A triangle has corners at #(-2 ,1 )#, #(6 ,-3 )#, and #(-1 ,4 )#. If the triangle is dilated by a factor of #5 # about point #(4 ,-6 ), how far will its centroid move?

Question

A triangle has corners at #(-2  ,1    )#, #(6  ,-3   )#, and #(-1  ,4    )#.  If the triangle is dilated by a factor of #5 # about point #(4  ,-6    ), how far will its centroid move?

Maverick Smith · Accepted Answer

The distance is #=29#

Let #ABC# be the triangle

#A=(-2,1)#

#B=(6,-3)#

#C=(-1,4)#

The centroid of triangle #ABC# is

#C_c=((-2+6-1)/3,(1+(-3)+4)/3)=(1,2/3)#

Let #A'B'C'# be the triangle after the dilatation

The center of dilatation is #D=(4,-6)#

#vec(DA')=5vec(DA)=5*<-6,7> = <-30,35>#

#A'=(-30+4,35-6)=(-26,29)#

#vec(DB')=5vec(DB)=5*<2,3> = <10,15>#

#B'=(10+4,15-6)=(14,9)#

#vec(DC')=5vec(DC)=5*<-5,10> = <-25,50>#

#C'=(-25+4,50-6)=(-19,44)#

The centroid #C_c'# of triangle #A'B'C'# is

#C_c'=((-26+14-19)/3,(29+9+44)/3)=(-31/3,82/3)#

The distance between the 2 centroids is

#C_cC_c'=sqrt((-31/3-1)^2+(82/3-2/3)^2)#

#=1/3sqrt(34^2+80^2)=86.93/3=29#

Violet Aldrich · Answer

undefined To find the centroid of the original triangle, we calculate the average of the x-coordinates and the average of the y-coordinates of its vertices.

Centroid of the original triangle:
x-coordinate = (-2 + 6 - 1) / 3 = 3/3 = 1
y-coordinate = (1 - 3 + 4) / 3 = 2/3

The centroid of the original triangle is at the point (1, 2/3).

After dilation by a factor of 5 about the point (4, -6), the distance between each vertex and the center of dilation increases by a factor of 5.

The distance between the original centroid and the center of dilation is:
sqrt((4 - 1)^2 + (-6 - 2/3)^2) = sqrt(9 + (192/9)) = sqrt(225/9 + 192/9) = sqrt(417/9)

After dilation, the new centroid will be 5 times this distance away from the center of dilation. Therefore, the distance the new centroid will move is:
5 * sqrt(417/9)