A triangle has corners at (3 ,4 ), (6 ,3 ), and (2 ,7 ). How far is the triangle's centroid from the origin?

Apr 4, 2016

≈ 5.935 units

Explanation:

The first step is to find the coordinates of the centroid.

If $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ and } \left({x}_{3} , {y}_{3}\right)$
are the coordinates of the vertices of a triangle , then

x-coord$\left({x}_{c}\right) = \frac{1}{3} \left({x}_{1} + {x}_{2} + {x}_{3}\right) \text{ and }$
y-coord$\left({y}_{c}\right) = \frac{1}{3} \left({y}_{1} + {y}_{2} + {y}_{3}\right)$

here ${x}_{c} = \frac{1}{3} \left(3 + 6 + 2\right) = \frac{11}{3} \text{ and } {y}_{c} = \frac{1}{3} \left(4 + 3 + 7\right) = \frac{14}{3}$

coords of centroid $= \left(\frac{11}{3} , \frac{14}{3}\right)$

To calculate the distance the centroid is from the origin use the $\textcolor{b l u e}{\text{ distance formula }}$

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coord points }$

The 2 points here are the centroid and the origin.
Since the origin is one of the points this simplifies the distance formula to :

$d = \sqrt{{\left(\frac{11}{3}\right)}^{2} + {\left(\frac{14}{3}\right)}^{2}} = \sqrt{\left(\frac{121}{9}\right) + \left(\frac{196}{9}\right)}$

 = sqrt(317/9) ≈ 5.935" units "