A triangle has corners at #(7 ,2 )#, #(6 ,7 )#, and #(3 ,5 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(7  ,2  )#, #(6  ,7  )#, and #(3   ,5  )#.  How far is the triangle's centroid from the origin?

Aurora Ayala · Accepted Answer

The centriod is the average of the coordinates:

#C=( {7+6+3}/3, {2+7+5}/3) = (16/3, 14/3)#

so its distance to the origin is

# \sqrt{(16/3)^2 + (14/3)^2}=sqrt{ (2/3)^2(8^2+7^2) } = 2/3 sqrt{113} #

Violet Aldrich · Answer

undefined To find the centroid of a triangle, you take the average of the coordinates of its vertices. Then, you calculate the distance from the centroid to the origin using the distance formula.

First, find the coordinates of the centroid:
$$ \text{Centroid} = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) $$

Given the vertices:
$$ (7, 2), (6, 7), (3, 5) $$

Average the x-coordinates:
$$ x_{\text{centroid}} = \frac{7 + 6 + 3}{3} = \frac{16}{3} $$

Average the y-coordinates:
$$ y_{\text{centroid}} = \frac{2 + 7 + 5}{3} = \frac{14}{3} $$

So, the centroid of the triangle is $\left( \frac{16}{3}, \frac{14}{3} \right)$.

Now, calculate the distance from the centroid to the origin:
$$ \text{Distance} = \sqrt{(\frac{16}{3})^2 + (\frac{14}{3})^2} $$
$$ = \sqrt{\frac{256}{9} + \frac{196}{9}} $$
$$ = \sqrt{\frac{452}{9}} $$
$$ = \frac{\sqrt{452}}{3} $$
$$ = \frac{2\sqrt{113}}{3} $$

Therefore, the distance from the centroid of the triangle to the origin is $\frac{2\sqrt{113}}{3}$.

Sadie Allen · Answer

undefined The centroid of a triangle is the average of its vertices' coordinates. Using the formula for the centroid, the coordinates of the centroid (G) can be found as follows:

$$G(x, y) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$$

Given the coordinates of the vertices:
$$A(7, 2), B(6, 7), C(3, 5)$$

Using the formula:
$$G(x, y) = \left(\frac{7 + 6 + 3}{3}, \frac{2 + 7 + 5}{3}\right)$$
$$G(x, y) = \left(\frac{16}{3}, \frac{14}{3}\right)$$

To find the distance between the centroid and the origin (0, 0), we use the distance formula:

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

Where $(x_0, y_0)$ is the origin (0, 0).

Substituting the values:
$$d = \sqrt{\left(\frac{16}{3} - 0\right)^2 + \left(\frac{14}{3} - 0\right)^2}$$
$$d = \sqrt{\left(\frac{16}{3}\right)^2 + \left(\frac{14}{3}\right)^2}$$
$$d = \sqrt{\frac{256}{9} + \frac{196}{9}}$$
$$d = \sqrt{\frac{452}{9}}$$
$$d \approx \frac{\sqrt{452}}{3} \approx \frac{2\sqrt{113}}{3}$$

So, the distance from the centroid to the origin is approximately $\frac{2\sqrt{113}}{3}$ units.