403] 
on pascal’s theorem. 
131 
The coordinates of the required point thus are 
/1 1 1 _ i _ i IN 
U-0’ 6-0’ c — 0’ f-d’ g — 0’ h-e)’ 
where 
_L + _L + 1 i i L__ 0 . 
a-e^b-e^o-t) f-e g-e h-e • 
and, the equation in 0 being of the fourth order, there are thus four points, say the 
points Oi, 0 2 , 0 2 , 0 4 , which have each of them the property in question. 
It will be convenient to designate the planes X = 0, Y= 0, Z = 0, TJ = 0, V=0, 
W = 0 as the planes a, b, c, f g, h respectively; the line of intersection of the planes 
X = 0, Y= 0 will then be the line ah, and the point of intersection of the planes 
X = 0, Y=0, Z =0 the point abc; and so in other cases. 
I say that from any one of the points 0 it is possible to draw 
a line 
meeting the lines af.bg. ch 
(1), 
„ ag .bh.cf 
(2), 
5? 
„ ah .bf .eg 
(3), 
99 
„ af .bli . eg 
(4), 
99 
„ ag.bf. ch 
(5), 
99 
,, ah .bg .cf 
(6), 
and consequently, that the four points 0 are the four common points of the six 
hyperboloids passing through these triads of lines respectively. 
In fact, considering 0 as determined by the foregoing quartic equation, and writing 
for shortness 
{a — 0) X — A, (f-0) U =F, 
(b — 0) Y = B, (g —0) V = G, 
(c-0)Z =C, (h-0) W=H, 
so that 
A + B + C + F+G + H = 0, 
the equations A + F = 0, B + G = 0, C + H = 0, are equivalent to two equations only, and 
it is at once seen, that these are in fact the equations of a line through the point 
O meeting the three lines af, bg, ch respectively. 
The equation A + F = 0, is in fact satisfied by the values X : U = —: ~~ f~Q * 
and by X = 0, U = 0; it is consequently the equation of the plane through O and 
the line af\ similarly, B+G = 0 is the equation of the plane through O and the line 
17—2