinates they are the parameters $[0:1:-1]$, $[1:0:-1]$, $[1:-1:0]$, giving respectively $x(z-y)=0$, $y(z-x)=0$, and $z(y-x)=0$.