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.

A. (1/3,0)
B. (0,1/3)
C. (1/3,1)
D. (1/6,3)

1 answer

To find the centroid of triangle ABC, we can use the formula for the centroid \(G\) of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

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

For triangle ABC, the vertices are:

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

Now we can plug in these coordinates into the centroid formula.

First, calculate the x-coordinate of the centroid:

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

Next, calculate the y-coordinate of the centroid:

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

Thus, the centroid \(G\) of triangle ABC is:

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

This means that the centroid and, therefore, the point where the medians of triangle ABC meet is \(\left( \frac{1}{3}, 0 \right)\).

The correct answer is:

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