Question #97211

What are the coordinates of the centroid of the triangle with vertices (X1,Y1), (X2,Y2), (X3 ,Y3)

Expert's answer

Let ABC be the triangle where A(x1, y1), B(x2, y2) and C(x3, y3):





Let G be the centroid of triangle ABC.

Let AD be the median of triangle ABC. So, D is the midpoint of BC.

Midpoint of B(x2, y2) and C(x3, y3) is

D=(x2+x32,y2+y32)D= (\frac {x_2 +x_3} {2},\frac {y_2 +y_3} {2})

We know that centroid divides median in the ratio 2:1. So, centroid G divides the median AD in the ratio 2:1.

We know that coordinate of point P(x, y) that divides the line segment joining  internally in the ratio m:n is

P(x,y)=(mx2+nx1m+n,my2+ny1m+n)P(x,y)= (\frac {mx_2 +nx_1} {m+n},\frac {my_2 +ny_1} {m+n})

Here,

x1=x1x_1=x_1 ,

y1=y1y_1=y_1 ,

x2=x2+x32x_2=\frac{x_2+x_3}{2} ,

y2=y2+y32y_2=\frac{y_2+y_3}{2} ,

m=2m=2 ,

n=1n=1

Coordinate of G are

G(x,y)=2(x2+x32)+1(x1)2+1,2(y2+y32)+1(y1)2+1=(x2+x3+x13,y2+y3+y13)G(x,y)= \frac {2 (\frac {x_2+x_3}{2}) +1(x_1)}{2+1},\frac {2 (\frac {y_2+y_3}{2}) +1(y_1)}{2+1} =(\frac {x_2+x_3+x_1}{3},\frac {y_2+y_3+y_1}{3})

Hence Coordinate of centroid are

G(x,y)=(x1+x2+x33,y1+y2+y33)G(x,y)=(\frac {x_1+x_2+x_3}{3},\frac {y_1+y_2+y_3}{3})


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