A line segment has endpoints at #(1 ,2 )# and #(3 ,4 )#. The line segment is dilated by a factor of #6 # around #(2 ,5 )#. What are the new endpoints and length of the line segment?

1 Answer
Jan 27, 2018

#(-4,-13),(8,-1),~~16.97#

Explanation:

#"label the endpoints "A(1,2)" and "B(3,4)#

#"label the centre of dilatation "C(2,5)#

#"let A' and B' be the images of A and B"#

#"then"#

#vec(CA')=color(red)(6)vec(CA)#

#rArrula'-ulc=6(ula-ulc)#

#rArrula'-ulc=6ula-6ulc#

#rArrula'=6ula-5ulc#

#color(white)(rArrula')=6((1),(2))-5((2),(5))#

#color(white)(rArrula')=((6),(12))-((10),(25))=((-4),(-13))#

#rArrA'=(-4,-13)#

#"and "#

#vec(CB')=color(red)(6)vec(CB)#

#rArrulb'-ulc=6(ulb-ulc)#

#rArrulb'=6ulb-5ulc#

#color(white)(rArrulb')=6((3),(4))-5((2),(5))#

#color(white)(rArrulb)=((18),(24))-((10),(25))=((8),(-1))#

#rArrB'=(8,-1)#

#"calculate the length using the "color(blue)"distance formula"#

#•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#"let "(x_1,y_1)=(-4,-13)" and "(x_2,y_2)=(8,-1)#

#d=sqrt((8+4)^2+(-1+13)^2)#

#color(white)(d)=sqrt(144+144)=sqrt288=12sqrt2~~16.97" 2 d.p"#