If A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.
Solution
It is given that A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD.
Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
\[\text{ ar } \left( ∆ ABC \right) = \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]
\[ = \frac{1}{2}\left| - 3\left[ - 7 - \left( - 8 \right) \right] + \left( - 2 \right)\left( - 8 - 5 \right) + 1\left[ 5 - \left( - 7 \right) \right] \right|\]
\[ = \frac{1}{2}\left| - 3 + 26 + 12 \right|\]
\[ = \frac{35}{2}\text{ square units } \]
\[\text{ ar } \left( ∆ ACD \right) = \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]
\[ = \frac{1}{2}\left| - 3\left( - 8 - 3 \right) + 1\left( 3 - 5 \right) + 6\left[ 5 - \left( - 8 \right) \right] \right|\]
\[ = \frac{1}{2}\left| 33 - 2 + 78 \right|\]
\[ = \frac{109}{2} \text{ square units } \]
∴ Area of the quadrilateral ABCD =\[\frac{35}{2} + \frac{109}{2} = \frac{144}{2} = 72\] square units
Hence, the area of the given quadrilateral is 72 square units.
