1. The problem asks to find the equations of lines symmetric to the given lines with respect to the line $y=1$. 2. The line of symmetry is $y=1$. For any point $(x,y)$ on the original line, its symmetric point $(x,y')$ satisfies: $$y' = 2 \cdot 1 - y = 2 - y$$ This means the symmetric line's equation can be found by replacing $y$ with $2 - y'$ and solving for $y'$. 3. For each given line $y = mx + b$, substitute $y$ with $2 - y'$: $$2 - y' = mx + b \implies y' = 2 - (mx + b) = -mx + (2 - b)$$ 4. Apply this to each line: (1) $y = 2x + 1$: $$y' = -2x + (2 - 1) = -2x + 1$$ (2) $y = x - 1$: $$y' = -x + (2 - (-1)) = -x + 3$$ (3) $y = 3x - 4$: $$y' = -3x + (2 - (-4)) = -3x + 6$$ (4) $y = -x - 6$: $$y' = x + (2 - (-6)) = x + 8$$ (5) $y = 0.5x + 6$: $$y' = -0.5x + (2 - 6) = -0.5x - 4$$ 5. Therefore, the symmetric lines with respect to $y=1$ are: 1) $y = -2x + 1$ 2) $y = -x + 3$ 3) $y = -3x + 6$ 4) $y = x + 8$ 5) $y = -0.5x - 4$