# A line segment has endpoints at (5 ,6 ) and (6 , 1). The line segment is dilated by a factor of 2  around (4 , 2). What are the new endpoints and length of the line segment?

Jan 14, 2018

$\left(6 , 10\right) , \left(8 , 0\right)$

#### Explanation:

$\text{let "A(5,6) ,B(6,1)" and } D \left(4 , 2\right)$

$\text{ then "A'" and "B'" are the images of A and B under}$
$\text{the dilatation}$

$\Rightarrow \vec{D A '} = \textcolor{red}{2} \vec{D A}$

$\Rightarrow \underline{a} ' - \underline{d} = 2 \left(\underline{a} - \underline{d}\right)$

$\Rightarrow \underline{a} ' - \underline{d} = 2 \underline{a} - 2 \underline{d}$

$\Rightarrow \underline{a} ' = 2 \underline{a} - \underline{d}$

$\textcolor{w h i t e}{\times \times} = 2 \left(\begin{matrix}5 \\ 6\end{matrix}\right) - \left(\begin{matrix}4 \\ 2\end{matrix}\right)$

$\textcolor{w h i t e}{\times \times} = \left(\begin{matrix}10 \\ 12\end{matrix}\right) - \left(\begin{matrix}4 \\ 2\end{matrix}\right) = \left(\begin{matrix}6 \\ 10\end{matrix}\right)$

$\Rightarrow A ' \left(6 , 10\right)$

$\text{similarly}$

$\vec{D B '} = \textcolor{red}{2} \vec{D B}$

$\Rightarrow \underline{b} ' - \underline{d} = 2 \left(\underline{b} - \underline{d}\right)$

$\Rightarrow \underline{b} ' = 2 \underline{b} - \underline{d}$

$\textcolor{w h i t e}{\times \times} = 2 \left(\begin{matrix}6 \\ 1\end{matrix}\right) - \left(\begin{matrix}4 \\ 2\end{matrix}\right) = \left(\begin{matrix}8 \\ 0\end{matrix}\right)$

$\Rightarrow B ' \left(8 , 0\right)$

$\text{to calculate length of segment use the "color(blue)"distance formula}$

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

$\text{let "(x_1,y_1)=(6,10)" and } \left({x}_{2} , {y}_{2}\right) = \left(8 , 0\right)$

$d = \sqrt{{\left(8 - 6\right)}^{2} + {\left(0 - 10\right)}^{2}} = \sqrt{104} \approx 10.2 \text{ 1 dec. place}$