Question

Triangle ABC is drawn on a coordinate plane with vertices A (-2, - 3)
, B (1, 0), and C (-2, 2) and medians indicating the midpoint of each of the line segments AB, BC', and C'A . Prove that the medians meet at a single point by finding the centroid. (1 point)
O (0, 0)
•（0，号）
0 (0,-)
〇（-1，）

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 the triangle:

- \( A(-2, -3) \)
- \( B(1, 0) \)
- \( C(-2, 2) \)

We can substitute these coordinates into the centroid formula.

Calculating the x-coordinate of the centroid:

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

Calculating the y-coordinate of the centroid:

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

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

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

This calculation confirms that the medians of triangle \( ABC \) do indeed meet at a single point, which is the centroid \( G \).

In summary, the centroid of triangle \( ABC \) is:

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