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To find the image of point Q after a rotation of [tex]\(-45^\circ\)[/tex] about the origin, we need to follow the steps involving the rotation formula for 2D coordinates. Given a point [tex]\((x, y)\)[/tex], the coordinates [tex]\((x', y')\)[/tex] after rotation by an angle [tex]\(\theta\)[/tex] are given by:
[tex]\[ x' = x \cos(\theta) - y \sin(\theta) \][/tex]
[tex]\[ y' = x \sin(\theta) + y \cos(\theta) \][/tex]
### Step-by-Step Solution:
1. Identify the coordinates of point Q:
Let’s say point Q is located at [tex]\((10, 0)\)[/tex].
2. Determine the rotation angle in degrees:
The rotation angle is [tex]\(-45^\circ\)[/tex].
3. Convert the angle from degrees to radians:
Since trigonometric functions in rotation formulae use radians, we convert [tex]\(-45^\circ\)[/tex] to radians.
[tex]\[ \theta = -45^\circ = -\frac{\pi}{4} \text{ radians} \][/tex]
4. Apply the rotation formulas to get the new coordinates:
Using the formulas:
[tex]\[ x' = 10 \cos\left(-\frac{\pi}{4}\right) - 0 \sin\left(-\frac{\pi}{4}\right) \][/tex]
[tex]\[ y' = 10 \sin\left(-\frac{\pi}{4}\right) + 0 \cos\left(-\frac{\pi}{4}\right) \][/tex]
Simplify the calculations:
[tex]\[ \cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \][/tex]
[tex]\[ \sin\left(-\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \][/tex]
5. Substitute the values into the formulas:
[tex]\[ x' = 10 \cdot \frac{\sqrt{2}}{2} - 0 \cdot \left(-\frac{\sqrt{2}}{2}\right) = 10 \cdot \frac{\sqrt{2}}{2} = 5\sqrt{2} \][/tex]
[tex]\[ y' = 10 \cdot \left(-\frac{\sqrt{2}}{2}\right) + 0 \cdot \frac{\sqrt{2}}{2} = -10 \cdot \frac{\sqrt{2}}{2} = -5\sqrt{2} \][/tex]
6. Approximate the numeric values:
[tex]\[ 5\sqrt{2} \approx 5 \cdot 1.414 \approx 7.071 \][/tex]
[tex]\[ -5\sqrt{2} \approx -5 \cdot 1.414 \approx -7.071 \][/tex]
Hence, the coordinates of the image of point Q after a [tex]\(-45^\circ\)[/tex] rotation about the origin are approximately [tex]\((7.071, -7.071)\)[/tex].
Thus, the correct point resulting from the rotation is closest to:
[tex]\[ (7.0710678118654755, -7.071067811865475) \][/tex]
[tex]\[ x' = x \cos(\theta) - y \sin(\theta) \][/tex]
[tex]\[ y' = x \sin(\theta) + y \cos(\theta) \][/tex]
### Step-by-Step Solution:
1. Identify the coordinates of point Q:
Let’s say point Q is located at [tex]\((10, 0)\)[/tex].
2. Determine the rotation angle in degrees:
The rotation angle is [tex]\(-45^\circ\)[/tex].
3. Convert the angle from degrees to radians:
Since trigonometric functions in rotation formulae use radians, we convert [tex]\(-45^\circ\)[/tex] to radians.
[tex]\[ \theta = -45^\circ = -\frac{\pi}{4} \text{ radians} \][/tex]
4. Apply the rotation formulas to get the new coordinates:
Using the formulas:
[tex]\[ x' = 10 \cos\left(-\frac{\pi}{4}\right) - 0 \sin\left(-\frac{\pi}{4}\right) \][/tex]
[tex]\[ y' = 10 \sin\left(-\frac{\pi}{4}\right) + 0 \cos\left(-\frac{\pi}{4}\right) \][/tex]
Simplify the calculations:
[tex]\[ \cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \][/tex]
[tex]\[ \sin\left(-\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \][/tex]
5. Substitute the values into the formulas:
[tex]\[ x' = 10 \cdot \frac{\sqrt{2}}{2} - 0 \cdot \left(-\frac{\sqrt{2}}{2}\right) = 10 \cdot \frac{\sqrt{2}}{2} = 5\sqrt{2} \][/tex]
[tex]\[ y' = 10 \cdot \left(-\frac{\sqrt{2}}{2}\right) + 0 \cdot \frac{\sqrt{2}}{2} = -10 \cdot \frac{\sqrt{2}}{2} = -5\sqrt{2} \][/tex]
6. Approximate the numeric values:
[tex]\[ 5\sqrt{2} \approx 5 \cdot 1.414 \approx 7.071 \][/tex]
[tex]\[ -5\sqrt{2} \approx -5 \cdot 1.414 \approx -7.071 \][/tex]
Hence, the coordinates of the image of point Q after a [tex]\(-45^\circ\)[/tex] rotation about the origin are approximately [tex]\((7.071, -7.071)\)[/tex].
Thus, the correct point resulting from the rotation is closest to:
[tex]\[ (7.0710678118654755, -7.071067811865475) \][/tex]
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