Question

An illustration of a coordinate plane with four quadrants shows the x-axis extending from negative 9 to 9 in increments of 1 and the y-axis extending from negative 11 to 11 in increments of 1. A triangle A B C is plotted on the graph. The triangle A B C has its vertices marked with closed points at A left parenthesis 2 comma negative 1 right parenthesis, B left parenthesis 6 comma negative 3 right parenthesis, and C left parenthesis 1 comma negative 3 right parenthesis.
Reflect triangle ABC
with respect to the x
-axis and the y
-axis. What are the vertices of triangle A′B′C′
?
The vertices of triangle A′B′C′
are A′(−2,1)
, B′(−6,3)
, and C′(−1,3)
.
The vertices of triangle upper A prime upper B prime upper C prime are upper A prime left parenthesis negative 2 comma 1 right parenthesis , upper B prime left parenthesis negative 6 comma 3 right parenthesis , and upper C prime left parenthesis negative 1 comma 3 right parenthesis .

The vertices of triangle A′B′C′
are A′(−2,1)
, B′(−1,3)
, and C′(−6,3)
.
The vertices of triangle upper A prime upper B prime upper C prime are upper A prime left parenthesis negative 2 comma 1 right parenthesis , upper B prime left parenthesis negative 1 comma 3 right parenthesis , and upper C prime left parenthesis negative 6 comma 3 right parenthesis .

The vertices of triangle A′B′C′
are A′(−2,−1)
, B′(−6,−3)
, and C′(−1,−3)
.
The vertices of triangle upper A prime upper B prime upper C prime are upper A prime left parenthesis negative 2 comma negative 1 right parenthesis , upper B prime left parenthesis negative 6 comma negative 3 right parenthesis , and upper C prime left parenthesis negative 1 comma negative 3 right parenthesis .

The vertices of triangle A′B′C′
are A′(2,1)
, B′(6,3)
, and C′(1,3)
.
The vertices of triangle upper A prime upper B prime upper C prime are upper A prime left parenthesis 2 comma 1 right parenthesis , upper B prime left parenthesis 6 comma 3 right parenthesis , and upper C prime left parenthesis 1 comma 3 right parenthesis .

Answer 1

To reflect triangle \( ABC \) with vertices \( A(2, -1) \), \( B(6, -3) \), and \( C(1, -3) \) first over the x-axis, and then over the y-axis, we follow these steps:

1. **Reflecting over the x-axis**:
- The reflection of a point \( (x, y) \) over the x-axis results in \( (x, -y) \).
- Therefore:
- \( A(2, -1) \) becomes \( A'(2, 1) \)
- \( B(6, -3) \) becomes \( B'(6, 3) \)
- \( C(1, -3) \) becomes \( C'(1, 3) \)

2. **Reflecting over the y-axis**:
- The reflection of a point \( (x, y) \) over the y-axis results in \( (-x, y) \).
- Applying this to our new points:
- \( A'(2, 1) \) becomes \( A''(-2, 1) \)
- \( B'(6, 3) \) becomes \( B''(-6, 3) \)
- \( C'(1, 3) \) becomes \( C''(-1, 3) \)

Thus, the vertices of triangle \( A''B''C'' \), which we can denote as triangle \( A'B'C' \) after both reflections, are:
- \( A'(-2, 1) \)
- \( B'(-6, 3) \)
- \( C'(-1, 3) \)

So the correct answer is:
**The vertices of triangle \( A'B'C' \) are \( A'(-2, 1) \), \( B'(-6, 3) \), and \( C'(-1, 3) \)**.