Question

Triangle ABC
is drawn on a coordinate plane with vertices A(−3,−3)
, B(0,6)
, and C(4,−3)
and medians indicating the midpoint of each of the lines AB
, BC
, and CA
. Prove the medians meet at a single point by finding the centroid.(1 point)
Responses

(13,1)
left parenthesis Start Fraction 1 over 3 End Fraction comma 1 right parenthesis

(0,13)
left parenthesis 0 comma Start Fraction 1 over 3 End Fraction right parenthesis

(13,0)
left parenthesis Start Fraction 1 over 3 End Fraction comma 0 right parenthesis

(16,3)

Answer 1

To find the centroid of triangle ABC with vertices A(-3, -3), B(0, 6), and C(4, -3), we can use the formula for the centroid \((G_x, G_y)\):

\[
G_x = \frac{x_1 + x_2 + x_3}{3}
\]
\[
G_y = \frac{y_1 + y_2 + y_3}{3}
\]

Substituting the coordinates of the vertices A, B, and C into the formulas:

- For \(G_x\):
\[
G_x = \frac{-3 + 0 + 4}{3} = \frac{1}{3}
\]

- For \(G_y\):
\[
G_y = \frac{-3 + 6 + (-3)}{3} = \frac{0}{3} = 0
\]

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

Given your options, the closest coordinate that represents the centroid of triangle ABC is:
\[
\left( \frac{1}{3}, 0 \right)
\]

Thus the valid response is \( \left( \frac{1}{3}, 0 \right) \) which matches one of the options you provided:

- (13,0)
- \( \left( \frac{1}{3}, 0 \right) \)

Therefore, the answer is \( \left( \frac{1}{3}, 0 \right) \).