A triangle has corners at #(7 ,1 )#, #(8 ,2 )#, and #(5 ,9 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(7  ,1  )#, #(8  ,2  )#, and #(5   ,9  )#.  How far is the triangle's centroid from the origin?

Caroline Aucoin · Accepted Answer

The distance of centroid from the origin is #7.77(2dp)# unit

The co-ordinate of centroid of the triangle is# x=(x_1+x_2+x_3)/3=(7+8+5)/3=20/3, y=(y_1+y_2+y_3)/3=(1+2+9)/3=4 #i.e #(20/3,4)#; The distance of centroid#(20/3,4)# from the origin#(0,0)# is #sqrt((20/3-0)^2+(4-0)^2)=sqrt(400/9+16)=sqrt(544/9)=7.77(2dp)#unit[Ans]

Noah Atkinson · Answer

undefined To find the centroid of a triangle, you take the average of the coordinates of its vertices. The coordinates of the centroid $ (x_c, y_c) $ are found by averaging the x-coordinates and y-coordinates of the vertices separately.

Given the vertices: A(7, 1), B(8, 2), and C(5, 9).

Average of x-coordinates = $ \frac{7 + 8 + 5}{3} = \frac{20}{3} $

Average of y-coordinates = $ \frac{1 + 2 + 9}{3} = \frac{12}{3} = 4 $

So, the centroid of the triangle is $ (\frac{20}{3}, 4) $.

Now, to find the distance between the centroid and the origin, we use the distance formula:

$$ \text{Distance} = \sqrt{(x_c - 0)^2 + (y_c - 0)^2} $$

$$ = \sqrt{\left(\frac{20}{3}\right)^2 + 4^2} $$

$$ = \sqrt{\frac{400}{9} + 16} $$

$$ = \sqrt{\frac{400 + 144}{9}} $$

$$ = \sqrt{\frac{544}{9}} $$

$$ = \frac{\sqrt{544}}{3} $$

$$ = \frac{8\sqrt{17}}{3} $$

So, the distance between the centroid of the triangle and the origin is $ \frac{8\sqrt{17}}{3} $ units.