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