Question

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

Answer 1

To determine the magnitude of dilation applied to triangle ABC to result in triangle A'B'C', we can use the coordinates of a corresponding point from triangle ABC and triangle A'B'C'.

The formula for dilation is given by:

\[
A' = k \cdot A
\]

where \(k\) is the scale factor (dilation magnitude), \(A\) is the original point, and \(A'\) is the dilated point. We can use the coordinates of point A and A' for our calculations:

1. **Calculate the dilation magnitude using A and A'**:
- Original A: \((4, 2)\)
- Dilated A': \((2, 1)\)

We can set up the following equations using the dilation formula:

\[
x' = k \cdot x
\]
\[
y' = k \cdot y
\]

From A to A':

\[
2 = k \cdot 4
\]
\[
1 = k \cdot 2
\]

Now, we can solve for \(k\) in each equation:

1. From the first equation:
\[
k = \frac{2}{4} = \frac{1}{2}
\]

2. From the second equation:
\[
k = \frac{1}{2} = \frac{1}{2}
\]

Both calculations give the same result, so we can conclude that the dilation factor \(k\) is \(\frac{1}{2}\).

To further ensure accuracy, we can verify with points B and B', and C and C':

**For B and B':**
- B: \((-2, 6)\)
- B': \((-1, 3)\)

\[
-1 = k \cdot (-2) \quad \Rightarrow \quad k = \frac{1}{2}
\]
\[
3 = k \cdot 6 \quad \Rightarrow \quad k = \frac{1}{2}
\]

**For C and C':**
- C: \((-6, -4)\)
- C': \((-3, -2)\)

\[
-3 = k \cdot (-6) \quad \Rightarrow \quad k = \frac{1}{2}
\]
\[
-2 = k \cdot (-4) \quad \Rightarrow \quad k = \frac{1}{2}
\]

Since all points confirm that the dilation magnitude is \(\frac{1}{2}\), the answer is:

\[
\text{Dilation magnitude} = \frac{1}{2}
\]