# What is the area of a triangle with vertices (x_1, y_1), (x_2, y_2), (x_3, y_3) ?

Oct 18, 2017

$\frac{1}{2} \left\mid {x}_{1} {y}_{2} + {x}_{2} {y}_{3} + {x}_{3} {y}_{1} - {x}_{1} {y}_{3} - {x}_{2} {y}_{1} - {x}_{3} {y}_{2} \right\mid$

#### Explanation:

Given three vertices: $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$, $\left({x}_{3} , {y}_{3}\right)$

Start by assuming:

${x}_{1} < {x}_{3} < {x}_{2}$

${y}_{1} < {y}_{2} < {y}_{3}$

The triangle can be drawn in a rectangle with vertices:

$\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{3}\right)$, $\left({x}_{1} , {y}_{3}\right)$

dividing it into $4$ triangles, the other $3$ of which are:

• $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$ with area: $\frac{1}{2} \left({x}_{2} - {x}_{1}\right) \left({y}_{2} - {y}_{1}\right)$

• $\left({x}_{2} , {y}_{2}\right)$, $\left({x}_{2} , {y}_{3}\right)$, $\left({x}_{3} , {y}_{3}\right)$ with area: $\frac{1}{2} \left({x}_{2} - {x}_{3}\right) \left({y}_{3} - {y}_{2}\right)$

• $\left({x}_{3} , {y}_{3}\right)$, $\left({x}_{1} , {y}_{3}\right)$, $\left({x}_{1} , {y}_{1}\right)$ with area: $\frac{1}{2} \left({x}_{3} - {x}_{1}\right) \left({y}_{3} - {y}_{1}\right)$

So the area of the given triangle is:

$\left({x}_{2} - {x}_{1}\right) \left({y}_{3} - {y}_{1}\right) - \frac{1}{2} \left({x}_{2} - {x}_{1}\right) \left({y}_{2} - {y}_{1}\right) - \frac{1}{2} \left({x}_{2} - {x}_{3}\right) \left({y}_{3} - {y}_{2}\right) - \frac{1}{2} \left({x}_{3} - {x}_{1}\right) \left({y}_{3} - {y}_{1}\right)$

$= \frac{1}{2} \left({x}_{1} {y}_{2} + {x}_{2} {y}_{3} + {x}_{3} {y}_{1} - {x}_{1} {y}_{3} - {x}_{2} {y}_{1} - {x}_{3} {y}_{2}\right)$

Note the symmetry of the final expression. It is symmetric in $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$, $\left({x}_{3} , {y}_{3}\right)$ except for its sign.

If we are only interested in the unsigned area of the triangle then we can ignore our initial assumptions and write the universal formula:

$\text{Area} = \frac{1}{2} \left\mid {x}_{1} {y}_{2} + {x}_{2} {y}_{3} + {x}_{3} {y}_{1} - {x}_{1} {y}_{3} - {x}_{2} {y}_{1} - {x}_{3} {y}_{2} \right\mid$

Jun 20, 2018

Just a small comment to the answer from George C.

#### Explanation:

The answer from George C. is 100% right, even for triangles with such vertices, that they do not fit on a rectangle. An example would be a triangle with the vertices
$\left(0 , 0\right) , \left(2 , 3\right) , \left(4 , 4\right)$

Even if the rectangle does not fit, we can still apply the formular and get the solution.

$A = \frac{1}{2} \left(| 0 \cdot 3 + 2 \cdot 4 + 4 \cdot 0 - 0 \cdot 4 - 4 \cdot 3 - 2 \cdot 0 |\right)$
$A = \frac{1}{2} \left(| 8 - 12 |\right)$
$A = \frac{1}{2} \left(| - 4 |\right)$
$A = \frac{4}{2} = 2 \text{ units}$ (confirmed by GeoGebra)