Answer:
The single transformation that takes shape 'p' onto shape 'q' is a dilation by a scale factor of 3 with the center point (1, 0)
Explanation:
The coordinates of the vertices of shape ΔP are;
A(2, 3), B(2, 1), C(3, 1)
The length AB = 3 - 1 = 2
The length CB = 3 - 2 = 1
∴ The length AC = √(2² + 1²) = √5
The coordinates of the vertices of shape ΔQ are;
A'(4, 9), B'(4, 3), C'(7, 3)
The length A'B' = 9 - 3 = 6
The length C'B' = 7 - 3 = 3
∴ The length A'C' = √(6² + 3²) = √(45) = 3·√5
∴ Figure Q ΔA'B'C' is a dilation of figure P ΔABC by a scale factor of A'B'/AB = 6/2 = 3
The distance of point B(2, 1) from the point (1, 0) = √((2 - 1)² + 1²) = √2
The distance of point B'(4, 3) from the point (1, 0) = √((4 - 1)² + 3²) = 3·√2
Therefore, the distances of points on figure Q from (1, 0) = 3 × the distances of points on figure from (1, 0)
Therefore, figure 'Q' can be obtained from figure 'P' by a dilation of figure 'P' by a scale factor of 3 with the center of dilation at point (1, 0).