# QUESTION OF FINDING CENTROID OF TRIANGLE

## About "Find Coordinates of Centroid of Triangle"

Find Coordinates of Centroid of Triangle :

Here we are going to see some example problems on finding 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 ]

## Questions on 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 - 3)/3  =  5h - 3  =  15h  =  15 + 3h  =  18 (-1 + k)/3  =  -1-1 + k  =  -3k  =  -3 + 1k  =  -2

(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

 (2a - 3)/3  =  32a - 3  =  92a  =  9 + 32a  =  12a  =  6 (2b+5)/3  =  32b + 5  =  92b  =  9 - 52b  =  4b  =  4/2  =  2

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)

(x+ x- x2, y+ y- y2)

(x+ x- x3, y+ y2 - y3)

C (x+ x- x1, y+ y3 - y1) Centroid  =  (1 + 2 + 12)/3 , (3 + 7 - 16)/3

=  (15/3), (-6/3)

=  (5, -2)

Hence the required centroid is (5, -2). After having gone through the stuff given above, we hope that the students would have understood, "Questions on Finding Centroid of Triangle"

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