A line segment has endpoints at #(3 ,2 )# and #(7 , 5)#. The line segment is dilated by a factor of #4 # around #(2 , 3)#. What are the new endpoints and length of the line segment?

1 Answer
Apr 21, 2017

#(6,-1),(22,11),20" units"#

Explanation:

let #A=(3,2),B=(7,5)" and " C=(2,3)#

#"and " A',B'" be the image of A and B under the dilatation"#

#vec(CA)=ula-ulc=((3),(2))-((2),(3))=((1),(-1))#

#rArrvec(CA')=4((1),(-1))=((4),(-4))#

#rArrA'=(2+4,3-4)=(6,-1)color(red)(larr)#

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

#rArrvec(CB')=4((5),(2))=((20),(8))#

#rArrB'=(2+20,3+8)=(22,11)color(red)(larr)#

To calculate 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)|)))#

#"let " (x_1,y_1)=(6,-1)" and " (x_2,y_2)=(22,11)#

#d_(A'B')=sqrt((22-6)^2+(11+1)^2)#

#color(white)(d_(A'B'))=sqrt(256+144)=sqrt400#

#rArrd_(A'B')=20" units"#