Formulas and examples for triangle

Area of the triangle?

The area of a triangle whose vertices are $A(x_A, y_A), B(x_B, y_B)$ and $C(x_C, y_C)$ is given by :

$$ {\color{blue}{ K = \frac12|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A-y_B)| }} $$

Example:

Find the area of the triangle whose vertices are $A(2, 4), B(3, -1)$ and $C(-3, 3)$.

Solution:

In this example we have: $ x_A = 2,~~ y_A = 4,~~ x_B = 3,~~ y_B = -1, x_C = -3,~~ y_C = 3$. So we have:

$$ \begin{aligned} K & = \frac12|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A-y_B)| \\ K & = \frac12|2(-1 - 3) + 3(3 - 4) -3(4-(-1))| \\ K & = \frac12|2\cdot(-4) + 3\cdot(-1) -3\cdot5| \\ K & = \frac12|-8 - 3 -15| \\ K & = \frac12|-26| \\ K & = \frac12\cdot26 \\ K & = 13 \end{aligned} $$

Centroid of the triangle?

The centroid of a triangle whose vertices are $A(x_A, y_A), B(x_B, y_B)$ and $C(x_C, y_C)$ is given by :

$$ {\color{blue}{ (x,y) = \left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\right) }} $$

Example:

Find the centroid of the triangle whose vertices are $A(2, 4), B(3, -1)$ and $C(-3, 3)$.

Solution:

Using the same $x_A, y_A, x_B, y_B, x_C, y_C$, as in previous example we have:

$$ \begin{aligned} (x,y) & = \left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\right) \\ (x,y) & = \left(\frac{2 + 3 - 3}{3}, \frac{4 - 1 + 3}{3}\right) \\ (x,y) & = \left(\frac{2}{3}, 2\right) \\ \end{aligned} $$