Question

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)

Answer 1

To find the centroid of a triangle, we need to find the average of the coordinates of the three vertices. Let's label the midpoint of AB as D, midpoint of BC as E, and midpoint of CA as F.

Finding the Midpoint of AB (D):
Using the midpoint formula:
x-coordinate of D = (x-coordinate of A + x-coordinate of B) / 2
= (4 + 9) / 2
= 13 / 2

y-coordinate of D = (y-coordinate of A + y-coordinate of B) / 2
= (4 + 1) / 2
= 5 / 2
Therefore, D = (13/2, 5/2).

Finding the Midpoint of BC (E):
Using the midpoint formula:
x-coordinate of E = (x-coordinate of B + x-coordinate of C) / 2
= (9 + 3) / 2
= 12 / 2
= 6

y-coordinate of E = (y-coordinate of B + y-coordinate of C) / 2
= (1 - 1) / 2
= 0 / 2
= 0
Therefore, E = (6, 0).

Finding the Midpoint of CA (F):
Using the midpoint formula:
x-coordinate of F = (x-coordinate of C + x-coordinate of A) / 2
= (3 + 4) / 2
= 7 / 2

y-coordinate of F = (y-coordinate of C + y-coordinate of A) / 2
= (-1 + 4) / 2
= 3 / 2
Therefore, F = (7/2, 3/2).

Now, we need to find the average of the x-coordinates and y-coordinates of points D, E, and F to find the centroid.

x-coordinate of centroid = (13/2 + 6 + 7/2) / 3
= (13/2 + 12/2 + 7/2) / 3
= (32/2) / 3
= 16/3

y-coordinate of centroid = (5/2 + 0 + 3/2) / 3
= (5/2 + 0/2 + 3/2) / 3
= (8/2) / 3
= 4/3.

Therefore, the centroid of triangle ABC is (16/3, 4/3), and the medians of the triangle meet at this point.