394 



clinant roots, then, we have F^O, and F a =0; whence we find, by 

 alternately eliminating s and r, F 3 (#, r)=0, and F 4 (#, s}= 3. 

 Now describe the loci of the points R and S, where 



OR=,r.OI + r.OB, F 3 (#, r) = 0, 

 and 



OS = #. OI + s. OB, F 4 (a, *) = 0. 



Then from the nature of the clinant roots, which will be in pairs of 

 the form r^ + i.s^ if we produce PQ to cut these curves, for every 

 point R' in which it cuts the first there will be two points S^, S' 2 , in 

 which it cuts the second. If, then, from the point R' we draw 

 si.PS^ and R'M' 2 =i . PS' 2 , two corresponding clinant roots 



of the equation will be i, 2, so that 

 O13 OB 



i i i OB OB 



and 



OB OB ' 



See the example and figure in (11). 



Hence we shall have for the geometrical representation of any 

 algebraical formation, by (2), 



f(x t/Y=X M ' Q MQ V MVQ M^Q M/)Q M,(-)Q 

 ' OB ' ' OB OB ' OB ' OB OB 



4. Next suppose that the origin of the coordinates be altered to 

 O f , so that OT=OI, and OT, when produced, passes through Q. 

 This will be equivalent to putting y + 6 instead of y, and the result 

 will not disturb the coefficient of y w +2. Also since one of the radical 

 loci for the scalar values remains unaltered, and the others (which 

 are the axes) are parallel to the former, the points M I} M 2 , . . . M n in 

 which PQ, when produced, will cut the curve, remain the same, and 

 consequently there are the same number of scalar roots as before. 



5. But as the terms themselves of the equation f(x, y)=0 are 

 altered, the radical loci of R and S in (3) will be altered, and as the 

 distances are now measured from Q and not from P, the lines 

 QR^-fi . QS' 3 , QR'. + i . QS' 4 will now be different and =Qw' 1 , Qm' 2 . 

 If then O'Q=* 2 . OT, and QP=y 2 . BO= -y 2 . OB, then 



" ~OB OB"" 



