The coordinates of the centroid (G) of a triangle with vertices
The centroid R of a triangle is two thirds of the distance from each vertex to the midpoint of the opposite side.
To find centroid of the triangle with three vertices, we use the formula
= (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3
Find the coordinates of the centroid of the triangle with the given vertices.
Problem 1 :
A(2, 3), B(8, 1), C(5, 7)
Solution :
Let the given points as (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{2}, y_{2}) are vertices of triangle.
Centroid = (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3
= (2 + 8 + 5)/3, (3 + 1 + 7)/3
= 15/3, 11/3
= (5, 11/3)
Problem 2 :
F(1, 5), G(−2, 7), H(−6, 3)
Solution :
Let the given points as (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{2}, y_{2}) are vertices of triangle.
Centroid = (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3
= (1 - 2 - 6)/3, (5 + 7 + 3)/3
= -7/3, 15/3
= (-7/3, 5)
Problem 3 :
S(5, 5), T(11, −3), U(−1, 1)
Solution :
Let the given points as (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{2}, y_{2}) are vertices of triangle.
Centroid = (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3
= (5 + 11 - 1)/3, (5 - 3 + 1)/3
= 15/3, 3/3
= (5, 1)
Problem 4 :
X(1, 4), Y(7, 2), Z(2, 3)
Solution :
Let the given points as (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{2}, y_{2}) are vertices of triangle.
Centroid = (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3
= (1 + 7 + 2)/3, (4 + 2 + 3)/3
= 10/3, 9/3
= (10/3, 3)
The point D is the centroid of △ABC. Find CD and CE.
Problem 5 :
Find the third vertex of triangle, if its two vertices are (–4, 1) and (5, 2) and its centroid is (1, 3).
Solution :
Let (x, y) be the third vertex.
Centroid = (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3
(–4, 1) and (5, 2) and (x, y)
(-4 + 5 + x)/3, (1 + 2 + y)/3 = (1, 3)
(1 + x)/3, (3 + y)/3 = (1, 3)
Equating x and y coordinates, we get
(1 + x)/3 = 1 1 + x = 3 x = 3 - 1 x = 2 |
(3 + y)/3 = 3 3 + y = 9 y = 9 - 3 y = 6 |
So, the missing point is (2, 6).
Problem 6 :
The co-ordinates of the mid points of the sides of a triangle are (1, 1) , (2, –3) and (3, 4). Find The co-ordinates of its centroid.
Solution :
Let A(x_{1}, y_{1}) B(x_{2}, y_{2}) and C(x_{3}, y_{3}) are three vertices of the triangle.
Midpoint of the side AB = (1, 1)
Midpoint of the side BC = (2, -3)
Midpoint of the side CA = (3, 4)
Midpoint of the side AB :
(x_{1 }+ x_{2})/2, (y_{1 }+ y_{2})/2 = (1, 1)
Equating x and y-coordinates, we get
(x_{1 }+ x_{2})/2 = 1 x_{1 }+ x_{2 }= 2 ----(1) |
(y_{1 }+ y_{2})/2 = 1 y_{1 }+ y_{2 }= 2----(2) |
Midpoint of the side BC :
(x_{2 }+ x_{3})/2, (y_{2 }+ y_{3})/2 = (2, -3)
Equating x and y-coordinates, we get
(x_{2 }+ x_{3})/2 = 2 x_{2 }+ x_{3 }= 4 ----(3) |
(y_{2 }+ y_{3})/2 = -3 y_{2 }+ y_{3 }= -6 ----(4) |
Midpoint of the side CA :
(x_{3 }+ x_{1})/2, (y_{3 }+ y_{1})/2 = (3, 4)
Equating x and y-coordinates, we get
(x_{3 }+ x_{1})/2 = 3 x_{3 }+ x_{1 }= 6 ----(5) |
(y_{3 }+ y_{1})/2 = 4 y_{3 }+ y_{1 }= 8 ----(6) |
(1) + (3) + (5)
x_{1 }+ x_{2 }+ x_{2 }+ x_{3 }+ x_{3 }+ x_{1} = 2 + 4 + 6
2x_{1 }+ 2x_{2 }+ 2x_{3} = 12
x_{1 }+ x_{2 }+ x_{3} = 6 ------(7)
Applying (1) in (7), we get
2 + x_{3} = 6
x_{3} = 6 - 2
x_{3} = 4
Applying the value of x_{3} in (3), we get
x_{2 }+ 4_{ }= 4
x_{2 }= 4 - 4
x_{2 }= 0
Applying the value of x_{2} in (1), we get
x_{1 }+ 0_{ }= 2
x_{1 }= 2
(2) + (4) + (6)
y_{1 }+ y_{2 }+ y_{2 }+ y_{3 }+ y_{3 }+ y_{1}= 2 - 6 + 8
2y_{1 }+ 2y_{2 }+ 2y_{3 }= 4
y_{1 }+ y_{2 }+ y_{3 }= 2 -----(8)
Applying (2) in (8), we get
2 + y_{3 }_{ }= 2
y_{3 }_{ }= 2 - 2
y_{3 }_{ }= 0
By applying the value of y_{3} in (4), we get
y_{2 }+ 0_{ }= -6
y_{2 }= -6
By applying the value of y_{2} and y_{3} in (8), we get
y_{1} + (-6) + 0 = 2
y_{1} = 2 + 6
y_{1} = 8
So, the required points are (2, 8) (0, -6) and (4, 0)
Centroid = (2 + 0 + 4)/3, (8 - 6 + 0)/3
= (6/3), (2/3)
= (2, 2/3)
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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