21 
1889 - 90 .] Dr T. Muir on Pascal’s Theorem. 
so that we are thus furnished with the useful identity 
| \Va^\ , fax y \ , hVe\ | = Kyp*s\ • \m^\ • 
Applying this four times, therefore, to our equation of condition, 
we have at once the surprisingly simple result 
- l^5%l-l*5^?2l-!*2% 2 4lrlw ! il= 0 ; 
or, as Cayley would write it, 
123.245 . 561 .634 - 456 . 5T2 . 234. 36l = 0 ; 
or, perhaps better still, 
T23 . 156 . 425 . 436 
-125.136.423.456 = 0. 
Now, if one of the points 1, 2, 3, 4, 5, 6, — say 1, — be viewed as 
current , this is evidently a cone of the second order, with its vertex 
at 0. And since 1 is associated in both of the terms with 2, 3, 
5, 6, both terms will vanish, and the equation therefore be satisfied 
when for (xtfft) we write (x 2 y 2 z 2 ) or (x 3 y&) or (xyy b z b ) or (x 6 y 6 z 6 ). 
Also it will be satisfied when (a? 4 y 4 z 4 ) is substituted, because then 
the two terms are alike and of opposite signs. The theorem in 
question is thus established. 
On comparison of this with Cayley’s, investigation, it will at once 
be seen that not only is the resulting equation here much simpler 
than his, but that it is also greatly superior for the purpose he had 
in view. 
There still remains the question as to the reconcilement of the 
two processes. Cayley’s resulting equation is 
145. 256. 423. MI + 245 . 123 . 456 . 36l 
- 245. 123. 356. 461 - 245.156.423.^1 = 0. 
Can it be simplified ? Observing that the two middle terms have 
more than one common factor, we write the aggregate of the two in 
the form 
245.123 (456. 361- 356. 461). 
The bracketed expression is then seen to be equal to 
346. 516, 
