Question

Triangle ABC with vertices at A(−3, −3), B(3, 3), C(0, 3) is dilated to create triangle A′B′C′ with vertices at A′(−1, −1), B′(1, 1), C′(0, 1). Determine the scale factor used.
1
one half
3
one third

Answer 1

To determine the scale factor of the dilation from triangle ABC to triangle A′B′C′, we can compare the distances between corresponding points.

For points A and A':
- A is at (-3, -3) and A' is at (-1, -1).
- To find the distance from the origin to A, we can use the distance formula:
\[
\text{Distance from origin to } A = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
\]
- The distance from the origin to A' is:
\[
\text{Distance from origin to } A' = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}
\]

Now we calculate the scale factor \( k \):
\[
k = \frac{\text{Distance from origin to A'}}{\text{Distance from origin to A}} = \frac{\sqrt{2}}{3\sqrt{2}} = \frac{1}{3}
\]

To verify, we'll check the other vertices:

For points B and B':
- B is at (3, 3) and B' is at (1, 1).
- The distance from the origin to B is:
\[
\text{Distance from origin to } B = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
\]
- The distance from the origin to B' is:
\[
\text{Distance from origin to } B' = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}
\]

The scale factor for B:
\[
k = \frac{\sqrt{2}}{3\sqrt{2}} = \frac{1}{3}
\]

For points C and C':
- C is at (0, 3) and C' is at (0, 1).
- The distance from the origin to C is:
\[
\text{Distance from origin to } C = \sqrt{0^2 + 3^2} = \sqrt{9} = 3
\]
- The distance from the origin to C' is:
\[
\text{Distance from origin to } C' = \sqrt{0^2 + 1^2} = \sqrt{1} = 1
\]

The scale factor for C:
\[
k = \frac{1}{3}
\]

Since all calculated scale factors are consistent and equal to \(\frac{1}{3}\), we conclude that the scale factor used to dilate triangle ABC to triangle A′B′C′ is:

\[
\boxed{\frac{1}{3}}
\]