94 
ON A LOCUS IN RELATION TO THE TRIANGLE. 
61 
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I suppose that two of the angles of the triangle are given, and I enquire into 
the locus of the remaining angle. To fix the ideas, let A, B, C be the angles of the 
triangle, A', B', C' those of the reciprocal triangle; and let the angles A and B be 
given. We have to find the locus of the point C: I observe however, that the lines 
AA', BB', CO meet in a point 0, and I conduct the investigation in such manner 
as to obtain simultaneously the loci of the two points C and 0. The lines C'B', C'A' 
are the polars of A, B respectively, let their equations be x = 0, and y — 0, and let the 
equation of the line AB be z = 0\ this being so, the equation of the given conic will 
be of the form 
(a, b, c, 0, 0, h\x, y, z) 2 = 0. 
I take (a, /3, 7) for the coordinates of 0 and (x, y, z) for those of O; the 
coordinates of either of these points being of course deducible from those of the other. 
Observing that the inverse coefficients are 
(be, ca, ab-h 2 , 0, 0, — ch), 
we find 
coordinates of A are ( b, — h, 0), 
„ B „ (- h, a, 0). 
The points A' and B' are then given as the intersections of AO with C'A'(y = 0) and 
of BO with C'B' (x = 0) ; we find 
coordinates of A' are (/¿a + b/3, 0 , /17), 
„ B' „ ( 0 , aa + h/3, hy). 
Moreover, coordinates of O are (0, 0, 1), 
» O „ (x, y, z). 
The six points A, B, C, A', B', C' are to lie in a conic; the equations of the 
lines C'A, C'B, AB are hX + bY= 0, aX+hY=0, Z= 0, and hence the equation of a 
conic passing through the points C, A, B is 
£ 
aX + hY 
+ 
M N_ 
hX + bY + Z ■ 
Hence, making the conic pass through the remaining points A, B, 6, we find 
L 
a (ha + b/3) 
L 
h (aa + hf3) 
M — = 0 
+ h (ha + b/3) + hy 
M X = 
+ b (aa + h/3) hy 
M w n 
hx +by z 
at 
lie 
L 
ax + hy