= 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 \b[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 \b[the unique <conic> is irreducible].
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