Question

Triangle ABC is dilated using a magnitude of , which results in triangle A'B'C'.
A=(4,2) B=(-2,6) C=(-6,-4)
A'=(2,1) B'=(-1,3) C=(-3,-2)

Answer 1

To solve this problem, we need to determine the scale factor of the dilation that transforms triangle ABC into triangle A'B'C'.

### Step 1: Calculate the distances of the original triangle ABC
We can find the distances from the origin (0, 0) to the points A, B, and C.

- For point A (4, 2):
\[
d_A = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}
\]

- For point B (-2, 6):
\[
d_B = \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}
\]

- For point C (-6, -4):
\[
d_C = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}
\]

### Step 2: Calculate the distances of the dilated triangle A'B'C'
Now, let’s calculate the distances from the origin to points A', B', and C'.

- For point A' (2, 1):
\[
d_{A'} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}
\]

- For point B' (-1, 3):
\[
d_{B'} = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}
\]

- For point C' (-3, -2):
\[
d_{C'} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]

### Step 3: Determine the scale factor of the dilation
The scale factor \(k\) can be found by dividing the distance of the dilated points by the original points.

- For point A:
\[
k = \frac{d_{A'}}{d_A} = \frac{\sqrt{5}}{2\sqrt{5}} = \frac{1}{2}
\]

- For point B:
\[
k = \frac{d_{B'}}{d_B} = \frac{\sqrt{10}}{2\sqrt{10}} = \frac{1}{2}
\]

- For point C:
\[
k = \frac{d_{C'}}{d_C} = \frac{\sqrt{13}}{2\sqrt{13}} = \frac{1}{2}
\]

### Conclusion
The dilation scale factor that transforms triangle ABC into triangle A'B'C' is \(k = \frac{1}{2}\).