# A triangle has corners at (1 ,9 ), (5 ,7 ), and (3 ,8 ). How far is the triangle's centroid from the origin?

Jan 10, 2018

$\sqrt{73}$ units

#### Explanation:

If,
$\left({x}_{1} , {y}_{1}\right) \implies \left(1 , 9\right)$
$\left({x}_{2} , {y}_{2}\right) \implies \left(5 , 7\right)$
$\left({x}_{3} , {y}_{3}\right) \implies \left(3 , 8\right)$
are the vertices of a triangle then,

Centroid (G) of a triangle is given by,
$G = \left(\frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} , \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right)$

So, substituting the above values we get,
$G = \left(3 , 8\right)$

Now, to find distance 'd' between $\left(0 , 0\right)$ and $\left(3 , 8\right)$ use distance formula

$d = \sqrt{{\left(8 - 0\right)}^{2} + {\left(3 - 0\right)}^{2}}$

$d = \sqrt{73}$ units

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Jan 10, 2018

Distance of centroid from origin is $\textcolor{red}{8.544}$

#### Explanation:

Formula to get centroid of a triangle, given the coordinates of three vertices is

$G \left(x , y\right) = \frac{x 1 + x 2 + x 3}{3} , \frac{y 1 + y 2 + y 3}{3}$

${G}_{x} = \frac{1 + 5 + 3}{3} = 3$

${G}_{y} = \frac{9 + 7 + 8}{3} = 8$

Coordinates of centroid $G \left(x , y\right) = \left(3 , 8\right)$

Coordinates of origin $O \left(x , y\right) = \left(0.0\right)$

Distance of centroid from origin is
$d = \sqrt{{\left(3 - 0\right)}^{2} + {\left(8 - 0\right)}^{2}} = \sqrt{{3}^{2} + {8}^{2}} = \ast 8.544 \ast$