A triangle has corners at $(1 ,3 )$ , $(7 ,7 )$ , and $(8 ,5 )$ . If the triangle is dilated by a factor of $2$ about point #(3 ,5 ), how far will its centroid move?

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Answer 1 · 2018-01-04T10:13:45

let the centroid of the triangle = $C(x,y)$
Then,
Co-ordinate of $C(x,y)$ = $((1+7+8)/3$ , $(3+7+5)/3)$
C $(x,y)$ = $(16/3, 5)$

Let us consider centroid $C'(x',y')$ after the triangle is dilated by a factor of 2 about point $D(3,5)$

Now, we can write,

$vec(DC')=2*vec(DC)$

$(x'-3 ,y'-5)$ = $2*(16/3-3 ,5-5)$

$x'-3=2*(16/3-3)$
$rArrx'=14/3+3$

$rArrx'=23/3$

Similarly
$y'-5=2*0$

$rArry'=5$

Therefore new centroid $C'(x',y')=(23/3,5)$
Now,
The distance between the centroids is
$CC'=sqrt((5-5)^2+(23/3-16/3)^2)$
$CC'=7/3$
$CC'=2.333$

A triangle has corners at (1 ,3 )(1 ,3 ), (7 ,7 )(7 ,7 ), and (8 ,5 )(8 ,5 ). If the triangle is dilated by a factor of 2 2 about point #(3 ,5 ), how far will its centroid move?