# A triangle has corners at (1 ,3 ), (7 ,7 ), and (8 ,5 ). If the triangle is dilated by a factor of 2  about point (3 ,5 ), how far will its centroid move?

Jan 4, 2018

let the centroid of the triangle = $C \left(x , y\right)$
Then,
Co-ordinate of $C \left(x , y\right)$= ((1+7+8)/3 , (3+7+5)/3)#
C$\left(x , y\right)$= $\left(\frac{16}{3} , 5\right)$

Let us consider centroid $C ' \left(x ' , y '\right)$ after the triangle is dilated by a factor of 2 about point $D \left(3 , 5\right)$

Now, we can write,

$\vec{D C '} = 2 \cdot \vec{D C}$

$\left(x ' - 3 , y ' - 5\right)$=$2 \cdot \left(\frac{16}{3} - 3 , 5 - 5\right)$

$x ' - 3 = 2 \cdot \left(\frac{16}{3} - 3\right)$
$\Rightarrow x ' = \frac{14}{3} + 3$

$\Rightarrow x ' = \frac{23}{3}$

Similarly
$y ' - 5 = 2 \cdot 0$

$\Rightarrow y ' = 5$

Therefore new centroid $C ' \left(x ' , y '\right) = \left(\frac{23}{3} , 5\right)$
Now,
The distance between the centroids is
$\mathbb{C} ' = \sqrt{{\left(5 - 5\right)}^{2} + {\left(\frac{23}{3} - \frac{16}{3}\right)}^{2}}$
$\mathbb{C} ' = \frac{7}{3}$
$\mathbb{C} ' = 2.333$