If

Question

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.

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