A triangle has corners at $(9 ,-5 )$ , $(-2 ,-1 )$ , and $(3 ,-4 )$ . If the triangle is dilated by a factor of $2/5$ about point #(1 ,4 ), how far will its centroid move?

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Answer 1 · 2017-06-10T06:39:21

The distance is $=4.62$

Let the centroid of the triangle be $C$

Then,

$C=((9-2+3)/3,(-5-1-4)/3)$

$=(10/3,-10/3)$

Let the new centroid be $C'=(x',y')$

The point $D=(1,4)$

So,

$vec(DC')=2/5vec(DC)$

$((x'-1),(y'-4))=2/5((10/3-1),(-10/3-4))$

$x'-1=2/5(10/3-1)=14/15$

$=>$ , $x'=14/15+1=29/15$

$y'-4=2/5(-10/3-4)=-44/15$

$y'=-44/15+4=16/15$

Therefore,

The new centroid is $C'=(29/15,16/15)$

The distance

$CC'=sqrt((10/3-29/15)^2+(-10/3-16/15)^2)$

$=sqrt((21/15)^2+(66/15)^2)$

$=1/15sqrt(21^2+66^2)$

$=4.62$

A triangle has corners at (9 ,-5 )(9 ,-5 ), (-2 ,-1 )(-2 ,-1 ), and (3 ,-4 )(3 ,-4 ). If the triangle is dilated by a factor of 2/5 2/5 about point #(1 ,4 ), how far will its centroid move?