Coordinate geometry application: Find the area (in square units) of the triangle formed by the three lines x = 4, y = 3, and 3x + 4y = 12. Clearly identify each pairwise intersection point and then compute the triangle's area.

Introduction / Context:This problem uses straight-line graphs to form a triangle. The goal is to find the three vertices by intersecting the given lines and then compute the triangle's area. It tests core coordinate geometry skills: finding intersections and applying a basic area formula for a right triangle.

Given Data / Assumptions:

• Lines: x = 4 (vertical), y = 3 (horizontal), and 3x + 4y = 12 (a slanted line).
• All intersections are within the real plane.
• Area is requested in square units.

Concept / Approach:The intersections of x = 4 and y = 3 with 3x + 4y = 12 give two vertices on the axes. The third vertex is simply the intersection of x = 4 and y = 3. The area of a right triangle is (1/2) * base * height; here the legs will be horizontal and vertical, making the calculation straightforward.

Step-by-Step Solution:

Verification / Alternative check:Plotting quickly shows the right angle at (4, 3) with legs along x and y directions. Shoelace formula also yields 6 if applied to (4,0), (0,3), (4,3).

Why Other Options Are Wrong:12 and 8 correspond to using full rectangle areas (4*3) or mis-halving; 10 is an arbitrary miscalculation. The only correct area is 6.

Common Pitfalls:Confusing which intersections lie on axes, or forgetting to take half the product of the perpendicular legs when computing the area.

Final Answer:6