The Imaginary in Geometry 



17 



the barb of one and only one arrow representing an imaginary 

 element of AB. Thus, knowing that the element belongs to AB, 

 we represent it perfectly by the point. If coordinates be rect- 



Fig. 17. 



angular, the centers of the /-line and the /-line through (x, y), 

 the barb and the butt of the arrow, are respectively 



O' — y", y' + x" ) and O' + y", y'—x" ) . 



Applying the ordinary formula to the distance between two 

 points in the plane to P, P x we get 



d^=(x l — xy+(y 1 -yy-. 



Working this out it will be found that 



d 2 = l l L_e i <t>, 



where l x is the length of the line joining the buts of the arrows, 

 /, the length of the line joining the barbs and <£ the angle de- 

 scribed by turning positively from the direction / x to the direc- 

 tion Z.,. 1 



Since the arrows for an /-ray all have their barbs at the center 



1 LaGuerre, Oevres, II, 97. 



17 



