= Solution
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 equations
$$
a yz+b zx+c xy=0,\qquad a+b+c=0,\qquad(a,b,c)\ne0.
$$
Coefficient vectors differing by a common nonzero scalar define the same <conic>. Their parameter space is the projectivization of a two-dimensional <vector space>, hence \b[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 \b[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 $z(y-x)=0$.
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