The centroid is said to the geometric center of a two-dimensional surface and is being used in physics and in mathematics. A centroid is a point which is located at average or arithmetic mean position of all points in a plane figure. We can say that the centroid is a point where the object could be balanced perfectly even on the tip of a pin. In mathematics, the centroid of different geometric shapes can be found in different ways. Let us talk about the centroid of a triangle.

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In a triangle, a centroid is defined as a point which is obtained by the intersection of its medians. A triangle has three medians which are the line segments joining the vertex to the midpoint of opposite side. All three medians of a triangle are concurrent (meet at one point) and their point of concurrency is referred as the centroid of the triangle.

In the diagram shown below, three medians of a triangle ABC intersect at centroid G.

In the diagram shown below, three medians of a triangle ABC intersect at centroid G.

The centroid is also known as the center of gravity. A triangular plate could be balanced at this point on the tip of a pencil or on your finger.

The properties of the centroid of a triangle are discussed below.

(iii)

(iv)

If the coordinates of three vertices of a triangle are given by ($x_{1}, y_{1}$), ($x_{2}, y_{2}$) and ($x_{3}, y_{3}$). Then the coordinates of centroid of a triangle can be determined using the following formula :

$(\frac{x_{1}+ x_{2} +x_{3}}{3}, \frac{y_{1}+ y_{2} +y_{3}}{3})$

Where, x coordinates of the vertices of a triangle are $x_{1}$, $x_{2}$ and $x_{3}$.

Likewise, y coordinates of the vertices of a triangle are $y_{1}$, $y_{2}$ and $y_{3}$. Assume that ABC be a triangle with coordinates of vertices given as A($x_{1}, y_{1}$), B($x_{2}, y_{2}$) and C($x_{3}, y_{3}$). The points D, E, and F are the midpoints of BC, AC, and AB respectively. The centroid is demonstrated by the point G.

$(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2})$

The point G divides the line segment AD in the ratio of 2 : 1. Therefore, the coordinates of G are calculated using section formula.

x = $\frac{2(\frac{x_{2}+x_{3}}{2}) + 1 . x_{1}}{2+1}$

= $\frac{x_{2}+x_{3} + x_{1}}{3}$

= $\frac{x_{1}+x_{2}+x_{3}}{3}$

y = $\frac{2(\frac{y_{2}+y_{3}}{2}) + 1 . y_{1}}{2+1}$

= $\frac{y_{2}+y_{3} + y_{1}}{3}$

= $\frac{y_{1}+y_{2}+y_{3}}{3}$

Thus, the coordinates of the centroid of a triangle are :

$(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3})$

Step 1 :

Step 3 :

Step 4 :

Solution :

$x_{2}$ = 18, $y_{2}$ = -3

$x_{3}$ = -4, $y_{3}$ = -2

Coordinates of centroid :

$(\frac{x_{1}+ x_{2} +x_{3}}{3}, \frac{y_{1}+ y_{2} +y_{3}}{3})$

= $(\frac{1+ 18 +(-4)}{3}, \frac{2+ (-3) +(-2)}{3})$

= $(\frac{15}{3}, \frac{-3}{3})$

= (5, -1)

Solution :

$x_{2}$ = 2, $y_{2}$ = 1

Let coordinates of centroid are (x, y). Then x = 3 and y = -2

x = $\frac{x_{1}+ x_{2} +x_{3}}{3}$

3 = $\frac{-1+ 2 +x_{3}}{3}$

9 = 1 + $x_{3}$

$x_{3}$ = 8

y = $\frac{y_{1}+ y_{2} +y_{3}}{3}$

-2 = $\frac{-3+ 1 +y_{3}}{3}$

-6 = -2 + $y_{3}$

$y_{3}$ = -4

The third vertex is located at (8, -4).

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