Triangle ABC

Question

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 which answer is correct and why

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

(16,3)
left parenthesis Start Fraction 1 over 6 End Fraction comma 3 right parenthesis

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

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

To find the centroid (the point where the medians of the triangle meet), we can use the formula for the centroid given by the coordinates of the triangle's vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \):

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

Substituting the coordinates of points A, B, and C:

• \( A(-3, -3) \)
• \( B(0, 6) \)
• \( C(4, -3) \)

Calculating the x-coordinate of the centroid:

\[ x_G = \frac{-3 + 0 + 4}{3} = \frac{1}{3} \]

Next, calculating the y-coordinate of the centroid:

\[ y_G = \frac{-3 + 6 - 3}{3} = \frac{0}{3} = 0 \]

Thus, the centroid \( G \) is at:

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

Now we compare this with the provided options:

1. (0, 13)
2. \( \left(0, \frac{1}{3}\right) \)
3. \( \left(\frac{1}{3}, 0\right) \)
4. \( \left(\frac{1}{3}, 1\right) \)

The correct answer is:

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

Reason why this is correct:

This result shows that the medians of triangle ABC indeed meet at the centroid, as the calculations align with the theoretical definition of a centroid for a triangle. The coordinates match exactly with option 3.