1) Find the image of the line \( l: y=2 x_{t} \) ) after reflection along the line \( L: y=2 x+6 \)

We are given the line \[ l: y=2x \] and the reflection line \[ L: y=2x+6. \] Since both lines have the same slope, they are parallel. In the case of parallel lines, the reflection of a line across another line will be a line with the same slope and with its vertical displacement “symmetrical” relative to the mirror line. 1. Write the image of \(l\) as a line with the same slope: \[ r: y=2x+b, \] where \(b\) is the new \(y\)-intercept to be determined. 2. The distance between two parallel lines of the form \[ y=2x+C \] and \[ y=2x+D \] is given by \[ \frac{|C-D|}{\sqrt{1+2^2}}=\frac{|C-D|}{\sqrt{5}}. \] 3. Compute the distance between \(l\) and \(L\). Their intercepts are \(0\) (for \(l\)) and \(6\) (for \(L\)), hence \[ d = \frac{|6-0|}{\sqrt{5}} = \frac{6}{\sqrt{5}}. \] 4. Since \(L\) is the mirror, the distance from \(L\) to the image \(r\) must also be \(\frac{6}{\sqrt{5}}\). So we require \[ \frac{|b-6|}{\sqrt{5}}=\frac{6}{\sqrt{5}}. \] Cancel \(\frac{1}{\sqrt{5}}\) on both sides: \[ |b-6|=6. \] 5. This equation gives two possibilities: - \(b-6=6\) which yields \(b=12\). - \(b-6=-6\) which gives \(b=0\). However, \(b=0\) corresponds to the original line \(l\) (which we are reflecting), so we select the other value that gives the new line on the opposite side of \(L\). 6. Thus, the image of the line \(l\) after reflection along \(L\) is: \[ r: y=2x+12. \]