# A line segment has endpoints at (9 ,2 ) and (7 , 4). The line segment is dilated by a factor of 3  around (1 , 5). What are the new endpoints and length of the line segment?

May 13, 2018

$\left(9 , 2\right) \to \left(25 , - 4\right)$

$\left(7 , 4\right) \to \left(19 , 2\right)$

New length $6 \sqrt{2}$

#### Explanation:

There is indeed a Bolivia, United States, in North Carolina, not too far from Myrtle Beach.

I did the general case of this question [here].(https://socratic.org/questions/a-line-segment-has-endpoints-at-2-4-and-5-3-the-line-segment-is-dilated-by-a-fac-1)

I got for endpoints $\left(a , b\right) , \left(c , d\right) ,$ and factor $r$ around dilation point $\left(p , q\right) :$

$\left(a , b\right) \to \left(\begin{matrix}1 - r p + r a \\ 1 - r q + r b\end{matrix}\right)$, similarly for $\left(c , d\right)$,

new length $l = r \setminus \sqrt{{\left(a - c\right)}^{2} + {\left(b - d\right)}^{2}}$

These are old problems that probably no one ever looks at that I think are just here to give the noobs something to do. I'm an old timer at 26 days; I only answered because of Bolivia, United States.

I will now mindlessly substitute.

$a = 9 , b = 2 , c = 7 , d = 4 , p = 1 , q = 5 , r = 3$

$\left(9 , 2\right) \to \left(\left(1 - 3\right) 1 + 3 \left(9\right) , \left(1 - 3\right) 5 + 3 \left(2\right)\right) = \left(25 , - 4\right)$

$\left(7 , 4\right) \to \left(\left(1 - 3\right) 1 + 3 \left(7\right) , \left(1 - 3\right) 5 + 3 \left(4\right)\right) = \left(19 , 2\right)$

new length $l = 3 \setminus \sqrt{{\left(9 - 7\right)}^{2} + {\left(2 - 4\right)}^{2}} = 3 \sqrt{8} = 6 \sqrt{2}$