First, let's find the intersection point of the mirror line y = 2x
and line L2 whose equation is y = (1/2)x+2.


For it, we should solve the system of two equations

    y = 2x,

    y = 0.5x + 2.


It quickly reduces to 

    2x = 0.5x + 2,

which gives the solution

    1.5x = 2,  x =  =  = .


Thus the mirror line and L2 intersect at the point  with x-coordinate    and y-coordinate   = .


Again, the intersection point of the mirror line and L2 is the point (x,y) = (,).


It means that line L1 also passes through this point (it is the reason why we found this point).


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    |      At this point, half of the problem is just solved.        |
    |    From this point, the other half of the solution starts.     |
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The mirror line y = 2x has the slope 2;       it means that its angle "a" with x-axis is tan(a) = 2.

Line L2 y = (1/2)x+2 has the slope 1/2 = 0.5; it means that its angle "b" with x-axis is tan(b) = 0.5.


Let's find the angle (a-b) between these lines.  We have 

    tan(a-b) =  =  =  = 0.75 = .



After mirroring about y = 2x, line L2 becomes L1 with the angle with x-axis a+(a-b) = 2a-b.

I want to calculate tan(2a-b), since it gives me the slope of line L1.


I calculate tan(2a) first: it is  tan(2a) =  =  = .

Next, I calculate tan(2a-b).  It is

    tan(2a-b) =  =  = .


In the denominator, we have 1-1 = 0; it means that line L1 is vertical.


Since line L1 is vertical and passes through the point  (,),  its equation is

    x = ,

or

    3x = 4.


ANSWER.  An equation of line L1 in the requested form is 3x = 4, or (which is the same) 3x + 0*y = 4.