394] 
ON A LOCUS IN RELATION TO THE TRIANGLE. 
55 
that is 
. (A B G \ (x y z \ A 
X V Z + \js yz + qt zx + ^ xyj (jp + p + jpJ =°, 
and the cubic will therefore break up into a line and conic if only 
^.(ABC-FGH)-^-^-A-0, 
and it is easy to see that conversely this is the necessary and sufficient condition in 
order that the cubic may so break up. 
The condition is 
H = 2F'G'H' (ABC-FGH) - AG'-II' 2 - BH' 2 F' 2 - CF' 2 G' 2 = 0, 
we have 
AA' + BB' + GG' = SABO - AF 2 - BG 2 - GIF, = K' + 2 (ABC - FGH), 
and thence 
n = F'G'H' (AA' + BB' + GG' - IF) - A G' 2 II' 2 - BH' 2 F 2 - GF' 2 G' 2 , 
that is 
n = -A G'H' (O'H' - A'F) - BIFF’ (H'F' - EG') - CFG' (F'G' - G'H') - IF F'G'H', 
= - A G' IF IFF - BH'F'K'G - GF’G'K'H - K'F'G'H', 
= -IF (AFG'IF + BGH'F + CII F'G' + F'G'H'), 
so that the condition O = 0 is satisfied if IF = 0, that is if the equation 
(A, B, G, F, G, H^, y, D 2 = 0, 
which is the line-equation of the Absolute breaks up into factors; that is, if the 
Absolute be a point-pair. 
In the case in question we may write 
(A, B, G, F, G, Af$f, v, f)> = 2(af + /3, + 7D(“'f + ^'>! + 7'D, 
that is 
(A, B, C, F, G, II) = (2aa, 2/3/3', 2 7 y', /87' +£7, 7“' + 7«> «£' + «'£), 
whence also, putting for shortness, 
(/3 7 ' -/3' 7 , yct-y'a, aft' — a'/3) = (X, /a, v), 
we have 
(A', B', G', F, G', IF) = - (X 2 , /a 2 , v 2 , fiv, v\, X/a), 
and also 
- v (ABC-FGH)-±-*° 
F'G'H 
F' 2 G' 2 IF 2 
IF = 0, 2 (A BG - FGH) = AA' + BB'+ GG’, = - 2 (aa'X 2 + /3/3> 2 + 77V).