On the Harmonic Relation of two Lines or two Points. 105 



\ 2 = m?(abc+pk0) 



+ m{—abc(p + k + O)+pk0{a+b-\-c)} 



+ {abc[k0 + 0p + kp)+pk0{bc + ca + ab)} 



2\p, = m 2 { — (be + ca + ab) + (k6 + Op +pk) } 



+ m{-abc—pk0+{bc + ca + ab)(p + k + 0) 



+ {k0 + 0p+pk){a + b + c)} 



+ {abc(p + k + 0)—pk0(a + b + c)} 



!#=. rn2{ a + b + c+p + k + 0} 



+ m{(bc + ca + ab) — (k0 + 0p+pk)} 



+ abc+pk0. 



And substituting in 4(\ 2 ./& 2 ) — (2\|*) 2 =0, we have the relation 

 required. To verify that the equation so obtained is in fact 

 the algebraical equivalent of the transcendental equation, it is 

 only necessary to remark, that the values of A, 2 , /j? are unaltered, 

 and that of \/uu only changes its sign when #,#,c,mand^,£,0,— m 

 are interchanged. Hence this change will not affect the equa- 

 tion obtained by substituting in the equation 4V 2 . /x 2 — (2a/a) 2 = 0. 

 Hence precisely the same equation would be obtained by elimi- 

 nating L, M from 



(h + a)(k + b) (k + c) = (L + M£) 2 (m -k) 



{p + a){p + b){p + c) = (L + Mp)%m-p) 



(0 + a)(0 + b)(0 + c) = (L + M0)*(0-p); 



or putting (L + Mk) (m — k) = a + /3k + yk* 2 , by eliminating a, /3, y 



from 



(m-k)(k + a)(k + b)(k + c) = {a + 0k + ytf)* 

 (m-p){p + a)(p + b){p + c)=:(ct + /3p + yp <2 )* 

 ( m _6>)(<9 + «)(^ + 6)(^ + c) = (a+^(9 + 7(9 2 ) 2 

 = (a + pm + ym*f } 



which by Abel's theorem show that p, k, are connected by the 

 transcendental equation above mentioned. 



2 Stone Buildings, 

 July 9, 1853. 



XIV. On the Harmonic Relation of two Lines or two Points. 

 By A. Cayley, Esq* 



THE ' harmonic relation of a point and line with respect to a 

 triangle' is well known and understoodf; but the analo- 



* Communicated by the Author. 



t The relation to which I refer is contained in the theorem, " If on each 

 side of a triangle there be taken two points harmonically related with 



