62 
ON A LOCUS IN RELATION TO THE TRIANGLE. 
[394 
and eliminating the L, M, X, we find 
1 1 
a ’ A 
1 1 
A ’ A ’ 
1 _ 1 
ax + %’ hx 4- by ’ 
or developing and reducing, this is 
ha + A/3 
= 0, 
aa + A/3 
hy 
(ab — h 2 )y^ 1 aa 4- A/3 1 ha + b/3 
hob z h ax + hy h hx + by 
1 aa + A/3 1 ha 4- A/3 _ ^ 
a hx 4- by b ax + hy 
We have still to find the relation between (a, /3, 7) and (x, y, z); this is obtained 
by the consideration that the line A'B', through the two points A', B' the coordinates 
of which are known in terms of (a, ¡3, 7), is the polar of the point C, the coordinates 
of which are (x, y, z). The equation of A'B' is thus obtained in the two forms 
and 
(aa + A/3) X 4- (ha + A/3) F - ( №a + A^) (Aa + A^) z = ^ 
(ax + hy)X+ (hx+ by) Z + cz Z= 0, 
and comparing these, we have 
x : y : z = a : (3 : 
or what is the same thing 
a : /3 : 7 = x : y : 
— (aa 4- A/3) (ha 4- A/3) 
chy 
— (ax 4- hy) (hx 4- by) 
chz 
(where it is to be observed that the equation a : ¡3 = x : y is the verification of the 
theorem that the lines AA', BB', CC' meet in a point 0). 
We may now from the above found relation eliminate either the (a, /3, 7) or the 
(x, y, z); first eliminating the (a, /3, 7), we find 
ab — A 2 Y 2 1 ax + hy 1 hx 4- by _ ^ 
hab Z A a hx + by b ax + by * 
where 
Y _ (ax 4- hy) (hx 4- by) 
Z chz 2 
or, completing the elimination, 
ab-h? _ ab ' halax + hy} hx+by y = 0t 
which is a quartic curve having a node at each of the points 
(z = 0, ax + hy = 0), (z — 0, hx 4- by = 0), (ax 4- hy = 0, hx 4- by = 0),