Answer:
[tex](\frac{1}{3}x - 3,-\frac{1}{3}y+2)[/tex]
Both triangles will be similar
Step-by-step explanation:
See comment for complete question.
Given
Let the coordinates of the triangle be T(x,y)
First transformation: Dilation by 1/3
The new points will be:
[tex]T' = \frac{1}{3}(x,y)[/tex]
[tex]T' = (\frac{1}{3}x,\frac{1}{3}y)[/tex]
Second: Reflection over the x-axis
When a point (x,y) is reflected over the x-axis, the new point is (x,-y).
So, we have:
[tex]T'' = (\frac{1}{3}x,-\frac{1}{3}y)[/tex]
Third: Translation 3 units left and 2 units up.
When a point (x,y) is translated b units left and h units up, the new point is (x - b,y+h).
In this case:
[tex]b = 3[/tex] and [tex]h = 2[/tex]
So, we have:
[tex]T''' = (\frac{1}{3}x - 3,-\frac{1}{3}y+2)[/tex]
Hence, the coordinate of the new triangle will be: [tex](\frac{1}{3}x - 3,-\frac{1}{3}y+2)[/tex]
Additionally, both triangles will be similar because all the transformation done are rigid transformations i.e. Dilation, Reflection and Translation