Question

△ABC
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.(1 point)

Answer 1

To prove that the medians of triangle \( ABC \) meet at a single point, we need to find the centroid of the triangle. The centroid \( G \) of a triangle with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is given by the formula:

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

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

Let's plug in the coordinates into the formula:

1. Calculate the x-coordinate of the centroid:
\[
x_G = \frac{1 + 6 + 3}{3} = \frac{10}{3}
\]

2. Calculate the y-coordinate of the centroid:
\[
y_G = \frac{3 + 6 + 1}{3} = \frac{10}{3}
\]

Thus, the coordinates of the centroid \( G \) are:
\[
G \left( \frac{10}{3}, \frac{10}{3} \right)
\]

Since the centroid is a single point determined by the vertices' coordinates, this shows that the medians meet at the centroid. Therefore, we have proved that the medians of triangle \( ABC \) intersect at a single point \( G \), which is \( \left( \frac{10}{3}, \frac{10}{3} \right) \).