Question: 5
8.1.10 Practice 5
Here is triangle XYZ:
Draw these three rotations of triangle XYZ together on the graph. Use the menu at the top of the graph. Select "Polygon" to begin drawing a polygon. A background image of XYZ has been provided.
1. Rotate triangle XYZ 90 degrees clockwise around Z.
2. Rotate triangle XYZ 180 degrees around Z.
3. Rotate triangle XYZ 270 degrees clockwise around Z.
Graph shape: a triangle with vertices at about (4,5), (4,3), and (7,4), with the left side vertical and the point facing right; position_hint=center
whats the answer
1. **Problem Statement:** We have triangle XYZ with vertices approximately at $X(4,5)$, $Y(4,3)$, and $Z(7,4)$. We need to perform three rotations of this triangle around point $Z$:
- Rotate 90 degrees clockwise around $Z$
- Rotate 180 degrees around $Z$
- Rotate 270 degrees clockwise around $Z$
2. **Rotation Formula:** To rotate a point $(x,y)$ around a pivot point $(x_0,y_0)$ by an angle $\theta$ clockwise, use:
$$
\begin{cases}
x' = x_0 + (x - x_0) \cos \theta + (y - y_0) \sin \theta \\
y' = y_0 - (x - x_0) \sin \theta + (y - y_0) \cos \theta
\end{cases}
$$
Here, angles are in radians, but since we have multiples of 90 degrees, we use:
- $90^\circ = \frac{\pi}{2}$
- $180^\circ = \pi$
- $270^\circ = \frac{3\pi}{2}$
3. **Coordinates of $Z$ (pivot):** $Z = (7,4)$
4. **Rotate 90 degrees clockwise ($\theta = 90^\circ$):**
For each point $P=(x,y)$, compute:
$$
x' = 7 + (x - 7) \cos 90^\circ + (y - 4) \sin 90^\circ = 7 + 0 + (y - 4)(1) = 7 + y - 4 = y + 3
$$
$$
y' = 4 - (x - 7) \sin 90^\circ + (y - 4) \cos 90^\circ = 4 - (x - 7)(1) + 0 = 4 - x + 7 = 11 - x
$$
Apply to each vertex:
- $X(4,5)$: $x' = 5 + 3 = 8$, $y' = 11 - 4 = 7$
- $Y(4,3)$: $x' = 3 + 3 = 6$, $y' = 11 - 4 = 7$
- $Z(7,4)$: pivot stays the same $(7,4)$
So rotated points: $X'=(8,7)$, $Y'=(6,7)$, $Z'=(7,4)$
5. **Rotate 180 degrees ($\theta = 180^\circ$):**
$$
x' = 7 + (x - 7) \cos 180^\circ + (y - 4) \sin 180^\circ = 7 + (x - 7)(-1) + (y - 4)(0) = 7 - x + 7 = 14 - x
$$
$$
y' = 4 - (x - 7) \sin 180^\circ + (y - 4) \cos 180^\circ = 4 - (x - 7)(0) + (y - 4)(-1) = 4 - y + 4 = 8 - y
$$
Apply to each vertex:
- $X(4,5)$: $x' = 14 - 4 = 10$, $y' = 8 - 5 = 3$
- $Y(4,3)$: $x' = 14 - 4 = 10$, $y' = 8 - 3 = 5$
- $Z(7,4)$: pivot stays $(7,4)$
Rotated points: $X'=(10,3)$, $Y'=(10,5)$, $Z'=(7,4)$
6. **Rotate 270 degrees clockwise ($\theta = 270^\circ$):**
$$
x' = 7 + (x - 7) \cos 270^\circ + (y - 4) \sin 270^\circ = 7 + (x - 7)(0) + (y - 4)(-1) = 7 - y + 4 = 11 - y
$$
$$
y' = 4 - (x - 7) \sin 270^\circ + (y - 4) \cos 270^\circ = 4 - (x - 7)(-1) + (y - 4)(0) = 4 + x - 7 = x - 3
$$
Apply to each vertex:
- $X(4,5)$: $x' = 11 - 5 = 6$, $y' = 4 - 3 = 1$ (correcting: $y' = 4 + 4 - 7 = 1$ but formula is $x - 3 = 4 - 3 = 1$)
- $Y(4,3)$: $x' = 11 - 3 = 8$, $y' = 4 - 3 = 1$ (correcting: $y' = 4 + 4 - 7 = 1$ but formula is $x - 3 = 4 - 3 = 1$)
- $Z(7,4)$: pivot stays $(7,4)$
Rotated points: $X'=(6,1)$, $Y'=(8,1)$, $Z'=(7,4)$
7. **Summary of rotated triangles:**
- 90° clockwise: $X'(8,7)$, $Y'(6,7)$, $Z'(7,4)$
- 180°: $X'(10,3)$, $Y'(10,5)$, $Z'(7,4)$
- 270° clockwise: $X'(6,1)$, $Y'(8,1)$, $Z'(7,4)$
These are the coordinates of the rotated triangles around point $Z$.