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Centroid of a Triangle

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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Definition

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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.

Centroid of a Triangle

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.

Properties

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

(i) The centroid is located at the intersecting point of three medians.

(ii) The centroid is one of the three points of concurrency (incenter, circumcenter, centroid) in a triangle.

(iii)
 It is situated in the interior of the triangle (not necessary with incenter and circumcenter).

(iv)
 At centroid, each median is divided in a ratio of 2 :1.

Formula

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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 $y_{3}$.

Proof

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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.

Poof of Centroid of a Triangle

As D is the midpoint of BC, the coordinates of D can be determined using midpoint formula :

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

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-coordinate of G

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

= $\frac{x_{2}+x_{3} +  x_{1}}{3}$

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

Similarly, y-coordinate of G

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 :

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

How To Find Centroid of a Triangle

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In order to find the coordinates of centroid of a triangle, following steps should be acquired.

Step 1 :
 Determine the coordinates of the vertices of a triangle.

Step 2 : Name them as $x_{1}$, $x_{2}$, $x_{3}$ and $y_{1}$, $y_{2}$, $y_{3}$.

Step 3 :
 Plug the values in the formula $(\frac{x_{1}+ x_{2} +x_{3}}{3}, \frac{y_{1}+ y_{2} +y_{3}}{3})$.

Step 4 :
 On solving, you obtain the coordinates of the centroid.

Examples

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The examples based on the centroid of a triangle are discussed below.
Example 1: Calculate the coordinates of the centroid of a triangle whose three vertices have the coordinates (1, 2), (18, -3) and (-4, -2).

Solution :
 $x_{1}$ = 1, $y_{1}$ = 2

$x_{2}$ = 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)
Example 2 : Two vertices of a triangle are (-1, -3) and (2, 1). If the coordinates of centroid are (3, -2), determine the third vertex.

Solution :
 
$x_{1}$ = -1, $y_{1}$ = -3

$x_{2}$ = 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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