A line segment has endpoints at (7 ,6 ) and (5 ,8 ). The line segment is dilated by a factor of 3  around (2 ,4 ). What are the new endpoints and length of the line segment?

May 7, 2017

The new emd points are $\left(17 , 10\right)$ and $\left(11 , 16\right)$
The length of the line segment is $= 8.49$

Explanation:

Let the end points be$A = \left(7 , 6\right)$ and $B = \left(5 , 8\right)$

and $C = \left(2 , 4\right)$

Let $A '$ and $B '$ be the new end points

Then,

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

$= 3 \cdot < 7 - 2 , 6 - 4 > = 3 \cdot < 5 , 2 > = < 15 , 6 >$

$A ' = \left(15 , 6\right) + \left(2 , 4\right) = \left(17 , 10\right)$

Similarly,

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

$= 3 \cdot < 5 - 2 , 8 - 4 > = 3 \cdot < 3 , 4 > = < 9 , 12 >$

$B ' = \left(9 , 12\right) + \left(2 , 4\right) = \left(11 , 16\right)$

The length of the line segment is

$A ' B ' = \sqrt{{\left(11 - 17\right)}^{2} + {\left(16 - 10\right)}^{2}}$

$= \sqrt{{\left(6\right)}^{2} + {\left(6\right)}^{2}}$

$= \sqrt{72}$

$= 8.49$

The length of the old line segment is

$A B = \sqrt{{\left(5 - 7\right)}^{2} + {\left(8 - 6\right)}^{2}}$

$= \sqrt{4 + 4}$

$= \sqrt{8}$

$= 2.83$

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