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

##### 1 Answer

#### Explanation:

Label the endpoints A(7 ,3) and B(1 ,2).

Label the centre of dilatation C(3 ,4)

#CtoA" is the translation" ((x_A-x_C),(y_A-y_C))#

#=((7-3),(3-4))=((4),(-1))#

#rArrCtoA'=3((4),(-1))=((12),(-3))# Now add the components of the translation to the coordinates of C to obtain the coordinates of the image A'.

#rArrA'=(3+12,4-3)=(15,1)# Repeat the process to find B'

#CtoB" is the translation " ((x_B-x_C),(y_B-y_C))#

#=((1-3),(2-4))=((-2),(-2))#

#rArrCtoB'=3((-2),(-2))=((-6),(-6))# Add the components of the translation to the coordinates of C to obtain the coordinates of the image B'.

#rArrB'=(3-6,4-6)=(-3,-2)# To calculate the length of the segment use the

#color(blue)"distance formula"#

#color(orange)"Reminder " 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)|)))#

#color(white)(xxxxx)" where" (x_1,y_1),(x_2,y_2)" are 2 coordinate points"# The 2 points here are (15 ,1) and (-3 ,-2)

let

# (x_1,y_1)=(15,1)" and " (x_2,y_2)=(-3,-2)#

#d=sqrt((-3-15)^2+(-2-1)^2)=sqrt(324+9)≈18.248#