132 
on pascal’s theorem. 
[403 
bg; and C + H = 0 is the equation of the plane through 0 and the line ch; and the 
three equations being equivalent to two equations only, the planes have a common 
line which is the line in 
question. 
The equations of the 
six lines thus 
are: 
(1) 
A +F =0, 
B+G = 0, 
G + H= 0, 
(2) 
A + G = 0, 
B + H = 0, 
G + F = 0, 
(3) 
A + H = 0, 
5 + ^=0, 
C + G = 0, 
(4) 
A + F =0, 
B + H = 0, 
C+G = 0, 
(5) 
A + G — 0, 
B + F =0, 
C + H = 0, 
(6) 
¿ + # = 0, 
B+G=0, 
C + F = 0. 
It is further to be noticed, that if in any one of these systems, for instance in the 
system A + F = 0, B + G = 0, G + H = 0, we consider 6 as an arbitrary quantity, then 
the equations are those of any line whatever cutting the lines af, bg, ch; and hence 
eliminating 6, we have the equation of the hyperboloid through the three lines 
af bg, ch; the equations of the six hyperboloids are thus found to be 
(1) 
ax +fu 
x + u 
by + gv 
y + v 
cz + hw 
z + w 
(2) 
ax + gv 
x 4- v 
by + hw 
y + w 
cz + fu 
z + u 
(3) 
ax + hw 
X + w 
by +fu 
y + u 
cz + gv 
z + v 
(4) 
ax + fu 
x + u 
by + hw 
y+iv 
cz + gv 
z + v 
(5) 
ax + gv 
x + v 
by +fu 
y+u 
cz + hw 
z + w 
(6) 
ax + hw 
by + gv 
cz + fu 
x+w 
y + v 
z + u 
respectively; the equations in the same line being of course equivalent to a single 
equation. 
For each one of the six lines we have 
(A, B, C) = (— F, -G, - H) 
in some order or other, and it is thus seen that the six lines lie on a cone of the 
second order, the equation whereof is 
A 2 + B 2 + C 2 - F 2 - G 2 -H 2 = 0.