Question

in exercises 1–4, trace the polygon and point p. then draw a rotation of the polygon about point p using the given number of degrees. example 1

1. 90°
2. 90°
3. 180°
4. 180°

in exercises 5–8, graph the polygon with the given vertices and its image after a rotation of the given number of degrees about the origin. example 2

1. a(-3, 2), b(2, 4), c(3, 1); 90°
2. d(-3, -1), e(-1, 2), f(4, -2); 180°
3. j(1, 4), k(5, 5), l(7, 2), m(2, 2); 180°
4. q(-6, -3), r(-5, 0), s(-3, 0), t(-1, -3); 270°

error analysis in exercises 9 and 10, the endpoints of \\(\overline{cd}\\) are c(-1, 1) and d(2, 3). describe and correct the error in finding the coordinates of the endpoints of the image after a rotation of 270° about the origin.

1. \\(c(-1, 1) \to c(-1, -1)\\), \\(d(2, 3) \to d(2, -3)\\)
2. \\(c(-1, 1) \to c(1, -1)\\), \\(d(2, 3) \to d(3, 2)\\)

in exercises 11–14, graph \\(\overline{xy}\\) with endpoints x(-3, 1) and y(4, -5) and its image after the composition. example 3

1. translation: \\((x, y) \to (x, y + 2)\\), rotation: 90° about the origin
2. rotation: 180° about the origin, translation: \\((x, y) \to (x - 1, y + 1)\\)
3. rotation: 270° about the origin, reflection: in the y-axis
4. reflection: in the line \\(y = x\\), rotation: 90° about the origin

in exercises 15 and 16, graph \\(\triangle lmn\\) with l(1, 6), m(-2, 4), and n(3, 2) and its image after the composition.

1. rotation: 90° about the origin, translation: \\((x, y) \to (x - 3, y + 2)\\)
2. reflection: in the x-axis, rotation: 270° about the origin

in exercises 17–20, determine whether the figure has rotational symmetry. if so, describe any rotations that map the figure onto itself. example 4
17.
18.
19.
20.

college prep in exercises 21–24, select all angle measures of a rotation about the center of the regular polygon that map the polygon onto itself.
a 30°
b 45°
c 60°
d
e 90°
f 120°
g 144°
h
21.
22.
23.
24.

chapter 4 transformations

Explanation:

Step1: Recall Rotation Rule for \(270^\circ\)

The rule for rotating a point \((x,y)\) \(270^\circ\) counterclockwise about the origin is \((x,y)\to(y, -x)\) (or equivalently, \(90^\circ\) clockwise: \((x,y)\to(y, -x)\)).

Step2: Apply Rule to Point \(C(-1,1)\)

For \(C(-1,1)\), using the rule \((x,y)\to(y, -x)\):
\(x = -1\), \(y = 1\)
So \(C'\) should be \((1, -(-1))=(1, 1)\)? Wait, no, wait: Wait, \(270^\circ\) counterclockwise: the formula is \((x,y)\to(y, -x)\). Wait, let's re - check. The standard rotation rules:
- \(90^\circ\) counterclockwise: \((x,y)\to(-y,x)\)
- \(180^\circ\) counterclockwise: \((x,y)\to(-x,-y)\)
- \(270^\circ\) counterclockwise: \((x,y)\to(y, -x)\) (which is the same as \(90^\circ\) clockwise)

So for \(C(-1,1)\):
\(x=-1\), \(y = 1\)
Applying \(270^\circ\) counterclockwise rule \((x,y)\to(y, -x)\):
\(y = 1\), \(-x=-(-1)=1\), so \(C'=(1,1)\)? Wait, no, that's not right. Wait, maybe I mixed up. Let's use the matrix method. The rotation matrix for \(270^\circ\) counterclockwise is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). So if we have a point \(\begin{pmatrix}x\\y\end{pmatrix}\), the rotated point is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\-x\end{pmatrix}\).

So for \(C(-1,1)\):
\(x=-1\), \(y = 1\)
Rotated point \(C'\): \(\begin{pmatrix}1\\-(-1)\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}\)? Wait, no, \(-x=-(-1) = 1\), so \((y,-x)=(1,1)\).

For point \(D(2,3)\):
Using the rule \((x,y)\to(y, -x)\)
\(x = 2\), \(y = 3\)
So \(D'\) should be \((3,-2)\)

Wait, the original error in problem 9: The student thought it was a reflection over the x - axis (rule \((x,y)\to(x,-y)\)), which is wrong. The correct rule for \(270^\circ\) rotation is different.

For problem 9:
The student's mistake: The student applied the reflection over the x - axis rule \((x,y)\to(x,-y)\) instead of the \(270^\circ\) rotation rule.
To correct:
For \(C(-1,1)\), using \(270^\circ\) rotation rule \((x,y)\to(y, -x)\): \(C(-1,1)\to(1,1)\) (Wait, no, wait, I think I made a mistake in the formula. Let's use the correct formula for \(270^\circ\) counterclockwise: \((x,y)\to(y, -x)\) is incorrect. Let's use the correct formula from the rotation matrix. The rotation matrix for \(\theta = 270^\circ=\frac{3\pi}{2}\) radians is \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). So when we multiply \(\begin{pmatrix}x\\y\end{pmatrix}\) by this matrix, we get \(\begin{pmatrix}y\\-x\end{pmatrix}\).

Wait, for \(C(-1,1)\): \(x=-1\), \(y = 1\). Then \(y = 1\), \(-x=-(-1)=1\), so \(C'=(1,1)\). For \(D(2,3)\): \(y = 3\), \(-x=-2\), so \(D'=(3,-2)\).

The original work in problem 9 used the reflection over x - axis rule \((x,y)\to(x,-y)\) (got \(C'(-1,-1)\) and \(D'(2,-3)\)), which is wrong. The correct transformation for \(270^\circ\) rotation is different.

For problem 10:
The student used the rule for reflection over \(y = x\) (which is \((x,y)\to(y,x)\)) for \(D(2,3)\to(3,2)\) and for \(C(-1,1)\to(1,-1)\) (which is a reflection over \(y=-x\) maybe? No, the rule for \(270^\circ\) rotation is not the reflection or the \(y = x\) reflection. The student's error is using the wrong transformation rule (reflection or \(90^\circ\) rotation in the wrong direction or something else) instead of the \(270^\circ\) rotation rule.

Let's re - do problem 9:

The correct rule for \(270^\circ\) counterclockwise rotation about the origin is \((x,y)\to(y, -x)\). Wait, no, let's check with an example. Take the point \((1,0)\). Rotating \(270^\circ\) counterclockwise about the origin should take it to \((0,1)\)? Wait, no. Wait, \((1,0)\) rotated \(90^\circ\) counterclockwise is \((0,1)\), \(180^\circ\) is \((-1,0)\), \(270^\circ\) is \((0,-1)\). Using the formula \((x,y)\to(y, -x)\) for \((1,0)\): \((0, - 1)\), which is correct. For \((0,1)\) rotated \(270^\circ\) counterclockwise: \((1,0)\), using the formula \((0,1)\to(1,0)\) (since \(y = 1\), \(-x = 0\)), which is correct.

So for \(C(-1,1)\):
\(x=-1\), \(y = 1\)
Applying \((x,y)\to(y, -x)\): \(y = 1\), \(-x=-(-1)=1\), so \(C'=(1,1)\)
For \(D(2,3)\):
\(x = 2\), \(y = 3\)
Applying \((x,y)\to(y, -x)\): \(y = 3\), \(-x=-2\), so \(D'=(3,-2)\)

The original work in problem 9 used the reflection over x - axis (\((x,y)\to(x,-y)\)) which is incorrect. The correct transformation for \(270^\circ\) rotation is as above.

Answer:

For problem 9:

• Error Description: The student used the rule for reflecting a point over the \(x\) - axis \((x,y)\to(x, - y)\) instead of the rule for rotating a point \(270^\circ\) about the origin.
• Correction:
• For \(C(-1,1)\), using the \(270^\circ\) rotation rule \((x,y)\to(y, - x)\), we have \(C(-1,1)\to(1,1)\).
• For \(D(2,3)\), using the \(270^\circ\) rotation rule \((x,y)\to(y, - x)\), we have \(D(2,3)\to(3,-2)\).

For problem 10:

• Error Description: The student used the wrong transformation rule (possibly the rule for reflecting over \(y = x\) or a wrong rotation rule) instead of the \(270^\circ\) rotation rule.
• Correction:
• For \(C(-1,1)\), using the \(270^\circ\) rotation rule \((x,y)\to(y, - x)\), \(C(-1,1)\to(1,1)\).
• For \(D(2,3)\), using the \(270^\circ\) rotation rule \((x,y)\to(y, - x)\), \(D(2,3)\to(3,-2)\).