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 point by finding the centroid.

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

1 answer

To find the centroid of triangle ABC with vertices A(-3, -3), B(0, 6), and C(4, -3), we use the formula for the centroid of a triangle, which is:

\[ G\left(x_G, y_G\right) = \left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\right) \]

where \((x_A, y_A)\), \((x_B, y_B)\), and \((x_C, y_C)\) are the coordinates of the vertices of the triangle.

Plugging in the coordinates of the vertices:

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

We find the x-coordinate of the centroid \(x_G\):

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

Next, we find the y-coordinate of the centroid \(y_G\):

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

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

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

Now, let's compare this result with the options given:

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

B. \((\frac{1}{6}, 3)\)

C. \((\frac{1}{3}, 0)\)

D. \((\frac{1}{3}, 1)\)

The correct answer is:

C. \((\frac{1}{3}, 0)\)

This proves that the medians of triangle ABC meet at a single point, which is the centroid \(G\).