For $v=48$, points either lie on four 9-lines (types A), or one 9-line
(types B).  If there are $M$ type A points, then divisibility
conditions for 9-lines imply that $M\equiv2$ mod 3, say $M=2+3m$, and
$N_9=6+m$; if there are at most 43 type B points, and at least 7
9-lines, then some 9-line contains at least 3 type A points, and so
$N_9\geq 3*(4-1)+1=10$. Repeating the above arguement with 34, 10, 6,
$N_9\geq19$, and with 7, 19, 9 $N_9\geq28$ eliminates all possible
point distributions except $M=2$ and $N_9=6$, but then these two type
A points have 8 incidences on the 6 9-lines, which causes too many
pairs, so there is no viable distribution.


-- Malcolm Greig
Jan. 1998