# A triangle has corners at (6 ,3 ), (4 ,-1 ), and (2 ,-9 ). If the triangle is dilated by a factor of 5  about point #(2 ,-4 ), how far will its centroid move?

It will move by distance$= \frac{4}{3} \sqrt{61} = 10.4137 \text{ }$units

#### Explanation:

We need to compute the old centroid and the new centroid first.

We compute the old centroid $\left({x}_{c} , {y}_{x}\right)$ first

${x}_{c} = \frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} = \frac{6 + 2 + 4}{3} = 4$

${y}_{c} = \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3} = \frac{3 + \left(- 9\right) + \left(- 1\right)}{3} = - \frac{7}{3}$

Old centroid $\left({x}_{c} , {y}_{x}\right) = \left(4 , - \frac{7}{3}\right)$

Now we compute the new centroid $\left({x}_{c} ' , {y}_{c} '\right)$

The given problem is about dilated by a factor of 5, therefore the ratio of the distance of the new centroid from the reference point $\left(2 , - 4\right)$ to the distance of the old centroid from the reference point is $5 : 1$

For ${x}_{c} '$

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

$\frac{{x}_{c} ' - 2}{4 - 2} = \frac{5}{1}$

${x}_{c} ' = 12$

For ${y}_{c} '$

$\frac{{y}_{c} ' - - 4}{{y}_{c} - - 4} = \frac{5}{1}$

$\frac{{y}_{c} ' - - 4}{- \frac{7}{3} - - 4} = \frac{5}{1}$

${y}_{c} ' = \frac{13}{3}$

New centroid $\left({x}_{c} ' , {y}_{c} '\right) = \left(12 , \frac{13}{3}\right)$

Distance $d$ from old centroid to the new centroid

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

$d = \sqrt{{\left(4 - 12\right)}^{2} + {\left(- \frac{7}{3} - \frac{13}{3}\right)}^{2}}$

$d = \sqrt{\frac{976}{9}} = \frac{4}{3} \sqrt{61} = 10.4137$

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