A triangle has corners at #(5 ,1 )#, #(7 ,2 )#, and #(6 ,7 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(5  ,1  )#, #(7  ,2  )#, and #(6   ,7  )#.  How far is the triangle's centroid from the origin?

Maverick Smith · Accepted Answer

The distance between the centroid and the origin is #d=2/3 sqrt106#

The centroid of a triangle can be found by averaging the #x# coordinates and #y# coordinates of the vertices.

#x=(5+7+6)/3= 18/3=6#

#y=(1+2+7)/3=10/3#

The coordinates of the centroid are #(6, 10/3)#

To find the distance from the origin, use the distance formula.

#d=sqrt((x_1 - x_2)^2 + (y_1 -y_2)^2)#

#d=sqrt((6-0)^2 +(10/3-0)^2)#

#d=sqrt(36+100/9)#

#d=sqrt(424/9)#

#d=sqrt((4 * 106)/9)#

#d=2/3 sqrt106#

Autumn Armstrong · Answer

undefined To find the centroid of a triangle, you take the average of the x-coordinates and the average of the y-coordinates of its vertices. Then, to find the distance between the centroid and the origin, you use the distance formula, considering the centroid's coordinates as one point and the origin as the other point.

Centroid coordinates:
$x_{\text{centroid}} = \frac{5 + 7 + 6}{3} = 6$  
$y_{\text{centroid}} = \frac{1 + 2 + 7}{3} = \frac{10}{3}$

Distance from the centroid to the origin:
$d = \sqrt{(x_{\text{centroid}} - 0)^2 + (y_{\text{centroid}} - 0)^2}$  
$d = \sqrt{(6 - 0)^2 + (\frac{10}{3} - 0)^2}$  
$d = \sqrt{6^2 + (\frac{10}{3})^2}$  
$d = \sqrt{36 + \frac{100}{9}}$  
$d = \sqrt{\frac{324}{9} + \frac{100}{9}}$  
$d = \sqrt{\frac{424}{9}}$  
$d = \frac{\sqrt{424}}{3}$  
$d \approx \frac{20.59}{3}$  
$d \approx 6.86$

So, the distance between the centroid of the triangle and the origin is approximately 6.86 units.