Question

(-4,-9) to (-9,4) rotation

Answer 1

To rotate the point \((-4, -9)\) to the point \((-9, 4)\), we first need to determine the angle and direction of rotation.

### Step-by-step approach to find the rotation:

1. **Determine the center of rotation**:
To find the angle of rotation, we'll first calculate the midpoint of the line segment connecting \((-4, -9)\) and \((-9, 4)\).

\[
\text{Midpoint} = \left( \frac{-4 + (-9)}{2}, \frac{-9 + 4}{2} \right) = \left( \frac{-13}{2}, \frac{-5}{2} \right) = \left(-6.5, -2.5\right)
\]

2. **Find the vector from the midpoint to both points**:
- For \((-4, -9)\) to the midpoint:
\[
V_1 = \left(-4 + 6.5, -9 + 2.5\right) = \left(2.5, -6.5\right)
\]
- For \((-9, 4)\) to the midpoint:
\[
V_2 = \left(-9 + 6.5, 4 + 2.5\right) = \left(-2.5, 6.5\right)
\]

3. **Calculate the angle of rotation**:
The angle \( \theta \) can be found using the dot product of the two vectors \( V_1 \) and \( V_2 \).

\[
V_1 \cdot V_2 = (2.5)(-2.5) + (-6.5)(6.5) = -6.25 - 42.25 = -48.5
\]
The magnitudes of each vector are:
\[
|V_1| = \sqrt{(2.5)^2 + (-6.5)^2} = \sqrt{6.25 + 42.25} = \sqrt{48.5}
\]
\[
|V_2| = \sqrt{(-2.5)^2 + (6.5)^2} = \sqrt{6.25 + 42.25} = \sqrt{48.5}
\]

The cosine of the angle \( \theta \) is:
\[
\cos(\theta) = \frac{V_1 \cdot V_2}{|V_1||V_2|} = \frac{-48.5}{\sqrt{48.5}\sqrt{48.5}} = -1
\]
Thus, \( \theta = 180^\circ \).

4. **Perform the rotation**:
To rotate the point \((-4, -9)\) by \(180^\circ\) around the midpoint \((-6.5, -2.5)\), we use the results from vector addition/subtraction:
\[
R = (-6.5, -2.5) + ((-6.5, -2.5) - (-4, -9))
\]
Calculating this gives:
\[
R = (-6.5, -2.5) + (-2.5, 6.5) = (-9, 4)
\]

The point \((-4, -9)\) is therefore rotated to \((-9, 4)\) around the midpoint \((-6.5, -2.5)\) by \(180^\circ\).