Question

Polygon ABCD with vertices at A(1, −2), B(3, −2), C(3, −4), and D(1, −4) is dilated to create polygon A′B′C′D′ with vertices at A′(4, −8), B′(12, −8), C′(12, −16), and D′(4, −16). Determine the scale factor used to create the image.
one fourth
one half
2
4

Answer 1

To find the scale factor used in the dilation from polygon ABCD to polygon A'B'C'D', we can compare the coordinates of the corresponding vertices.

First, let's find the distances from the origin (0,0) to the vertices of the original polygon ABCD:

1. For vertex A(1, -2):
\[
d_A = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}
\]
2. For vertex B(3, -2):
\[
d_B = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]
3. For vertex C(3, -4):
\[
d_C = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
4. For vertex D(1, -4):
\[
d_D = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}
\]

Next, let's find the distances from the origin to the vertices of the dilated polygon A'B'C'D':

1. For vertex A'(4, -8):
\[
d_{A'} = \sqrt{4^2 + (-8)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}
\]
2. For vertex B'(12, -8):
\[
d_{B'} = \sqrt{12^2 + (-8)^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}
\]
3. For vertex C'(12, -16):
\[
d_{C'} = \sqrt{12^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20
\]
4. For vertex D'(4, -16):
\[
d_{D'} = \sqrt{4^2 + (-16)^2} = \sqrt{16 + 256} = \sqrt{272} = 4\sqrt{17}
\]

To find the scale factor, we can compare one pair of distances. Let's use vertex C and C':

\[
\text{Scale factor} = \frac{d_{C'}}{d_C} = \frac{20}{5} = 4
\]

Thus, the scale factor used to create the image A'B'C'D' from ABCD is:

\[
\boxed{4}
\]