Question

If
𝐴
𝐵
‾

=
AB
=
15
15 inches then find the distance of
𝐴
′
𝐵
′
‾
A
′
B
′

after a reflection over the x-axis and a translation of
(
𝑥
+
2
,

𝑦
−
3
)
(x+2, y−3) Click Here for Help Video




𝐴
′
𝐵
′
‾

=
A
′
B
′

=

Answer 1

To find the distance \( A'B' \) after a reflection over the x-axis and a translation, let's break down the steps.

1. **Reflection over the x-axis**:
- When a point \( (x, y) \) is reflected over the x-axis, it transforms to \( (x, -y) \).
- Therefore, if \( A \) has coordinates \( (x_A, y_A) \) and \( B \) has coordinates \( (x_B, y_B) \), after reflection:
- \( A' = (x_A, -y_A) \)
- \( B' = (x_B, -y_B) \)

2. **Translation**:
- The translation is given as \( (x + 2, y - 3) \), which means we add 2 to the x-coordinates and subtract 3 from the y-coordinates of the reflected points:
- After translation:
- \( A' = (x_A + 2, -y_A - 3) \)
- \( B' = (x_B + 2, -y_B - 3) \)

3. **Calculating the distance \( A'B' \)**:
- The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
- For our points \( A' \) and \( B' \):
- \( x_1 = x_A + 2 \)
- \( y_1 = -y_A - 3 \)
- \( x_2 = x_B + 2 \)
- \( y_2 = -y_B - 3 \)

- The distance \( A'B' \) becomes:
\[
A'B' = \sqrt{((x_B + 2) - (x_A + 2))^2 + ((-y_B - 3) - (-y_A - 3))^2}
\]
- Simplifying:
\[
A'B' = \sqrt{(x_B - x_A)^2 + (-y_B + y_A)^2}
\]
- We know that the original distance \( AB \) is 15 inches:
\[
AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = 15
\]
- Therefore, the distance \( A'B' \) is:
\[
A'B' = \sqrt{(x_B - x_A)^2 + (-(y_B - y_A))^2} = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = 15
\]

Thus, the distance after the reflection and translation is also 15 inches.

\[
\boxed{15 \text{ inches}}
\]