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

May 25, 2018

$\textcolor{b l u e}{\frac{\sqrt{265}}{3} \text{ units}}$

#### Explanation:

If all vertices of a triangle are dilated by a factor $\boldsymbol{a}$ then the centroid will also be dilated by a factor $\boldsymbol{a}$

Let $A$ be the triangle in its original position.

Let $A '$ be the dilated triangle.

Centroid of $A$

The coordinates of the centroid can be found by taking the arithmetic mean of the x coordinates and the y coordinates.

$\left(\frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} , \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right)$

$\left(\frac{1 - 6 + 1}{3} , \frac{9 - 1 - 2}{3}\right) \implies \left(- \frac{4}{3} , 2\right)$

Using vectors:

Position vector of centre of dilation:

$\vec{O D} = \left(\begin{matrix}- 5 \\ - 2\end{matrix}\right)$

Let $C$ be the centroid of $A '$.

Vector from D to C:

$\vec{D C} = \left(\begin{matrix}\frac{11}{3} \\ 4\end{matrix}\right)$

Scale factor = 2:

Position vector of new centroid $C '$

$\vec{O D} + 3 \vec{D C} = \left(\begin{matrix}- 5 \\ - 2\end{matrix}\right) + 2 \left(\begin{matrix}\frac{11}{3} \\ 4\end{matrix}\right) = \left(\begin{matrix}\frac{7}{3} \\ 6\end{matrix}\right)$

Distance the centroid has moved can be found using the distance formula:

$d = \sqrt{{\left(- \frac{4}{3} - \frac{7}{3}\right)}^{2} + {\left(2 - 6\right)}^{2}} = \frac{\sqrt{265}}{3}$ units