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#