Final answer:
The area of the region in the first quadrant between the lines x + 3y = 12 and x + 3y = 18 is calculated as the area of a trapezoid and is found to be 12 square units.
Step-by-step explanation:
Calculating Area Between Lines
The question involves finding the area of a region in the first quadrant that lies between two parallel lines, x + 3y = 12 and x + 3y = 18. The region between these lines forms a strip that is parallel to the x-axis.
To start, let's identify the points where each line intersects the x and y axes, which will give us the corners of the strip. For x + 3y = 12, the x-intercept is (12, 0) when y = 0, and the y-intercept is (0, 4) when x = 0. For x + 3y = 18, the x-intercept is (18, 0) when y = 0, and the y-intercept is (0, 6) when x = 0.
Next, we can calculate the area of the region. This strip can be viewed as a trapezoid in the first quadrant. Since the two lines are parallel, the height (h) of the trapezoid is the difference in y-intercepts, which is 6 - 4 = 2 units. The top base (b1) is the portion of the x-axis from x = 12 to x = 18, and the bottom base (b2) is along the x-axis from x = 0. The length of each base is the distance between the x-intercepts, which is 18 - 12 = 6 units.
The area (A) of a trapezoid is given by the formula A = \(\frac{1}{2}(b1 + b2) \times h\), so substituting our values gives:
A = \(\frac{1}{2}(6 + 6) \times 2\)
A = 6 \times 2
A = 12 square units
Therefore, the area of the region is 12 square units.