name
geometry
1 write a single translation that is equivalent to ( t _ { 3, - 1 } ) followed by ( t _ { - 5,5 } ).
2 if the coordinates of point ( p ) are ( ( 2, - 3 ) ), then ( ( r _ { 90 } circ r _ { 180 } ) ( p ) ) is
1) ( ( - 2,3 ) ) 2) ( ( - 2, - 3 ) ) 3) ( ( 3, - 2 ) ) 4) ( ( - 3, - 2 ) )
3 what is the image of ( p ( - 4,6 ) ) under the composite ( r _ { x - 2 } circ r _ { y - ext { axis } } )?
1) ( ( - 8,6 ) ) 2) ( ( 4, - 2 ) ) 3) ( ( 6,0 ) ) 4) ( ( 0,6 ) )
4 find the coordinates of ( r _ { y - ext { axis } } circ r _ { y - pi } ( a ) ) if the coordinates of ( a ) are ( ( 6,1 ) ).
5 find the coordinates of the image of ( ( 2,4 ) ) under the transformation ( r _ { y - ext { axis } } circ t _ { 3, - 5 } ).
6 find the coordinates of point ( n ( - 1,3 ) ) under the composite ( r _ { y - ext { axis } } circ r _ { 90 ^ { circ } } ).

Question

name
geometry
1 write a single translation that is equivalent to ( t _ { 3, - 1 } ) followed by ( t _ { - 5,5 } ).
2 if the coordinates of point ( p ) are ( ( 2, - 3 ) ), then ( ( r _ { 90 } circ r _ { 180 } ) ( p ) ) is
1) ( ( - 2,3 ) ) 2) ( ( - 2, - 3 ) ) 3) ( ( 3, - 2 ) ) 4) ( ( - 3, - 2 ) )
3 what is the image of ( p ( - 4,6 ) ) under the composite ( r _ { x - 2 } circ r _ { y - 	ext { axis } } )?
1) ( ( - 8,6 ) ) 2) ( ( 4, - 2 ) ) 3) ( ( 6,0 ) ) 4) ( ( 0,6 ) )
4 find the coordinates of ( r _ { y - 	ext { axis } } circ r _ { y - pi } ( a ) ) if the coordinates of ( a ) are ( ( 6,1 ) ).
5 find the coordinates of the image of ( ( 2,4 ) ) under the transformation ( r _ { y - 	ext { axis } } circ t _ { 3, - 5 } ).
6 find the coordinates of point ( n ( - 1,3 ) ) under the composite ( r _ { y - 	ext { axis } } circ r _ { 90 ^ { circ } } ).

Accepted Answer

### Question 1
# Explanation:
## Step1: Recall translation composition rule
For translations $ T_{a,b} $ (which translates a point $(x,y)$ to $(x + a,y + b)$) and $ T_{c,d} $, the composition $ T_{c,d}\circ T_{a,b} $ is equivalent to $ T_{a + c,b + d} $. Here we have $ T_{3,-1} $ followed by $ T_{-5,5} $, so we apply the rule for the composition of translations (note that the order is first $ T_{3,-1} $ then $ T_{-5,5} $, so the composition is $ T_{-5,5}\circ T_{3,-1} $).
## Step2: Calculate the new translation vector
For the $ x $-component: $ 3+(-5)=3 - 5=-2 $
For the $ y $-component: $ -1 + 5 = 4 $
So the single translation equivalent is $ T_{-2,4} $
# Answer:
$ T_{-2,4} $

### Question 2
# Explanation:
## Step1: Recall rotation rules
First, recall the rule for $ R_{180^\circ} $: $ R_{180^\circ}(x,y)=(-x,-y) $
Then, recall the rule for $ R_{90^\circ} $: $ R_{90^\circ}(x,y)=(-y,x) $
We need to compute $ (R_{90^\circ}\circ R_{180^\circ})(P) $, which means we first apply $ R_{180^\circ} $ to $ P(2,-3) $, then apply $ R_{90^\circ} $ to the result.
## Step2: Apply $ R_{180^\circ} $ to $ P $
$ R_{180^\circ}(2,-3)=(-2,3) $
## Step3: Apply $ R_{90^\circ} $ to the result
Now apply $ R_{90^\circ} $ to $(-2,3)$. Using $ R_{90^\circ}(x,y)=(-y,x) $, we get $ R_{90^\circ}(-2,3)=(-3,-2) $? Wait, no, wait: Wait, $ R_{90^\circ}(x,y)=(-y,x) $. So for $(x=-2,y = 3)$, $ -y=-3 $, $ x=-2 $? Wait, no, I made a mistake. Wait, $ R_{90^\circ} $ is counterclockwise rotation by $ 90^\circ $, the formula is $ (x,y)\to(-y,x) $. So for point $(-2,3)$, $ x=-2 $, $ y = 3 $. Then $ -y=-3 $, $ x=-2 $? No, wait, no: Wait, let's re - check. The standard $ R_{90^\circ} $ (counterclockwise) is $ (x,y)\to(-y,x) $. So if we have a point $(a,b)$, after $ R_{90^\circ} $ it is $(-b,a)$. So for $(-2,3)$, $ a=-2 $, $ b = 3 $, so $ -b=-3 $, $ a=-2 $? Wait, that can't be. Wait, no, I think I messed up the order of rotation. Wait, the problem is $ (R_{90^\circ}\circ R_{180^\circ})(P) $, which is $ R_{90^\circ}(R_{180^\circ}(P)) $. So first $ R_{180^\circ}(2,-3)=(-2,3) $. Then $ R_{90^\circ}(-2,3) $: using $ (x,y)\to(-y,x) $, so $ x=-2 $, $ y = 3 $, so $ -y=-3 $, $ x=-2 $? No, that's wrong. Wait, no, the formula is $ (x,y)$ rotated $ 90^\circ $ counterclockwise is $ (-y,x) $. So for $(-2,3)$, $ x=-2 $, $ y = 3 $, so $ -y=-3 $, $ x=-2 $? No, that would be $(-3,-2)$? Wait, no, let's take a simple point. Let's take $(1,0)$, $ R_{90^\circ}(1,0)=(0,1) $? Wait, no, $ (1,0) $ rotated $ 90^\circ $ counterclockwise is $ (0,1) $? Wait, no, the correct formula for $ 90^\circ $ counterclockwise rotation is $ (x,y)\to(-y,x) $. So $ (1,0) $: $ -y = 0 $, $ x = 1 $, so $(0,1)$, which is correct. $ (0,1) $ rotated $ 90^\circ $ counterclockwise is $(-1,0)$, using the formula: $ -y=-1 $, $ x = 0 $, so $(-1,0)$, correct. So for $(-2,3)$, $ x=-2 $, $ y = 3 $, so $ -y=-3 $, $ x=-2 $? Wait, that gives $(-3,-2)$? But let's check the options. Wait, maybe I mixed up the order of rotation. Wait, the problem is $ (R_{90^\circ}\circ R_{180^\circ})(P) $, which is $ R_{90^\circ}(R_{180^\circ}(P)) $. Wait, maybe I should recall the formula for $ R_{180^\circ} $ is $(x,y)\to(-x,-y)$, so $ R_{180^\circ}(2,-3)=(-2,3) $. Then $ R_{90^\circ}(-2,3) $: using $ (x,y)\to(-y,x) $, so $ -y=-3 $, $ x=-2 $, so $(-3,-2)$? But let's check the options. Option 4 is $(-3,-2)$. Wait, but let's verify again. Alternatively, maybe the rotation is clockwise? Wait, the notation $ R_{90^\circ} $ usually means counterclockwise, but sometimes clockwise. Wait, the formula for $ 90^\circ $ clockwise rotation is $(x,y)\to(y,-x)$. Let's try that. If $ R_{90^\circ} $ is clockwise, then $ R_{90^\circ}(-2,3)=(3,2) $, which is not in the options. So it must be counterclockwise. So $ R_{90^\circ}(-2,3)=(-3,-2) $, which is option 4.
# Answer:
4) $(-3,-2)$

### Question 3
# Explanation:
## Step1: Recall reflection rules
First, recall the rule for reflection over the $ y $-axis: $ r_{y - \text{axis}}(x,y)=(-x,y) $
Then, recall the rule for reflection over the line $ x = 2 $: $ r_{x = 2}(x,y)=(4 - x,y) $ (since the distance from $ x $ to $ 2 $ is $ |x - 2| $, so the reflected $ x $-coordinate is $ 2+(2 - x)=4 - x $)
We need to compute $ (r_{x = 2}\circ r_{y - \text{axis}})(P) $, which means we first apply $ r_{y - \text{axis}} $ to $ P(-4,6) $, then apply $ r_{x = 2} $ to the result.
## Step2: Apply $ r_{y - \text{axis}} $ to $ P $
$ r_{y - \text{axis}}(-4,6)=(4,6) $
## Step3: Apply $ r_{x = 2} $ to the result
Now apply $ r_{x = 2} $ to $(4,6)$. Using the formula $ r_{x = 2}(x,y)=(4 - x,y) $, for $ x = 4 $, $ 4-4 = 0 $, so the point is $(0,6)$
# Answer:
4) $(0,6)$

### Question 4
# Explanation:
## Step1: Recall reflection rules
First, recall the rule for reflection over the line $ y=x $: $ r_{y = x}(x,y)=(y,x) $
Then, recall the rule for reflection over the $ y $-axis: $ r_{y - \text{axis}}(x,y)=(-x,y) $
We need to compute $ (r_{y - \text{axis}}\circ r_{y = x})(A) $, which means we first apply $ r_{y = x} $ to $ A(6,1) $, then apply $ r_{y - \text{axis}} $ to the result.
## Step2: Apply $ r_{y = x} $ to $ A $
$ r_{y = x}(6,1)=(1,6) $
## Step3: Apply $ r_{y - \text{axis}} $ to the result
$ r_{y - \text{axis}}(1,6)=(-1,6) $
Wait, wait, the problem is $ r_{y - \text{axis}}\circ r_{y = x}(A) $, so first $ r_{y = x}(6,1)=(1,6) $, then $ r_{y - \text{axis}}(1,6)=(-1,6) $? But let's check the problem statement again. Wait, the problem says $ r_{y - \text{axis}}\circ r_{y = x}(A) $ with $ A(6,1) $. Wait, maybe I misread the reflection line. Wait, the problem says $ r_{y - \text{axis}}\circ r_{y = \pi} $? No, it's $ r_{y - \text{axis}}\circ r_{y = x} $? Wait, the original problem says " $ r_{y - \text{axis}}\circ r_{y = \pi}(A) $"? No, looking back, the user's image shows " $ r_{y - \text{axis}}\circ r_{y = x}(A) $"? Wait, the user's question 4: "Find the coordinates of $ r_{y - \text{axis}}\circ r_{y = x}(A) $ if the coordinates of $ A $ are $ (6,1) $." Wait, maybe it's a typo, and it's $ r_{y = x} $. So proceeding with that, we have:
After $ r_{y = x}(6,1)=(1,6) $, then $ r_{y - \text{axis}}(1,6)=(-1,6) $. But let's check the reflection over $ y = x $ and $ y $-axis. Alternatively, maybe the second reflection is over $ y = 2 $? No, the problem says $ r_{y - \text{axis}}\circ r_{y = x} $. Wait, maybe I made a mistake. Wait, the formula for reflection over $ y = x $ is $ (x,y)\to(y,x) $, so $ (6,1)\to(1,6) $. Then reflection over $ y $-axis: $ (1,6)\to(-1,6) $. But let's see if that makes sense. Alternatively, maybe the problem is $ r_{y - \text{axis}}\circ r_{x = y} $, same as $ r_{y = x} $. So the answer would be $(-1,6)$. But let's check again.
Wait, maybe the second reflection is over $ y = \pi $? No, that doesn't make sense. It's probably a typo, and it's $ y = x $. So we proceed with that.
# Answer:
$(-1,6)$ (assuming the second reflection is over $ y = x $; if there is a different intended reflection, the answer may change, but based on the given problem and common notations, this is the solution)

### Question 5
# Explanation:
## Step1: Recall transformation composition order
The transformation is $ r_{y - \text{axis}}\circ T_{3,-5} $, which means we first apply the translation $ T_{3,-5} $ to the point $(2,4)$, then apply the reflection over the $ y $-axis to the result.
## Step2: Apply the translation $ T_{3,-5} $
For a translation $ T_{a,b} $, the point $(x,y)$ becomes $(x + a,y + b)$. So for $ T_{3,-5} $ and point $(2,4)$:
$ x $-coordinate: $ 2+3 = 5 $
$ y $-coordinate: $ 4+( - 5)=4 - 5=-1 $
So after translation, the point is $(5,-1)$
## Step3: Apply the reflection over the $ y $-axis
The rule for reflection over the $ y $-axis is $ r_{y - \text{axis}}(x,y)=(-x,y) $. So for the point $(5,-1)$:
$ x $-coordinate: $ - 5 $
$ y $-coordinate: $ -1 $
So the image is $(-5,-1)$
# Answer:
$(-5,-1)$

### Question 6
# Explanation:
## Step1: Recall transformation composition order
The transformation is $ r_{y - \text{axis}}\circ R_{90^\circ} $, which means we first apply the rotation $ R_{90^\circ} $ to the point $ N(-1,3) $, then apply the reflection over the $ y $-axis to the result.
## Step2: Apply the rotation $ R_{90^\circ} $
Recall the rule for $ 90^\circ $ counterclockwise rotation: $ R_{90^\circ}(x,y)=(-y,x) $. For the point $ (-1,3) $:
$ x=-1 $, $ y = 3 $
So $ R_{90^\circ}(-1,3)=(-3,-1) $ (using $ -y=-3 $, $ x=-1 $)
## Step3: Apply the reflection over the $ y $-axis
The rule for reflection over the $ y $-axis is $ r_{y - \text{axis}}(x,y)=(-x,y) $. For the point $ (-3,-1) $:
$ x=-3 $, so $ -x = 3 $
$ y=-1 $
So the image is $(3,-1)$
Wait, but let's verify the rotation again. If $ R_{90^\circ} $ is clockwise, the formula is $ (x,y)\to(y,-x) $. Then $ R_{90^\circ}(-1,3)=(3,1) $, then reflection over $ y $-axis is $(-3,1)$, which is not in the expected form. So we assume counterclockwise rotation. So $ R_{90^\circ}(-1,3)=(-3,-1) $, then reflection over $ y $-axis is $(3,-1)$.
# Answer:
$(3,-1)$

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Recall translation composition rule

Step2: Calculate the new translation vector

Step1: Recall rotation rules

Step2: Apply \( R_{180^\circ} \) to \( P \)

Step3: Apply \( R_{90^\circ} \) to the result

Step1: Recall reflection rules

Step2: Apply \( r_{y - \text{axis}} \) to \( P \)

Step3: Apply \( r_{x = 2} \) to the result

Answer:

Question 2