### Theory:

Consider a triangle $$ABC$$ whose vertices are $$A (x_1, y_1)$$, $$B (x_2, y_2)$$ and $$C (x_3, y_3)$$.

Draw the perpendiculars $$AP$$, $$BQ$$ and $$CR$$ from $$A$$, $$B$$ and $$C$$ respectively to the $$x -$$ axis. From the figure, it is clear that $$ABQP$$, $$APRC$$ and $$BQRC$$ are all trapezoids.

Therefore, we have:
Area of $$\Delta ABC$$ $$=$$ Area of trapezium $$ABQP$$ $$+$$ Area of trapezium $$APRC$$ $$–$$ Area of trapezium $$BQRC$$.

We know that the area of a trapezium is given by:
Area of a trapezium $$=$$ $$\frac{1}{2} \times \text{sum of parallel sides} \times \text{distance between the parallel sides}$$
Therefore, Area of $$\Delta ABC = \frac{1}{2}(BQ + AP) QP + \frac{1}{2}(AP + CR) PR – \frac{1}{2}(BQ + CR) QR$$

Rewrite the above equation in terms of coordinates as follows:

Area of $$\Delta ABC$$ $$=$$ $$\frac{1}{2}\left(y_2 + y_1\right)\left(x_1 - x_2\right)$$ $$+$$ $$\frac{1}{2}\left(y_1 + y_3\right)\left(x_3 - x_1\right)$$ $$-$$ $$\frac{1}{2}\left(y_2 + y_3\right)\left(x_3 - x_2\right)$$

Factor out the common terms and rearrange the terms.

Thus, we have:

Area of $$\Delta ABC$$ $$=$$ $$\frac{1}{2}\left[x_1 \left(y_2 - y_3\right) + x_2 \left(y_3 - y_1\right) + x_3 \left(y_1 - y_2\right)\right]$$ square units.

Another form of the formula:

The above given formula can also be written as follows:

Area of $$\Delta ABC$$ $$=$$ $$\frac{1}{2}\left[ \left(x_1y_2 + x_2y_3 + x_3y_1\right) - \left(x_2y_1 + x_3y_2 + x_1y_3\right)\right]$$ square units.

The above formula can be easily remembered using the following pictorial representation. Example:
Find the area of the triangle joining the vertices $$(- 2, 1)$$, $$(- 11, -9)$$ and $$(6, 7)$$.

Solution:

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

Area of the triangle $$=$$ $$\frac{1}{2}\left[- 2 \left(-9 - 7\right) + (- 11) \left(7 - 1\right) + 6 \left(1 - (-9)\right)\right]$$

$$=$$ $$\frac{1}{2}\left[- 2 \left(-16\right) + (- 11) \left(6\right) + 6 \left(10\right)\right]$$

$$=$$ $$\frac{1}{2}\left[32 - 66 + 60 \right]$$

$$=$$ $$\frac{26}{2}$$

$$=$$ $$13$$

Therefore, area of the triangle joining the vertices $$(2, 1)$$, $$(11, -9)$$ and $$(6, 7)$$ is $$13$$ square units.
Important!
To calculate the area of a polygon, divide it into triangular regions, which have no common area, and add the areas of these regions.