I haven't done math for many years. I don't want the middle school exam to be so difficult now. There seems to be no need for students to have pressure. I don't know if it is mainstream.

1, quadrilateral ABDC perimeter =AB (fixed) +BD+DC (fixed a+3-a=3)+CA.

That is to say, the shortest BD distance of BD+CA is the distance from (a+3-4,0-1) to the origin, that is, point E (a- 1,1), BD=OE, similarly AC = of (a-2,3), so

2. If there are A'(-2, -3) and B'(4, 1), then Na' and MB = MB', we can know that ABMN must have the shortest perimeter, that is, the four points of A'NMB' are on a straight line.

The equation of the straight line A'B' is y=kx+b 2x-3y-5=0, and the focus of the equation and the XY axis is m=5/2 and n=-5/3.

I hope it helps you.