The area of the given triangle is one-half base times height: (1/2)(9)(4) = 18. So we want the vertical line to divide the triangle into two pieces each with area 9.
Any point on segment AC will have coordinates of the form (2a,a); the vertical line through (2a,a) will intersect AB at (2a,0).
The area of the triangle formed by the vertical line and point A will have area one-half base times height: (1/2)(2a)(a) = a^2.
Since we want the area of that triangle to be 9, a^2=9 --> a = 3.