# A triangle has corners at (6 ,7 ), (2 ,6 ), and (1 ,2 ). How far is the triangle's centroid from the origin?

Jun 20, 2016

≈ 5.83 units

#### Explanation:

The first step here is to calculate the coordinates of the centroid.

Given 3 vertices $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) , \left({x}_{3} , {y}_{3}\right)$ then the coordinates of the centroid are found as follows.

x-coordinate ${x}_{c} = \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{1}{3} \left({x}_{1} + {x}_{2} + {x}_{3}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and y-coordinate ${y}_{c} = \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{1}{3} \left({y}_{1} + {y}_{2} + {y}_{3}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Using the 3 vertices given here.

${x}_{c} = \frac{1}{3} \left(6 + 2 + 1\right) = \frac{1}{3} \times 9 = 3$

and ${y}_{c} = \frac{1}{3} \left(7 + 6 + 2\right) = \frac{1}{3} \times 15 = 5$

Hence coordinates of centroid = (3 ,5)

To calculate the distance between (3 ,5) and (0 ,0) use $\textcolor{b l u e}{\text{Pythagoras' theorem}}$

d=sqrt(3^2+5^2)=sqrt(9+25)=sqrt34≈5.83" units"