Question

16. (02.05 MC)Two similar polygons are shown below:A coordinate grid is shown from positive 6 to negative 6 on the x-axis and from positive 6 to negative 6 on the y-axis. A polygon P prime Q prime R prime S prime is shown with vertex P prime on ordered pair 2, negative 4, vertex Q prime on ordered pair 6, negative 4, vertex R prime on ordered pair 4, negative 2 and vertex S prime on ordered pair 2, negative 2. A polygon PQRS is shown with vertex P on ordered pair 1, negative 2, vertex Q on ordered pair 3, negative 2, vertex R on ordered pair 2, negative 1 and vertex S on ordered pair 1, negative 1.Which transformation was performed on PQRS to form P′Q′R′S′? (1 point)A dilation factor of 2A dilation factor of 4A dilation factor of 1 over 2A dilation factor of 1 over 4

Answer 1

a dilation of 2 (option A)

Step-by-step explanation:

To determine the transformation, we will use the vertices of PQRS and P'Q'R'S' and compare them

P (1, -2), Q (3, -2), R (2, -1) and S(1, -1)

P' (2, -4) Q' (6, -4), R' (4, -2) and S' (2, -2)

`\begin{gathered} \text{From P to P'} \\ (1,\text{ -2) }\rightarrow\text{ (2, -4)} \\ \lbrack\text{2(1), 2(-2)\rbrack = (2, -4)} \\ \\ \text{From Q to Q'} \\ (3,\text{ -2) }\rightarrow\text{ (6, -4)} \\ \lbrack2(3),\text{ 2(-2)\rbrack = (6, -4)} \end{gathered}`
`\begin{gathered} \text{From R to R'} \\ (2,\text{ -1) }\rightarrow\text{ (4, -2)} \\ \lbrack2(2),\text{ 2(-1)\rbrack = (4, -2)} \\ \\ \text{From S to S}^(\prime) \\ (1,\text{ -1) }\rightarrow\text{ (}2,\text{ -2)} \\ \lbrack2(1),\text{ 2(-1)\rbrack = (2, -2)} \end{gathered}`

Since we multiplied the vertices of PQRS to vertices of P'Q'R'S', the transformation is a dilation of 2 (option A)