A triangle has corners at `(2, 1 )`, `( 1, 3 )`, and `(5 , 5 )`. If the triangle is dilated by `3 x` around `(1, 6)`, what will the new coordinates of its corners be?

The vector equation of a line from `(1, 6)` to `(2,1)` is:

`(x,y)=(1,6)+t((2-1)hati+(1-6)hatj)`

Perform the subtraction within the vector:

`(x,y)=(1,6)+t(hati-5hatj)" [1]"`

The vector equation of a line from `(1, 6)` to `(1,3)` is:

`(x,y)=(1,6)+t((1-1)hati+(3-6)hatj)`

Perform the subtraction within the vector:

`(x,y)=(1,6)+t(-3hatj)" [2]"`

The vector equation of a line from `(1, 6)` to `(5,5)` is:

`(x,y)=(1,6)+t((5-1)hati+(5-6)hatj)`

Perform the subtraction within the vector:

`(x,y)=(1,6)+t(4hati-hatj)" [3]"`

Here are the 3 vector equations:

`(x,y)=(1,6)+t(hati-5hatj)" [1]"`
`(x,y)=(1,6)+t(-3hatj)" [2]"`
`(x,y)=(1,6)+t(4hati-hatj)" [3]"`

To obtain the new points, evaluate equations [1], [2], and [3] at t = 3

`(x,y)=(1,6)+3(hati-5hatj)`
`(x,y)=(1,6)+3(-3hatj)`
`(x,y)=(1,6)+3(4hati-hatj)`

`(x,y)=(4,-9)`
`(x,y)=(1,-3)`
`(x,y)=(13,3)`

`"let " A=(2,1),B=(1,3),C=(5,5)" and " D=(1,6)`

`"and " A',B',C'" be the images of A,B" and "C`
`"under the dilatation"`

`vec(DA)=ula-uld=((2),(1))-((1),(6))=((1),(-5))`

`rArrvec(DA')=3((1),(-5))=((3),(-15))`

`rArrA'=(1+3,6-15)=(4,-9)`
`color(blue)"--------------------------------------------------"`

`vec(DB)=ulb-uld=((1),(3))-((1),(6))=((0),(-3))`

`rArrvec(DB')=3((0),(-3))=((0),(-9))`

`rArrB'=(1+0,6-9)=(1,-3)`
`color(blue)"--------------------------------------------------"`

`vec(DC)=ulc-uld=((5),(5))-((1),(6))=((4),(-1))`

`rArrvec(DC')=3((4),(-1))=((12),(-3))`

`rArrC'=(1+12,6-3)=(13,3)`
`color(blue)"-------------------------------------------------"`