The centroid of a triangle lies at the intersection of the triangle's medians, because it lies one-third of the way from each
side towards the opposite vertex. Use this result to find the centroid of the triangle whose vertices appear as following.
-(0, 0), (4, 0), (2, 8)
A) $\left( 2 , \frac { 8 } { 3 } \right)$
B) $\left( \frac { 4 } { 3 } , 4 \right)$
C) $\left( 0 , \frac { 8 } { 3 } \right)$
D) $\left( 2 , \frac { 16 } { 3 } \right)$

Question

The centroid of a triangle lies at the intersection of the triangle's medians, because it lies one-third of the way from each
side towards the opposite vertex. Use this result to find the centroid of the triangle whose vertices appear as following.
-(0, 0), (4, 0), (2, 8) 
A) $\left( 2 , \frac { 8 } { 3 } \right)$
B) $\left( \frac { 4 } { 3 } , 4 \right)$
C) $\left( 0 , \frac { 8 } { 3 } \right)$
D) $\left( 2 , \frac { 16 } { 3 } \right)$

Quizplus · Accepted Answer

To find the centroid of the triangle, we need to find the average of the x-coordinates and the average of the y-coordinates of the three vertices. Using the formula for the midpoint of a line segment, we can find the midpoint of each side of the triangle:

Midpoint of AB: $\left( \frac { 0 + 4 } { 2 } , \frac { 0 + 0 } { 2 } \right) = (2,0)$
Midpoint of BC: $\left( \frac { 4 + 2 } { 2 } , \frac { 0 + 8 } { 2 } \right) = (3,4)$
Midpoint of AC: $\left( \frac { 0 + 2 } { 2 } , \frac { 0 + 8 } { 2 } \right) = (1,4)$

These midpoints are also the endpoints of two of the medians of the triangle. To find the third median, we need to find the midpoint of the third side, which is:

Midpoint of CA: $\left( \frac { 0 + 2 } { 2 } , \frac { 0 + 8 } { 2 } \right) = (1,4)$

This midpoint happens to be the same as the midpoint of side AC, so the third median passes through it. The intersection of the three medians is therefore the average of the three midpoints:

Centroid: $\left( \frac { 2+3+1 } { 3 } , \frac { 0+4+4 } { 3 } \right) = \left( 2 , \frac { 8 } { 3 } \right)$

Therefore, the answer is A.

The Centroid of a Triangle Lies at the Intersection of the Triangle's