Answer :
Solution:
If a triangle has vertices (a,b), (p,q) and (c,d) then it's
Area= [tex]\frac{1}{2}[/tex][a(q-d)-p(d-b)+c(b-q)] Square Units
Vertices of Preimage i.e Original Triangle = (0,0), (2,0) , (2,-3).
Area of Original Triangle = [tex]\frac{1}{2}[0(0+3)-2(-3-0)+2(0-0)]=\frac{1}{2}\times 6=3[/tex] square units
As , given Dilation Factor = [tex]\frac{1}{3}[/tex]
Now, Dilation Factor < 1
So, Image will be smaller than pre-image.
Now, the area of Image, that is new Triangle= [tex]\frac{1}{3} \times 3=1[/tex]
Now, we will check by looking at the options Which Triangle has area of 1 Units.
Option A: Triangle with vertices: (0, 0), (2, 0), and (2, 3).
Area of Triangle = [tex]\frac{1}{2}[0(0-3)-2(3-0)+2(0-0)]=\frac{1}{2}\times -6=3[/tex] square units(Area is a positive Quantity)
Option B: Triangle with vertices: (0, 0), (0.5, 0), and (0.5, negative 1).
Area of Triangle = [tex]\frac{1}{2}[0(0+1)-0.5(-1-0)+0.5(0-0)]=\frac{1}{2}\times 0.5=0.25[/tex] square units
Option C :Triangle with vertices: (0, 0), (6, 0), and (6, 9).
Area of Triangle = [tex]\frac{1}{2}[0(0-9)-6(9-0)+6(0-0)]=\frac{1}{2}\times -54=27[/tex] square units(Area is a positive Quantity)
Option D: Triangle with vertices: (0, 0), (6, 0), and (6, negative 9).
Area of Triangle = [tex]\frac{1}{2}[0(0+9)-6(-9-0)+6(0-0)]=\frac{1}{2}\times 54=27[/tex] square units
None of the option matches the answer.
Now , it is not clear in the question whether original triangle or Image has been dilated by a factor of [tex]\frac{1}{3}[/tex].
Original Triangle Vertices : (0,0), (2,0) (2, -3)
Dilation factor of image = [tex]\frac{1}{3}[/tex]
So, Dilation Factor of Preimage =3
So, to get the vertices of image i.e new triangle we have to multiply the vertices of original triangle by 3.
So, Coordinates of Vertices of Image = (0,0),(6,0),(6,-9)
Option C. Triangle with vertices: (0, 0), (6, 0), and (6, 9) is true.