256 MR CHARLES TWEEDIE ON THE 



Now it is possible to determine a and b such that 



a{k(l)+l(2)}+b{m(l)-k(2)}=a(l)+b{2), 



a'{k{l) + l(2)} + b'{vi(l)-k(2)}=-{a'(l) + b'(2)}; 



for these lead to 



ak + bm = a \ , . . a'k + b'm = —a' } ,r>\ 



al-blc=br } ' a'l-b'k=-b'r h 



pairs of equations which are consistent, since k 2 + Im = 1. Also —f can not be equal to . r 



Hence the original equations may be replaced by 



a (l)+6(2) = 



a'(l) + b'(2) = 0, 



so that the discussion again reduces to that of Case (1). 



§ 5. If Z = 0, m = 0, then k = ±1 ; n = =hl, a case already discussed. 



If Z=j=0, m= 0, there result, when k= +1, 



(l)=(l) + /(2) 

 (2)= -(2). 



From the identity (2)'= — (2) it follows that (2) has the form 



x— g + K(y — >/) + L(s;j; — yg) = 0, (K and L constants). 



The other identity leads to 



£( A^x + B# + C x ) + }] ( A 2 x + B 2 y + C 2 ) + A. A x + B 3 t/ + C 3 

 ^x(A,i+ B lV +C 1 ) + l[x- i+ K(y - ,,) + U(xq - yg)]. 



The comparison of coefficients shows that the corresponding equation is 



(1) g(A 1 x+B 1 y+CJ+*i(B 1 x+B0+C 2 )+C 1 x+C 2 y+C 3 +l{lMv+x+Ky) = O; 



while (2) is 



Now, so far as solutions of (1) and (2) are concerned, 



x + Ky + La;,, = g+ K,, + Lyg 



= h { * +£+ K(y + n) + U%>i + vB}- 

 Therefore for (1) may be substituted 



i(A,x+ .... )+etc. + |[z+£+K(y+^)+L(^+yf)] = 0, 



an equation collaterally symmetrical in xy | fy, and therefore of a type already discussed. 

 Finally, if Z=|=0, m = 0, k= — 1, we obtain 



(iy=-(i)+«(2) 



(2)'= +(2), 

 the solutions for which are the same as for 



(1)-A( 2 ) = 



(2) = 0, 



which are of a type already discussed. 



