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\(\bullet\) Interactive Triangle Centroid (geogebra.org)

 

Lesson

 

Notes

 

Questions

Find the centroid of the following triangles with the following vertices.

\(\textbf{1)}\) \((1,2), (5,4), (3,6)\)

Show Centroid

The Centroid is \((3,4)\)

Show Work (Centroid Formula)

\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{1+5+3}{3},\frac{2+4+6}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{9}{3},\frac{12}{3} \right)\)
\(\,\,\,\,\,\,\)The Centroid is \((3,4)\)

Show Work (Equation Method)

First, we need to find the equations of the medians of the triangle.
A median of a triangle is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.
\(\text{Midpoint} = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
Let’s find the equation of the median connecting vertex A to the midpoint of side \(\overline{BC}\):
M is the midpoint of side \(\overline{BC}\),
M \( = \left(\frac{5+3}{2}, \frac{4+6}{2}\right) = (4, 5)\)
The slope of the median \(\overline{AM}\) is
\(\frac{y_2-y_1}{x_2-x_1}=\frac{5-2}{4-1} = \frac{3}{3}=1\)
Use this slope and the coordinates of point A to find the equation of the median in point-slope form:
\(y – 2 = 1(x – 1)\)
Solving for y, we get:
\(y = 1x +1 \)
Use the same process to find the the median connecting vertex B to the midpoint of side \(\overline{AC}\).
\(y = 4\)
We can find the intersection of these medians to find the coordinates of the centroid.
\(4 = 1x +1 \)
\(x=3 \)
The Centroid is \((3,4)\)

 

\(\textbf{2)}\) \((-8,0), (4,7), (10,8)\)

Show Centroid

The Centroid is \((2,5)\)

Show Work (Centroid Formula)

\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{-8+4+10}{3},\frac{0+7+8}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{6}{3},\frac{15}{3} \right)\)
\(\,\,\,\,\,\,\)The Centroid is \((2,5)\)

Show Work (Equation Method)

First, we need to find the equations of the medians of the triangle.
A median of a triangle is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.
\(\text{Midpoint} = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
Let’s find the equation of the median connecting vertex A to the midpoint of side \(\overline{BC}\):
M is the midpoint of side \(\overline{BC}\),
M \( = \left(\frac{4+10}{2}, \frac{7+8}{2}\right) = (7, 7.5)\)
The slope of the median \(\overline{AM}\) is
\(\frac{y_2-y_1}{x_2-x_1}=\frac{7.5-0}{7-(-8)} = \frac{7.5}{15} = 0.5\)
Use this slope and the coordinates of point A to find the equation of the median in point-slope form:
\(y – 0 = 0.5(x – (-8))\)
Solving for y, we get:
\(y = 0.5x + 4\)
Use the same process to find the median connecting vertex B to the midpoint of side \(\overline{AC}\).
\(y = 5\)
We can find the intersection of these medians to find the coordinates of the centroid.
\(5 = 0.5x + 4\)
\(x = 2\)
The Centroid is \((2,5)\)

\(\textbf{3)}\) \((1,-2), (1,2), (-2,3)\)

Show Centroid

The Centroid is \((0,1)\)

Show Work

\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{1+1+(-2)}{3},\frac{-2+2+3}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{0}{3},\frac{3}{3} \right)\)
\(\,\,\,\,\,\,\)The Centroid is \((0,1)\)

 

\(\textbf{4)}\) \((1,-2), (2,2), (3,0)\)

Show Centroid

The Centroid is \((2,0)\)

Show Work

\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{1+2+3}{3},\frac{-2+2+0}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{6}{3},\frac{0}{3} \right)\)
\(\,\,\,\,\,\,\)The Centroid is \((2,0)\)

 

\(\textbf{5)}\) \((0,6), (4,0), (5,0)\)

Show Centroid

The Centroid is \((3,2)\)

Show Work

\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{0+4+5}{3},\frac{6+0+0}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{9}{3},\frac{6}{3} \right)\)
\(\,\,\,\,\,\,\)The Centroid is \((3,2)\)

 

\(\textbf{6)}\) \((-8,9), (0,2), (2,4)\)

Show Centroid

The Centroid is \((-2,5)\)

Show Work

\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{-8+0+2}{3},\frac{9+2+4}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{-6}{3},\frac{15}{3} \right)\)
\(\,\,\,\,\,\,\)The Centroid is \((-2,5)\)

 

\(\textbf{7)}\) \((-2,-2), (-2,7), (4,7)\)

Show Centroid

The Centroid is \((0,4)\)

Show Work

\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{-2-2+4}{3},\frac{-2+7+7}{3} \right)\)
\(\,\,\,\,\,\,\text{Centroid}=\left(\frac{0}{3},\frac{12}{3} \right)\)
\(\,\,\,\,\,\,\)The Centroid is \((0,4)\)

 

 

See Related Pages\(\)

\(\bullet\text{ Geometry Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Orthocenter}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Circumcenter}\)
\(\,\,\,\,\,\,\,\,\)

 

In Summary

The centroid is the point where a shape’s mass is evenly distributed. If you suspended the shape from that point, it would be perfectly balanced. In triangles the centroid can be found using different methods.

If the coordinates of the points of the vertices are known, the coordinates of the vertex can be found by averaging the x values and the y values of the vertices, \(\text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)\). In the case of 3 dimensions, you would also average the z values, \(\text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}\right)\).

The centroid is also the point of concurrency of the medians of a triangle. A median of a triangle can be found by locating the midpoint of one side and connecting the segment from that midpoint to the opposite vertex. If you repeat this for all 3 medians of any triangle, the point where the medians meet, known as the point of concurrency, will be the centroid of that triangle.