DM and CN intersect at point F.
Then △ ABO△ CDF
∴DF=2,CF= 1
The area of ∫ quadrilateral BCDE is five times that of △ABE.
∴(BC+AD)=5AE
∴DE=2AE
∴MO=2AO
The abscissa of point d is 2,
The abscissa of point C is 3.
The coordinate of the set point is (2, m)
The coordinate of point ∴c is (3, m-2)
∵C and d are both on the image of function y = k/X.
∴k=2m=3(m-2)
The solution is m = 6 and k = 12.