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# Triangle calculator

This calculator finds all the main triangle parameters, such as area, medians, altitudes, centroid and incenter. The calculator shows a formula and an explanation for each parameter of a triangle.

Triangle in coordinate geometry
Input vertices and choose one of seven triangle characteristics to compute.
help ↓↓ examples ↓↓
 Area (default) Medians Altitudes Centroid (intersection of medians) Incenter (center of the incircle) Circumcenter (center of circumscribed circle) Orthocenter (intersection of the altitudes)
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examples
example 1:ex 1:
Find area of a triangle whose vertices are $(4, 4), (-2, 3)$ and $(-4, -5)$.
example 2:ex 2:
Find area of a triangle whose vertices are $\left(\frac{4}{3}, 3.5\right), \left(8, -\frac{1}{2}\right)$ and $(-7, 5.2)$.
example 3:ex 3:
Find altitudes of a triangle whose vertices are $(1,1), (3,5)$ and $(-10, 8)$.
example 4:ex 4:
Find circumcenter of a triangle whose vertices are $(-2,-5), (3,4)$ and $(10, -3)$.
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# 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}
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