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