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

##### 1 Answer

The displacement from old centroid to the new centroid is
$d = \sqrt{596}$
$d = 24.413 \text{ }$units

#### Explanation:

From the given, let ${P}_{1} \left({x}_{1} , {y}_{1}\right) = \left(3 , 1\right)$ and ${P}_{2} \left({x}_{2} , {y}_{2}\right) = \left(5 , 2\right)$ and ${P}_{3} \left({x}_{3} , {y}_{3}\right) = \left(9 , 4\right)$ and $R \left(1 , 9\right)$

Compute the centroid ${C}_{1} \left({x}_{c} , {y}_{c}\right)$ of the triangle first

${x}_{c} = \frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} = \frac{3 + 5 + 9}{3} = \frac{17}{3}$
${y}_{c} = \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3} = \frac{1 + 2 + 4}{3} = \frac{7}{3}$

The centroid ${C}_{1} \left({x}_{c} , {y}_{c}\right) = \left(\frac{17}{3} , \frac{7}{3}\right)$

Determine the new centroid ${C}_{2} \left({x}_{c} ' , {y}_{c} '\right)$

The ratio:
$\frac{{C}_{2} R}{{C}_{1} R} = \frac{4}{1}$
Find ${x}_{c} '$
$\frac{{x}_{c} ' - 1}{\frac{17}{3} - 1} = \frac{4}{1}$

${x}_{c} ' = 1 + 4 \left(\frac{17}{3} - 1\right)$

${x}_{c} ' = 1 + \frac{56}{3}$

${x}_{c} ' = \frac{59}{3}$

Find ${y}_{c} '$

$\frac{{y}_{c} ' - 9}{\frac{7}{3} - 9} = \frac{4}{1}$

${y}_{c} ' = 9 + 4 \left(\frac{7}{3} - 9\right)$

${y}_{c} ' = 9 + \frac{4 \left(7 - 27\right)}{3}$
${y}_{c} ' = 9 - \frac{80}{3}$
${y}_{c} ' = \frac{27 - 80}{3}$

${y}_{c} ' = - \frac{53}{3}$

The New Centroid is at ${C}_{2} \left(\frac{59}{3} , - \frac{53}{3}\right) = \left(19.67 , - 17.67\right)$

Let us determine the displacement $d$

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

$d = \sqrt{{\left(\frac{59}{3} - \frac{17}{3}\right)}^{2} + {\left(- \frac{53}{3} - \frac{7}{3}\right)}^{2}}$

$d = \sqrt{{\left(\frac{59}{3} - \frac{17}{3}\right)}^{2} + {\left(- \frac{53}{3} - \frac{7}{3}\right)}^{2}}$

$d = \sqrt{{14}^{2} + {\left(- 20\right)}^{2}}$
$d = \sqrt{596}$
$d = 24.413 \text{ }$units

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