A triangle has corners at #(2 ,4 )#, #(8 ,6 )#, and #(7 ,9 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(2  ,4  )#, #(8   ,6  )#, and #(7   ,9  )#.  How far is the triangle's centroid from the origin?

Claire Akin · Accepted Answer

The distance is #sqrt(650)/3#

Let point O = the centroid = #(O_x, O_y)#

#O_x = (2 + 8 + 7)/3 = 17/3#

#O_y = (4 + 6 + 9)/3 = 19/3#

The distance, d, from the origin is:

#d = sqrt((17/3 - 0)^2 + (19/3 - 0)^2) = sqrt(650)/3#

Savannah Aguilar · Answer

undefined To find the centroid of a triangle with vertices at (2, 4), (8, 6), and (7, 9), we can use the following formula:

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

Substituting the coordinates of the vertices, we get:

$$ \text{Centroid} = \left( \frac{2 + 8 + 7}{3}, \frac{4 + 6 + 9}{3} \right) $$
$$ \text{Centroid} = \left( \frac{17}{3}, \frac{19}{3} \right) $$

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

$$ d = \sqrt{(x - x_0)^2 + (y - y_0)^2} $$

Substituting the coordinates of the centroid and the origin (0, 0), we get:

$$ d = \sqrt{\left(\frac{17}{3} - 0\right)^2 + \left(\frac{19}{3} - 0\right)^2} $$
$$ d = \sqrt{\left(\frac{17}{3}\right)^2 + \left(\frac{19}{3}\right)^2} $$
$$ d = \sqrt{\frac{289}{9} + \frac{361}{9}} $$
$$ d = \sqrt{\frac{650}{9}} $$
$$ d = \frac{\sqrt{650}}{3} $$

So, the distance between the triangle's centroid and the origin is $ \frac{\sqrt{650}}{3} $ units.