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

Write $P_5=[u:v:w]$ in the same normalized coordinates. Its coordinates are all nonzero, because the coordinate lines join pairs of the first three points. Evaluation at $P_5$ gives the additional linear equation $avw+bwu+cuv=0$ on the two-dimensional coefficient plane $a+b+c=0$. It is not identically zero there: that would imply $vw=wu=uv$, hence $u=v=w$ and $P_5=P_4$, contradicting the hypothesis. Its kernel is therefore one-dimensional. Thus there is exactly one projective conic through all five points. If it were reducible, $P_5$ would lie on a line joining two of the other points, contrary to no three being collinear. Hence the unique conic is irreducible.