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 (0,−13) open paren 0 comma negative 1 third close paren (−1, 53) open paren negative 1 comma 5 thirds close paren (0, 0) open paren 0 comma 0 close paren (0, 53)

Answer 1

To find the centroid \( G \) of triangle \( ABC \), we use the formula for the centroid which is given by:

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

where \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) are the coordinates of the vertices of the triangle.

Given the vertices:

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

We will substitute these coordinates into the formula:

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

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

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

\[ G(0, -\frac{1}{3}) \]

This result shows that the medians of triangle \( ABC \) intersect at a single point, specifically at \( G(0, -\frac{1}{3}) \).

The closest option given in your choices is: \[ (0, -\frac{1}{3}) \]

Therefore, the answer is:

\[ \boxed{(0, -\frac{1}{3})} \]