Triangle ABC is drawn on a coordinate plane with vertices A(4,4), B(9,1), 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)

Question

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Answer 1

To find the centroid of triangle ABC with vertices A(4,4), B(9,1), and C(3,−1), we use the formula for the centroid \((G)\) of a triangle, which is given by:

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

where \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are the coordinates of the vertices.

For triangle ABC, we have:

• \(A(4, 4)\), so \(x_1 = 4\) and \(y_1 = 4\)
• \(B(9, 1)\), so \(x_2 = 9\) and \(y_2 = 1\)
• \(C(3, -1)\), so \(x_3 = 3\) and \(y_3 = -1\)

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

\[ G_x = \frac{4 + 9 + 3}{3} = \frac{16}{3} \]

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

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

\[ G\left(\frac{16}{3}, \frac{4}{3}\right) \]

Therefore, the centroid is:

\[ \boxed{\left(\frac{16}{3}, \frac{4}{3}\right)} \]

This proves that the three medians of triangle ABC meet at the point \(\left(\frac{16}{3}, \frac{4}{3}\right)\).