A triangle has corners at #(2 ,4 )#, #(8 ,6 )#, and #(7 ,1 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(2  ,4  )#, #(8   ,6  )#, and #(7   ,1  )#.  How far is the triangle's centroid from the origin?

Claire Akin · Accepted Answer

The distance is#=sqrt410/3# Let the corners of a triangle be #(x_1,y_1),(x_2,y_2) and(x_3,y_3)# And the centrid #(x_c, y_c)# Then #x_c=(x_1+x_2+x_3)/3=(2+8+7)/3=17/3# #y_c=(y_1+y_2+y_3)/3=(4+6+1)/3=11/3# The distance of the centroid from the origin is #d=sqrt(x_c^2+y_c^2)=sqrt((17/3)^2+(11/3)^2)=sqrt410/3#

Jayden Jackson · Answer

undefined The distance between the centroid of the triangle and the origin is approximately 5.196 units.

Allison Allen · Answer

undefined To find the centroid of a triangle with vertices at coordinates (x1, y1), (x2, y2), and (x3, y3), you can use the formula:

$$ \text{Centroid} = \left( \frac{x1 + x2 + x3}{3}, \frac{y1 + y2 + y3}{3} \right) $$

Given the coordinates of the vertices of the triangle:
Vertex 1: (2, 4)
Vertex 2: (8, 6)
Vertex 3: (7, 1)

Using the centroid formula:

$$ \text{Centroid} = \left( \frac{2 + 8 + 7}{3}, \frac{4 + 6 + 1}{3} \right) $$

$$ \text{Centroid} = \left( \frac{17}{3}, \frac{11}{3} \right) $$

To find the distance from the centroid to the origin, you can use the distance formula, which states that the distance between two points (x1, y1) and (x2, y2) is given by:

$$ \text{Distance} = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} $$

In this case, one point is the centroid $(\frac{17}{3}, \frac{11}{3})$ and the other is the origin (0, 0):

$$ \text{Distance} = \sqrt{(\frac{17}{3} - 0)^2 + (\frac{11}{3} - 0)^2} $$

$$ \text{Distance} = \sqrt{(\frac{17}{3})^2 + (\frac{11}{3})^2} $$

$$ \text{Distance} = \sqrt{\frac{289}{9} + \frac{121}{9}} $$

$$ \text{Distance} = \sqrt{\frac{410}{9}} $$

$$ \text{Distance} = \frac{\sqrt{410}}{3} $$

So, the distance from the centroid of the triangle to the origin is $ \frac{\sqrt{410}}{3} $ units.