118 
APPLICATION OF THE NEWTON-FOURIER METHOD 
[736 
where of course the equation with k shows that k is equal to the ratio of the 
distances of the point from the points N, N' respectively, and the equation in cf>, 
taken in the second form, shows that <£ is the angle subtended at the point by N, N\ 
It is sometimes convenient to write heke~ 2i(l> =p, q respectively; we then have 
70 . • 1 -P p • 1 + q 
pq = K\ x+iy = YZp’ x ~ i y = Y^q' 
Suppose for a moment that we have (p 1} q 2 ), (p 2 , q 2 ), (p 3 , q 3 ) as the (p, q) coordinates 
of any three points, the condition that these three points may lie in a line, is given 
in the form, determinant = 0, where each line of the determinant is of the form 
1+p 1+ q 
1-p’ T~—’ L ’ 
or, what is the same thing, it is 
1 -pq+p-q, 1-pq-p + q, 1+pq-p-q, 
pq-1, p-q, p + q- 2, 
0. 
or, again 
viz. the condition is 
Pi9i-1, Pi-qi, Pi + qi-2 
p 2 q 2 - 1, p, - q 2 , p 2 + q 2 -2 
p s q 3 -1, p s ~ q 3 , p 3 + q 3 -2 
Suppose the (k, <fr) coordinates of the three points are (l, a), (m, /3), (n, 7) respectively; 
then this equation is 
P — 1, l sin a, l cos a — 1 
w 2 —1, m sin ß, mcos/3— 1 
V. 2 — 1, n sin 7, n cos 7 - 1 
= 0, 
viz. it is 
p -1, 
l sin a, 
l cos a 
- 
P 
-l, 
l sin a, 
1 
m 2 — 1, 
m sin /3, m cos ß 
m 2 
-1, 
m sin /3, 
1 
n 2 — 1, 
n sin 7, 
n cos 7 
n 2 
-1, 
n sin 7, 
1 
= 0, 
or, what is the same thing, it is 
[(P — 1) mn sin (/3 - 7) + (m 2 — 1) nl sin (7 — a) + (w 2 - 1) bn sin (a — /3)] 
+ [(m 2 — n 2 ) l sin a + (n 2 — P) m sin /3 + (P — m 2 ) n sin 7] = 0. 
If in this equation 7 is put =7r, and ¡3 = 2a, so that sin (a - /3) = - sin a, the equation 
will contain only terms in sin a, and sin 2a, viz. it will be 
[ (m 2 — n 2 ) l + (m 2 — 1) nl — (n 2 — 1) bn] sin a 
+ [- (P - 1) mn + m (n 2 - P) ] sin 2a = 0, 
that is, 
l (m — 1) (n + 1) (m — n) sin ct + m (m + 1) (n - P) sin 2a = 0,