Question

When the three medians of the triangle are drawn, they meet at a single point what is the point of centroid? leave all results in fractions
a,(2,3)
b,(-3,-1)
c,(2,-1)

Answer 1

To find the centroid (the point where the three medians of a triangle meet) of a triangle with vertices at points \( A(2, 3) \), \( B(-3, -1) \), and \( C(2, -1) \), we can use the formula for calculating the centroid \( G \) of a triangle given by the coordinates of its vertices:

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

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

Substituting in the given coordinates:

- \( A(2, 3) \) so \( x_1 = 2 \), \( y_1 = 3 \)
- \( B(-3, -1) \) so \( x_2 = -3 \), \( y_2 = -1 \)
- \( C(2, -1) \) so \( x_3 = 2 \), \( y_3 = -1 \)

Now calculate the \( x \) and \( y \) coordinates of the centroid:

\[
G_x = \frac{2 + (-3) + 2}{3} = \frac{2 - 3 + 2}{3} = \frac{1}{3}
\]

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

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

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

So, the centroid of triangle \( ABC \) is \( \left( \frac{1}{3}, \frac{1}{3} \right) \).