# A line segment has endpoints at (5 ,6 ) and (2 , 1). The line segment is dilated by a factor of 3  around (6 , 2). What are the new endpoints and length of the line segment?

Oct 19, 2016

$\textcolor{b l u e}{\text{Point A'} \to \left(x , y\right) = \left(- 6 , - 1\right)}$
$\textcolor{b l u e}{\text{Point B'} \to \left(x , y\right) = \left(3 , 14\right)}$
$\textcolor{b l u e}{L = \sqrt{{9}^{2} + {15}^{2}} = \sqrt{306}}$

#### Explanation:

Dilation is a fancy word for scale. As soon as you talk about scale your are dealing with ratios. Dilation that is positive increases the magnitude. Assuming positive dilation as not indicated otherwise.

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color(blue)("Determine point B' ")" "larr B prime

$\frac{O B}{O B '} = \frac{1}{3}$

$\frac{{x}_{O} - {x}_{B}}{{x}_{O} - {x}_{B} '} = \frac{1}{3}$

$\implies 3 \left({x}_{O} - {x}_{B}\right) = {x}_{O} - {x}_{B} '$

$\implies 3 \left(6 - 5\right) = 6 - {x}_{B} '$

${x}_{B} ' = 3$
.......................................................................................

$\frac{{y}_{O} - {y}_{B}}{{y}_{O} - {y}_{B} '} = \frac{1}{3}$

$\implies 3 \left({y}_{O} - {y}_{B}\right) = {y}_{O} - {y}_{B} '$

$\implies 3 \left(2 - 6\right) = 2 - {y}_{B} '$

${y}_{B} ' = 14$

$\textcolor{b l u e}{\text{Point B'} \to \left(x , y\right) = \left(3 , 14\right)}$
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color(blue)("Determine point A' ")" "larr A prime

$\frac{O A}{O A '} = \frac{1}{3}$

$\frac{{x}_{O} - {x}_{A}}{{x}_{O} - {x}_{A} '} = \frac{1}{3}$

$\implies 3 \left({x}_{O} - {x}_{A}\right) = {x}_{O} - {x}_{A} '$

$\implies 3 \left(6 - 2\right) = 6 - {x}_{A} '$

${x}_{A} ' = - 6$
............................................................................

$\frac{{y}_{O} - {y}_{A}}{{y}_{O} - {y}_{A} '} = \frac{1}{3}$

$\implies 3 \left({y}_{O} - {y}_{A}\right) = {y}_{O} - {y}_{A} '$

$\implies 3 \left(2 - 1\right) = 2 - {y}_{A} '$

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

$\textcolor{b l u e}{\text{Point A'} \to \left(x , y\right) = \left(- 6 , - 1\right)}$
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$\textcolor{b l u e}{\text{Determine length of A'B'}}$

$L = \sqrt{{\left({x}_{B} ' - {x}_{A} '\right)}^{2} + {\left({y}_{B} ' - {y}_{B} '\right)}^{2}}$

$\textcolor{b l u e}{L = \sqrt{{9}^{2} + {15}^{2}} = \sqrt{306}}$
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$\textcolor{b l u e}{\text{Check}}$

$\frac{A B}{A ' B '} = \frac{1}{3}$

Length $A B = \sqrt{{\left(5 - 2\right)}^{2} + {\left(6 - 1\right)}^{2}} = \sqrt{34}$

$L H S \to R H S$

$\frac{1}{3} \to \frac{\sqrt{34}}{\sqrt{306}} = \frac{1}{3}$

$L H S = R H S \textcolor{red}{\leftarrow \text{ True}}$