A triangle has corners at $(-1 ,2 )$ , $(3 ,-5 )$ , and $(7 ,4 )$ . If the triangle is dilated by a factor of $5$ about point #(-2 ,6 ), how far will its centroid move?

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Answer 1 · 2018-04-26T11:52:08

Given vertices $(a,b),(c,d),(e,f)$ , dilation point $(p,q)$ , dilation factor $r$ , the distance the centroid moves after dilation is:

${r-1}/3 sqrt{ (a+c+e-3p)^2 + (b+d+e-3q)^2 }$

$= 4/3 sqrt( (-1+3+7-3(-2))^2 + (2+-5+4-3(6))^2 } = 4/3 sqrt{514}$

I've answered one or two of these before. Let's do this one in general:

Given a triangle with vertices $(a,b),(c,d),(e,f)$ and dilation point $(p,q)$ determine how far the centroid moves when dilated by a factor of $r$ .

The centroid is $(u,v)=(1/3(a+c+e), 1/3(b+d+e))$

The distance $s$ from the centroid to the dilation point is before dilation is

$s = sqrt{(u-p)^2 + (v-q)^2}$

After dilation it will be $rs$ so the total distance moved is:

$m= (r-1) s = (r-1)sqrt{(u-p)^2 + (v-q)^2}$

$m = (r-1) sqrt { (1/3(a+c+e)-p)^2 + (1/3(b+d+e)-q)^2 }$

$m = {r-1}/3 sqrt{ (a+c+e-3p)^2 + (b+d+e-3q)^2 }$

A triangle has corners at (-1 ,2 )(-1 ,2 ), (3 ,-5 )(3 ,-5 ), and (7 ,4 )(7 ,4 ). If the triangle is dilated by a factor of 5 5 about point #(-2 ,6 ), how far will its centroid move?