A triangle has corners at #(4 ,4 )#, #(8 ,6 )#, and #(7 ,1 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(4  ,4  )#, #(8   ,6  )#, and #(7   ,1  )#.  How far is the triangle's centroid from the origin?

Avery Allen · Accepted Answer

8.945(approx)

centroid of a triangle with given points [ (x1+x2+x3)/3 , (y1+y2+y3)/3] where x1,x2,x3,y1,y2,y3 are the coordinates of the vertices o the triangle. substituting will get u the centroid as (19/3,11/3) now the distance from the origin (0,0) will be #sqrt((x1-x2)^2+(y1-y2)^2)# X1=19/3,X2=0,Y1=11/3,Y2=0 substitute to get the answer as 8.945(approx)

Carson Arnold · Answer

undefined To find the centroid of the triangle, you first calculate the average of the x-coordinates and the average of the y-coordinates of the vertices. Then, you use these averages as the coordinates of the centroid.

Given the vertices are (4, 4), (8, 6), and (7, 1), you can find the centroid by averaging the x-coordinates and the y-coordinates:

Average of x-coordinates = $\frac{4 + 8 + 7}{3} = \frac{19}{3}$

Average of y-coordinates = $\frac{4 + 6 + 1}{3} = \frac{11}{3}$

So, the centroid has coordinates $\left(\frac{19}{3}, \frac{11}{3}\right)$.

To find the distance between the centroid and the origin, you can use the distance formula:

Distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Where $(x_1, y_1)$ are the coordinates of the origin (0, 0), and $(x_2, y_2)$ are the coordinates of the centroid.

Substituting the values, you get:

Distance = $\sqrt{\left(\frac{19}{3} - 0\right)^2 + \left(\frac{11}{3} - 0\right)^2}$

Calculating this expression will give you the distance from the origin to the centroid of the triangle.