Question: Reflect the figure over the line $y = x - 1$. Plot all of the points of the reflected figure. You may click a plotted point to delete it. Graph: The line $y = x - 1$ is a diagonal line with positive slope crossing the y-axis at $-1$ and the x-axis at $1$, extending from bottom-left to top-right; the polygon is in the lower-right quadrant with vertices at $(1,-1)$, $(1,-5)$, $(8,-8)$, $(9,-3)$, and $(5,-3)$.
1. **State the problem:** We need to reflect a polygon with vertices $(1,-1)$, $(1,-5)$, $(8,-8)$, $(9,-3)$, and $(5,-3)$ over the line $y = x - 1$.
2. **Reflection formula:** To reflect a point $(x,y)$ over the line $y = x - 1$, first translate the line to $y = x$ by shifting coordinates by $-1$ in $y$-direction: define $y' = y + 1$. Then reflect over $y = x$ by swapping coordinates: $(x,y') \to (y',x)$. Finally, translate back by subtracting $1$ from $y$ coordinate: $(y',x) \to (y' , x - 1)$. So the reflected point $(x_r,y_r)$ is:
$$
(x_r,y_r) = (y + 1, x - 1)
$$
3. **Apply reflection to each vertex:**
- For $(1,-1)$:
$$
x_r = -1 + 1 = 0, \quad y_r = 1 - 1 = 0
$$
Reflected point: $(0,0)$
- For $(1,-5)$:
$$
x_r = -5 + 1 = -4, \quad y_r = 1 - 1 = 0
$$
Reflected point: $(-4,0)$
- For $(8,-8)$:
$$
x_r = -8 + 1 = -7, \quad y_r = 8 - 1 = 7
$$
Reflected point: $(-7,7)$
- For $(9,-3)$:
$$
x_r = -3 + 1 = -2, \quad y_r = 9 - 1 = 8
$$
Reflected point: $(-2,8)$
- For $(5,-3)$:
$$
x_r = -3 + 1 = -2, \quad y_r = 5 - 1 = 4
$$
Reflected point: $(-2,4)$
4. **Summary:** The reflected polygon vertices are $(0,0)$, $(-4,0)$, $(-7,7)$, $(-2,8)$, and $(-2,4)$.
5. **Plotting:** Plot these points and connect them in the same order to visualize the reflected polygon.
This completes the reflection over the line $y = x - 1$.