# A line segment has endpoints at (a, b) and (c, d ). The line segment is dilated by a factor of r around (p,q ). What are the new endpoints and length of the line segment?

May 1, 2018

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

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

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

#### Explanation:

I have a theory all these questions are here so the there's something for newbies to do. I'll do the general case here and see what happens.

We translate the plane so the dilation point P maps to the origin. Then the dilation scales the coordinates by a factor of $r$. Then we translate the plane back:

$A ' = r \left(A - P\right) + P = \left(1 - r\right) P + r A$

That's the parametric equation for a line between P and A, with $r = 0$ giving P, $r = 1$ giving A, and $r = r$ giving A', the image of A under dilation by $r$ around P.

The image of $A \left(a , b\right)$ under dilation by $r$ around $P \left(p , q\right)$ is thus

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

Similarly, the image of $\left(c , d\right)$ is

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

The new length is $r$ times the original length.

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