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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

1 answer

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:

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