Since this holds for all x in the plane, we conclude that (σ_L)^2 is the identity map, so it has order at most two.
Since the only points fixed by σ_L are those on L itself, and there are points not on L, we note that σ_L doesn't have order one. Therefore it has order two.
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Every vector x in the plane uniquely decomposes as
x = y + z
where y lies on L and z is orthogonal to L.
With this decomposition, you can define the reflection by
σ_L(x) = y-z
and therefore, (noting if z is orthogonal to L, then so is -z)
σ_L( σ_L(x) ) = σ_L( y-z ) = σ_L(y+(-z)) = y-(-z) = y+z = x
Since this holds for all x in the plane, we conclude that (σ_L)^2 is the identity map, so it has order at most two.
Since the only points fixed by σ_L are those on L itself, and there are points not on L, we note that σ_L doesn't have order one. Therefore it has order two.