The area of a triangle whose vertices are (-4, 0), (2, 4) and (4, 0) is

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Solution

The correct option is C 16
$L e t u s c o n s i d e r t h e Δ a s A B C w i t h A = (x_{1}, y_{1}) = (- 4, 0), B = (x_{2}, y_{2}) = (2, 4) a n d C = (x_{3}, y_{3}) = (4, 0) . T h e | a r Δ A B C | = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} - 4 & 0 & 1 2 & 4 & 1 4 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ ⟹ a r Δ A B C = \frac{1}{2} [- 4 (4 - 0) + 2 (0 - 0) + 4 (1 - 4)] = | - 16 | S i n c e i t i s a n a r e a w e t a k e t h e p o s i t i v e v a l u e o n l y . ∴ a r Δ A B C = 16 s q u n i t s . A n s - O p t i o n C$

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