**Question: **Find the area of the parallelogram with vertices A (1, -2), B(2, -2), C(5, 3), D(-4, 1)

** Solution: **

For a parallelogram ABCD, firstly find the area of two triangles ABC and ACD.

To find the area of triangle ABC

Let ($x_{1}, y_{1}$) = A (1, -2)

($x_{2}, y_{2}$) = B (2, -2)

($x_{3}, y_{3}$) = C (5, 3)

Area of $\triangle$ ABC = $\frac{1}{2}$ [$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}$)]

= $\frac{1}{2}$ [1(- 2 - 3) + 2 (3 + 2 ) + 5 ( - 2 + 2)]

Area of $\triangle$ ABC = $\frac{5}{2}$

The area of triangle ACD

Let ($x_{1}, y_{1}$) = A (1, -2)

Let ($x_{2}, y_{2}$) = C (5 , 3)

Let ($x_{3}, y_{3}$) = D (-4, 1)

Area of $\triangle$ ACD = $\frac{1}{2}$ [$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}$)]

= $\frac{1}{2}$ [ (1 (3 - 1) + 5 (1 + 2) - 4 ( -2 - 3)

Area of $\triangle$ ACD = $\frac{37}{2}$

Therefore Area of parallelogram ABCD = (Area of triangle ABC + Area of triangle ACD)

Area of parallelogram ABCD = $\frac{5}{2}$ + $\frac{37}{2}$

Area of parallelogram ABCD = 21 square units.