A triangle has corners at #(4 ,3 )#, #(2 ,6 )#, and #(1 ,5 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(4  ,3  )#, #(2   ,6  )#, and #(1   ,5  )#.  How far is the triangle's centroid from the origin?

Mason Adams · Accepted Answer

The distance is #=5.2#

The centroid of the triangle, #A=(4,3)#, #B=(2,6)# and #C=(1,5)# is

#C_c=((x_A+x_B+x_C)/3,(y_A+y_B+y_C)/3)=((4+2+1)/3,(3+6+5)/3)#

#=(7/3,14/3)#

The distance of the centroid from the origin is

#=sqrt((7/3)^2+(14/3)^2)#

#=sqrt(49/9+196/9)#

#=sqrt(245)/3#

#=5.2#

Sophia Alfred · Answer

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

Centroid coordinates:
$x_{centroid} = \frac{(x_1 + x_2 + x_3)}{3}$
$y_{centroid} = \frac{(y_1 + y_2 + y_3)}{3}$

Given coordinates of the triangle's vertices:
$P_1(4, 3)$
$P_2(2, 6)$
$P_3(1, 5)$

Calculate centroid coordinates:
$x_{centroid} = \frac{(4 + 2 + 1)}{3} = \frac{7}{3}$
$y_{centroid} = \frac{(3 + 6 + 5)}{3} = \frac{14}{3}$

Now, calculate the distance from the centroid to the origin:
$distance = \sqrt{(x_{centroid})^2 + (y_{centroid})^2}$
$distance = \sqrt{(\frac{7}{3})^2 + (\frac{14}{3})^2}$
$distance = \sqrt{\frac{49}{9} + \frac{196}{9}}$
$distance = \sqrt{\frac{245}{9}}$
$distance \approx 5.78$

Therefore, the distance from the triangle's centroid to the origin is approximately 5.78 units.