Question

Question Triangle ABC is drawn on a coordinate plane with vertices A(4,4) , B(9,1) , and C(3,−1) and with medians indicating the midpoint of each of the lines AB , BC , and CA . Prove that the medians meet at a single point by finding the centroid. Express all results in fractions.(1 point) centroid= (,) Remaining Attempts : 3 Skip to navigation

Answer 1

To find the centroid of triangle ABC, we can use the formula for the centroid (G) of a triangle with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \):

\[
G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
\]

Given the coordinates for the vertices:
- \( A(4, 4) \)
- \( B(9, 1) \)
- \( C(3, -1) \)

Plugging in the values:

1. For the x-coordinate of the centroid \( G \):

\[
x_G = \frac{4 + 9 + 3}{3} = \frac{16}{3}
\]

2. For the y-coordinate of the centroid \( G \):

\[
y_G = \frac{4 + 1 + (-1)}{3} = \frac{4}{3}
\]

Thus, the centroid \( G \) of triangle ABC is:

\[
G = \left( \frac{16}{3}, \frac{4}{3} \right)
\]

This result shows that the medians of triangle ABC indeed meet at a single point, which is the centroid \((\frac{16}{3}, \frac{4}{3})\).

So, the final answer is:

**Centroid = \(\left( \frac{16}{3}, \frac{4}{3} \right)\)**.