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

The centroid will move by about $d = \frac{4}{3} \sqrt{233} = 20.35245 \text{ }$units

#### Explanation:

We have a triangle with vertices or corners at the points $A \left(- 6 , 3\right)$and $B \left(3 , - 2\right)$ and $C \left(5 , 4\right)$.

Let $F \left({x}_{f} , {y}_{f}\right) = F \left(- 2 , 6\right) \text{ }$the fixed point

Compute the centroid $O \left({x}_{g} , {y}_{g}\right)$ of this triangle, we have

${x}_{g} = \frac{{x}_{a} + {x}_{b} + {x}_{c}}{3} = \frac{- 6 + 3 + 5}{3} = \frac{2}{3}$
${y}_{g} = \frac{{y}_{a} + {y}_{b} + {y}_{c}}{3} = \frac{3 + \left(- 2\right) + 4}{3} = \frac{5}{3}$

Centroid $O \left({x}_{g} , {y}_{g}\right) = O \left(\frac{2}{3} , \frac{5}{3}\right)$

Compute the centroid of the bigger triangle (scale factor =5)

Let $O ' \left({x}_{g} ' , {y}_{g} '\right) =$the centroid of the bigger triangle

the working equation:

$\frac{F O '}{F O} = 5$

solve for ${x}_{g} '$:

$\frac{{x}_{g} ' - - 2}{\frac{2}{3} - - 2} = 5$

$\left({x}_{g} ' + 2\right) = 5 \cdot \frac{8}{3}$
${x}_{g} ' = \frac{40}{3} - 2$
${x}_{g} ' = \frac{34}{3}$

solve for ${y}_{g} '$

$\frac{{y}_{g} ' - 6}{\frac{5}{3} - 6} = 5$

${y}_{g} ' = 6 + 5 \left(- \frac{13}{3}\right) = \frac{18 - 65}{3}$

${y}_{g} ' = - \frac{47}{3}$

Compute now the distance from centroid O(2/3, 5/3) to new centroid O'(34/3, -47/3).

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

$d = \sqrt{{\left(\frac{2}{3} - \frac{34}{3} '\right)}^{2} + {\left(\frac{5}{3} - - \frac{47}{3}\right)}^{2}}$

$d = \sqrt{{\left(- \frac{32}{3}\right)}^{2} + {\left(\frac{52}{3}\right)}^{2}}$

$d = \sqrt{{\left(\frac{- 4 \cdot 8}{3}\right)}^{2} + {\left(\frac{4 \cdot 13}{3}\right)}^{2}}$

$d = \frac{4}{3} \cdot \sqrt{64 + 169}$

$d = \frac{4}{3} \cdot \sqrt{233} = 20.35245$

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