1. Area of the triangle

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:

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.

 

 

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.