Two of the lines forming the triangle are fixed (y=0, y=2x), while one is variable. One of the vertices is thus fixed at the intersection of y=0 and y=2x, which is the origin. The remaining vertices occur where y=2x intersects y= -(x/2) + k, and also where y = -(x/2) + k intersects y=0. Let's call these A and B, respectively.
The base length of the triangle is the x-coordinate of B, and the height is given by the y-coordinate of A. Make a sketch, including y=0, y=2x, and the remaining line for a few positive values of k. Everything stated above will become crystal clear if you do.
To find the x-coordinate of A, set 2x = -(x/2) + k and solve for x:
(2+1/2)x = k
(5/2)x = k
Since we need the y-coordinate of A as the height of the triangle, substitute this into either equation. y=2x provides the least difficulty: y = 2*(2/5)k = (4/5)k.
To find the x-coordinate of B, set y = -(x/2) + k to 0 and solve for x:
-(x/2) + k = 0
-(x/2) = -k
x = 2k
So we have as the height of the triangle (4/5)k, and as the base length 2k. Substitute into the equation for the area of a triangle, and set this area to 80:
(1/2)b*h = A
(1/2)(2k)(4/5)k = 80
(4/5) k^2 = 80
k^2 = 400/4 = 100
k = ± 10. Since the problem states that k > 0, retain k = 10.