QUESTION IMAGE
Question
list the coordinates of the figure.
then list the coordinates after a reflection
over the x-axis.
r: b: g: y:
r’: b’: g’: y’:
list the coordinates of the figure.
then list the coordinates after a reflection
over the y-axis.
r: b: g: y:
r’: b’: g’: y’:
list the coordinates of the figure.
then list the coordinates after a reflection
over the x-axis.
r: b: y:
r’: b’: y’:
list the coordinates of the figure.
then list the coordinates after a reflection
over the y-axis.
r: b: y:
r’: b’: y’:
Top - Left (Reflection over x - axis)
Step 1: Determine original coordinates
- Let's assume each grid square has a side length of 1 unit. For point \(R\), looking at the grid, if we consider the origin (intersection of x - axis and y - axis) as a reference, \(R\) is at \((0,1)\) (since it's on the y - axis, x - coordinate is 0 and y - coordinate is 1).
- Point \(B\): Let's count the grid. If we move left 2 units from \(R\) (x - coordinate: \(0 - 2=- 2\)) and y - coordinate remains 1 (since it's on the same horizontal line as \(R\) initially), so \(B=(-2,1)\)
- Point \(G\): Let's count the grid. From the origin, moving left 10 units (x - coordinate: \(- 10\)) and up 6 units (y - coordinate: \(6\)), so \(G = (-10,6)\)
- Point \(Y\): From the origin, moving left 8 units (x - coordinate: \(-8\)) and up 2 units (y - coordinate: \(2\)), so \(Y=(-8,2)\)
Step 2: Apply reflection over x - axis rule
The rule for reflection over the x - axis is \((x,y)\to(x, - y)\)
- For \(R=(0,1)\), \(R'=(0,-1)\)
- For \(B = (-2,1)\), \(B'=(-2,-1)\)
- For \(G=(-10,6)\), \(G'=(-10,-6)\)
- For \(Y = (-8,2)\), \(Y'=(-8,-2)\)
Top - Right (Reflection over y - axis)
Step 1: Determine original coordinates
- Point \(R\): Let's assume the origin. If we move right 1 unit (x - coordinate: \(1\)) and up 1 unit (y - coordinate: \(1\)), so \(R=(1,1)\)
- Point \(B\): Move right 2 units from \(R\) (x - coordinate: \(1 - 2=-1\))? Wait, no. Wait, looking at the grid, if \(R\) is at \((1,1)\), \(B\) is at \((-1,1)\) (since it's to the left of \(R\) on the same horizontal line).
- Point \(G\): Let's count. From the origin, moving left 10 units (x - coordinate: \(-10\)) and up 6 units (y - coordinate: \(6\)), so \(G = (-10,6)\)
- Point \(Y\): From the origin, moving left 8 units (x - coordinate: \(-8\)) and up 1 unit (y - coordinate: \(1\)), so \(Y=(-8,1)\)
Step 2: Apply reflection over y - axis rule
The rule for reflection over the y - axis is \((x,y)\to(-x,y)\)
- For \(R=(1,1)\), \(R'=(-1,1)\)
- For \(B=(-1,1)\), \(B'=(1,1)\)
- For \(G = (-10,6)\), \(G'=(10,6)\)
- For \(Y=(-8,1)\), \(Y'=(8,1)\)
Bottom - Left (Reflection over x - axis)
Step 1: Determine original coordinates
- Point \(R\): Let's assume the origin. \(R\) is at \((0,-1)\) (on the y - axis, below the x - axis)
- Point \(B\): Moving left 8 units (x - coordinate: \(-8\)) and down 3 units (y - coordinate: \(-4\)), so \(B=(-8,-4)\)
- Point \(Y\): Moving left 6 units (x - coordinate: \(-6\)) and down 2 units (y - coordinate: \(-2\)), so \(Y=(-6,-2)\)
Step 2: Apply reflection over x - axis rule
The rule for reflection over the x - axis is \((x,y)\to(x, - y)\)
- For \(R=(0,-1)\), \(R'=(0,1)\)
- For \(B=(-8,-4)\), \(B'=(-8,4)\)
- For \(Y=(-6,-2)\), \(Y'=(-6,2)\)
Bottom - Right (Reflection over y - axis)
Step 1: Determine original coordinates
- Point \(R\): Let's assume the origin. Moving right 1 unit (x - coordinate: \(1\)) and up 1 unit (y - coordinate: \(1\)), so \(R=(1,1)\)
- Point \(B\): Moving left 6 units (x - coordinate: \(-6\)) and up 5 units (y - coordinate: \(5\)), so \(B=(-6,5)\)
- Point \(Y\): Moving left 8 units (x - coordinate: \(-8\)) and up 0 units (on the x - axis, y - coordinate: \(0\)), so \(Y=(-8,0)\)
Step 2: Apply reflection over y - axis rule
The rule for reflection over the y - axis is \((x,y)\to(-x,y)\)
- For \(R=(1,1)\), \(R'=(-1,1)\)
- For \(B=(-6,5)\), \(B'=(6,5)\)
- For \(Y=(-8,0)\), \(Y'=(8,0)\)
Filling the tables:
Top - Left Table
| \(R\) | \(B\) | \(G\) | \(Y\) |
|---|
| \((0,1)\) | \((-2,1)\)…
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Top - Left (Reflection over x - axis)
Step 1: Determine original coordinates
- Let's assume each grid square has a side length of 1 unit. For point \(R\), looking at the grid, if we consider the origin (intersection of x - axis and y - axis) as a reference, \(R\) is at \((0,1)\) (since it's on the y - axis, x - coordinate is 0 and y - coordinate is 1).
- Point \(B\): Let's count the grid. If we move left 2 units from \(R\) (x - coordinate: \(0 - 2=- 2\)) and y - coordinate remains 1 (since it's on the same horizontal line as \(R\) initially), so \(B=(-2,1)\)
- Point \(G\): Let's count the grid. From the origin, moving left 10 units (x - coordinate: \(- 10\)) and up 6 units (y - coordinate: \(6\)), so \(G = (-10,6)\)
- Point \(Y\): From the origin, moving left 8 units (x - coordinate: \(-8\)) and up 2 units (y - coordinate: \(2\)), so \(Y=(-8,2)\)
Step 2: Apply reflection over x - axis rule
The rule for reflection over the x - axis is \((x,y)\to(x, - y)\)
- For \(R=(0,1)\), \(R'=(0,-1)\)
- For \(B = (-2,1)\), \(B'=(-2,-1)\)
- For \(G=(-10,6)\), \(G'=(-10,-6)\)
- For \(Y = (-8,2)\), \(Y'=(-8,-2)\)
Top - Right (Reflection over y - axis)
Step 1: Determine original coordinates
- Point \(R\): Let's assume the origin. If we move right 1 unit (x - coordinate: \(1\)) and up 1 unit (y - coordinate: \(1\)), so \(R=(1,1)\)
- Point \(B\): Move right 2 units from \(R\) (x - coordinate: \(1 - 2=-1\))? Wait, no. Wait, looking at the grid, if \(R\) is at \((1,1)\), \(B\) is at \((-1,1)\) (since it's to the left of \(R\) on the same horizontal line).
- Point \(G\): Let's count. From the origin, moving left 10 units (x - coordinate: \(-10\)) and up 6 units (y - coordinate: \(6\)), so \(G = (-10,6)\)
- Point \(Y\): From the origin, moving left 8 units (x - coordinate: \(-8\)) and up 1 unit (y - coordinate: \(1\)), so \(Y=(-8,1)\)
Step 2: Apply reflection over y - axis rule
The rule for reflection over the y - axis is \((x,y)\to(-x,y)\)
- For \(R=(1,1)\), \(R'=(-1,1)\)
- For \(B=(-1,1)\), \(B'=(1,1)\)
- For \(G = (-10,6)\), \(G'=(10,6)\)
- For \(Y=(-8,1)\), \(Y'=(8,1)\)
Bottom - Left (Reflection over x - axis)
Step 1: Determine original coordinates
- Point \(R\): Let's assume the origin. \(R\) is at \((0,-1)\) (on the y - axis, below the x - axis)
- Point \(B\): Moving left 8 units (x - coordinate: \(-8\)) and down 3 units (y - coordinate: \(-4\)), so \(B=(-8,-4)\)
- Point \(Y\): Moving left 6 units (x - coordinate: \(-6\)) and down 2 units (y - coordinate: \(-2\)), so \(Y=(-6,-2)\)
Step 2: Apply reflection over x - axis rule
The rule for reflection over the x - axis is \((x,y)\to(x, - y)\)
- For \(R=(0,-1)\), \(R'=(0,1)\)
- For \(B=(-8,-4)\), \(B'=(-8,4)\)
- For \(Y=(-6,-2)\), \(Y'=(-6,2)\)
Bottom - Right (Reflection over y - axis)
Step 1: Determine original coordinates
- Point \(R\): Let's assume the origin. Moving right 1 unit (x - coordinate: \(1\)) and up 1 unit (y - coordinate: \(1\)), so \(R=(1,1)\)
- Point \(B\): Moving left 6 units (x - coordinate: \(-6\)) and up 5 units (y - coordinate: \(5\)), so \(B=(-6,5)\)
- Point \(Y\): Moving left 8 units (x - coordinate: \(-8\)) and up 0 units (on the x - axis, y - coordinate: \(0\)), so \(Y=(-8,0)\)
Step 2: Apply reflection over y - axis rule
The rule for reflection over the y - axis is \((x,y)\to(-x,y)\)
- For \(R=(1,1)\), \(R'=(-1,1)\)
- For \(B=(-6,5)\), \(B'=(6,5)\)
- For \(Y=(-8,0)\), \(Y'=(8,0)\)
Filling the tables:
Top - Left Table
| \(R\) | \(B\) | \(G\) | \(Y\) |
|---|---|---|---|
| \(R'\) | \(B'\) | \(G'\) | \(Y'\) |
| \((0,-1)\) | \((-2,-1)\) | \((-10,-6)\) | \((-8,-2)\) |
Top - Right Table
| \(R\) | \(B\) | \(G\) | \(Y\) |
|---|---|---|---|
| \(R'\) | \(B'\) | \(G'\) | \(Y'\) |
| \((-1,1)\) | \((1,1)\) | \((10,6)\) | \((8,1)\) |
Bottom - Left Table
| \(R\) | \(B\) | \(Y\) |
|---|---|---|
| \(R'\) | \(B'\) | \(Y'\) |
| \((0,1)\) | \((-8,4)\) | \((-6,2)\) |
Bottom - Right Table
| \(R\) | \(B\) | \(Y\) |
|---|---|---|
| \(R'\) | \(B'\) | \(Y'\) |
| \((-1,1)\) | \((6,5)\) | \((8,0)\) |