A triangle has corners at $(9, 5)$ $(9 ,5 )$ , $(4, - 9)$ $(4 ,-9 )$ , and $(2, - 4)$ $(2 ,-4 )$ . If the triangle is dilated by a factor of $\frac{7}{4}$ $7/4$ about point #(1 ,4 ), how far will its centroid move?

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Answer 1 · 2017-12-21T14:35:14

The centroid will move by $= 8.6 u$ $=8.6u$

The corners of the triangle are $A = (9, 5)$ $A=(9,5)$ , $B = (4, - 9)$ $B=(4,-9)$ and $C = (2, - 4)$ $C=(2,-4)$

The centroid of triangle $A B C$ $ABC$ is

$C = (\frac{9 + 4 + 2}{3}, \frac{5 - 9 - 4}{3}) = (\frac{15}{3}, - \frac{8}{3})$ $C=((9+4+2)/3, (5-9-4)/3)=(15/3,-8/3)$

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

Let the fixed point be $D=(1,4)$

Let the coefficient of dilatation be $k=7/4$

Therefore, in vector notation

$vec(DC')=kvec(DC)$

$((x-1),(y-4))=7/4((15/3-1),(-8/3-4))$

$x-1=7/4(*4)$ , $=>$ , $x=8$

$y-4=7/4*(-20/3)$ , $=>$ , $y=-35/3+4=-23/3$

The new coordinates of the centroid are $C'=(8,-23/3)$

The centroid will move by

$=sqrt((8-1)^2+(-23/3+8/3)^2)$

$=sqrt(49+25)$

$=sqrt(74)$

$=8.6u$

A triangle has corners at (9 ,5 )(9,5)(9 ,5 ), (4 ,-9 )(4,−9)(4 ,-9 ), and (2 ,-4 )(2,−4)(2 ,-4 ). If the triangle is dilated by a factor of 7/4 747/4 about point #(1 ,4 ), how far will its centroid move?