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)

Question

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Answer 1

The centroid (geometric center) of a triangle can be found using the coordinates of its vertices. Given the vertices of triangle ABC:

• \( A(4, 4) \)
• \( B(9, 1) \)
• \( C(3, -1) \)

The formula for the centroid \( G(x, y) \) of a triangle with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is:

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

Substituting the coordinates of points A, B, and C into the formula:

1. \( x \)-coordinate: \[ x = \frac{4 + 9 + 3}{3} = \frac{16}{3} \]

2. \( y \)-coordinate: \[ y = \frac{4 + 1 - 1}{3} = \frac{4}{3} \]

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

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

Therefore, the centroid of triangle ABC is:

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