A line segment has endpoints at #(3 ,2 )# and #(5 ,4 )#. The line segment is dilated by a factor of #4 # around #(2 ,3 )#. What are the new endpoints and length of the line segment?

2 Answers
P dilip_k
Oct 13, 2016

drawn
We know,if a point P of coordinate (a,b) be dilated by a factor n around the point of coordinate (h,k), then after dilation the new position of point will be #P'equiv(n(a-h)+h,n(b-k)+k)#

This means

#P(a,b)stackrel("dilated nX arnd"(h,k))->P^'(n(a-h)+h,n(b-k)+k)#

Using this formula we get

#A(3,2)stackrel("dilated 4X arnd"(2,3))->A^'(4(3-2)+2,4(2-3)+3)=A^'(6,-1)#

#B(5,4)stackrel("dilated 4X arnd"(2,3))->B^'(4(5-2)+2,4(4-3)+3)=B^'(14,7)#

Length of

#AB=sqrt((3-5)^2+(2-4)^2)=2sqrt2#

Length of

#A'B'=sqrt((14-6)^2+(7+1)^2)=8sqrt2#

Cesareo R.
Oct 13, 2016

#(6,-1)# and #(14,7)#
length = #8sqrt(2)#

Explanation:

If #p_0=(2,3)# is the dilation center, then #p_1=(3,2)# and #p_2 = (5,4)# after the dilation by a factor #lambda# their position will be

#p'_1=p_0+lambda(p_1-p_0)#
#p'_2=p_0+lambda(p_2-p_0)#

so if #lambda= 4#

#p'_1=(2,3)+4(3-2,2-3)=(6,-1)#
#p'_2=(2,3)+4(5-2,4-3)=(14,7)#

and

#norm(p'_1-p'_2)=8sqrt(2)#

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