FIND COORDINATES OF CENTROID OF TRIANGLE

The centroid G of the triangle with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is

G [ (x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3 ]

Question 1 :

Find the centroid of the triangle whose vertices are

(i)  (2, −4), (−3, −7) and (7, 2)

Solution :

Let the points be A (2, −4) B (−3, −7) and C (7, 2)

=  G [ (x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3 ]

=  (2 - 3 + 7)/3, (-4 - 7 + 2)/3

=  (6/3, -9/3)

=  (2, -3)

(ii)  (−5,−5),  (1,−4) and (−4,−2)

Solution :

Let the points be A (−5,−5) B (1,−4) and C (-4, -2)

=  G [ (x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3 ]

=  (-5 + 1 - 4)/3, (-5 - 4 - 2)/3

=  (-9 + 1)/3, (-11/3)

=  (-8/3, -11/3)

Question 2 :

If the centroid of a triangle is at (4,−2) and two of its vertices are (3,−2) and (5,2) then find the third vertex of the triangle

Solution :

Let the given vertices be A(3, -2) and B (5, 2)

Centroid of the triangle  =  (4, -2)

Let the third vertex be C (a, b)

=  G [ (x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3 ]

(3 + 5 + a)/3, (-2 + 2 + b)/3  =  (4, -2)

(8 + a)/3, (b/3)  =  (4, -2)

By equating x and y-coordinates, we get

Hence the required vertex is (4, -6).

Question 3 :

Find the length of median through A of a triangle whose vertices are A(−1, 3), B(1, −1) and C(5, 1).

Solution :

If we draw a median through A, it will intersect the side BC exactly at middle.

Mid point of the side BC  =  D

  =  (x₁ + x₂)/2, (y₁ + y₂)/2

=  (1 + 5)/2, (-1 + 1)/2

=   (6/2, 0/2)

=  (3, 0)

Length of median AD  =   √(x₂ - x₁)² + (y₂ - y₁)²

=   √(3 + 1)² + (0 - 3)²

=   √4² + (-3)²

=   √(16 + 9)

=   √25

=  5 units.

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