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

Question

1472 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-03T06:33:34

The new segments are #(5/2,3)# and #(7/2,11/2)#
The new length of the segment is #=sqrt29/2#

Let the segment be #A=(1,4)# and #B=(3,9)#

After dilatation, the segment becomes #(A'=(x_A,y_A))# and #B'=(x_B,y_B)#

Let the center of dilatation be #D=(4,2)#

Factor of dilatation is #=1/2#

We do this calcutation with vectors

#vec(DA')=1/2vec(DA)#

#((x_A-4),(y_A-2))=1/2((1-4),(4-2))#

#((x_A),(y_A))=((4-3/2),(2+1))=((5/2),(3))#

Similarly,

#vec(DB')=1/2vec(DB)#

#((x_B-4),(y_B-2))=1/2((3-4),(9-2))#

#((x_B),(y_B))=((4-1/2),(2+7/2))=((7/2),(11/2))#

Original length of line segment is

#AB=sqrt((3-1)^2+(9-4)^2)#

#=sqrt(4+25)#

#=sqrt29#

New length of ine segment is

#A'B'=sqrt((7/2-5/2)^2+(11/2-3)^2)#

#=sqrt(1+25/4)#

#=sqrt29/2#