Question

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

Answer 1

`color(blue)("Point A'"->(x,y)=(-6,-1))`
`color(blue)("Point B'"->(x,y)=(3,14))`
`color(blue)(L=sqrt(9^2+15^2) = sqrt(306))`

Explanation:

Dilation is a fancy word for scale. As soon as you talk about scale your are dealing with ratios. Dilation that is positive increases the magnitude. Assuming positive dilation as not indicated otherwise.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine point B' ")" "larr` B prime

`(OB)/(OB') = 1/3`

`(x_O-x_B)/ (x_O-x_B') =1/3`

`=>3(x_O-x_B)=x_O-x_B'`

`=>3(6-5)=6-x_B'`

`x_B'=3`
.......................................................................................

`(y_O-y_B)/ (y_O-y_B') =1/3`

`=>3(y_O-y_B)=y_O-y_B'`

`=>3(2-6)=2-y_B'`

`y_B'=14`

`color(blue)("Point B'"->(x,y)=(3,14))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine point A' ")" "larr` A prime

`(OA)/(OA') = 1/3`

`(x_O-x_A)/ (x_O-x_A') =1/3`

`=>3(x_O-x_A)=x_O-x_A'`

`=>3(6-2)=6-x_A'`

`x_A'=-6`
............................................................................

`(y_O-y_A)/ (y_O-y_A') =1/3`

`=>3(y_O-y_A)=y_O-y_A'`

`=>3(2-1)=2-y_A'`

`y_A'=-1`

`color(blue)("Point A'"->(x,y)=(-6,-1))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine length of A'B'")`

`L=sqrt((x_B'-x_A')^2+(y_B'-y_B')^2)`

`color(blue)(L=sqrt(9^2+15^2) = sqrt(306))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Check")`

`(AB)/(A'B') = 1/3`

Length `AB = sqrt((5-2)^2+(6-1)^2) =sqrt( 34)`

`LHS ->RHS`

`1/3->sqrt(34)/(sqrt(306))=1/3`

`LHS=RHS color(red)(larr" True")`