A line segment has endpoints at #(7 ,6 )# and #(5 ,8 )#. The line segment is dilated by a factor of #4 # around #(2 ,1 )#. What are the new endpoints and length of the line segment?

Question

A line segment has endpoints at #(7  ,6  )# and #(5 ,8   )#.  The line segment is dilated by a factor of #4  # around #(2  ,1  )#.  What are the new endpoints and length of the line segment?

Theodore Abad · Accepted Answer

#(22,21),(14,29)" and "8sqrt2#

Let the endpoints be A (7 ,6) and B (5 ,8) and their images be A' and B', respectively, under the dilation. Let the centre of dilatation be C (2 ,1)

#vec(CA)=ula-ulc=((7),(6))-((2),(1))=((5),(5)) #

#rArrvec(CA')=4((5),(5))=((20),(20))#

#rArrA'=(2+20,1+20)=(color(red)(22,21))#

Similar process to obtain coordinates of B'

#vec(CB)=ulb-ulc=((5),(8))-((2),(1))=((3),(7))#

#rArrvec(CB')=4((3),(7))=((12),(28))#

#rArrB'=(2+12,1+28)=(color(red)(14,29))#

#"new endpoints are "(22,21)" and " (14,29)#

To calculate the length, use the #color(blue)"distance formula"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(2/2)|)))# where # (x_1,y_1),(x_2,y_2)" are 2 coordinate points"#

The 2 points here are (22 ,21) and (14 ,29)

let # (x_1,y_1)=(22,21)" and " (x_2,y_2)=(14,29)#

#d=sqrt((14-22)^2+(29-21)^2)=sqrt(64+64)=sqrt128#

#"length of line segment" =sqrt128=8sqrt2≈11.31#

Violet Aldrich · Answer

undefined The new endpoints of the dilated line segment are (18, 17) and (10, 23). The length of the dilated line segment is 14 units.