Respuesta :
Vertices of triangle dilated about the origin by a scale factor of 1/3, and then translated 2 units left and 3 units down
coordinates of A'B'C' are
A' (-1,-1)
B'(0,0)
C' (1,-2)
Given :
The vertices of a triangle are A(3,6), B(6,9), and C(9.3).
When the vertices by dilated by origin by scale factor 'k', then we multiply 'k' with each vertices
It is dilated about the origin by a scale factor of 1/3
So we multiply (x,y) by scale factor
[tex](x,y) - (\frac{1}{3}x,\frac{1}{3}y)\\A(3,6) - (\frac{1}{3}(3),\frac{1}{3}(6))->(1,2)\\\\B(6,9) - (\frac{1}{3}(6),\frac{1}{3}(9))->(2,3)\\\\C(9,3) - (\frac{1}{3}(9),\frac{1}{3}(3))->(3,1)\\\\[/tex]
Now we translate 2 units left and 3 units down
For 2 units left , we subtract 2 from x
For 3 units down , we subtract 3 from y
So (x,y) becomes (x-2, y-3)
Lets apply this rule to our new vertices
[tex](1,2) \; becomes \; (-1,-1)\\(2,3) \; becomes \; (0,0)\\(3,1) \; becomes \; (1,-2)\\[/tex]
coordinates are
A' (-1,-1)
B'(0,0)
C' (1,-2)
Learn more : brainly.com/question/23375150
