# A line segment has endpoints at (7 ,3 ) and (1 ,2 ). The line segment is dilated by a factor of 3  around (3 ,4 ). What are the new endpoints and length of the line segment?

Dec 17, 2016

$\left(7 , 3\right) \to \left(15 , 1\right) , \left(1 , 2\right) \to \left(- 3 , - 2\right)$

$\text{length " ≈18.248" to 3 decimal places}$

#### Explanation:

Label the endpoints A(7 ,3) and B(1 ,2).

Label the centre of dilatation C(3 ,4)

$C \to A \text{ is the translation} \left(\begin{matrix}{x}_{A} - {x}_{C} \\ {y}_{A} - {y}_{C}\end{matrix}\right)$

$= \left(\begin{matrix}7 - 3 \\ 3 - 4\end{matrix}\right) = \left(\begin{matrix}4 \\ - 1\end{matrix}\right)$

$\Rightarrow C \to A ' = 3 \left(\begin{matrix}4 \\ - 1\end{matrix}\right) = \left(\begin{matrix}12 \\ - 3\end{matrix}\right)$

Now add the components of the translation to the coordinates of C to obtain the coordinates of the image A'.

$\Rightarrow A ' = \left(3 + 12 , 4 - 3\right) = \left(15 , 1\right)$

Repeat the process to find B'

$C \to B \text{ is the translation } \left(\begin{matrix}{x}_{B} - {x}_{C} \\ {y}_{B} - {y}_{C}\end{matrix}\right)$

$= \left(\begin{matrix}1 - 3 \\ 2 - 4\end{matrix}\right) = \left(\begin{matrix}- 2 \\ - 2\end{matrix}\right)$

$\Rightarrow C \to B ' = 3 \left(\begin{matrix}- 2 \\ - 2\end{matrix}\right) = \left(\begin{matrix}- 6 \\ - 6\end{matrix}\right)$

Add the components of the translation to the coordinates of C to obtain the coordinates of the image B'.

$\Rightarrow B ' = \left(3 - 6 , 4 - 6\right) = \left(- 3 , - 2\right)$

To calculate the length of the segment use the $\textcolor{b l u e}{\text{distance formula}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\textcolor{w h i t e}{\times \times x} \text{ where" (x_1,y_1),(x_2,y_2)" are 2 coordinate points}$

The 2 points here are (15 ,1) and (-3 ,-2)

let $\left({x}_{1} , {y}_{1}\right) = \left(15 , 1\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 3 , - 2\right)$

d=sqrt((-3-15)^2+(-2-1)^2)=sqrt(324+9)≈18.248