# A line segment has endpoints at (1 ,6 ) and (6 ,7 ). The line segment is dilated by a factor of 4  around (4 ,3 ). What are the new endpoints and length of the line segment?

May 3, 2018

$\left(1 , 6\right) \to \left(- 8 , 15\right)$

$\left(6 , 7\right) \to \left(12 , 19\right)$

$l = 4 \sqrt{26}$

#### Explanation:

I did the general case here.

$\left(p , q\right) = \left(4 , 3\right) , \quad r = 4 , \quad \left(a , b\right) = \left(1 , 6\right) , \quad \left(c , d\right) = \left(6 , 7\right)$

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

$\left(1 , 6\right) \to \left(- 3 \left(4\right) + 4 \left(1\right) , - 3 \left(3\right) + 4 \left(6\right)\right) = \left(- 8 , 15\right)$

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

$\left(6 , 7\right) \to \left(- 3 \left(4\right) + 4 \left(6\right) , - 3 \left(3\right) + 4 \left(7\right)\right) = \left(12 , 19\right)$

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

$l = 4 \setminus \sqrt{{\left(6 - 1\right)}^{2} + {\left(7 - 6\right)}^{2}} = 4 \sqrt{26}$