A triangle has corners at $(8 ,3 )$ , $(4 ,-5 )$ , and $(2 ,1 )$ . If the triangle is dilated by a factor of $5$ about point #(1 ,-3 ), how far will its centroid move?

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Answer 1 · 2017-07-28T07:11:50

The centroid will move by $=18.1u$

Let the corners of the triangle be $(x_1,y_1)$ , $(x_2,y_2)$ and $(x_3,y_3)$

The coordinates of the centroid are

$C=((x_1+x_2+x_3)/2,(y_1+y_2+y_3)/3)$

Here, we have $(8,3)$ , $(4,-5)$ , and $(2,1)$

So,

The coordinates of the centroid are $C=((8+4+2)/3,(3-5+1)/3)=(14/3,-1/3)$

Let the coordinates of the centroid after dilatation be $C'=(x,y)$

The fixed point is $D=(1,-3)$

Therefore,

$vec(DC')=5vec(DC)$

$((x-1),(y+3))=5((14/3-1),(-1/3+3))=5((11/3),(8/3))=((55/3),(40/3))$

So,

$x-1=55/3$ , $=>$ , $x=55/3+1=58/3$

$y+3=40/3$ , $=>$ , $y=40/3-3=31/3$

So, the coordinates of $C'=(58/3,31/3)$

The distance between the centroids is

$C C'=sqrt((58/3-14/3)^2+(31/3+1/3)^2)$

$=sqrt((44/3)^2+(32/3)^2)$

$=sqrt(44^2+32^2)/3$

$=18.1u$

A triangle has corners at (8 ,3 )(8 ,3 ), (4 ,-5 )(4 ,-5 ), and (2 ,1 )(2 ,1 ). If the triangle is dilated by a factor of 5 5 about point #(1 ,-3 ), how far will its centroid move?