Question

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

Answer 1

To find the centroid of triangle ABC with vertices \( A(-2, -3) \), \( B(4, 0) \), and \( C(-2, 2) \), we use the formula for the centroid, which is given by:

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

Here, \( (x_1, y_1) = A(-2, -3) \), \( (x_2, y_2) = B(4, 0) \), and \( (x_3, y_3) = C(-2, 2) \).

First, we calculate the x-coordinates of the centroid:

\[
x_G = \frac{-2 + 4 - 2}{3} = \frac{0}{3} = 0
\]

Next, we calculate the y-coordinates of the centroid:

\[
y_G = \frac{-3 + 0 + 2}{3} = \frac{-1}{3}
\]

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

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

The centroid \( G(0, -\frac{1}{3}) \) is indeed a single point where the three medians of the triangle intersect.

Comparing with the given options:
- (−1, \( \frac{5}{3} \))
- (0, \( \frac{5}{3} \))
- (0, −13)
- (0, −\( \frac{1}{3} \))

The correct answer is:

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