### Theory:

Consider a quadrilateral $$ABCD$$ whose vertices are $$A (x_1, y_1)$$, $$B (x_2, y_2)$$, $$C (x_3, y_3)$$ and $$D (x_4, y_4)$$ taken in order.

Draw a diagonal $$BD$$ and split the quadrilateral into two triangles $$ABD$$ and $$BCD$$. Therefore, we have:
Area of the quadrilateral $$ABCD$$ $$=$$ Area of triangle $$ABD$$ $$+$$ Area of triangle $$BCD$$.

We know that the area of a triangle having its vertices $$(x_1, y_1)$$, $$(x_2, y_2)$$, $$(x_3, y_3)$$is
given by:
Area of the triangle $$=$$ $$\frac{1}{2}\left[ \left(x_1y_2 + x_2y_3 + x_3y_1\right) - \left(x_2y_1 + x_3y_2 + x_1y_3\right)\right]$$ square units.
Therefore, Area of the quadrilateral $$ABCD$$ $$=$$ $$\frac{1}{2}\left[ \left(x_1y_2 + x_2y_4 + x_4y_1\right) - \left(x_2y_1 + x_4y_2 + x_1y_4\right)\right]$$ $$+$$ $$\frac{1}{2}\left[ \left(x_2y_3 + x_3y_4 + x_4y_2\right) - \left(x_3y_2 + x_4y_3 + x_2y_4\right)\right]$$

Factor out the common terms and rearrange the terms.

Thus, we have:

Area of the quadrilateral $$ABCD$$ $$=$$ $$\frac{1}{2}\left[\left(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1\right) - \left(x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4\right)\right]$$

The formula can be easily remembered using the following pictorial representation. Another form of the formula:

Area of the quadrilateral $$ABCD$$ $$=$$ $$\frac{1}{2}\left[\left(x_1 - x_3\right)\left(y_2 - y_4\right) - \left(x_2 - x_4\right)\left(y_1 - y_3\right)\right]$$ square units.
Example:
Find the area of the quadrilateral joining the vertices $$(- 3, 4)$$, $$(- 5, - 6)$$, $$(4, - 1)$$ and $$(1, 2)$$.

Solution:

By the formula, we have:
Area of the quadrilateral $$=$$ $$\frac{1}{2}\left[\left(x_1 - x_3\right)\left(y_2 - y_4\right) - \left(x_2 - x_4\right)\left(y_1 - y_3\right)\right]$$ square units
Substitute the required values in the above formula.

Area of the quadrilateral $$=$$ $$\frac{1}{2}\left[\left(-3 - 4\right)\left(-6 - 2\right) - \left(-5 - 1\right)\left(4 - (-1)\right)\right]$$

$$=$$ $$\frac{1}{2}\left[(-7)(- 8) - (-6)(4)\right]$$

$$=$$ $$\frac{1}{2}\left[56 + 24\right]$$

$$=$$ $$\frac{80}{2}$$

$$=$$ $$40$$

Therefore, area of the quadrilateral joining the vertices $$(- 3, 4)$$, $$(- 5, - 6)$$, $$4, - 1)$$ and $$(1, 2)$$ is $$40$$ square units