Question

Find the area of the triangle with vertices at (0, 0), (3, 0), and (0, 4).

• A. 6
• B. 8
• C. 10
• D. 12

Hint:

For a triangle with vertices on the axes, use \( \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} \), or apply the determinant formula for general vertices.

Answer 1

The Correct Option is A

The vertices of the triangle are (0, 0), (3, 0), and (0, 4). This forms a right-angled triangle with the right angle at (0, 0). The base lies along the x-axis from (0, 0) to (3, 0) with length 3, and the height lies along the y-axis from (0, 0) to (0, 4) with length 4. The area of a right-angled triangle is: \[ \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 3 \cdot 4 = 6 \] Alternatively, use the determinant formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \): \[ \text{Area} = \frac{1}{2} \left| x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) \right| \] Substitute (0, 0), (3, 0), (0, 4): \[ \text{Area} = \frac{1}{2} \left| 0 \cdot (0 - 4) + 3 \cdot (4 - 0) + 0 \cdot (0 - 0) \right| = \frac{1}{2} \left| 0 + 12 + 0 \right| = \frac{12}{2} = 6 \] The area is: \[ \boxed{6} \]