# A triangle has corners at (1 ,3 ), (9 ,4 ), and (2 ,2 ). If the triangle is dilated by a factor of 2  about point #(9 ,5 ), how far will its centroid move?

distance $d = 5.3851648 \text{ }$units

#### Explanation:

Compute for the centroid $\left({x}_{c} , {y}_{c}\right)$

${x}_{c} = \frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} = \frac{1 + 9 + 2}{3} = 4$
${y}_{c} = \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3} = \frac{3 + 4 + 2}{3} = 3$

Centroid $\left({x}_{c} , {y}_{c}\right) = \left(4 , 3\right)$

Factor of 2 about the point $\left(9 , 5\right)$

Let $\left({x}_{c} ' , {y}_{c} '\right)$ be the new centroid

Solve for ${x}_{c} '$

$\frac{{x}_{c} ' - 9}{{x}_{c} - 9} = \frac{2}{1}$

$\frac{{x}_{c} ' - 9}{4 - 9} = \frac{2}{1}$

$\frac{{x}_{c} ' - 9}{- 5} = 2$

${x}_{c} ' = - 10 + 9$

${x}_{c} ' = - 1$

Solve for ${y}_{c} '$

$\frac{{y}_{c} ' - 5}{{y}_{c} - 5} = \frac{2}{1}$

$\frac{{y}_{c} ' - 5}{3 - 5} = \frac{2}{1}$

$\frac{{y}_{c} ' - 5}{- 2} = 2$

${y}_{c} ' = - 4 + 5$

${y}_{c} ' = 1$

The new centroid $\left({x}_{c} ' , {y}_{c} '\right) = \left(- 1 , 1\right)$

Solve for the distance between the two centroids

$d = \sqrt{{\left({x}_{c} - {x}_{c} '\right)}^{2} + {\left({y}_{c} - {y}_{c} '\right)}^{2}}$

$d = \sqrt{{\left(4 - - 1\right)}^{2} + {\left(3 - 1\right)}^{2}}$

$d = \sqrt{{\left(5\right)}^{2} + {\left(2\right)}^{2}}$

$d = \sqrt{25 + 4}$

$d = \sqrt{29}$

$d = 5.3851648 \text{ }$units

KIndly see the drawings of the old and new triangles

God bless....I hope the explanation is useful.