24 
ON A PROBLEM OF ELIMINATION. 
[490 
this is right, for writing P = ax + (3y + yz, Q = clx + ¡3'y + y'z, V — bx + b'y + b"z, then 
if two of the intersections of the curve TJ = 0 with the line V= 0 lie in a line 
with the point P = 0, Q = 0, then the point in question, that is the point 
(/3y — /3''7, ya' —y'a, a/3' — a/3), must lie in the line F=0; and the condition reduces 
itself to 
{(/3y' — /3y, yd — y'a, a/3' — a'0^b, V, 6")p TO (m_1) = 0, 
where the index \m (m — 1) is accounted for as denoting the number of pairs of 
points out of the m intersections of the curve U = 0 with the line V = 0. 
If in general Jc = 1, then writing as before P = ax + /3y + 70, Q = dx + /3'y + 7'z, 
we have 
f! = (jSy — (mn—i) ^ '' ysn (n 1) (am—1) ^ # (m-i) (27i-iq 
where 0 = 0 is the condition in order that the point (¡3y — (3'y, ...) may lie in lined 
with two of the intersections of the curves U=0, V=0. Or writing (X, Y, Z) for 
the coordinates of the given point, the condition is 
O = (a, .. №-l) (2m—1) ^ (m—l) (271-1) y ^Mmn (mn-l) = 0. 
I have found that if 
TJ = (a, ...Jx, y, z) m , V=(b, ...][x, y, z) n , 
W = (c, y, z)p, T = (d, y, zf, 
the condition in order that the point (X, F, Z) may lie in lined with one of the 
intersections of the curves U = 0, V= 0, and one of the intersections of the curves 
W = 0, T= 0, is 
n = (a, ...) n w(b, ...) m ^(c, ...) mn P(X, Y, Z) mn P^ = 0. 
Supposing that the curves IF = 0, T = 0 become identical with the curves U=0, 
V=0 respectively, this becomes 
fl = (a,...) n2 - 2m (b,...) m *- 2n (X, F, Z) mn ‘ mn = 0, 
and the variation from the correct form given above is what might have been expected.