# A line segment has endpoints at (2 ,8 ) and (3 , 1). The line segment is dilated by a factor of 3  around (1 , 4). What are the new endpoints and length of the line segment?

Apr 29, 2017

The new end points are $\left(4 , 16\right)$ and $\left(7 , - 5\right)$
The length of the new line segment is $= 21.2$

#### Explanation:

Let the end points be$A = \left(2 , 8\right)$ and $B = \left(3 , 1\right)$

and $C = \left(1 , 4\right)$

Let $A '$ and $B '$ bethe new end points

Then,

$\vec{C A '} = 3 \vec{C A}$

$= 3 \cdot < 2 - 1 , 8 - 4 > = 3 \cdot < 1 , 4 > = < 3 , 12 >$

$A ' = \left(3 , 12\right) + \left(1 , 4\right) = \left(4 , 16\right)$

Similarly,

$\vec{C B '} = 3 \vec{C B}$

$= 3 \cdot < 3 - 1 , 1 - 4 > = 3 \cdot < 2 , - 3 > = < 6 , - 9 >$

$B ' = \left(6 , - 9\right) + \left(1 , 4\right) = \left(7 , - 5\right)$

The length of the line segment is

$A ' B ' = \sqrt{{\left(7 - 4\right)}^{2} + {\left(- 5 - 16\right)}^{2}}$

$= \sqrt{{\left(3\right)}^{2} + {\left(- 21\right)}^{2}}$

$= \sqrt{450}$

$= 21.2$

The length of the olb line segment is

$A B = \sqrt{{\left(3 - 2\right)}^{2} + {\left(1 - 8\right)}^{2}}$

$= \sqrt{1 + 49}$

$= \sqrt{50}$

$= 7.07$

$A ' B ' = 3 \cdot A B$