I assume you mean x-2y-2=0 and 4x-y-1=0. (Later I'll generalise the second equation to 4x-y-a=0 where a is an arbitrary constant.)
From the second equation, y=4x-1, and substituting for y in the first equation:
x-2(4x-1)-2=0, x-8x+2-2=0, -7x=0, so x=0⇒y=-1, so (0,-1) is the intersection point of the two lines.
The lines intersect the x-axis when y=0, so the intersections are at x=2 and x=¼, (x-2=0 and 4x-1=0).
This line can be the base of the inverted triangle so its length is 2-¼=1¾ or 7/4.
The height of the triangle is 1 because the lines intersect at 1 unit (y=-1) vertically below the origin (0,0).
The area of the triangle=½base×height=½(7/4)(1)=⅞ sq units.
Now consider the equation 4x-y-a=0 as the second equation. y=4x-a, so we can substitute for y in the first equation:
x-2(4x-a)-2=0, x-8x+2a-2=0, 7x=2a-2, x=(2a-2)/7; y=4(2a-2)/7-a=(8a-8-7a)/7=(a-8)/7. Intersection point is ((2a-2)/7,(a-8)/7).
The lines intersect the x-axis when y=0, when x=2 and x=a/4. The base is 2-a/4 units long.
The lines intersect at ((2a-2)/7,(a-8)/7) so the vertical height is the y-coordinate (a-8)/7.
Area of the triangle ½base×height=½(2-a/4)(a-8)/7=⅛(8-a)(a-8)/7=-⅛(a-8)2/7. It's negative because the triangle is below the x-axis. In the earlier calculation the height was 1 below the x-axis at y=-1, but it was treated as being magnitude 1 which is -(-1). So the actual area of the triangle is (1/56)(a-8)2.