A triangle has corners at #(1 ,6 )#, #(8 ,2 )#, and #(5 ,9 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(1  ,6  )#, #(8  ,2  )#, and #(5   ,9  )#.  How far is the triangle's centroid from the origin?

Jayden Jackson · Accepted Answer

Triangle's centroid is #7.341# units away from the origin.

Centroid of a triangle, whose corners are #(x_1,y_1)#, #(x_2,y_2)# and #(x_3,y_3)#, is given by #(1/3(x_1+x_2+x_3),1/3(y_1+y_2+y_3))# Hence centroid of the triangle whose corners are #(1,6)#, #(8,2)# and #(5,9)# is #(1/3(1+8+5),1/3(6+2+9))# or #(14/3,17/3)# And its distance from origin #(0,0)# is #sqrt((14/3-0)^2+(17/3-0)^2)=sqrt(196/9+289/9)# = #1/3sqrt485=1/3xx22.023=7.341#

James Atkins · Answer

undefined The centroid of a triangle is the average of its vertices' coordinates. Given the vertices (1, 6), (8, 2), and (5, 9), the centroid's coordinates can be found by averaging the x-coordinates and y-coordinates separately.

Centroid's x-coordinate = (1 + 8 + 5) / 3 = 14 / 3
Centroid's y-coordinate = (6 + 2 + 9) / 3 = 17 / 3

Now, using the centroid's coordinates (14/3, 17/3) and the origin (0, 0), you can apply the distance formula:

Distance = √[(x2 - x1)^2 + (y2 - y1)^2]
Distance = √[(14/3 - 0)^2 + (17/3 - 0)^2]
Distance ≈ √[(14/3)^2 + (17/3)^2]
Distance ≈ √[(196/9) + (289/9)]
Distance ≈ √[(196 + 289) / 9]
Distance ≈ √[485 / 9]
Distance ≈ √(485) / √(9)
Distance ≈ √(485) / 3

Therefore, the distance from the centroid to the origin is approximately √(485) / 3.