Find the area of a triangle whose vertices are $A (- 8, - 2), B (- 4, - 6),$ and $C (- 1, 5)$

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Solution

$A = \frac{1}{2} ∣ ∣ ∣ \begin{matrix} (x_{1} - x_{2}) & (x_{2} - x_{3}) (y_{1} - y_{2}) & (y_{2} - y_{3}) \end{matrix} ∣ ∣ ∣$
$\Rightarrow A = \frac{1}{2} ∣ ∣ ∣ \begin{matrix} (- 8 - (- 4)) & (- 4 - (- 1)) (- 2 - (- 6)) & (- 6 - 5) \end{matrix} ∣ ∣ ∣$
$\Rightarrow A = \frac{1}{2}$ $∣ ∣ ∣ \begin{matrix} - 4 & - 3 4 & - 11 \end{matrix} ∣ ∣ ∣$ $= \frac{44 + 12}{2}$
$\Rightarrow A = 28$

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