Question

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?

Answer 1

`(7,3)to(15,1), (1,2)to(-3,-2)`

`"length " ≈18.248" to 3 decimal places"`

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`