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

Oct 13, 2016

We know,if a point P of coordinate (a,b) be dilated by a factor n around the point of coordinate (h,k), then after dilation the new position of point will be $P ' \equiv \left(n \left(a - h\right) + h , n \left(b - k\right) + k\right)$

This means

$P \left(a , b\right) \stackrel{\text{dilated nX arnd} \left(h , k\right)}{\to} {P}^{'} \left(n \left(a - h\right) + h , n \left(b - k\right) + k\right)$

Using this formula we get

$A \left(3 , 2\right) \stackrel{\text{dilated 4X arnd} \left(2 , 3\right)}{\to} {A}^{'} \left(4 \left(3 - 2\right) + 2 , 4 \left(2 - 3\right) + 3\right) = {A}^{'} \left(6 , - 1\right)$

$B \left(5 , 4\right) \stackrel{\text{dilated 4X arnd} \left(2 , 3\right)}{\to} {B}^{'} \left(4 \left(5 - 2\right) + 2 , 4 \left(4 - 3\right) + 3\right) = {B}^{'} \left(14 , 7\right)$

Length of

$A B = \sqrt{{\left(3 - 5\right)}^{2} + {\left(2 - 4\right)}^{2}} = 2 \sqrt{2}$

Length of

$A ' B ' = \sqrt{{\left(14 - 6\right)}^{2} + {\left(7 + 1\right)}^{2}} = 8 \sqrt{2}$

Oct 13, 2016

$\left(6 , - 1\right)$ and $\left(14 , 7\right)$
length = $8 \sqrt{2}$

#### Explanation:

If ${p}_{0} = \left(2 , 3\right)$ is the dilation center, then ${p}_{1} = \left(3 , 2\right)$ and ${p}_{2} = \left(5 , 4\right)$ after the dilation by a factor $\lambda$ their position will be

$p {'}_{1} = {p}_{0} + \lambda \left({p}_{1} - {p}_{0}\right)$
$p {'}_{2} = {p}_{0} + \lambda \left({p}_{2} - {p}_{0}\right)$

so if $\lambda = 4$

$p {'}_{1} = \left(2 , 3\right) + 4 \left(3 - 2 , 2 - 3\right) = \left(6 , - 1\right)$
$p {'}_{2} = \left(2 , 3\right) + 4 \left(5 - 2 , 4 - 3\right) = \left(14 , 7\right)$

and

$\left\lVert p {'}_{1} - p {'}_{2} \right\rVert = 8 \sqrt{2}$