420 Mr. A. Cayley on the Cubic Centres of a Line 



Then, first, in order to find the curve which is the locus of f, 

 the coordinates of the point XL are given by x : y : z=0 : v : —fi ; 

 or if, as it is convenient to do, we take X, Y, Z (instead of x, y, z) 

 for current coordinates, by X : Y : Z = : v : — fi. Hence taking 

 x y y, z as the coordinates of f , the equation of the line through 

 XL, f is 



X, Y, Z =0, 



0, v, -ft 



VIZ. 



X{fiy + vz) -x(jiY + vL) = 0. 



And at the point where this line meets the line I, the equation 

 whereof is 



X + Y + Z=0, 

 we have 



(Y + Z)(^ + v^)+^Y + vZ)=0; 

 that is, 



Y(fix + fiy + vz) + Z(vx + fiy + vz) = 0. 



Hence this line, and the line 



Yz-Zy=0, 

 with the lines 



Y=0, Z=0, 



are the lines which pass through the point YZ and the four har- 

 monic points, and they form therefore a harmonic pencil ; or we 

 have 



y(jjux+fiy + vz)— z{vx+fiy + vz)=0, 



or what is the same thing, 



(fiy— vz){x + y+z) + 2yz(v— fi) = 0, 



as the locus of the point f : the locus is therefore a conic pass- 

 ing through the points YI, ZI, YZ. 



The equations of the conies which are the loci of X, Y, Z 

 respectively, are therefore 



V = (fiy— vz){x + y + z)+2yz(v— fi) =0, 



Y=(vz— \x)(x + y + z)+2zx{\— v)=0, 



W={\x-fjLy){x + y + z) + 2xy(p-\)=0; 



and the identical equation, 



UKx + Vpy + Wvz=0, 



shows that these conies have three points of intersection in 

 common. The three equations, and a fourth one to which they 

 give rise, may be written 



