# A line segment has endpoints at (7 ,2 ) and (1 ,8 ). The line segment is dilated by a factor of 6  around (4 ,5 ). What are the new endpoints and length of the line segment?

Jan 28, 2018

$\left(22 , - 13\right) , \left(- 14 , 23\right) , \approx 37.4$

#### Explanation:

$\text{label the endpoints "A(7,2)" and } B \left(1 , 8\right)$

$\text{let A' and B' be the images of A and B}$

$\text{let the centre of dilatation be C}$

$\Rightarrow \vec{C A '} = \textcolor{red}{6} \vec{C A}$

$\Rightarrow \underline{a} ' - \underline{c} = 6 \left(\underline{a} - \underline{c}\right)$

$\Rightarrow \underline{a} ' - \underline{c} = 6 \underline{a} - 6 \underline{c}$

$\Rightarrow \underline{a} ' = 6 \underline{a} - 5 \underline{c}$

$\textcolor{w h i t e}{\Rightarrow \underline{a} '} = 6 \left(\begin{matrix}7 \\ 2\end{matrix}\right) - 5 \left(\begin{matrix}4 \\ 5\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow \underline{a} '} = \left(\begin{matrix}42 \\ 12\end{matrix}\right) - \left(\begin{matrix}20 \\ 25\end{matrix}\right) = \left(\begin{matrix}22 \\ - 13\end{matrix}\right)$

$\Rightarrow A ' = \left(22 , - 13\right)$

$\Rightarrow \vec{C B '} = \textcolor{red}{6} \vec{C B}$

$\Rightarrow \underline{b} ' - \underline{c} = 6 \underline{b} - 6 \underline{c}$

$\Rightarrow \underline{b} ' = 6 \underline{b} - 5 \underline{c}$

$\textcolor{w h i t e}{\Rightarrow \underline{b} '} = 6 \left(\begin{matrix}1 \\ 8\end{matrix}\right) - 5 \left(\begin{matrix}4 \\ 5\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow \underline{b} '} = \left(\begin{matrix}6 \\ 48\end{matrix}\right) - \left(\begin{matrix}20 \\ 25\end{matrix}\right) = \left(\begin{matrix}- 14 \\ 23\end{matrix}\right)$

$\Rightarrow B ' = \left(- 14 , 23\right)$

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

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

$\text{let "(x_1,y_1)=(22,-13)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 14 , 23\right)$

$d = \sqrt{{\left(- 14 - 22\right)}^{2} + {\left(23 - 13\right)}^{2}} = \sqrt{1296 + 100}$

$\Rightarrow d = \sqrt{1396} \approx 37.4 \text{ to 1 dec. place}$