Question: Reflect the figure over the line $y = x + 3$. Plot all of the points of the reflected figure. You may click a plotted point to delete it. Graph shapes and positions: A blue diagonal line with positive slope passes through $(0, 3)$ and $(7, 10)$, from bottom-left to top-right, position_hint = center. A pink pentagon is in the top-left, with vertices at approximately $(-8, 6)$, $(-7, 9)$, $(-2, 6)$, $(-4, 2)$, and $(-7, 4)$, position_hint = top-left. A light-blue pentagon is in the lower-right, with vertices at approximately $(-1, -1)$, $(3, 1)$, $(6, -4)$, $(2, -5)$, and $(1, -4)$, position_hint = bottom-right.
1. **State the problem:** We need to reflect a figure over the line $$y = x + 3$$ and plot the reflected points.
2. **Reflection formula:** To reflect a point $(x, y)$ over the line $$y = x + c$$, we use the formula:
$$
\text{Reflected point } (x', y') = \left( y - c, x + c \right)
$$
where $c = 3$ in this case.
3. **Apply the formula to each vertex of the pink pentagon:**
- For $(-8, 6)$:
$$x' = 6 - 3 = 3, \quad y' = -8 + 3 = -5$$
- For $(-7, 9)$:
$$x' = 9 - 3 = 6, \quad y' = -7 + 3 = -4$$
- For $(-2, 6)$:
$$x' = 6 - 3 = 3, \quad y' = -2 + 3 = 1$$
- For $(-4, 2)$:
$$x' = 2 - 3 = -1, \quad y' = -4 + 3 = -1$$
- For $(-7, 4)$:
$$x' = 4 - 3 = 1, \quad y' = -7 + 3 = -4$$
4. **Apply the formula to each vertex of the light-blue pentagon:**
- For $(-1, -1)$:
$$x' = -1 - 3 = -4, \quad y' = -1 + 3 = 2$$
- For $(3, 1)$:
$$x' = 1 - 3 = -2, \quad y' = 3 + 3 = 6$$
- For $(6, -4)$:
$$x' = -4 - 3 = -7, \quad y' = 6 + 3 = 9$$
- For $(2, -5)$:
$$x' = -5 - 3 = -8, \quad y' = 2 + 3 = 5$$
- For $(1, -4)$:
$$x' = -4 - 3 = -7, \quad y' = 1 + 3 = 4$$
5. **Summary:** The reflected pink pentagon vertices are approximately $(3, -5)$, $(6, -4)$, $(3, 1)$, $(-1, -1)$, and $(1, -4)$. The reflected light-blue pentagon vertices are approximately $(-4, 2)$, $(-2, 6)$, $(-7, 9)$, $(-8, 5)$, and $(-7, 4)$.
6. **Plotting:** Plot these reflected points to visualize the reflected figure over the line $y = x + 3$.