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A line segment has endpoints at #(2 ,8 )# and #(3 , 1)#. The line segment is dilated by a factor of #3 # around #(1 , 4)#. What are the new endpoints and length of the line segment?

Answer 1

Allison Allen

The new end points are #(4,16)# and #(7,-5)#
The length of the new line segment is #=21.2#

Let the end points be#A=(2,8)# and #B=(3,1)#

and #C=(1,4)#

Let #A'# and #B'# bethe new end points

Then,

#vec(CA')=3vec(CA)#

#=3*<2-1,8-4> =3*<1,4> = <3,12>#

#A'=(3,12)+(1,4)=(4,16)#

Similarly,

#vec(CB')=3vec(CB)#

#=3*<3-1,1-4> =3*<2,-3> = <6,-9>#

#B'=(6,-9)+(1,4)=(7,-5)#

The length of the line segment is

#A'B'=sqrt((7-4)^2+(-5-16)^2)#

#=sqrt((3)^2+(-21)^2)#

#=sqrt450#

#=21.2#

The length of the olb line segment is

#AB=sqrt((3-2)^2+(1-8)^2)#

#=sqrt(1+49)#

#=sqrt(50)#

#=7.07#

#A'B'=3*AB#

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Answer 2

Maverick Smith

The new endpoints of the dilated line segment are (7, 16) and (10, -11). The length of the dilated line segment is approximately 21.931 units.

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