Question

The area of the triangle whose vertices are \( (a, a) \), \( (a + 1, a + 1) \) and \( (a + 2, a) \) is:

• A. \( a^3 \)
• B. \( a^2 \)
• C. \( 2a \)
• D. \( 2a^2 \)

Hint:

Use the general area formula for triangles to simplify calculations involving specific vertices.

Answer 1

The Correct Option is D

Using the formula for the area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \): \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substituting the given vertices \( (a, a) \), \( (a + 1, a + 1) \), and \( (a + 2, a) \), we get: \[ \text{Area} = \frac{1}{2} \left| a(a + 1 - a) + (a + 1)(a - a) + (a + 2)(a - (a + 1)) \right| \] Simplifying: \[ \text{Area} = \frac{1}{2} \left| a \times 1 + 0 + (a + 2)(-1) \right| \] \[ \text{Area} = \frac{1}{2} \left| a - (a + 2) \right| = \frac{1}{2} \times 2a = a^2 \] Thus, the area is \( 2a^2 \).