A triangle has corners at #(2 ,4 )#, #(7 ,6 )#, and #(4 ,5 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(2  ,4  )#, #(7   ,6  )#, and #(4   ,5  )#.  How far is the triangle's centroid from the origin?

Savannah Aguilar · Accepted Answer

The triangle's centroid is #6.62(2dp)# unit from the origin.

Coordinates of the vertices of the triangle are #A(2,4),B(7,6),C(4,5)#. The coordinates of centroid #(x,y)# of triangle is the average of the x-coordinate's value and the average of the y-coordinate's value of all the vertices of the triangle. #:.x= (2+7+4)/3=13/3=4 1/3 , y= (4+6+5)/3=5 # . So centroid is at #(4 1/3,5)# , Its distance from the origin #(0,0)# is #D= sqrt((x-0)^2+(y-0)^2) = sqrt((13/3-0)^2+(5-0)^2) # or #D=sqrt((13/3)^2+5^2) ~~6.62(2dp)# unit. The triangle's centroid is #6.62(2dp)# unit from the origin [Ans]

Parker Allen · Answer

undefined To find the centroid of a triangle with vertices at (2, 4), (7, 6), and (4, 5), follow these steps:

1. Find the coordinates of the centroid, which is the average of the coordinates of the vertices.
   Centroid's x-coordinate = (2 + 7 + 4) / 3
   Centroid's y-coordinate = (4 + 6 + 5) / 3

2. Once you have the coordinates of the centroid, use the distance formula to find the distance between the centroid and the origin (0, 0).
   Distance = sqrt((x_centroid)^2 + (y_centroid)^2)

Calculating:

Centroid's x-coordinate = (2 + 7 + 4) / 3 = 13 / 3
Centroid's y-coordinate = (4 + 6 + 5) / 3 = 15 / 3 = 5

Distance = sqrt((13/3)^2 + (5)^2)
         = sqrt((169/9) + 25)
         = sqrt((169 + 225) / 9)
         = sqrt(394 / 9)
         ≈ sqrt(43.78)
         ≈ 6.62 units

Therefore, the distance from the centroid of the triangle to the origin is approximately 6.62 units.