A triangle has corners at #(4 ,1 )#, #(8 ,3 )#, and #(5 ,8 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(4  ,1  )#, #(8   ,3  )#, and #(5   ,8  )#.  How far is the triangle's centroid from the origin?

Theodore Abad · Accepted Answer

The distance is #=6.9#

If we have a triangle has corners #(x_1,y_1)# ; #(x_2, y_2)# and #(x_3,y_3)#

The coordinates of the centroid are

#x_c=(x_1+x_2+x_3)/3#

and

#y_c=(y_1+y_2+y_3)/3#

and the distance from the orogin is

#OC=sqrt((x_c)^2+(y_c)^2)#

Here, we have

#x_c=(4+8+5)/3=17/3#

#y_c=(1+3+8)/3=4#

Therefore,

#OC=sqrt((17/3)^2+(4^2))=sqrt48.11=6.9#

Claire Akin · Answer

undefined To find the centroid of a triangle with vertices at (4, 1), (8, 3), and (5, 8), you can use the formula:

$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$

Substitute the coordinates of the vertices:

$$ x_1 = 4, \quad y_1 = 1 $$
$$ x_2 = 8, \quad y_2 = 3 $$
$$ x_3 = 5, \quad y_3 = 8 $$

$$ \text{Centroid} = \left( \frac{4 + 8 + 5}{3}, \frac{1 + 3 + 8}{3} \right) $$
$$ \text{Centroid} = \left( \frac{17}{3}, \frac{12}{3} \right) $$
$$ \text{Centroid} = \left( \frac{17}{3}, 4 \right) $$

The distance from the origin to the centroid can be found using the distance formula:

$$ d = \sqrt{(x_c - 0)^2 + (y_c - 0)^2} $$

Where $ (x_c, y_c) $ are the coordinates of the centroid.

Substitute the coordinates of the centroid:

$$ d = \sqrt{\left(\frac{17}{3}\right)^2 + 4^2} $$
$$ d = \sqrt{\frac{289}{9} + 16} $$
$$ d = \sqrt{\frac{289}{9} + \frac{144}{9}} $$
$$ d = \sqrt{\frac{433}{9}} $$

So, the distance from the origin to the centroid of the triangle is $ \sqrt{\frac{433}{9}} $ units.