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On the over-fixed point in compulsory mathematics. . .
In fact, the straight line through the fixed point is like this:

The standard formula is as follows: M(Ax+By+C)+N(A'x+B'y+C')=0. Pass through the intersection of the line Ax+By+C=0 and the line A'x+B'y+C'=0.

The constant intersection (x0, y0) with the shape y-y0=k(x-x0) is a special case of the above standard formula. After a deformation, it becomes: k(x-x0)+(y0-y)=0, that is, it is the intersection of straight lines y-y0=k(x-x0) = 0 and y0-y=0.

The more essential thing is that this thing is sorted by parameters, and then each item is set to 0 to see if there is such a solution (x0, y0). The straight line you gave (M- 1)X-Y+2M+ 1=0, sorted according to the parameter m (put the items containing m together and propose m), becomes M(x+2)+(-x-y+ 1)=0. If there is, the solution is a fixed point; Without it, there is no constant point.

This thing can also be extended to other kinds of curves, such as circles. You can try!