Question

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

Answer 1

`"Distance" = sqrt277/3`

Explanation:

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`