🚀 go-pugleaf

No it's not the easiest.  The a simpler (and not that inefficient) would 
be to test the slopes of the 2 lines in question and if m1 <= m2 then the 
segment is on the right...the only catch is if the slope is negative; is 
it the numerator positive and the denominator negative or vic versa?  To 
elimate this (and to eliminate a nasty divide) do if (dx1*dy2) <= 
(dy2*dx2) then the line is on the right of the partion line. EX:

int     OnRight( short Index){
 long   x1 = Nodes[Index].x;
 long   y1 = Nodes[Index].y;
 long   x2 = x1 + Nodes[Index].dx;
 long   y2 = y1 + Nodes[Index].dy;
        if ( ((x1-x2)*(py-y2)) <= ((y1-y2)*(px-x2)) )
           return 1;
        else return 0;
}

more intuative that dot or cross product in my opinion but almost 
certainly not the best.  I remember glancing at a recent issue of DD 
sourcebook and Abrash mentioning a *very* simple test but I may be 
mistaken. 

-- 
Those who danced were thought to be quite insane by those who could not 
hear the music.