Question

The three medians of ABC, A (2, 2), B (8, 2), C(4, 7). what is the point of the centroid? express all results in fractions.

Answer 1

To find the centroid (the point of concurrency of the medians) of a triangle with vertices \( A(2, 2) \), \( B(8, 2) \), and \( C(4, 7) \), we can use the centroid formula:

\[
G\left(x_G, y_G\right) = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right)
\]

Now we substitute the coordinates of points \( A \), \( B \), and \( C \):

- Coordinates of \( A \) are \( (2, 2) \)
- Coordinates of \( B \) are \( (8, 2) \)
- Coordinates of \( C \) are \( (4, 7) \)

Now calculate \( x_G \) and \( y_G \):

1. **Calculate \( x_G \)**:

\[
x_G = \frac{x_A + x_B + x_C}{3} = \frac{2 + 8 + 4}{3} = \frac{14}{3}
\]

2. **Calculate \( y_G \)**:

\[
y_G = \frac{y_A + y_B + y_C}{3} = \frac{2 + 2 + 7}{3} = \frac{11}{3}
\]

So, the coordinates of the centroid \( G \) are:

\[
G\left(\frac{14}{3}, \frac{11}{3}\right)
\]

Thus, the point of the centroid is:

\[
\boxed{\left( \frac{14}{3}, \frac{11}{3} \right)}
\]