963] 
NOTE ON DR MUIRS PAPER ON ELIMINATION. 
547 
which are cK — 0, fK = 0, gK=0, all satisfied by If = 0; and similarly for the first 
and third, and the second and third equations. It will be remembered that the 
true form of the result is not K= 0 but K 2 = 0, and this seems to be an indication 
that the three equations should be, as they have been found to be, equivalent to a 
single equation. 
The problem may be further illustrated as follows: instead of the original 
equations TJ = 0, V=0, W= 0, consider the like equations with the inverse coefficients 
(A, B, G, F, G, H), viz. 
Bz 2 — 2 Fyz + Gy 1 — 0, 
Ox 2 — 2 Gzx + Az 2 = 0, 
Ay 2 — 2 Hxy + Bx 2 = 0, 
so that the result of the elimination should be 
(ABC - AF 2 - BG 2 - CH 2 + 2FGH) 2 = 0. 
Here considering in connexion with the triangle x = 0, y = 0, z = 0 (say the vertices 
hereof are the points A, B, C) the conic 
(a, b, c, /, g, h) (x, y, z) 2 = 0, 
the first equation represents the pair of tangents from the point A to the conic, 
the second the pair of tangents from the point B to the conic, and the third the 
pair of tangents from the point C to the conic. The first and second pairs of 
tangents intersect in four points, and if one of the third pair of tangents passes 
through one of the four points, then it is at once seen that the conic must touch 
one of the sides x = 0, y = 0, z — 0 of the triangle, viz. we must have be —f 2 = 0, 
ca —g 2 = 0, or ab — h 2 = 0. But we have a = BG — F 2 , &c., or writing 
K, = ABC - AF 2 - BG 2 - CH 2 + 2FGH, 
then these equations are K 1 A = 0, K 1 B = 0, K^C = 0, all satisfied by 1^ = 0. We may 
regard K 1 — 0 as the condition in order that the conic (a, b, c, f g, h) (x, y, z) 2 = 0 
may be a point-pair: the analytical reason for this is not clear, but we see at once 
that, if the conic be a point-pair, then the three pairs of tangents are the lines 
drawn from the points A, B, G respectively to the two points of the point-pair, so 
that the three pairs of tangents have in common these two points. Regarding 
K 1 = 0 as the condition in order to the existence of a single common point, and 
recollecting that the true result of the elimination is K 1 2 =0, the form perhaps 
indicates what we have just seen is the case, that there are in fact two common 
points of intersection: but at any rate the foregoing geometrical considerations lead 
to K! = 0, as the condition for the coexistence of the three equations. 
I remark in conclusion that I do not know that there is any general theory of 
the case where a result of elimination presents itself in the form il 2 = 0, as distinguished 
from the ordinary form il = 0. 
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