144 ON THE AREA OF THE CONIC SECTION REPRESENTED BY THE [192 
The expression for the distance of the two points 0, P is readily obtained in 
terms of the relative coordinates, viz. calling this distance r, we have 
r 2 = Lp 4- Mrf + Nip, 
where, if l, m, n are the sides of the triangle ABC, we have 
L = \ (m 2 + n" + l 2 ), 
M = \ (n 2 + l 2 — m 2 ), 
N = | (l 2 + m 2 — n 2 ) ; 
and it is to be remarked that these values give 
MN + NL + LM = (2m 2 w 2 + 2n 2 l 2 + 2l 2 m 2 — l 4 — to 4 — n 4 ), — 4A 2 , 
if A denote the area of the triangle ABC. 
Consider now a conic 
(a, b, c, f g, K$se, y, z) 2 , 
and suppose as usual that 51, 33, (£, 8, ©, are the inverse coefficients and that K 
is the discriminant, suppose also for shortness 
P = (21, 33, 6, g, @, ££1, 1, l) 2 . 
The coordinates of the centre being a, /3, y, we have 
«= p (a, •§, @$i, i, i), 
/3 = p(-&, 9, S $1, 1. 1), 
7=p(®, S$1, 1, 1), 
and writing as before £,77, £ for x — cl, y — ¡3, z — y, so that f, 77, £ are the coordinates 
of a point P of the conic, in relation to the centre, we have x, y, z respectively 
equal to f + a, 77 -h /3, £+y, and the equation of the conic gives 
(a, ...$£ + a, 77+73, £+y) 2 = 0, 
(a, ...$£ 77, £) 2 
+ 2 (a, /3, y) (f, 77, £) 
+ (a, /3, y) 2 = 0. 
which may be written