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 $\frac{2}{5}$ $2/5$ about point #(1 ,4 ), how far will its centroid move?

Question

1363 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-10T06:39:21

The distance is $= 4.62$ $=4.62$

Let the centroid of the triangle be $C$ $C$

Then,

$C = (\frac{9 - 2 + 3}{3}, \frac{- 5 - 1 - 4}{3})$ $C=((9-2+3)/3,(-5-1-4)/3)$

$= (\frac{10}{3}, - \frac{10}{3})$ $=(10/3,-10/3)$

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)(9 ,-5 ), (-2 ,-1 )(−2,−1)(-2 ,-1 ), and (3 ,-4 )(3,−4)(3 ,-4 ). If the triangle is dilated by a factor of 2/5 252/5 about point #(1 ,4 ), how far will its centroid move?