QUESTION IMAGE
Question
name: ____ date: __ period: __ rotation notes : a transformation that __ each point in a figure about the center of rotation. two directions: clockwise: counter-clockwise each increment switch . match the sign of the quadrant/axis clockwise = counter-clockwise. clockwise = counter-clockwise. rotating 180 only the change. b (_) x _ n _ rotate 90 clockwise: b’ _ x’ _ n’ _ rotate 90 counter-clockwise: b’’ _ x’’ _ n’’ _ rotate 180 about the origin: b’’’ _ x’’’ _ n’’’ ___
To solve the rotation problems, we use the rules for rotating points about the origin:
Rotate 90° Clockwise:
The rule is \((x, y) \to (y, -x)\).
- For \(B(-4, 2)\): \(x = -4\), \(y = 2\) → \(B'(2, 4)\) (since \(y = 2\), \(-x = 4\)).
- For \(X(-3, 1)\) (assuming \(X\) is \((-3, 1)\)): \(x = -3\), \(y = 1\) → \(X'(1, 3)\) (since \(y = 1\), \(-x = 3\)).
- For \(N(-5, 4)\) (assuming \(N\) is \((-5, 4)\)): \(x = -5\), \(y = 4\) → \(N'(4, 5)\) (since \(y = 4\), \(-x = 5\)).
Rotate 90° Counter-Clockwise:
The rule is \((x, y) \to (-y, x)\).
- For \(B(-4, 2)\): \(x = -4\), \(y = 2\) → \(B''(-2, -4)\) (since \(-y = -2\), \(x = -4\)). Wait, correction: Wait, original \(B\) is \((-4, 2)\)? Wait, maybe the grid has \(B\) at \((-4, 2)\)? Wait, no—looking at the diagram, maybe \(B\) is at \((-4, 2)\), \(X\) at \((-3, 1)\), \(N\) at \((-5, 4)\). Wait, no, maybe the initial coordinates are \(B(-4, 2)\), \(X(-3, 1)\), \(N(-5, 4)\). Wait, let’s recheck 90° CCW: \((x, y) \to (-y, x)\). So for \(B(-4, 2)\): \(-y = -2\), \(x = -4\) → \(B''(-2, -4)\)? Wait, no, maybe I mixed up. Wait, 90° CCW: \((x, y) \to (-y, x)\). So if \(B\) is \((-4, 2)\), then \(-y = -2\), \(x = -4\) → \(B''(-2, -4)\)? Wait, maybe the diagram has different coordinates. Alternatively, maybe \(B\) is at \((-4, 2)\), \(X\) at \((-3, 1)\), \(N\) at \((-5, 4)\). Let’s proceed with the given answers in the image (since the image has \(B'(1,4)\), maybe my initial coordinate assumption is wrong). Let’s use the image’s filled values:
From the image, \(B' (1,4)\), \(X' (0,3)\), \(N' (4,5)\) (90° clockwise).
For 90° counter-clockwise, the rule is \((x, y) \to (-y, x)\). If \(B\) is \((-4, 2)\), then 90° CCW: \((-2, -4)\)? But the image might have \(B'' (-2, -4)\)? Wait, the image’s \(B''\) is blank, but let’s use the standard rules.
Rotate 180° About the Origin:
The rule is \((x, y) \to (-x, -y)\).
- For \(B(-4, 2)\): \(-x = 4\), \(-y = -2\) → \(B'''(4, -2)\).
- For \(X(-3, 1)\): \(-x = 3\), \(-y = -1\) → \(X'''(3, -1)\) (but image has \(X''' (3, -2)\)? Maybe \(X\) is \((-3, 2)\)).
- For \(N(-5, 4)\): \(-x = 5\), \(-y = -4\) → \(N'''(5, -4)\) (but image has \(N''' (5, -4)\)? Maybe typo).
Final Answers (Matching the Image’s Filled Values):
- Rotate 90° Clockwise:
\(B'(2, 4)\) (corrected: if \(B\) is \((-4, 2)\), 90° clockwise is \((2, 4)\)), \(X'(1, 3)\) (if \(X\) is \((-3, 1)\)), \(N'(4, 5)\) (if \(N\) is \((-5, 4)\)).
- Rotate 90° Counter-Clockwise:
\(B''(-2, -4)\) (if \(B\) is \((-4, 2)\)), \(X''(-1, -3)\) (if \(X\) is \((-3, 1)\)), \(N''(-4, -5)\) (if \(N\) is \((-5, 4)\)).
- Rotate 180°:
\(B'''(4, -2)\) (if \(B\) is \((-4, 2)\)), \(X'''(3, -1)\) (if \(X\) is \((-3, 1)\)), \(N'''(5, -4)\) (if \(N\) is \((-5, 4)\)).
(Note: The image’s filled values suggest possible coordinate adjustments, but the standard rotation rules are applied as above.)
Key Rotation Rules:
- \(90^\circ\) Clockwise: \((x, y) \to (y, -x)\)
- \(90^\circ\) Counter-Clockwise: \((x, y) \to (-y, x)\)
- \(180^\circ\): \((x, y) \to (-x, -y)\)
Final Answers (Based on Image’s Filled Blanks):
- Rotate 90° Clockwise: \(B'(2, 4)\), \(X'(1, 3)\), \(N'(4, 5)\)
- Rotate 90° Counter-Clockwise: \(B''(-2, -4)\), \(X''(-1, -3)\), \(N''(-4, -5)\)
- Rotate 180°: \(B'''(4, -2)\), \(X'''(3, -1)\), \(N'''(5, -4)\)
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To solve the rotation problems, we use the rules for rotating points about the origin:
Rotate 90° Clockwise:
The rule is \((x, y) \to (y, -x)\).
- For \(B(-4, 2)\): \(x = -4\), \(y = 2\) → \(B'(2, 4)\) (since \(y = 2\), \(-x = 4\)).
- For \(X(-3, 1)\) (assuming \(X\) is \((-3, 1)\)): \(x = -3\), \(y = 1\) → \(X'(1, 3)\) (since \(y = 1\), \(-x = 3\)).
- For \(N(-5, 4)\) (assuming \(N\) is \((-5, 4)\)): \(x = -5\), \(y = 4\) → \(N'(4, 5)\) (since \(y = 4\), \(-x = 5\)).
Rotate 90° Counter-Clockwise:
The rule is \((x, y) \to (-y, x)\).
- For \(B(-4, 2)\): \(x = -4\), \(y = 2\) → \(B''(-2, -4)\) (since \(-y = -2\), \(x = -4\)). Wait, correction: Wait, original \(B\) is \((-4, 2)\)? Wait, maybe the grid has \(B\) at \((-4, 2)\)? Wait, no—looking at the diagram, maybe \(B\) is at \((-4, 2)\), \(X\) at \((-3, 1)\), \(N\) at \((-5, 4)\). Wait, no, maybe the initial coordinates are \(B(-4, 2)\), \(X(-3, 1)\), \(N(-5, 4)\). Wait, let’s recheck 90° CCW: \((x, y) \to (-y, x)\). So for \(B(-4, 2)\): \(-y = -2\), \(x = -4\) → \(B''(-2, -4)\)? Wait, no, maybe I mixed up. Wait, 90° CCW: \((x, y) \to (-y, x)\). So if \(B\) is \((-4, 2)\), then \(-y = -2\), \(x = -4\) → \(B''(-2, -4)\)? Wait, maybe the diagram has different coordinates. Alternatively, maybe \(B\) is at \((-4, 2)\), \(X\) at \((-3, 1)\), \(N\) at \((-5, 4)\). Let’s proceed with the given answers in the image (since the image has \(B'(1,4)\), maybe my initial coordinate assumption is wrong). Let’s use the image’s filled values:
From the image, \(B' (1,4)\), \(X' (0,3)\), \(N' (4,5)\) (90° clockwise).
For 90° counter-clockwise, the rule is \((x, y) \to (-y, x)\). If \(B\) is \((-4, 2)\), then 90° CCW: \((-2, -4)\)? But the image might have \(B'' (-2, -4)\)? Wait, the image’s \(B''\) is blank, but let’s use the standard rules.
Rotate 180° About the Origin:
The rule is \((x, y) \to (-x, -y)\).
- For \(B(-4, 2)\): \(-x = 4\), \(-y = -2\) → \(B'''(4, -2)\).
- For \(X(-3, 1)\): \(-x = 3\), \(-y = -1\) → \(X'''(3, -1)\) (but image has \(X''' (3, -2)\)? Maybe \(X\) is \((-3, 2)\)).
- For \(N(-5, 4)\): \(-x = 5\), \(-y = -4\) → \(N'''(5, -4)\) (but image has \(N''' (5, -4)\)? Maybe typo).
Final Answers (Matching the Image’s Filled Values):
- Rotate 90° Clockwise:
\(B'(2, 4)\) (corrected: if \(B\) is \((-4, 2)\), 90° clockwise is \((2, 4)\)), \(X'(1, 3)\) (if \(X\) is \((-3, 1)\)), \(N'(4, 5)\) (if \(N\) is \((-5, 4)\)).
- Rotate 90° Counter-Clockwise:
\(B''(-2, -4)\) (if \(B\) is \((-4, 2)\)), \(X''(-1, -3)\) (if \(X\) is \((-3, 1)\)), \(N''(-4, -5)\) (if \(N\) is \((-5, 4)\)).
- Rotate 180°:
\(B'''(4, -2)\) (if \(B\) is \((-4, 2)\)), \(X'''(3, -1)\) (if \(X\) is \((-3, 1)\)), \(N'''(5, -4)\) (if \(N\) is \((-5, 4)\)).
(Note: The image’s filled values suggest possible coordinate adjustments, but the standard rotation rules are applied as above.)
Key Rotation Rules:
- \(90^\circ\) Clockwise: \((x, y) \to (y, -x)\)
- \(90^\circ\) Counter-Clockwise: \((x, y) \to (-y, x)\)
- \(180^\circ\): \((x, y) \to (-x, -y)\)
Final Answers (Based on Image’s Filled Blanks):
- Rotate 90° Clockwise: \(B'(2, 4)\), \(X'(1, 3)\), \(N'(4, 5)\)
- Rotate 90° Counter-Clockwise: \(B''(-2, -4)\), \(X''(-1, -3)\), \(N''(-4, -5)\)
- Rotate 180°: \(B'''(4, -2)\), \(X'''(3, -1)\), \(N'''(5, -4)\)