∫A(4，3)，

∴3=4k 1，

The solution is k 1=34,

The analytical formula of the straight line where OA is located is: y=34x,

Similarly, the analytical formula of straight line AB can be obtained as follows: y=-32x+9,

∫MN∨AB，

∴ Let the analytical formula of straight line MN be y=-32x+b and substitute it into m (1, 0).

Get: b=32,

The analytical formula of line MN is y=-32x+32,

Solution y = 34xy =? 32x+32，

X = 23y = 12，

∴N(23, 12).

(2) As shown in Figure 2, if NH⊥OB is in H and AG⊥OB is in G, then Ag = 3.

∫MN∨AB，

∴△ area of ∴△MBN =△ area =△PMN =S,

∴△OMN∽△OBA，

∴NH:AG=OM:OB，

∴ NH: 3 = x: 6, that is, NH= 12x.

∴S= 12MB？ NH = 12×(6-x)× 12x =- 14(x-3)2+94(0 < x < 6)，

When x=3, s has a maximum value of 94.

(3) As shown in Figure 2, ∫MN∨AB,

∴△AMB area =△ANB area =S△ANB, △NMB area =△NMP area = S.

∫S:S△ANB = 2:3，

∴ 12MB？ NH: 12MB？ Ag = 2: 3, that is, NH: Ag = 2: 3,

∴ON:OA=NH:AG=2:3，

∫MN∨AB，

∴OM:OB=ON:OA=2:3，

OA = 6，

∴OM6=23，

∴OM=4，

∴M(4,0)

The analytical formula of ∫ straight line AB is: y=-32x+9,

The analytic formula y =-32x+b of the yield line MN.

Substitute point m into 0=-32×4+b,

The solution is b=6,

The analytical formula of line MN is y=-32x+6,

Solution y = 34xy =? 32x+6，

X = 83y = 2，

∴N(83,2).