48 operators

from fractions import Fraction
def C(Ax,Ay,Bx,By,Cx,Cy,Dx,Dy):
 Ax,Ay,Bx,By,Cx,Cy,Dx,Dy=map(Fraction,(Ax,Ay,Bx,By,Cx,Cy,Dx,Dy))#convert to Fraction
 dBx=Bx-Ax;dBy=Ay-By#4
 Bx_=dBx*dBx+dBy*dBy#4
 Cx_=(Cx-Ax)*dBx-(Cy-Ay)*dBy#6
 Cy_=(Cx-Ax)*dBy+(Cy-Ay)*dBx#6
 Dx_=(Dx-Ax)*dBx-(Dy-Ay)*dBy#6
 Dy_=(Dx-Ax)*dBy+(Dy-Ay)*dBx#6
 if Cy_*Dy_>0:return 0#2
 if Cy_==Dy_==0:return(Cx_>=0 or Dx_>=0)and(Cx_<=Bx_ or Dx_<=Bx_)#6
 return 0<=(Cx_*Dy_-Dx_*Cy_)/(Dy_-Cy_)<=Bx_#7

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42 operators if assigning to already existing variables is allowed

def R(Ax,Ay,Bx,By,Cx,Cy,Dx,Dy):
 Ax,Ay,Bx,By,Cx,Cy,Dx,Dy=map(Fraction,(Ax,Ay,Bx,By,Cx,Cy,Dx,Dy))#convert to Fraction
 if Ax==Bx:Ax=Ax+Ay;Bx=Bx+By;Cx=Cx+Cy;Dx=Dx+Dy#9
 dBx=Bx-Ax;dBy=By-Ay#4
 Cx_m=(Cx-Ax)/dBx#3
 Cy_=Cy-Ay-Cx_m*dBy#4
 Dx_m=(Dx-Ax)/dBx#3
 Dy_=Dy-Ay-Dx_m*dBy#4
 if Cy_*Dy_>0:return 0#2
 if Cy_==Dy_==0:return(Cx_m>=0 or Dx_m>=0)and(Cx_m<=1 or Dx_m<=1)#6
 return 0<=(Cx_m*Dy_-Dx_m*Cy_)/(Dy_-Cy_)<=1#7

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Uses Fraction only to convert to fraction so that division is exact.

Python, 88 58 74

I originally used bool-int conversions in place of if/else/and/or, and then left it as is once the spec was clarified because l4m2's answer used way fewer operations even if I made the switch so it didn't feel worth it, but apparently it was wrong (at time of writing) so I might as well update mine.

Implements this algorithm with all subroutines inlined and slightly streamlined to improve op count as follows:

1. If three points are colinear we don't actually need to check if the y-value is in range. nvm
2. Stolen from l4m2's answer: Instead of evaluating the sign of each orientation and comparing those to each other, multiply them, and if ≤ 0 either the signs are different or they're both 0. For the double 0 case, an inequality check is added to prevent the problem l4m2 had with colinear line segments. If I just compared the signs I'd be doing something like (1 if o1 > 0 else -1 if o1 < 0 else 0) != (1 if o2 > 0 else -1 if o2 < 0 else 0) which is 5 operations vs o1*o2 <= 0 and o1 != o2 having 3.

def intersect(x1,y1,x2,y2,x3,y3,x4,y4):
 o1 = (y2-y1)*(x3-x2) - (x2-x1)*(y3-y2)                    # Cost 8
 o2 = (y2-y1)*(x4-x2) - (x2-x1)*(y4-y2)                    # Cost 8
 o3 = (y4-y3)*(x1-x4) - (x4-x3)*(y1-y4)                    # Cost 8
 o4 = (y4-y3)*(x2-x4) - (x4-x3)*(y2-y4)                    # Cost 8
 if o1*o2 <= 0 and o1 != o2 and o3*o4 <= 0 and o3 != o4:   # Cost 6
  return True
 elif o1 == 0 and (x1<=x3<=x2 or x1>=x3>=x2) and (y1<=y3<=y2 or y1>=y3>=y2):
  return True                                              # Cost 9
 elif o2 == 0 and (x1<=x4<=x2 or x1>=x4>=x2) and (y1<=y4<=y2 or y1>=y4>=y2):
  return True                                              # Cost 9
 elif o3 == 0 and (x3<=x1<=x4 or x3>=x1>=x4) and (y3<=y1<=y4 or y3>=y1>=y4):
  return True                                              # Cost 9
 else:                                                     # Cost 9
  return o4 == 0 and (x3<=x2<=x4 or x3>=x2>=x4) and (y3<=y2<=y4 or y3>=y2>=y4)

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