# A line segment has endpoints at (1 ,2 ) and (3 ,4 ). The line segment is dilated by a factor of 6  around (2 ,5 ). What are the new endpoints and length of the line segment?

Jan 27, 2018

$\left(- 4 , - 13\right) , \left(8 , - 1\right) , \approx 16.97$

#### Explanation:

$\text{label the endpoints "A(1,2)" and } B \left(3 , 4\right)$

$\text{label the centre of dilatation } C \left(2 , 5\right)$

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

$\text{then}$

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

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

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

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

$\textcolor{w h i t e}{\Rightarrow \underline{a} '} = 6 \left(\begin{matrix}1 \\ 2\end{matrix}\right) - 5 \left(\begin{matrix}2 \\ 5\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow \underline{a} '} = \left(\begin{matrix}6 \\ 12\end{matrix}\right) - \left(\begin{matrix}10 \\ 25\end{matrix}\right) = \left(\begin{matrix}- 4 \\ - 13\end{matrix}\right)$

$\Rightarrow A ' = \left(- 4 , - 13\right)$

$\text{and }$

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

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

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

$\textcolor{w h i t e}{\Rightarrow \underline{b} '} = 6 \left(\begin{matrix}3 \\ 4\end{matrix}\right) - 5 \left(\begin{matrix}2 \\ 5\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow \underline{b}} = \left(\begin{matrix}18 \\ 24\end{matrix}\right) - \left(\begin{matrix}10 \\ 25\end{matrix}\right) = \left(\begin{matrix}8 \\ - 1\end{matrix}\right)$

$\Rightarrow B ' = \left(8 , - 1\right)$

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

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

$\text{let "(x_1,y_1)=(-4,-13)" and } \left({x}_{2} , {y}_{2}\right) = \left(8 , - 1\right)$

$d = \sqrt{{\left(8 + 4\right)}^{2} + {\left(- 1 + 13\right)}^{2}}$

$\textcolor{w h i t e}{d} = \sqrt{144 + 144} = \sqrt{288} = 12 \sqrt{2} \approx 16.97 \text{ 2 d.p}$