The Imaginary m Geometry 1 1 



opposite, until the black line P x ' P 2 ' is met. Divide the blue 

 line i 3 /' P 2 " in the same ratio as the black line is thus divided. 

 From the point of division of the blue line draw a parallel to the 

 black to a point such that a black vector from it equal to that 1 

 from the arbitrary point will just reach the blue line. This vec- 

 tor reversed will give the blue point that goes with the arbitrary 

 black point to determine the red vector required. Fig. II illus- 

 trates this construction. Similarly, we can start with any point 

 in the plane for a blue point. 



Return to the equation ax -f- by— i. This is equivalent to the 

 pair 



a'x' + b'y' — a"x" — b"y" = i, 



a'x" -f b'y" + a"x' + b"y' =o. 



These show very plainly that there is a one-to-one correspond- 

 ence between black points and blue points, provided 



V 



b" 



H=o. 



In particular, let x" = y" = o, so that the blue point coincides 

 with the black. This will give a point (;r , y ) =P > whose 

 red vector vanishes. We get, in fact, 



b" —a" 



3'o 



~o a'b" — a"b' ' -° a'b" — a"b' ' 



When the denominator vanishes, but at least one numerator 

 does not, P goes to infinity. We have in fact the case of a line 

 of real slope. If a" and b" both vanish the line is real and P 

 is indeterminately any point whatever on the line. Note in these 

 cases, that, although x" = y" =0 gives the point P with co- 

 ordinates as above, yet to assume coordinates as above for a 

 black point leaves the blue point to go with it indeterminate. 



The point P t> we call the center of the line. When it goes to 

 infinity or becomes indeterminate we say that the line is non- 

 central. 



Given two red vectors P 1 and P. 2 (fig. 12) to construct the 

 center P of the line to which they belong, proceed as follows : 

 Draw the black line and the blue line joining them. The vector 



IT 



