Triangle ABC is drawn on a coordinate plane with vertices A(−2,−3) , B(4, 0) , and C(−2, 2) and medians indicating the midpoint of each of the line segments AB¯¯¯¯¯¯¯¯ , BC¯¯¯¯¯¯¯¯ , and CA¯¯¯¯¯¯¯¯ . Prove that the medians meet at a single point by finding the centroid. (1 point) Responses (−1, 53) open paren negative 1 comma 5 thirds close paren (0,−13) open paren 0 comma negative 1 third close paren (0, 53) open paren 0 comma 5 thirds close paren (0, 0)

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 vertices of triangle \( ABC \):

• \( A(-2, -3) \)
• \( B(4, 0) \)
• \( C(-2, 2) \)

We can substitute the coordinates of the vertices into the centroid formula:

1. Calculate the x-coordinate of the centroid: \[ x_G = \frac{-2 + 4 - 2}{3} = \frac{0}{3} = 0 \]

2. Calculate the y-coordinate of the centroid: \[ y_G = \frac{-3 + 0 + 2}{3} = \frac{-1}{3} \]

Thus, the coordinates of the centroid \( G \) are: \[ G \left( 0, -\frac{1}{3} \right) \]

Hence, we can conclude that the medians of triangle \( ABC \) meet at the centroid \( G \) which is at the point \( \left( 0, -\frac{1}{3} \right) \).

The correct answer from the given options is: \[ (0, -\frac{1}{3}) \quad \text{(0,−13)} \]

This indicates the medians do indeed meet at a single point, confirming that the centroid serves as the point of intersection for the triangle's medians.