Solution: Let the coordinates of point A be (a, k/a) and point B be (2a, k/2a).

The coordinates of c are set to (x, 0). Because of the ABC line.

So there is (0-k/a)/(x-a)=(k/2a? -? k/a)？ /(2a-a)

The solution is x=3a.

S△AOC= 1/2*？ Supervisor *(k/a)

= 1/2*3a*k/a=6

So k=4