56 
ON A LOCUS IN RELATION TO THE TRIANGLE. 
[394 
The original cubic equation is 
(olol'X I 2 + (3(3'y 2 + 77 V) xyz + OioiXyz (/iy + vz) + (3(3'yzx (Xx + vz) + y y vxy (Xx + yy) = 0, 
and this in fact is 
(iOLoiXyz + (3/3'yzx + 77'vxy) (Xx + yy + vz) = 0. 
The equation Xx + yy + vz = 0 is that of the line through the two points which 
constitute the Absolute; the other factor gives 
aa'Xyz + (3(3'yzx + 77 vxy = 0, 
which is the equation of a conic through the angles of the triangle (x = 0, y = 0, 0 = 0), 
and which also passes through the two points of the Absolute; in fact, writing (a, /3, 7) 
for (x, y, z) the equation becomes a(3y (a'X + (3'y + y'v) = 0, and so also writing (a', /9', y) 
for (x, y, z) it becomes a (3'y' (aX + (3y + yv) = 0, which relations are identically satisfied 
by the values of (X, y, v). Hence we see that the Absolute being a point-pair, the 
locus is the conic passing through the angles of the triangle, and the two points of 
the Absolute; that is, it is the circle passing through the angles of the triangle. 
But assuming that K' is not = 0, or that the Absolute is a proper conic, the 
equation 0=0 will be satisfied if 
AFG'H' + BGH'F' + CHF'G' -f F'G'H’ = 0, 
we have F', G', H’ = Kf, Kg, Kh respectively, or omitting the factor K-, the equation 
becomes 
AFgh + BGhf+ GHfg + Kfgh = 0, 
f-g-fi 2 — bcg-h 2 — cah-f - — abf 2 g 2 + labcfgh = 0, 
which is 
or, as it may also be written, 
I remark that we have ABC — FGH = K(abc —fgh); substituting also for F, G', H' 
the values Kf Kg, Kh, the equation of the cubic curve is 
2 (abc — fgh) xyz + Ayz (hy + gz) + Bzx (hx +fy) + Cxy (gx + fy) = 0, 
and the transformed form is 
we have 
so that the foregoing condition 
1 _ 1_ 1 _ 1 2__ 
abc a/ 2 bg 2 ch 2 + fgh ’