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

##### 1 Answer
Apr 3, 2017

The new segments are $\left(\frac{5}{2} , 3\right)$ and $\left(\frac{7}{2} , \frac{11}{2}\right)$
The new length of the segment is $= \frac{\sqrt{29}}{2}$

#### Explanation:

Let the segment be $A = \left(1 , 4\right)$ and $B = \left(3 , 9\right)$

After dilatation, the segment becomes $\left(A ' = \left({x}_{A} , {y}_{A}\right)\right)$ and $B ' = \left({x}_{B} , {y}_{B}\right)$

Let the center of dilatation be $D = \left(4 , 2\right)$

Factor of dilatation is $= \frac{1}{2}$

We do this calcutation with vectors

$\vec{D A '} = \frac{1}{2} \vec{D A}$

$\left(\begin{matrix}{x}_{A} - 4 \\ {y}_{A} - 2\end{matrix}\right) = \frac{1}{2} \left(\begin{matrix}1 - 4 \\ 4 - 2\end{matrix}\right)$

$\left(\begin{matrix}{x}_{A} \\ {y}_{A}\end{matrix}\right) = \left(\begin{matrix}4 - \frac{3}{2} \\ 2 + 1\end{matrix}\right) = \left(\begin{matrix}\frac{5}{2} \\ 3\end{matrix}\right)$

Similarly,

$\vec{D B '} = \frac{1}{2} \vec{D B}$

$\left(\begin{matrix}{x}_{B} - 4 \\ {y}_{B} - 2\end{matrix}\right) = \frac{1}{2} \left(\begin{matrix}3 - 4 \\ 9 - 2\end{matrix}\right)$

$\left(\begin{matrix}{x}_{B} \\ {y}_{B}\end{matrix}\right) = \left(\begin{matrix}4 - \frac{1}{2} \\ 2 + \frac{7}{2}\end{matrix}\right) = \left(\begin{matrix}\frac{7}{2} \\ \frac{11}{2}\end{matrix}\right)$

Original length of line segment is

$A B = \sqrt{{\left(3 - 1\right)}^{2} + {\left(9 - 4\right)}^{2}}$

$= \sqrt{4 + 25}$

$= \sqrt{29}$

New length of ine segment is

$A ' B ' = \sqrt{{\left(\frac{7}{2} - \frac{5}{2}\right)}^{2} + {\left(\frac{11}{2} - 3\right)}^{2}}$

$= \sqrt{1 + \frac{25}{4}}$

$= \frac{\sqrt{29}}{2}$