A triangle has corners at $(7, 9)$ $(7 ,9 )$ , $(1, 7)$ $(1 ,7 )$ , and $(6, 2)$ $(6 ,2 )$ . How far is the triangle's centroid from the origin?

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A triangle has corners at $(7, 9)$ $(7 ,9 )$ , $(1, 7)$ $(1 ,7 )$ , and $(6, 2)$ $(6 ,2 )$ . How far is the triangle's centroid from the origin?

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Answer 1 · 2016-06-26T08:00:58

Triangle's centroid is $7.6$ $7.6$ units away from the origin.

Explanation:

Centroid of a triangle, whose corners are $(x_{1}, y_{1})$ $(x_1,y_1)$ , $(x_{2}, y_{2})$ $(x_2,y_2)$ and $(x_{3}, y_{3})$ $(x_3,y_3)$ , is given by $(\frac{1}{3} (x_{1} + x_{2} + x_{3}), \frac{1}{3} (y_{1} + y_{2} + y_{3}))$ $(1/3(x_1+x_2+x_3),1/3(y_1+y_2+y_3))$

Hence centroid of the triangle whose corners are $(7, 9)$ $(7,9)$ , $(1.7)$ $(1.7)$ and $(6, 2)$ $(6,2)$ is

$(\frac{1}{3} (7 + 1 + 6), \frac{1}{3} (9 + 7 + 2))$ $(1/3(7+1+6),1/3(9+7+2))$ or $(\frac{14}{3}, \frac{18}{3})$ $(14/3,18/3)$

And its distance from origin $(0, 0)$ $(0,0)$ is

$\sqrt{{(\frac{18}{3} - 0)}^{2} + {(\frac{14}{3} - 0)}^{2}} = \sqrt{\frac{324}{9} + \frac{196}{9}}$ $sqrt((18/3-0)^2+(14/3-0)^2)=sqrt(324/9+196/9)$

= $\frac{1}{3} \sqrt{520} = \frac{1}{3} \times 22.8 = 7.6$ $1/3sqrt520=1/3xx22.8=7.6$

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A triangle has corners at $(7, 9)$ $(7 ,9 )$ , $(1, 7)$ $(1 ,7 )$ , and $(6, 2)$ $(6 ,2 )$ . How far is the triangle's centroid from the origin?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

A triangle has corners at (7 ,9 )(7,9)(7 ,9 ), (1 ,7 )(1,7)(1 ,7 ), and (6 ,2 )(6,2)(6 ,2 ). How far is the triangle's centroid from the origin?

1 Answer

Explanation:

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A triangle has corners at $(7, 9)$ $(7 ,9 )$ , $(1, 7)$ $(1 ,7 )$ , and $(6, 2)$ $(6 ,2 )$ . How far is the triangle's centroid from the origin?