Question

given \\( \triangle xyz \\), line \\( n \\) with equation \\( x=-3 \\), and line \\( p \\) with equation \\( x = 4 \\), write a translation that is equivalent to \\( r_{n}\circ r_{p} \\).

Explanation:

Step1: Calculate the distance between the two lines

The distance \(d\) between two vertical lines \(x = a\) and \(x = b\) is given by \(|a - b|\). Here, \(a=-3\) and \(b = 4\), so \(d=| - 3-4|=7\).
Since we are doing \(R_{n}\circ R_{p}\) (reflection over \(p\) first then over \(n\)), the translation is to the left.

Step2: Write the translation vector

A translation in the coordinate plane is of the form \((x,y)\to(x - k,y)\) for a horizontal translation. Since the distance between the lines is \(7\), the translation vector is \((x,y)\to(x-14,y)\). The reason is that when you reflect over two parallel lines, the translation distance is \(2\times\) the distance between the lines. The formula for the translation \(\vec{v}\) when reflecting over two parallel vertical lines \(x = x_1\) and \(x=x_2\) (and the order of reflections \(R_{x_1}\circ R_{x_2}\)) is \(\vec{v}=(2(x_1 - x_2),0)\). Substituting \(x_1=-3\) and \(x_2 = 4\), we get \(\vec{v}=(2(-3 - 4),0)=(-14,0)\)

Answer:

\((x,y)\to(x - 14,y)\)