A line segment has endpoints at #(9 ,7 )# and #(1 ,2 )#. The line segment is dilated by a factor of #4 # around #(3 ,3 )#. What are the new endpoints and length of the line segment?

Answer 1

James Atkins

#color(blue)((27,19) \ \ "and" \ \(-5,-1)#

One method to perform these is by using vectors.

Let:

#bba# be position vector #((9),(7))#

#bb(b)# be position vector #((1),(2))#

#bbd# be position vector #((3),(3))#

#vec(bb(da))=((6),(4))#

Dilations by a factor 4:

#4vec(bb(da))=4((6),(4))=((24),(16))#

Position vector of the image of #bba#:

We now just add:

#a'=bbd+4vec(bb(da))=((3),(3))+4((6),(4))=((27),(19))#

#:.#

#(9,7)->(27,19)#

We do the same for point b:

#vec(bb(db))=((-2),(-1))#

Position vector of the image of #bb(b)#

#bb(b)+4vec(bb(db))=((3),(3))+4((-2),(-1))=((-5),(-1))#

#(1,2)->(-5,-1)#

So new endpoints are:

#(27,19),(-5,-1)#

PLOT:

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Answer 2

Aaliyah Anderson

The new endpoints of the line segment after dilation by a factor of 4 around (3, 3) are (21, 19) and (-3, -5). The length of the line segment is the distance between these two points, which can be calculated using the distance formula. The length is found to be 28 units.

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