设此直线是y-1=k(x-2),代入双曲线并整理得到(3-k^2)x^2+2k(2k-1)x-4(k^2-k+1)=0所以x1+x2=-2k(2k-1)/(3-k^2)因为M为AB的中点,所以(x1x+2)/2=2-2k(2k-1)/(3-k^2)＝4-2k^2+k=6-2k^2k=6所以直线是y=6x-11...