Coordinates of A'B'C':
[tex]\begin{gathered} A^{\prime}(-9,3) \\ B^{\prime}(-3,0) \\ C^{\prime}(0,6) \end{gathered}[/tex]
To get coordinates of triangle ABC:
Dilate the triangle A'B'C' by a facor of 1/3:
Rule of dilation:
[tex]P(x,y)\rightarrow P^{\prime}(kx,ky)[/tex]
k is the factor of dilation:
Then, you get;
[tex]\begin{gathered} A^{\prime}(-9,3)\rightarrow A^{\doubleprime}(\frac{1}{3}(-9),\frac{1}{3}(3)) \\ A^{\prime}(-9,3)\rightarrow A^{\prime}^{\prime}(-3,1) \end{gathered}[/tex]
Relect over x-axis:
Rule of relection:
[tex]P(x,y)\rightarrow P^{\prime}(x,-y)[/tex]
Then, coordiate of point A is:
[tex]A^{\doubleprime}(-3,1)\rightarrow A(-3,-1)[/tex]Then, coordinates of A are (-3,-1)
