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

Question

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

Mason Adams · Accepted Answer

#2/3\sqrt{106}=6.864\ text{unit##

The coordinates of centroid of given triangle with vertices at #(x_1, y_1)\equiv(5, 6)#, #(x_2, y_2)\equiv(2, 7)# & #(x_3, y_3)\equiv(3, 5)# are given as #(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})# #\equiv(\frac{5+2+3}{3}, \frac{6+7+5}{3})# #\equiv(10/3, 6)# hence the distance of centroid of triangle #(10/3, 6) # from the origin #(0, 0)# is given by using distance formula #\sqrt{(10/3-0)^2+(6-0)^2}# #=2/3\sqrt{106}# #=6.864\ text{unit#

Evelyn Arboleda · Answer

undefined To find the centroid of a triangle with vertices at (5, 6), (2, 7), and (3, 5), follow these steps:

1. Find the coordinates of the centroid, which is the average of the x-coordinates and the average of the y-coordinates of the vertices.

2. Average of x-coordinates:
   $$ \text{Average of } x = \frac{5 + 2 + 3}{3} = \frac{10}{3} $$

3. Average of y-coordinates:
   $$ \text{Average of } y = \frac{6 + 7 + 5}{3} = 6 $$

4. So, the coordinates of the centroid are $ \left(\frac{10}{3}, 6\right) $.

5. To find the distance between the centroid and the origin, use the distance formula:
   $$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

6. Plug in the coordinates of the centroid $ (\frac{10}{3}, 6) $ and the origin $ (0, 0) $:
   $$ \text{Distance} = \sqrt{\left(\frac{10}{3} - 0\right)^2 + (6 - 0)^2} $$

7. Simplify:
   $$ \text{Distance} = \sqrt{\left(\frac{10}{3}\right)^2 + 6^2} $$
   $$ = \sqrt{\frac{100}{9} + 36} $$
   $$ = \sqrt{\frac{100}{9} + \frac{324}{9}} $$
   $$ = \sqrt{\frac{424}{9}} $$
   $$ = \frac{\sqrt{424}}{3} $$

8. Approximate the square root of 424:
   $$ \sqrt{424} \approx 20.59 $$

9. So, the distance between the centroid and the origin is approximately:
   $$ \frac{20.59}{3} \approx \boxed{6.86} $$