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

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Answer 1 · 2018-04-25T01:14:55

$\frac{10 \sqrt{5}}{3}$ ${10 \sqrt{5}}/3$

The centroid is the average of the coordinates:

$C = (\frac{1 + 4 + 2}{3}, \frac{7 + 3 + 6}{3}) = (\frac{7}{3}, \frac{16}{3})$ $C = ( (1+4+2)/3, (7+3+6)/3) = (7/3, 16/3)$

That's a distance from the dilation point $D (4, 2)$ $D(4,2)$ of

$d = \sqrt{{(4 - \frac{7}{3})}^{2} + {(2 - \frac{16}{3})}^{2}} = \frac{5 \sqrt{5}}{3}$ $d = sqrt{(4-7/3)^2 + (2-16/3)^2 } = { 5 sqrt{5} }/3$

$C$ $C$ is currently $1 d$ $1 d$ away from $D$ $D$ , after dilation it will be $3 d$ $3d$ away from $D$ $D$ , so it will have moved by

$2 d = \frac{10 \sqrt{5}}{3}$ $2d = {10 \sqrt{5}}/3$

A triangle as corners at (1 ,7 )(1,7)(1 ,7 ), (4 ,3 )(4,3)(4 ,3 ), and (2 ,6 )(2,6)(2 ,6 ). If the triangle is dilated by a factor of 3 33 about #(4 ,2 ), how far will its centroid move?