# A line segment has endpoints at (9 ,1 ) and (1 ,2 ). The line segment is dilated by a factor of 4  around (3 ,3 ). What are the new endpoints and length of the line segment?

The new endpoints are $A ' \left({x}_{a} ' , {y}_{a} '\right) = \left(27 , - 5\right)$ and $B ' \left({x}_{b} ' , {y}_{b} '\right) = \left(- 5 , - 1\right)$

#### Explanation:

Let $A \left({x}_{a} , {y}_{a}\right) = \left(9 , 1\right)$ and $B \left({x}_{b} , {y}_{b}\right) = \left(1 , 2\right)$

Dilated by a factor of $4$

Let $A ' \left({x}_{a} ' , {y}_{a} '\right)$ and
Let $B ' \left({x}_{b} ' , {y}_{b} '\right)$ be the new points

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Let us solve the new point $A ' \left({x}_{a} ' , {y}_{a} '\right)$

Working equation to solve ${x}_{a} '$

$\frac{{x}_{a} ' - 3}{{x}_{a} - 3} = \frac{4}{1}$

$\frac{{x}_{a} ' - 3}{9 - 3} = \frac{4}{1}$

$\frac{{x}_{a} ' - 3}{6} = 4$

${x}_{a} ' = 27$

Working equation to solve ${y}_{a} '$

$\frac{{y}_{a} ' - 3}{{y}_{a} - 3} = \frac{4}{1}$

$\frac{{y}_{a} ' - 3}{1 - 3} = \frac{4}{1}$

$\frac{{y}_{a} ' - 3}{- 2} = 4$

${y}_{a} ' = - 5$

the new point $A ' \left({x}_{a} ' , {y}_{a} '\right) = \left(27 , - 5\right)$

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Let us solve the new point $B ' \left({x}_{b} ' , {y}_{b} '\right)$

Working equation to solve ${x}_{b} '$

$\frac{{x}_{b} ' - 3}{{x}_{b} - 3} = \frac{4}{1}$

$\frac{{x}_{b} ' - 3}{1 - 3} = \frac{4}{1}$

$\frac{{x}_{b} ' - 3}{-} 2 = 4$

${x}_{b} ' = - 5$

Working equation to solve ${y}_{b} '$

$\frac{{y}_{b} ' - 3}{{y}_{b} - 3} = \frac{4}{1}$

$\frac{{y}_{b} ' - 3}{2 - 3} = \frac{4}{1}$

$\frac{{y}_{b} ' - 3}{- 1} = 4$

${y}_{b} ' = - 1$

the new point $B ' \left({x}_{b} ' , {y}_{b} '\right) = \left(- 5 , - 1\right)$

Kindly see the graph of the segments

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