1. **State the problem:** We need to find the image of triangle $\triangle EFG$ after reflecting it over the line $y=1$. The vertices are $E(0,5)$, $F(1,5)$, and $G(1,3)$. 2. **Reflection formula:** When reflecting a point $(x,y)$ over the horizontal line $y=k$, the image point $(x',y')$ is given by: $$y' = 2k - y$$ and $x' = x$ (since reflection is vertical). 3. **Apply the formula to each vertex:** - For $E(0,5)$: $$y'_E = 2(1) - 5 = 2 - 5 = -3$$ So, $E' = (0, -3)$. - For $F(1,5)$: $$y'_F = 2(1) - 5 = -3$$ So, $F' = (1, -3)$. - For $G(1,3)$: $$y'_G = 2(1) - 3 = 2 - 3 = -1$$ So, $G' = (1, -1)$. 4. **Final answer:** The reflected triangle $\triangle E'F'G'$ has vertices: $$E'(0, -3), F'(1, -3), G'(1, -1)$$ This completes the reflection over the line $y=1$.