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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

(0, 0)
open paren 0 comma 0 close paren

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

(0, 53)
open paren 0 comma 5 thirds close paren

1 answer

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

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

Substituting in the coordinates of the vertices:

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

Now, we calculate:

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

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

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

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

Therefore, the centroid of the triangle \(ABC\) is at the point \((0, -\frac{1}{3})\).

Among the provided responses, the correct one is:

(0, −1/3)