Question

The area of the triangle whose vertices are (a, a),

(a + 1, a + 1), (a + 2, a) is-

• A. a²
• B. 2a
• C. 1
• D. $\sqrt{2}$

Answer 1

Answer: (C)

1

Answer 2

DABC = $\frac{1}{2}$ $\begin{Bmatrix} a & a + 1 & a + 2 & a \\ a & a + 1 & a & a \end{Bmatrix}$

= $\frac{1}{2}${a(a + 1) – a(a + 1) + a(a + 1) – (a + 1) (a + 2) + a(a + 2) –a²}

=$\frac{1}{2}$ {a² + a – a² – 3a – 2 + a² + 2a – a²}

= $\frac{1}{2}${–2} = – 1 = 1