Question

1. how many units apart are parallel lines m and n such that ( t_{(0, -12)}(x, y) = (r_n circ r_m)(x, y) )?

Explanation:

Step1: Recall Reflection Over Parallel Lines

When reflecting a point over two parallel lines \( m \) and \( n \) (in that order, \( r_n \circ r_m \)), the composition of the two reflections is a translation. The vector of the translation \( T_{(a,b)} \) has a magnitude related to the distance between the parallel lines. For two parallel lines, the distance between them, \( d \), and the translation vector \( (a, - 12) \) (here the vertical component is - 12, assuming the lines are horizontal or vertical; if they are horizontal, the vertical translation is related to twice the distance between the lines) satisfy the relationship: if the lines are horizontal (parallel to the \( x \)-axis), the vertical component of the translation is \( 2d \) (since reflecting over two parallel horizontal lines will move a point twice the distance between the lines vertically).

Step2: Solve for Distance

Given the translation vector \( T_{(a, - 12)} \), and knowing that for reflections over two parallel lines, the magnitude of the translation in the direction perpendicular to the lines is \( 2d \), where \( d \) is the distance between the lines. Here, the vertical component of the translation is \( - 12 \), so \( 2d=\vert - 12\vert = 12 \). Then, solving for \( d \), we divide both sides by 2: \( d=\frac{12}{2}=6 \).

Answer:

\( 6 \)