# A line segment has endpoints at (9 ,3 ) and (5 ,4 ). The line segment is dilated by a factor of 3  around (4 ,6 ). What are the new endpoints and length of the line segment?

Apr 26, 2018

Endpoints $\left(19 , - 3\right) \mathmr{and} \left(7 , 0\right)$ and length $3 \sqrt{17} .$

#### Explanation:

Let $D = \left(4.6\right)$ be the dilation point, $r$ be the dilation factor, and $P$ be the point being dilated.

I always think of translating the dilation point to the origin, dilating, then translating back. So $P '$, the image of $P$, is

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

That's the standard parametric form for a line from $D$
to $P$,

$l \left(t\right) = \left(1 - t\right) D + t P$.

We have $l \left(0\right) = D$, $l \left(1\right) = P$ and our image $P ' = l \left(r\right)$. That makes sense; dilation sends points off along a ray from the dilation point.

We can precompute

$\left(1 - r\right) D = \left(1 - 3\right) D = - 2 \left(4 , 6\right) = \left(- 8 , - 12\right) .$

The image of $\left(9 , 3\right)$ is $3 \left(9 , 3\right) + \left(- 8 , - 12\right) = \left(19 , - 3\right)$

The image of $\left(5 , 4\right)$ is $3 \left(5 , 4\right) + \left(- 8 , - 12\right) = \left(7 , 0\right)$

The length will be three times the original length,

$l = 3 \setminus \sqrt{{\left(9 - 5\right)}^{2} + {\left(3 - 4\right)}^{2}} = 3 \sqrt{17}$

We can check that.

sqrt{ (19-7)^2 + (-3)^2}= sqrt{144+9}=sqrt{153}=3 sqrt{17} quad sqrt