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

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Answer 1 · 2018-02-16T09:04:25

Distance moved by centroid is

$vec(GG’) ~~ color (green)(2.5$

Coordinates of centroid is obtained using the formula,

$G (x,y) = G ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3)$

Given : $A (2,-5), B(-1,2), C(3, -3)$

Centroid $G ((x),(y)) = (((2-1+3)/3),((-5+2-3)/3)) = color(brown)(((4/3),(-2))$

Triangle is dilated about point D(-1,3) by a factor of 2/5

$bar(A’D) = (2/5) bar(AD)$

$a’ - d = (2/5) (a - d) = 0.4 a - 0.4d$

$a’((x),(y)) = 0.4((2),(-5)) + 0.6 ((-1),(2)) = color(blue)(((0.2),(-0.8))$

$b’((x),(y)) = 0.4((-1),(2)) + 0.6 ((-1),(2)) = color(blue)(((1),(2))$

$c’((x),(y)) = 0.4((3),(-3)) + 0.6 ((-1),(2)) = color(blue)(((0.6),(0))$

New centroid $G’((x),(y)) = (((0.2+1+0.6)/3),((-0.8+2+0)/3)) = color(brown)(((0.6),(0.4))$

Using distance formula, we can calculate the distance moved by centroid

$vec(GG’) = sqrt((4/3 - 0.6)^2 + (0.4+2)^2) ~~ color (green)(2.5$

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