6. If when $m$ takes any real number not equal to 0 and 1, the parabola $y=\frac{m-1}{m} x^{2}+\frac{2}{m} x-\frac{m-3}{m}$ passes through two fixed points in the Cartesian coordinate system, then, the distance between these two fixed points is $\qquad$
$6.4 \sqrt{5}$.
Taking $m=\frac{1}{2}$ yields the parabola equation $y=-x^{2}+4 x+5$;
Taking $m=\frac{1}{4}$ yields the parabola equation $y=-3 x^{2}+8 x+11$.
Solving simultaneously, the coordinates of the two fixed points are $M(-1,0), N(3,8)$.