To prove that the medians of triangle ABC meet at a single point (the centroid), we can find the coordinates of the midpoint of each side and then find the intersection point of these medians.
Let's find the coordinates of the midpoints first:
Midpoint of AB:
x-coordinate = (x1 + x2) / 2 = (-3 + 0) / 2 = -3/2
y-coordinate = (y1 + y2) / 2 = (-3 + 6) / 2 = 3/2
So, the midpoint of AB is (-3/2, 3/2).
Midpoint of BC:
x-coordinate = (x2 + x3) / 2 = (0 + 4) / 2 = 2
y-coordinate = (y2 + y3) / 2 = (6 + -3) / 2 = 3/2
So, the midpoint of BC is (2, 3/2).
Midpoint of CA:
x-coordinate = (x3 + x1) / 2 = (4 + -3) / 2 = 1/2
y-coordinate = (y3 + y1) / 2 = (-3 + -3) / 2 = -3
So, the midpoint of CA is (1/2, -3).
Now, let's find the equation of the line passing through the midpoints of AB and BC:
Slope of the line passing through the midpoints of AB and BC = (3/2 - 3/2) / (-3/2 - 2) = 0 / (-3/2 - 4/2) = 0 / (-7/2) = 0
Since the slope is 0, the line is horizontal.
The equation of the line passing through the midpoint of AB and BC is y = 3/2.
Next, let's find the equation of the line passing through the midpoints of BC and CA:
Slope of the line passing through the midpoints of BC and CA = (3/2 - -3) / (2 - 1/2) = (3/2 + 3) / (4/2 - 1/2) = (9/2) / (3/2) = 9/3 = 3
So, the equation of the line passing through the midpoints of BC and CA is y = 3x - 3/2.
To find the intersection point of these two lines, we can solve the system of equations y = 3/2 and y = 3x - 3/2:
3/2 = 3x - 3/2
4/2 = 3x
2 = 3x
x = 2/3
Substituting this value of x back into either equation, we find y = 3/2.
Therefore, the intersection point of the medians is (2/3, 3/2), which is the centroid of triangle ABC.
Hence, we have proved that the medians of triangle ABC meet at a single point, which is the centroid.
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 by finding the centroid.
3 answers
the answers provided are
(0,1/3)
(1/6,3)
(1/3,0)
(1/3,1)
(0,1/3)
(1/6,3)
(1/3,0)
(1/3,1)
Apologies for the error in the previous response. Let's recalculate the coordinates of the centroid.
To find the centroid, we need to find the average of the coordinates of the vertices of the triangle.
Let's find the x-coordinate of the centroid:
x-coordinate = (x1 + x2 + x3) / 3 = (-3 + 0 + 4) / 3 = 1/3
Now, let's find the y-coordinate of the centroid:
y-coordinate = (y1 + y2 + y3) / 3 = (-3 + 6 - 3) / 3 = 0
Therefore, the coordinates of the centroid are (1/3, 0).
To find the centroid, we need to find the average of the coordinates of the vertices of the triangle.
Let's find the x-coordinate of the centroid:
x-coordinate = (x1 + x2 + x3) / 3 = (-3 + 0 + 4) / 3 = 1/3
Now, let's find the y-coordinate of the centroid:
y-coordinate = (y1 + y2 + y3) / 3 = (-3 + 6 - 3) / 3 = 0
Therefore, the coordinates of the centroid are (1/3, 0).
1 answer