Question
The point Q(−5,−9)
is rotated −270°
about the origin. Select the location of Q′
. (1 point)
Responses
Q′(9,−5)
upper Q prime left parenthesis 9 comma negative 5 right parenthesis
Q′(9,5)
upper Q prime left parenthesis 9 comma 5 right parenthesis
Q′(−5,9)
upper Q prime left parenthesis negative 5 comma 9 right parenthesis
Q′(5,9)
upper Q prime left parenthesis 5 comma 9 right parenthesis
is rotated −270°
about the origin. Select the location of Q′
. (1 point)
Responses
Q′(9,−5)
upper Q prime left parenthesis 9 comma negative 5 right parenthesis
Q′(9,5)
upper Q prime left parenthesis 9 comma 5 right parenthesis
Q′(−5,9)
upper Q prime left parenthesis negative 5 comma 9 right parenthesis
Q′(5,9)
upper Q prime left parenthesis 5 comma 9 right parenthesis
Answers
GPT-4o mini
To rotate a point \((x, y)\) about the origin by an angle θ, you can use the following formulas:
- When rotating counterclockwise by an angle θ:
- \(x' = x \cdot \cos(\theta) - y \cdot \sin(\theta)\)
- \(y' = x \cdot \sin(\theta) + y \cdot \cos(\theta)\)
- When rotating clockwise by an angle θ (which is equivalent to rotating counterclockwise by \(-\theta\)):
- \(x' = x \cdot \cos(-\theta) - y \cdot \sin(-\theta)\)
- \(y' = x \cdot \sin(-\theta) + y \cdot \cos(-\theta)\)
In this case, we need to rotate the point \(Q(-5, -9)\) by \(-270^\circ\). A rotation of \(-270^\circ\) is equivalent to a rotation of \(90^\circ\) clockwise.
Using the rotation formulas for \(90^\circ\) clockwise, we have:
- \(\cos(90^\circ) = 0\)
- \(\sin(90^\circ) = 1\)
Now, apply the formulas:
- \(x' = -5 \cdot 0 - (-9) \cdot 1 = 0 + 9 = 9\)
- \(y' = -5 \cdot 1 + (-9) \cdot 0 = -5 + 0 = -5\)
Thus, after the rotation, point \(Q'\) is at \((9, -5)\).
The correct response is:
**Q′(9,−5)**
- When rotating counterclockwise by an angle θ:
- \(x' = x \cdot \cos(\theta) - y \cdot \sin(\theta)\)
- \(y' = x \cdot \sin(\theta) + y \cdot \cos(\theta)\)
- When rotating clockwise by an angle θ (which is equivalent to rotating counterclockwise by \(-\theta\)):
- \(x' = x \cdot \cos(-\theta) - y \cdot \sin(-\theta)\)
- \(y' = x \cdot \sin(-\theta) + y \cdot \cos(-\theta)\)
In this case, we need to rotate the point \(Q(-5, -9)\) by \(-270^\circ\). A rotation of \(-270^\circ\) is equivalent to a rotation of \(90^\circ\) clockwise.
Using the rotation formulas for \(90^\circ\) clockwise, we have:
- \(\cos(90^\circ) = 0\)
- \(\sin(90^\circ) = 1\)
Now, apply the formulas:
- \(x' = -5 \cdot 0 - (-9) \cdot 1 = 0 + 9 = 9\)
- \(y' = -5 \cdot 1 + (-9) \cdot 0 = -5 + 0 = -5\)
Thus, after the rotation, point \(Q'\) is at \((9, -5)\).
The correct response is:
**Q′(9,−5)**