The area (in sq. units) of a quadrilateral whose vertices are $A (5, 4),$ $B (- 3, 2),$ $C (- 5, - 4)$ and $D (7, - 6)$ taken in order, is

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Solution

Given vertices are $A (5, 4), B (- 3, 2), C (- 5, - 4)$ and $D (7, - 6)$
So, the area of quadrilateral is
$Δ = \frac{1}{2} ∣ ∣ ∣ \begin{matrix} 5 & - 3 & - 5 & 7 & 5 4 & 2 & - 4 & - 6 & 4 \end{matrix} ∣ ∣ ∣ \Rightarrow Δ = \frac{1}{2} | 22 + 22 + 58 + 58 | ∴ Δ = 80 sq. units$

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