QUESTION OF FINDING CENTROID OF TRIANGLE 

Practice Questions

Question 1 :

Th e vertices of a triangle are (1, 2), (h, −3) and (−4, k). If the centroid of the triangle is at the point (5, −1) then find the value of √(h + k)² + (h + 3k)².

Solution :

Let A (1, 2) B (h, −3) and C (−4, k) be the vertices of the triangle.

Centroid of the triangle  =  (5, -1)

G [ (x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3 ]  =  (5, -1)

(1 + h - 4)/3, (2 - 3 + k)/3  =  (5, -1)

(h - 3)/3, (-1 + k)/3  =  (5, -1)

 √(h + k)² + (h + 3k)²  =   √(18 - 2)² + (18 - 6)²

   =   √16² + 12²

   =   √256 + 144

  =  √400

  =  20

Question 2 :

Orthocentre and centroid of a triangle are A(−3, 5) and B(3, 3) respectively. If C is the circumcentre and AC is the diameter of this circle, then find the radius of the circle.

Solution  :

The distance between the centroid and the orthocenter is twice the distance between the centroid and  he circumcenter. The centroid is dividing the othocenter and circumcenter in the ratio 2 : 1.

  =  (mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)

Let C be (a, b)

(2a+1(-3))/(2+1), (2b+1(5))/(2+1)  =  (3, 3)

(2a-3)/3, (2b+5)/3  =  (3, 3)

Equating the x and y coordinates, we get

Hence the circumcenter is (6, 2)

AC  =  diameter of the circle 

A(-3, 5) C(6, 2)

  =   √(6+3)² + (2-5)²

  =   √9² + (-3)²

  =   √(81+9)

  =   √90

  =  3√10

Radius  =  3√10/2  =  3√(5/2)

Question 3 :

ABC is a triangle whose vertices are A(3, 4), B(−2, −1) and C(5, 3) . If G is the centroid and BDCG is a parallelogram then find the coordinates of the vertex D.

Solution :

So, midpoint of diagonal BD = midpoint of diagonal CG 

let D = (a, b) 

midpoint of BD = [(-2+a)/2, (-1+b)/2 ] 

midpoint of CG = [(5+2)/2,(3 + 2)/2 ] = (7/2, 5/2) 

hence, [(-2+x)/2, (-1+y)/2] = (7/2, 5/2)

(-2+a)/2 = 7/2 => a = 9

(-1+b)/2 = 5/2 => b = 6

So the required vertex is D (9,6)

Question 4 :

If (3/2, 5) (7, -9/2) and (13/2, -13/2) are mid-points of the sides of a triangle, then find the centroid of the triangle.

Solution :

Let the given midpoints be (x₁, y₁) (x₂,  y₂) and (x₃, y₃)

A (x₁ + x₃ - x_2, y₁ + y₃ - y₂)

B (x₁ + x₂ - x_3, y₁ + y₂ - y₃)

C (x₂ + x₃ - x_1, y₂ + y₃ - y₁)

Centroid  =  (1 + 2 + 12)/3 , (3 + 7 - 16)/3

  =  (15/3), (-6/3)

  =  (5, -2)

Hence the required centroid is (5, -2).

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