Question

AABC is drawn on a coordinate plane with vertices A (1, 3), B (6, 6), and C (3, 1) and with medians indicating the midpoint of each of the lines AB, BC, and CA. Prove that the medians meet at a single point by finding the centroid. Express all results in fractions.

Answer 1

To find the centroid of triangle \(AABC\), we can use the formula for the centroid, which is given by

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

where \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) are the vertices of the triangle.

Given the vertices:
- \(A(1, 3)\)
- \(B(6, 6)\)
- \(C(3, 1)\)

We can identify the coordinates of the points:
- \(x_1 = 1\), \(y_1 = 3\)
- \(x_2 = 6\), \(y_2 = 6\)
- \(x_3 = 3\), \(y_3 = 1\)

Now, we calculate the coordinates of the centroid \(G\):

1. **Calculating \(G_x\)**:
\[
G_x = \frac{x_1 + x_2 + x_3}{3} = \frac{1 + 6 + 3}{3} = \frac{10}{3}
\]

2. **Calculating \(G_y\)**:
\[
G_y = \frac{y_1 + y_2 + y_3}{3} = \frac{3 + 6 + 1}{3} = \frac{10}{3}
\]

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

\[
G\left(\frac{10}{3}, \frac{10}{3}\right)
\]

### Conclusion

The centroid \(G\) of triangle \(AABC\) is \(\left(\frac{10}{3}, \frac{10}{3}\right)\). Since the centroid is the point where all three medians of the triangle intersect, we have thus shown that the medians of triangle \(AABC\) meet at the single point \(G\) with coordinates \(\left(\frac{10}{3}, \frac{10}{3}\right)\).

This confirms that the medians indeed meet at a single point.