# A line segment has endpoints at (9 ,7 ) and (1 ,2 ). The line segment is dilated by a factor of 4  around (3 ,3 ). What are the new endpoints and length of the line segment?

May 17, 2018

color(blue)((27,19) \ \ "and" \ \(-5,-1)

#### Explanation:

One method to perform these is by using vectors.

Let:

$\boldsymbol{a}$ be position vector $\left(\begin{matrix}9 \\ 7\end{matrix}\right)$

$\boldsymbol{b}$ be position vector $\left(\begin{matrix}1 \\ 2\end{matrix}\right)$

$\boldsymbol{d}$ be position vector $\left(\begin{matrix}3 \\ 3\end{matrix}\right)$

$\vec{\boldsymbol{\mathrm{da}}} = \left(\begin{matrix}6 \\ 4\end{matrix}\right)$

Dilations by a factor 4:

$4 \vec{\boldsymbol{\mathrm{da}}} = 4 \left(\begin{matrix}6 \\ 4\end{matrix}\right) = \left(\begin{matrix}24 \\ 16\end{matrix}\right)$

Position vector of the image of $\boldsymbol{a}$:

$a ' = \boldsymbol{d} + 4 \vec{\boldsymbol{\mathrm{da}}} = \left(\begin{matrix}3 \\ 3\end{matrix}\right) + 4 \left(\begin{matrix}6 \\ 4\end{matrix}\right) = \left(\begin{matrix}27 \\ 19\end{matrix}\right)$

$\therefore$

$\left(9 , 7\right) \to \left(27 , 19\right)$

We do the same for point b:

$\vec{\boldsymbol{\mathrm{db}}} = \left(\begin{matrix}- 2 \\ - 1\end{matrix}\right)$

Position vector of the image of $\boldsymbol{b}$

$\boldsymbol{b} + 4 \vec{\boldsymbol{\mathrm{db}}} = \left(\begin{matrix}3 \\ 3\end{matrix}\right) + 4 \left(\begin{matrix}- 2 \\ - 1\end{matrix}\right) = \left(\begin{matrix}- 5 \\ - 1\end{matrix}\right)$

$\left(1 , 2\right) \to \left(- 5 , - 1\right)$

So new endpoints are:

$\left(27 , 19\right) , \left(- 5 , - 1\right)$

PLOT: