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 shape: a quadrilateral with vertices at $(-9, 4)$, $(-8, 2)$, $(-5, 1)$, and $(-1, 8)$, reflected over the line $y = x + 1$; position_hint = center
1. **State the problem:**
We need to reflect a quadrilateral with vertices at $(-9, 4)$, $(-8, 2)$, $(-5, 1)$, and $(-1, 8)$ over the line $$y = x + 1$$.
2. **Formula and approach:**
To reflect a point $(x, y)$ over the line $$y = x + 1$$, we use the following steps:
- Translate the line to the origin by subtracting 1 from the $y$-coordinate: consider the line $$y' = x$$ where $y' = y - 1$.
- Reflect the point $(x, y')$ over the line $$y' = x$$ by swapping coordinates: $(x, y') \to (y', x)$.
- Translate back by adding 1 to the $y$-coordinate: $(y', x) \to (y', x + 1)$.
So the reflection of $(x, y)$ over $$y = x + 1$$ is:
$$
(x, y) \to (y - 1, x + 1)
$$
3. **Apply the formula to each vertex:**
- For $(-9, 4)$:
$$
(x, y) = (-9, 4) \to (4 - 1, -9 + 1) = (3, -8)
$$
- For $(-8, 2)$:
$$
(x, y) = (-8, 2) \to (2 - 1, -8 + 1) = (1, -7)
$$
- For $(-5, 1)$:
$$
(x, y) = (-5, 1) \to (1 - 1, -5 + 1) = (0, -4)
$$
- For $(-1, 8)$:
$$
(x, y) = (-1, 8) \to (8 - 1, -1 + 1) = (7, 0)
$$
4. **Final reflected vertices:**
The reflected quadrilateral has vertices at:
$$
(3, -8), (1, -7), (0, -4), (7, 0)
$$
These points can be plotted to visualize the reflected figure.