A triangle is a plane figure with three sides and three angles. Triangles can be of many types – acute, obtuse, right angles, isosceles, equilateral and scalene. Median of a triangle is a line segment joining a vertex to mid point of the opposite side. Centroid is where medians meet.

All triangles have three medians. Medians of a triangle meet at a point called centroid. The centroid is the center of a triangle which means the triangle is a center of gravity is located at the centroid. It is also known as barycenter of a triangle. It is one of the points of concurrency of a triangle. Centroid is usually denoted by $G$.
Properties of a Centroid:
1. The centroid is always located inside of a triangle (as opposed to orthocenter and circumcenter which may or may not be located inside the triangle).
2. The triangle is bisected into two triangles of equal area by each median.
3. The centroid divides each median into two segments whose lengths are in the ratio $2:1$ with the segment from vertex to median being the longer one.
$ABC$ is a triangle with vertices $A, B, C$ located at $(x1, y1),\ (x2, y2),\ (x3, y3)$ respectively.
Centroid of $\Delta ABC$ is located at = $\frac{(x1 + x2 + x3)}{3}$ and $\frac{(y1 + y2 + y3)}{3}$
Example:
Let have vertices at $A (2, 6),\ B (3, 2),\ C (4, 1)$
Centroid of $\Delta ABC$ is located at = $\frac{(2 + 3 + 4)}{3}$ and $\frac{(6 + 2 + 1)}{3}$ = $(3, 1)$
If $h$ be the altitude and $b$ be the base of a right angled triangle, then the centroid $G$ of a right angled triangle is located at $(\frac{b}{3}$, $\frac{h}{3})$ as shown in the diagram below An equilateral triangle is a triangle all of whose three sides have the same length. Each of the vertex angles of an equilateral triangle is equal to $60^{\circ}$. It is the simplest regular polygon.
The centroid of an equilateral triangle coincides with orthocenter and circumcenter. There is no Euler line in an equilateral triangle (Euler line joins the orthocenter , triangle centroid , circumcenter and other centers of a triangle).
The line joining the centroid with vertex of an equilateral triangle bisects the $60^{\circ}$ vertex angle. The median is also perpendicular bisector of sides in an equilateral triangle. Therefore in an equilateral triangle when the medians meet at centroid they form six $30  60  90$ triangles. See diagram below for better understanding.
$\Delta ABC$ is equilateral triangle. $AD$ is median from vertex $A$ to side $BC$.
Example 1:
Find the centroid of a triangle whose vertices are located at $(5, 4),\ (2, 1)$ and $(8, 3)$.
Solution:
Write the coordinates in a table as $x1, x2 ,x3$ and $y1 ,y2 ,y3$
$x1$ = $5, y1$ = $4$
$x2$ = $2, y2$ = $1$
$x3$ = $8, y3$ = $3$
Formula for centroid is $\frac{(x1 + x2 + x3)}{3}$ and $\frac{(y1 + y2 + y3)}{3}$
$\frac{(5 + 2 + 8)}{3}$ = $\frac{15}{5}$ = $3$
$(4 + 1 + 3)$ = $\frac{6}{3}$ = $2$
Centroid is located at $(3, 2)$
Example 2:
Find coordinates for the centroid of $\Delta PQR$ whose vertices are located at $P (2, 6),\ Q (2, 9)$ and $R (4, 3)$
Solution:
$x1$ = $2, y1$ = $6$
$x2$ = $2, y2$ = $9$
$x3$ = $4, y3$ = $3$
Formula for centroid is $\frac{(x1 + x2 + x3)}{3}$ and $\frac{(y1 + y2 + y3)}{3}$
$\frac{(2 + 2 + 4)}{3}$ = $\frac{0}{3}$ = $0$
$(6 + 9 + 3)$ = $\frac{15}{3}$ = $5$
Centroid of $\Delta PQR$ is located at = $(0, 5)$
Example 3:
Show that the centroid of the $\Delta KLM$ formed by the points $K (x  y, y  z),\ L (x, y)$ and $M (y, z)$ is the origin.
Solution:
$x1$ = $x  y,\ y1$ = $y  z$
$x2$ = $x,\ y2$ = $y$
$x3$ = $y,\ y3$ = $z$
Formula for centroid is $\frac{(x1 + x2 + x3)}{3}$ and $\frac{(y1 + y2 + y3)}{3}$
$\frac{[(x – y + (x) + y)]}{3}$ = $\frac{(x – y – x + y)}{3}$ = $0$
$\frac{[(y – z + (y) + z)]}{3}$ = $\frac{(y – z – y + z)}{3}$ = $0$
It is proved that centroid of the $\Delta KLM$ is located at origin $(0, 0)$
Example 4:
In $\Delta ABC,\ A$ is located at $(6, 2)$ and $B$ at $(8, 3)$. If centroid is located at $(4, 5)$ what are the coordinates of vertex $C$?
Solution:
Let coordinates of vertex $C$ be at $(x3, y3)$
Formula for centroid is $\frac{(x1 + x2 + x3)}{3}$ and $\frac{(y1 + y2 + y3)}{3}$
$4$ = $\frac{(6 + 8 + x3)}{3}$
Multiplying both sides by $3$
$12$ = $6 + 8 + x3$
$14 + x3$ = $12$
$x3$ = $2$
$5$ = $\frac{[(2) + (3) + y3]}{3}$
Multiplying both sides by $3$
$15$ = $2 3 + y3$
$y3 – 5$ = $15$
$y3$ = $10$
$C$ is located at $(2,\ 10)$