Answer:
B
Step-by-step explanation:
Quadrilateral ABCD has its vertices at points A(1,7), B(4,6), C(6,1) and D(1,4).
Consider dilation centered at point A(1,7) with a scale factor [tex]\frac{2}{3}.[/tex] This dilation has the rule
[tex](x,y)\rightarrow \left(\dfrac{2}{3}x+\dfrac{1}{3},\dfrac{2}{3}y+\dfrac{7}{3}\right)[/tex]
Thus,
- [tex]A(1,7)\rightarrow A'(1,7)\ \left[\dfrac{2}{3}\cdot 1+\dfrac{1}{3}=1,\ \dfrac{2}{3}\cdot 7+\dfrac{7}{3}=7\right];[/tex]
- [tex]B(4,6)\rightarrow B'\left(3,6\dfrac{1}{3}\right)\ \left[\dfrac{2}{3}\cdot 4+\dfrac{1}{3}=3,\ \dfrac{2}{3}\cdot 6+\dfrac{7}{3}=\dfrac{19}{3}=6\dfrac{1}{3}\right];[/tex]
- [tex]C(6,1)\rightarrow C'\left(4\dfrac{1}{3},3\right)\ \left[\dfrac{2}{3}\cdot 6+\dfrac{1}{3}=\dfrac{13}{3}=4\dfrac{1}{3},\ \dfrac{2}{3}\cdot 1+\dfrac{7}{3}=3\right];[/tex]
- [tex]D(1,4)\rightarrow D'(1,5)\ \left[\dfrac{2}{3}\cdot 1+\dfrac{1}{3}=1,\ \dfrac{2}{3}\cdot 4+\dfrac{7}{3}=5\right];[/tex]
These are exactly coordinates of quadrilateral A'B'C'D'
