Question

.

An illustration shows a coordinate plane with four quadrants. The x-axis ranges from negative 9 to 9 in one unit increments. The y-axis ranges from negative 11 to 11 in one unit increments. Four points are plotted. A square is graphed using the four plotted points as vertices. Point A is plotted at coordinates left parenthesis 3 comma negative 4 right parenthesis. Point B is plotted at coordinates left parenthesis 7 comma negative 4 right parenthesis. Point C is plotted at coordinates left parenthesis 7 comma negative 8 right parenthesis. Point D is plotted at coordinates left parenthesis 3 comma negative 8 right parenthesis. Points A, B, C, and D are connected by solid lines to form square A B C D.

Reflect square ABCD
first across the x
-axis, and then across the y
-axis to create square A′B′C′D′
. What are the vertices of square A′B′C′D′
?

(1 point)
Responses

Square A′B′C′D′
has vertices A′(−3,−4)
, B′(−7,−4)
, C′(−7,−8)
, and D′(−3,−8)
.
Square upper A prime upper B prime upper C prime upper D prime has vertices upper A prime left parenthesis negative 3 comma negative 4 right parenthesis , upper B prime left parenthesis negative 7 comma negative 4 right parenthesis , upper C prime left parenthesis negative 7 comma negative 8 right parenthesis , and upper D prime left parenthesis negative 3 comma negative 8 right parenthesis .

Square A′B′C′D′
has vertices A′(−3,4)
, B′(−7,4)
, and C′(−7,8)
.
Square upper A prime upper B prime upper C prime upper D prime has vertices upper A prime left parenthesis negative 3 comma 4 right parenthesis , upper B prime left parenthesis negative 7 comma 4 right parenthesis , and upper C prime left parenthesis negative 7 comma 8 right parenthesis .

Square A′B′C′D′
has vertices A′(−3,4)
, B′(−7,4)
, C′(−7,8)
, and D′(−3,8)
.
Square upper A prime upper B prime upper C prime upper D prime has vertices upper A prime left parenthesis negative 3 comma 4 right parenthesis , upper B prime left parenthesis negative 7 comma 4 right parenthesis , upper C prime left parenthesis negative 7 comma 8 right parenthesis , and upper D prime left parenthesis negative 3 comma 8 right parenthesis .

Square A′B′C′D′
has vertices A′(3,4)
, B′(7,4)
, C′(7,8)
, and D′(3,8)
.

Answer 1

To find the vertices of square \( A'B'C'D' \), we will reflect the vertices of square \( ABCD \) first across the x-axis, and then across the y-axis.

1. **Identify the original vertices of square \( ABCD \)**:
- \( A(3, -4) \)
- \( B(7, -4) \)
- \( C(7, -8) \)
- \( D(3, -8) \)

2. **Reflect across the x-axis**:
- To reflect a point \( (x, y) \) across the x-axis, the new coordinates become \( (x, -y) \).
-
- \( A(3, -4) \) becomes \( A'(3, 4) \)
- \( B(7, -4) \) becomes \( B'(7, 4) \)
- \( C(7, -8) \) becomes \( C'(7, 8) \)
- \( D(3, -8) \) becomes \( D'(3, 8) \)

So, after reflecting across the x-axis, we have:
- \( A'(3, 4) \)
- \( B'(7, 4) \)
- \( C'(7, 8) \)
- \( D'(3, 8) \)

3. **Reflect across the y-axis**:
- Now, to reflect a point \( (x, y) \) across the y-axis, the new coordinates become \( (-x, y) \).
-
- \( A'(3, 4) \) becomes \( A''(-3, 4) \)
- \( B'(7, 4) \) becomes \( B''(-7, 4) \)
- \( C'(7, 8) \) becomes \( C''(-7, 8) \)
- \( D'(3, 8) \) becomes \( D''(-3, 8) \)

So, the final vertices after both reflections are:
- \( A''(-3, 4) \)
- \( B''(-7, 4) \)
- \( C''(-7, 8) \)
- \( D''(-3, 8) \)

Thus, the vertices of square \( A'B'C'D' \) are:

- \( A'(-3, 4) \)
- \( B'(-7, 4) \)
- \( C'(-7, 8) \)
- \( D'(-3, 8) \)

In response format, this corresponds to:

"Square \( A'B'C'D' \) has vertices \( A'(-3, 4) \), \( B'(-7, 4) \), \( C'(-7, 8) \), and \( D'(-3, 8) \)."