The area of (1) RT△AOB is 3: 1/2 * ob * AB = 3, 1/2 * 2 * m=3, m=3, a (-2,3).

The image with Y=K/X passes through point A (-2,3): 3 = k/(-2), k=-6, and the inverse scaling function is y =-6/x.

Y = ax+b passes through point a:3 = a *(2)+b, b-2a = 3...( 1).

Another point C(n, -3/2) on the image with the inverse scaling function Y = K/X:-3/2 =-6/n n = 4c(4, -3/2).

Y = ax+b passes through point c:-3/2 = 4a+b...(2)

The joint solution of (1)(2) is: a=-3/4, b=3/2. The linear function is: y=-3/4x+3/2, that is 3x+4y-6=0.

(2) Distance from O to 3x+4y-6=0: |3*0+4*0-6|/√(3? +4? )=6/5

Ac distance: √((-2-4))? +(3-(-3/2))? )= 15/2

△AOC area:1/2 * (15/2) * (6/5) = 9/2.

(3) Let P(k, 0) and the lengths of the three sides of the triangle POA are:

AO = √(-2-0)？ +(3-0)? )=√ 13 AP=√((-2-k)？ +(3-0)? )=√((2+k)？ +9)

OP=√((k-0)？ +(0-0)? )=|k|

Suppose there is an isosceles triangle POA, there are three situations: one is AP=AO, the other is AP=OP, and the third is ao = op.

AP=AO:√((2+k)？ +9)=√ 13 k=0(P and o overlap, so they cannot form a triangle, so they are discarded) or k=-4, so P(-4, 0).

AP=OP:√((2+k)？ +9)= | k | k =- 13/4 P(- 13/4，0)

Ao = op: √13 = | k | k = √13 or k =-√13, 0) or P(-√ 13, 0).

To sum up, there is an isosceles triangle POA, and the coordinates of point P are as above.