# A line segment has endpoints at (5 ,9 ) and (6 ,7 ). The line segment is dilated by a factor of 2  around (5 ,3 ). What are the new endpoints and length of the line segment?

Jul 27, 2018

$\left(5 , 15\right) , \left(7 , 11\right) , 2 \sqrt{5}$

#### Explanation:

$\text{let "A=(5,9),B=(6,7)" and "C=(5,3)" then}$

$\vec{C A '} = 2 \vec{C A} \text{ where A' is the image of A}$

$\underline{a} ' - \underline{c} = 2 \underline{a} - 2 \underline{c}$

$\underline{a} ' = 2 \underline{a} - \underline{c}$

$\textcolor{w h i t e}{\underline{a} '} = 2 \left(\begin{matrix}5 \\ 9\end{matrix}\right) - \left(\begin{matrix}5 \\ 3\end{matrix}\right)$

$\textcolor{w h i t e}{\underline{a} '} = \left(\begin{matrix}10 \\ 18\end{matrix}\right) - \left(\begin{matrix}5 \\ 3\end{matrix}\right) = \left(\begin{matrix}5 \\ 15\end{matrix}\right)$

$\Rightarrow A ' = \left(5 , 15\right)$

$\text{Similarly }$

$\underline{b} ' = 2 \underline{b} - \underline{c}$

$\textcolor{w h i t e}{\underline{b} '} = 2 \left(\begin{matrix}6 \\ 7\end{matrix}\right) - \left(\begin{matrix}5 \\ 3\end{matrix}\right)$

$\textcolor{w h i t e}{\underline{b} '} = \left(\begin{matrix}12 \\ 14\end{matrix}\right) - \left(\begin{matrix}5 \\ 3\end{matrix}\right) = \left(\begin{matrix}7 \\ 11\end{matrix}\right)$

$\Rightarrow B ' = \left(7 , 11\right)$

$\text{calculate the length using the "color(blue)"distance formula}$

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

$\text{let "(x_1,y_1)=(7,11)" and } \left({x}_{2} , {y}_{2}\right) = \left(5 , 15\right)$

d=sqrt((5-7)^2+(15-11)^2

$\textcolor{w h i t e}{d} = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5}$