AC:y=x+ 1

(2)D( 1，4) N(0，3)

The value of MN+MD is the smallest.

Then do point D2 (5,4), which is the symmetrical point of point D with respect to the straight line x=3.

Then the straight line nd2: y = 1/5x+3.

So m is in a straight line, and m= 18/5.

(3) B( 1, 2) is BD=2.

Let e (xe, xe+ 1) and f (xe, -xe2+2xe+3).

If it is a parallelogram, EF=BD=2.

So Xe+ 1-(-Xe2+2Xe+3)=2.

Or Xe+ 1-(-Xe2+2Xe+3)=-2 (note that point e here may be on both sides of the symmetry axis).

Solve two equations to get three solutions, namely 1, add and subtract the root sign 17, and divide by 2 and 0 (omit 1).

Like the third question, it becomes

Let P(X, -X2+2X+3) Q(X, X+ 1).

s△APC = pqx3x 1/2 = pqx 1.5

That is, find the maximum PQ.

The maximum value of -X2+2X+3-(X+ 1) (because p must be above q).

Formula -(X2- 1/2)2+9/4

So the maximum X=0.5.

PQ=2.25 at this time.

S△APC=3.375