A triangle has corners at $(6 ,3 )$ , $(2 ,6 )$ , and $(1 ,5 )$ . How far is the triangle's centroid from the origin?

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Answer 1 · 2018-07-13T20:34:27

$"Distance" = sqrt277/3$

The centroid of a triangle with vertices at $(x_A,y_A)$ , $(x_B,y_B)$ and $(x_C,y_C)$ has coordinates

$O((x_A+x_B+x_C)/3,(y_A+y_B+y_C)/3)$

Hence:

${(A(6,3)),(B(2,6)),(C(1,5)) :}=>O((6+2+1)/3,(3+6+5)/3) -= O(3,14/3)$

The distance between two points $(x_1,y_1)$ , $(x_2,y_2)$ is given by the formula

$"Distance" = sqrt((x_1-x_2)^2+(y_1-y_2)^2)$

The distance between the centroid, $O$ , and the origin, $(0,0)$ is going to be:

$"Distance" = sqrt((color(red)3-0)^2+(color(red)(14/3)-0)^2)$

$=sqrt(9+196/9)=sqrt(277/9)=sqrt227/3$

A triangle has corners at (6 ,3 )(6 ,3 ), (2 ,6 )(2 ,6 ), and (1 ,5 )(1 ,5 ). How far is the triangle's centroid from the origin?