clicliclic@freenet.de schrieb:   
   >   
   > Here's that solution by Derive 6.10. It's actually four closely   
   > related solutions, and I wouldn't call them big:   
   >   
   > [...]   
   >   
   > The factor SIGN(m0-m1) can be set to 1 throughout, I think, as the   
   > sign of the accompanying square root alternates betwen solutions.   
   >   
      
   This and other kinds of massaging give the simplified solution:   
      
   x0=(b0-b1)/(m1-m0)+r*(SQRT(m0^2+1)*SQRT(m1^2+1)+m0*m1+1)/(SQRT(m~   
   0^2+1)*(m1-m0)) AND x1=(b0-b1)/(m1-m0)+r*(SQRT(m0^2+1)*SQRT(m1^2~   
   +1)+m0*m1+1)/((m1-m0)*SQRT(m1^2+1)) AND xc=(b0-b1)/(m1-m0)+r*(SQ~   
   RT(m0^2+1)+SQRT(m1^2+1))/(m1-m0) AND y0=(b0*m1-b1*m0)/(m1-m0)+m0~   
   *r*(SQRT(m0^2+1)*SQRT(m1^2+1)+m0*m1+1)/((m1-m0)*SQRT(m0^2+1)) AN~   
   D y1=(b0*m1-b1*m0)/(m1-m0)+m1*r*(SQRT(m0^2+1)+SQRT(m1^2+1))*(m1*~   
   SQRT(m0^2+1)+m0*SQRT(m1^2+1))/((m1^2-m0^2)*SQRT(m1^2+1)) AND yc=~   
   (b0*m1-b1*m0)/(m1-m0)+r*(m1*SQRT(m0^2+1)+m0*SQRT(m1^2+1))/(m1-m0)   
      
   For other solutions just flip the sign of SQRT(m0^2+1) and/or   
   SQRT(m1^2+1).   
      
   Martin.   
      
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