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

New Endpoints are $\left(- 2 , - 1\right)$ and $\left(6 , - 5\right)$
Length $l = 8.94427 \text{ }$

#### Explanation:

Let $A \left({x}_{a} , {y}_{a}\right) = \left(1 , 2\right)$
Let $B \left({x}_{b} , {y}_{b}\right) = \left(3 , 1\right)$
Let $R \left({x}_{r} , {y}_{r}\right) = \left(2 , 3\right)$

Let the new endpoints be
$C \left({x}_{c} , {y}_{c}\right)$ and $D \left({x}_{d} , {y}_{d}\right)$

Let $k = 4$

Solve the new endpoints by ratio and proportion

$\frac{R C}{R A} = k$

$\frac{{x}_{c} - {x}_{r}}{{x}_{a} - {x}_{r}} = 4$
$\frac{{x}_{c} - 2}{1 - 2} = 4$

${x}_{c} = - 2$

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

${y}_{c} = - 1$

$\frac{R D}{R B} = k$
$\frac{{x}_{d} - {x}_{r}}{{x}_{b} - {x}_{r}} = 4$
$\frac{{x}_{d} - 2}{3 - 2} = 4$
${x}_{d} = 6$

$\frac{{y}_{d} - {y}_{r}}{{y}_{b} - {y}_{r}} = 4$
$\frac{{y}_{d} - 3}{1 - 3} = 4$
${y}_{d} = - 5$

Length

$l = \sqrt{{\left({x}_{d} - {x}_{c}\right)}^{2} + {\left({y}_{d} - {y}_{c}\right)}^{2}}$
$l = 4 \sqrt{5} = 8.94427 \text{ }$

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