Find the area of the quadrilateral whose vertices are A (5, 3), B (6, - 4), C (- 3, - 2) and D (- 4, 7) ?

CONCEPT:

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = 

\(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

CALCULATION:

Given: A (5, 3), B (6, - 4), C (- 3, - 2) and D (- 4, 7) are the vertices of a quadrilateral ABCD

Here, we have to find the area of quadrilateral ABCD

Area of quadrilateral ABCD = Area of ΔABC + Area of Δ ACD

Let's find out the area of ΔABC

∵ A (5, 3), B (6, - 4), C (- 3, - 2) are the vertices of ΔABC

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

⇒ Area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{5}}&{{3}}&1\\ {{6}}&{{-4}}&1\\ {{-3}}&{{-2}}&1 \end{array}} \right|\)