A triangle has corners at #(1 ,4 )#, #(9 ,6 )#, and #(4 ,5 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(1  ,4  )#, #(9  ,6  )#, and #(4   ,5  )#.  How far is the triangle's centroid from the origin?

Noah Atkinson · Accepted Answer

The distance of triangle's centroid from the origin is #6.839# units

The centroid of a triangle whose vertices are #(x_1,y_1)#, #(x_2,y_2)# and (x_3,y_3)# is given by #((x_1+x_2+x_3)/3,(y_1+y_2+y_3)/3)# Hence centrid of given triangle is #((1+9+4)/3,(4+6+5)/3)# or #(14/3,5)# and its distance from origin is #sqrt((14/3)^2+5^2)=sqrt(196/9+25)=sqrt(421/9)=1/3xx20.518=6.839#

James Atkins · Answer

undefined The centroid of a triangle is the point of intersection of its medians. The coordinates of the centroid can be found by taking the average of the coordinates of the vertices. The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.

For the given triangle with vertices at (1, 4), (9, 6), and (4, 5), the centroid's coordinates are:

$$(\frac{1+9+4}{3}, \frac{4+6+5}{3}) = (\frac{14}{3}, \frac{15}{3}) = (\frac{14}{3}, 5)$$

The distance from the origin to the centroid is the length of the line segment connecting the origin to the centroid. This distance can be found using the distance formula:

$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

Substituting the coordinates of the centroid and the origin (0, 0) into the formula:

$$d = \sqrt{(\frac{14}{3}-0)^2 + (5-0)^2} = \sqrt{(\frac{14}{3})^2 + 5^2} = \sqrt{\frac{196}{9} + 25} = \sqrt{\frac{196+225}{9}} = \sqrt{\frac{421}{9}} = \frac{\sqrt{421}}{3}$$

So, the distance from the centroid to the origin is $\frac{\sqrt{421}}{3}$ units.

Sadie Allen · Answer

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

Average the x-coordinates and y-coordinates of the vertices:
$$ \text{Average of } x \text{-coordinates} = \frac{1 + 9 + 4}{3} = \frac{14}{3} $$
$$ \text{Average of } y \text{-coordinates} = \frac{4 + 6 + 5}{3} = \frac{15}{3} = 5 $$

So, the centroid is at $ (\frac{14}{3}, 5) $.

Now, calculate the distance from this centroid to the origin:
$$ \text{Distance} = \sqrt{(\frac{14}{3})^2 + 5^2} $$
$$ = \sqrt{\frac{196}{9} + 25} $$
$$ = \sqrt{\frac{196}{9} + \frac{225}{9}} $$
$$ = \sqrt{\frac{421}{9}} $$

Therefore, the distance from the triangle's centroid to the origin is $ \frac{\sqrt{421}}{3} $ units.