In triangle ABC, A (o,o) , <BAC = 30 degree , AC =2 and midpoint of segment AB is (4,0) and if B is on the x axis, find the centroid of triangle ABC
The line drawn from each vertex of a
triangle to the midpoint of the opposite side will intersect at the center of the triangle. The point of
intersection is called the Centroid.
A(0,0),B(8,0),C(Xc,Yc).
Xc = 2cos30 = 1.732.
Yc = 2sin30 = 1.
A(0,0),B(8,0),C(1.73,1).
Xo = (Xa + Xb + Xc) / 3,
Xo = (0 + 8 + 1.73) / 3 = 3.24.
Yo = (Ya + yB + Yc) / 3,
Yo = (0 + 0 + 1) / 3 = 1/3 = 0.3333.
Co-ordinates of the Centroid:
C(Xo,Yo),
C(3.24,0.333).
To find the centroid of triangle ABC, we need to find the coordinates of point B and point C first.
Given that point A is at (0, 0) and the midpoint of segment AB is at (4, 0), we can find the coordinates of point B. Since point B is on the x-axis, its y-coordinate will be 0.
Let's assume the x-coordinate of point B as 'b'. Since the midpoint of segment AB is (4, 0), we can use the midpoint formula to find the x-coordinate of point B:
Midpoint formula: ((x1 + x2) / 2, (y1 + y2) / 2)
(4, 0) = ((0 + b) / 2, (0 + 0) / 2)
(4, 0) = (b / 2, 0)
From this equation, we can see that b / 2 = 4. By multiplying both sides of the equation by 2, we find b = 8.
So, point B is located at (8, 0).
Now, let's find the coordinates of point C.
Since AC = 2 and triangle ABC is isosceles, we can conclude that BC will also have a length of 2 units.
Let's assume the x-coordinate of point C as 'c'. Since point C lies on the line segment AC, the y-coordinate of point C will be the same as that of point A, which is 0.
So, point C is located at (c, 0), and BC = 2 means the distance between point B and point C is 2 units. We can use the distance formula to find the value of 'c':
Distance formula: √((x2 - x1)^2 + (y2 - y1)^2) = BC
√((c - 8)^2 + (0 - 0)^2) = 2
√((c - 8)^2) = 2
(c - 8)^2 = 2^2
(c - 8)^2 = 4
Taking the square root of both sides and solving for 'c', we get:
c - 8 = ±2
c = 8 ± 2
So, point C can be located at (6, 0) or (10, 0).
Now that we have the coordinates of all three vertices of triangle ABC:
A = (0, 0)
B = (8, 0)
C = (6, 0) or (10, 0)
To find the centroid of a triangle, we can find the average of the x-coordinates and the average of the y-coordinates of the vertices.
Centroid formula: ( (x1+x2+x3) / 3 , (y1+y2+y3) / 3 )
Using the coordinates of the vertices, we can calculate the centroid:
For point C at (6, 0):
Centroid = ( (0 + 8 + 6) / 3 , (0 + 0 + 0) / 3 )
= (14 / 3, 0)
For point C at (10, 0):
Centroid = ( (0 + 8 + 10) / 3 , (0 + 0 + 0) / 3 )
= (18 / 3, 0)
= (6, 0)
Therefore, the centroid of triangle ABC can be either (14/3, 0) or (6, 0) depending on the location of point C.