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?

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Answer 1 · 2018-04-26T14:16:21

Endpoints #(19,-3) and (7,0)# and length #3sqrt{17}.#

Let #D=(4.6)# 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 (P-D) + D = rP + (1-r)D #

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

#l(t)=(1-t)D+tP#.

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

We can precompute

#(1-r)D = (1-3)D=-2(4,6)=(-8,-12).#

The image of #(9,3)# is #3(9,3)+(-8,-12)=(19,-3)#

The image of #(5,4)# is #3(5,4)+(-8,-12)=(7,0)#

The length will be three times the original length,

# l = 3\sqrt{ (9-5)^2+(3-4)^2} = 3sqrt{17}#

We can check that.

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