B' (-2,6) B (-1,3) A' (4,2) A (2,1) C' (-6, -4) C (-3, -2)

Use the image to answer the question. In the image, triangle ABC is dilated with reference to the origin to create triangle A'B'C'. What is the magnitude of dilation?

1 answer

To find the magnitude of dilation from triangle ABC to triangle A'B'C', you can use the formula for the scale factor of dilation, which is given by the ratio of the distances from the origin to corresponding points in the triangles.

Here's how to calculate it step by step:

  1. Calculate the distance from the origin to point A (2, 1): \[ d_A = \sqrt{(2 - 0)^2 + (1 - 0)^2} = \sqrt{4 + 1} = \sqrt{5} \]

  2. Calculate the distance from the origin to point A' (4, 2): \[ d_{A'} = \sqrt{(4 - 0)^2 + (2 - 0)^2} = \sqrt{16 + 4} = \sqrt{20} \]

  3. Calculate the scale factor (magnitude of dilation): The scale factor \( k \) is the ratio of \( d_{A'} \) to \( d_A \): \[ k = \frac{d_{A'}}{d_A} = \frac{\sqrt{20}}{\sqrt{5}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2 \]

  4. Verification with another point (B and B'):

    • Distance to B (-1, 3): \[ d_B = \sqrt{(-1 - 0)^2 + (3 - 0)^2} = \sqrt{1 + 9} = \sqrt{10} \]

    • Distance to B' (-2, 6): \[ d_{B'} = \sqrt{(-2 - 0)^2 + (6 - 0)^2} = \sqrt{4 + 36} = \sqrt{40} \]

    • Scale factor: \[ k = \frac{d_{B'}}{d_B} = \frac{\sqrt{40}}{\sqrt{10}} = \sqrt{\frac{40}{10}} = \sqrt{4} = 2 \]

  5. Final Verification with point C and C':

    • Distance to C (-3, -2): \[ d_C = \sqrt{(-3 - 0)^2 + (-2 - 0)^2} = \sqrt{9 + 4} = \sqrt{13} \]

    • Distance to C' (-6, -4): \[ d_{C'} = \sqrt{(-6 - 0)^2 + (-4 - 0)^2} = \sqrt{36 + 16} = \sqrt{52} \]

    • Scale factor: \[ k = \frac{d_{C'}}{d_C} = \frac{\sqrt{52}}{\sqrt{13}} = \sqrt{\frac{52}{13}} = \sqrt{4} = 2 \]

Since in all cases (points A, B, and C), the scale factor calculated is the same, we confirm that the magnitude of dilation is:

\[ \boxed{2} \]