130 
on pascal’s theorem. 
[403 
respectively, and we have to find (x, y, z, u, v, w), such that these two equations 
may represent the same plane, or that the two equations may in virtue of the linear 
relations between (X, Y, Z, U, V, W) be the same equation. 
The ordinary process by indeterminate multipliers gives 
1 . 
- + X + aa = 0, 
x 
- -I - A -f- ul) — 0, 
y 
—I- A 4- fie — 0, 
z 
!+x+ M /=o, 
k , 
- + A + yg — 0, 
—h A -f- fill — 0, 
w 
and we have the before-mentioned linear relations between (x, y, z, u, v, w) ; these 
last are satisfied by the values 
1 1 1 
1 
1 
r-‘ 
1 ' 
,a-0’ b-0’ 6-0’ 
1 
X 
1 
9-0’ 
i^t> 
! i 
i 
111 
a — 6^ b—0 + c — 0 
1 
1 
rH 
f-e 
9-0 
h-0 
if only 
in fact, 6 satisfying this equation, the relation 
x + y + z + u + v + iu = 0 
is obviously satisfied; and observing that we have 
= 0; 
n e , -/ , 0 
Q/X ^ — 1 -f- ^, . ., Jll j, p 1 /. /}>••> 
a — 6 a — 6 
we have 
f-e f-e 
ax +%+ cz + fa + gv + hw 
= s (' + a^) _s ( 1 + f=e)' 
= 9 ( s ^-V-«)' =0 - 
so that the relation ax + by + cz + fa + gv + hw = 0 is also satisfied. Substituting the 
foregoing values of (x, y, z, u, v, w) the six equations containing k, A, g, will be all 
of them satisfied if only 
= A = 6, k = — 1.