A triangle has corners at #(1 ,6 )#, #(7 ,4 )#, and #(5 ,9 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(1  ,6  )#, #(7  ,4  )#, and #(5   ,9  )#.  How far is the triangle's centroid from the origin?

Wyatt Anderson · Accepted Answer

the centroid is #58.89# when 3 vertices of a triangle #(xunits away from the origin.

when 3 vertices of a triangle are #(x_1,y_1),(x_2,y_2),(x_3,y_3)# then the centroid of the triangle is #G=((x_1+x_2+x_3)/3,(y_1+y_2+y_3)/3)# substituting the values we get the centroid as #G=(13/3,19/3)# the distance between #G(13/3,19/3) and C(0,0)# is #sqrt[(x_2-x_1)^2+(y_2-y_1)^2# substituting in the formula we get distance as #sqrt(530/9)~~58.89#

Autumn Armstrong · Answer

undefined To find the centroid of a triangle, you calculate the average of the x-coordinates and the average of the y-coordinates of its vertices. Then, the distance from the centroid to the origin can be found using the distance formula.

First, calculate the centroid:

Average x-coordinate: $(1 + 7 + 5) / 3 = 13/3$

Average y-coordinate: $(6 + 4 + 9) / 3 = 19/3$

So, the centroid is at $(13/3, 19/3)$.

Now, find the distance from the centroid to the origin:

$$ d = \sqrt{(13/3)^2 + (19/3)^2} $$

$$ d = \sqrt{(169/9) + (361/9)} $$

$$ d = \sqrt{530/9} $$

$$ d = \sqrt{\frac{530}{9}} $$

$$ d = \frac{\sqrt{530}}{3} $$

Therefore, the distance from the centroid to the origin is $ \frac{\sqrt{530}}{3} $ units.