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

Answer 1

Maverick Smith

The centroid will move by #=sqrt68/3=2.75#

The centroid of the triangle, #A=(5,2)#, #B=(-3,-3)# and #C=(1,-2)# is

#C_c=((5-3+1)/3,(2-3-2)/3)#

#=(1,-1)#

Let the point #D=(5,-2)#

Let the centroid be #C_c'# after the dilatation

#vec(DC_c')=1/3vec(DC_c)#

#=1/3<1-5,-1+2>#

#=<-4/3,1/3>#

#C_c'=(-4/3,1/3)+(5,-2)#

#=(11/3,-5/3)#

The distance between the centroids is

#d=sqrt((11/3-1)^2+(-5/3+1)^2)#

#=sqrt(64/9+4/9)#

#=sqrt68/3#

#=2.75#

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Answer 2

Allison Allen

The centroid of a triangle is the point of intersection of its medians. When a triangle is dilated by a factor of $k$ about a point, the centroid moves in the same direction as the original centroid, but the distance is scaled by $k$ .

To find the centroid of the original triangle, we calculate the average of the coordinates of its vertices.

Centroid of the original triangle:
$x_{\text{centroid}} = \frac{5 - 3 + 1}{3} = \frac{3}{3} = 1$
$y_{\text{centroid}} = \frac{2 - 3 - 2}{3} = \frac{-3}{3} = -1$

The centroid of the original triangle is at (1, -1).

When dilated by a factor of $\frac{1}{3}$ about point (5, -2), the centroid will move $\frac{1}{3}$ of the distance from the point of dilation to the original centroid.

Distance from (1, -1) to (5, -2):
$\sqrt{(5 - 1)^2 + (-2 - (-1))^2} = \sqrt{16 + 1} = \sqrt{17}$

So, the distance the centroid moves is:
$\frac{1}{3} \times \sqrt{17}$

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