A line segment goes from #(4 ,1 )# to #(2 ,3 )#. The line segment is dilated about #(2 ,2 )# by a factor of #2#. Then the line segment is reflected across the lines #x = -2# and #y=4#, in that order. How far are the new endpoints form the origin?

Answer 1

I get # 4 sqrt{2}# for the first endpoint and #0# for the second.

Yikes, that's a lot of transformations. We're just interested in the image of each endpoint.

To do the dilation we start by getting a direction vector from the dilation point to each endpoint, essentially translating the dilation point to the origin.

# (4,1) - (2,2) = (2,-1) quad quad quad quad (2,3)-(2,2)=(0,1)#

We dilate each direction vector by a factor of two and translate back:

# (2,2)+2(2,-1)=(6,0) quad quad quad quad (2,2)+2(0,1)=(2,4)#
Reflecting through #x=-2# leaves the #y# coordinate alone:
# (6,0) to (2-6,0)=(-4,0) quad quad quad quad (2,4) to (2-2,4)=(0,4)#
Reflecting through #y=4# leaves the #x# coordinate alone:
# (-4,0) to (-4,4-0)=(-4,4) quad quad quad quad (0,4) to(0,0) #
If I did that right the first endpoint is #\sqrt{4^2+4^2}=4\sqrt{2}# from the origin and the second endpoint is the origin, so a distance of zero.
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Answer 2

To find the new endpoints of the line segment after dilation, reflection across the lines ( x = -2 ) and ( y = 4 ), we'll follow these steps:

  1. Dilate the line segment about the point ( (2, 2) ) by a factor of 2.
  2. Reflect the dilated line segment across the line ( x = -2 ).
  3. Reflect the result across the line ( y = 4 ).

First, let's dilate the line segment from ( (4, 1) ) to ( (2, 3) ) about ( (2, 2) ) by a factor of 2.

Dilation formula: [ (x', y') = (2 + 2(x - 2), 2 + 2(y - 2)) ]

For ( (4, 1) ): [ x' = 2 + 2(4 - 2) = 6 ] [ y' = 2 + 2(1 - 2) = 0 ]

For ( (2, 3) ): [ x' = 2 + 2(2 - 2) = 2 ] [ y' = 2 + 2(3 - 2) = 4 ]

Now, let's reflect the dilated line segment across the line ( x = -2 ).

Reflecting across ( x = -2 ): [ x'' = -(-2) - (x' + 2) = 2 - (x' + 2) ] [ y'' = y' ]

For ( (6, 0) ): [ x'' = 2 - (6 + 2) = -6 ] [ y'' = 0 ]

For ( (2, 4) ): [ x'' = 2 - (2 + 2) = -2 ] [ y'' = 4 ]

Finally, let's reflect the result across the line ( y = 4 ).

Reflecting across ( y = 4 ): [ x''' = x'' ] [ y''' = 8 - y'' ]

For ( (-6, 0) ): [ x''' = -6 ] [ y''' = 8 - 0 = 8 ]

For ( (-2, 4) ): [ x''' = -2 ] [ y''' = 8 - 4 = 4 ]

So, the new endpoints after dilation and reflection are ( (-6, 8) ) and ( (-2, 4) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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