# centriod of trianlge

• January 28th 2010, 12:23 AM
yshridhar
centriod of trianlge
Hi everybody

Show that the centroid of a triangle coincides with that of the triangle formed by the mid points of the sides.

Thank you all
Sridhar
• January 28th 2010, 02:39 AM
HallsofIvy
For a triangle, the centriod is just the "average" of the vertices. That is, if the vertices are $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, in some coordinate system, then the centroid is at $\left(\frac{x_1+ x_2+ x_3}{3}, \frac{y_1+ y_2+ y_3}{3}\right)$.

Of course, the midpoints of the sides are $\left(\frac{x_1+ x_2}{2}, \frac{y_1+ y_2}{2}\right)$, $\left(\frac{x_2+ x_3}{2}, \frac{y_2+ y_3}{2}\right)$, and $\left(\frac{x_1+ x_3}{2}, \frac{y_1+ y_3}{2}\right)$. Use the formula above to find the centroid of that triangle.