# geometry

• July 21st 2009, 07:26 PM
sunnyyueliang
geometry
i need help with a question ...
ΔPQR has vertices P(1, 3), Q(-1, 1), and R(5, 1). Determine the coordinates of the centroid of ΔABC.

What goes centroid mean and how do you solve it!
(Bow)(Wait)(Itwasntme)(Smirk)
• July 21st 2009, 07:43 PM
VonNemo19
Quote:

Originally Posted by sunnyyueliang
i need help with a question ...
ΔPQR has vertices P(1, 3), Q(-1, 1), and R(5, 1). Determine the coordinates of the centroid of ΔABC.

What goes centroid mean and how do you solve it!
(Bow)(Wait)(Itwasntme)(Smirk)

Centroid means center.

Triangle Centroid -- from Wolfram MathWorld
• July 21st 2009, 07:52 PM
Goharm
it is the center of the triangle.

find the midpoint of all the segments. the point of intersection of lines formed from each vertex to the midpoint of each opposite segment is the centroid.
hope that helps!
• July 21st 2009, 07:59 PM
galactus
Don't get the centroid mixed up with the incenter.
• July 21st 2009, 08:08 PM
Goharm
centroid is located at intersection of medians where as the incenter is located at intersection of the angle bisectors
• July 22nd 2009, 12:15 AM
earboth
Quote:

Originally Posted by sunnyyueliang
i need help with a question ...
ΔPQR has vertices P(1, 3), Q(-1, 1), and R(5, 1). Determine the coordinates of the centroid of ΔABC.

What goes centroid mean and how do you solve it!
(Bow)(Wait)(Itwasntme)(Smirk)

If you know the coordinates of the vertices then the coordinates of the centroid are calculated by a simple formula:

If $P(x_P, y_P)$, $Q(x_Q, y_Q)$ and $R(x_R, y_R)$ then the centroid

$C\left(\dfrac13(x_P + x_Q + x_R)\ ,\ \dfrac13(y_P + y_Q + y_R)\right)$

With your question: $C\left(\dfrac53\ ,\ \dfrac53 \right)$
• July 22nd 2009, 06:43 AM
alunw
There is no such thing as "the" center of a general triangle. A triangle has many different kinds of center just as a collection of numbers has many different kinds of average. There is a very extensive online encycolpedia about triangle centers here:
ENCYCLOPEDIA OF TRIANGLE CENTERS
The formulas it gives use either trilinear or barycentric coordinates which are both quite different from ordinary Cartesian coordinates.