for items 7–10, use the figure shown. find the coordinates of the vertices of each image. 7. a reflection across ( y = 0 ), then a translation 1 unit left and 2 units up ( s : (square, square) ) 8. a rotation ( 180^circ ) about the origin, then a reflection across ( x = 0 ) ( r : (square, square) ) 9. a translation 2 units left and 3 units down, then a rotation ( 90^circ ) about the origin ( t : (square, square) ) 10. a translation 3 units right, then a reflection across ( x = 0 ) ( q : (square, square) )

Question

Accepted Answer

### Item 7:
# Explanation:
## Step1: Find original coordinates of S
From the graph, $ S $ is at $ (0, -1) $.
## Step2: Reflect across $ y = 0 $ (x - axis)
The rule for reflection across $ y = 0 $ is $ (x, y) \to (x, -y) $. So $ S $ becomes $ (0, 1) $.
## Step3: Translate 1 unit left and 2 units up
The rule for translation 1 unit left (subtract 1 from x) and 2 units up (add 2 to y) is $ (x, y) \to (x - 1, y + 2) $. So $ (0 - 1, 1 + 2)=(-1, 3) $.
# Answer:
$ S' : (-1, 3) $

### Item 8:
# Explanation:
## Step1: Find original coordinates of R
From the graph, $ R $ is at $ (3, -2) $.
## Step2: Rotate $ 180^\circ $ about the origin
The rule for $ 180^\circ $ rotation about the origin is $ (x, y) \to (-x, -y) $. So $ R $ becomes $ (-3, 2) $.
## Step3: Reflect across $ x = 0 $ (y - axis)
The rule for reflection across $ x = 0 $ is $ (x, y) \to (-x, y) $. So $ (-3, 2) \to (3, 2) $.
# Answer:
$ R' : (3, 2) $

### Item 9:
# Explanation:
## Step1: Find original coordinates of T
From the graph, $ T $ is at $ (-2, 2) $.
## Step2: Translate 2 units left and 3 units down
The rule for translation 2 units left (subtract 2 from x) and 3 units down (subtract 3 from y) is $ (x, y) \to (x - 2, y - 3) $. So $ (-2 - 2, 2 - 3)=(-4, -1) $.
## Step3: Rotate $ 90^\circ $ about the origin (counter - clockwise, assuming standard rotation)
The rule for $ 90^\circ $ counter - clockwise rotation about the origin is $ (x, y) \to (-y, x) $. So $ (-4, -1) \to (1, -4) $. (If clockwise, the rule is $ (x, y)\to(y, -x) $, but usually $ 90^\circ $ rotation in coordinate geometry is counter - clockwise unless specified otherwise. Let's re - check: For $ 90^\circ $ counter - clockwise, $ (x,y)\to(-y,x) $. For $ (-4,-1) $, $ -y = 1 $, $ x=-4 $, so $ (1, - 4) $)
# Answer:
$ T' : (1, -4) $ (If clockwise rotation, the answer would be different. But following standard $ 90^\circ $ counter - clockwise rotation)

### Item 10:
# Explanation:
## Step1: Find original coordinates of Q
From the graph, $ Q $ is at $ (1, 4) $.
## Step2: Translate 3 units right
The rule for translation 3 units right (add 3 to x) is $ (x, y) \to (x + 3, y) $. So $ (1+3, 4)=(4, 4) $.
## Step3: Reflect across $ x = 0 $ (y - axis)
The rule for reflection across $ x = 0 $ is $ (x, y) \to (-x, y) $. So $ (-4, 4) $.
# Answer:
$ Q' : (-4, 4) $

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QUESTION IMAGE

Question

Explanation:

Item 7:

Step1: Find original coordinates of S

Step2: Reflect across \( y = 0 \) (x - axis)

Step3: Translate 1 unit left and 2 units up

Step1: Find original coordinates of R

Step2: Rotate \( 180^\circ \) about the origin

Step3: Reflect across \( x = 0 \) (y - axis)

Step1: Find original coordinates of T

Step2: Translate 2 units left and 3 units down

Step3: Rotate \( 90^\circ \) about the origin (counter - clockwise, assuming standard rotation)

Answer:

Item 8: