Question

rotations on the coordinate plane
use the figure to answer the questions below. match your answers in the table to solve the riddle.
grid with points a, b, c, d

1. find a after a is rotated 90° clockwise.
2. find b after b is rotated 270° clockwise.
3. find c after c is rotated 180° clockwise.
4. find d after d is rotated 90° clockwise.
5. find a after a is rotated 180° counterclockwise.
6. find b after b is rotated 270° counterclockwise.
7. find c after c is rotated 90° counterclockwise.
8. find d after d is rotated 180° counterclockwise.
9. find a after a is rotated 270° clockwise.
10. find b after b is rotated 180° clockwise.
11. find c after c is rotated 90° clockwise.
12. find d after d is rotated 90° counterclockwise.

s (7, -7) l (1, -7) c (-2, -7)
b (-7, -7) o (-7, -1) v (2, 1)
i (-7, 2) r (2, 1) (1, -2)
u (-2, -1) e (-1, 2) p (-1, 7)
n (-7, 7) a (7, -2) d (2, 1)
why does nobody talk to circles?
__ __
10 12 5 1 8 6 12 9 4 6 7 3 9 2 4 11 12 6 6
©maneuvering the middle llc, 2017

Answer:

To solve these rotation problems, we first need the coordinates of points \( A \), \( B \), \( C \), and \( D \) from the graph. Let's assume the coordinates (by analyzing the grid):

• \( A \): Let's say \( A = (2, 7) \) (since it's 2 units right on the x - axis and 7 units up on the y - axis)
• \( B=(7, 7) \)
• \( C=(7, 1) \)
• \( D=(2, 1) \)

Problem 1: Find \( A' \) after \( A(2,7) \) is rotated \( 90^{\circ} \) clockwise.

Step 1: Recall the rotation rule for \( 90^{\circ} \) clockwise.

The rule for rotating a point \( (x,y) \) \( 90^{\circ} \) clockwise is \( (x,y)\to(y, - x) \)

Step 2: Apply the rule to \( A(2,7) \)

For \( x = 2 \) and \( y=7 \), the new coordinates \( A'=(7,- 2) \)

Problem 2: Find \( B' \) after \( B(7,7) \) is rotated \( 270^{\circ} \) clockwise.

Step 1: Recall the rotation rule for \( 270^{\circ} \) clockwise (or \( 90^{\circ} \) counter - clockwise).

The rule for rotating a point \( (x,y) \) \( 270^{\circ} \) clockwise is \( (x,y)\to(y, - x) \) (same as \( 90^{\circ} \) counter - clockwise: \( (x,y)\to(-y,x) \)? Wait, correction: The correct rule for \( 270^{\circ} \) clockwise rotation of a point \( (x,y) \) is \( (x,y)\to(y, - x) \)? No, let's derive it. A \( 90^{\circ} \) clockwise rotation: \( (x,y)\to(y, - x) \), \( 180^{\circ} \) rotation: \( (x,y)\to(-x,-y) \), \( 270^{\circ} \) clockwise rotation: \( (x,y)\to(-y,x) \)

Let's re - derive: A full rotation is \( 360^{\circ} \). \( 270^{\circ} \) clockwise is equivalent to \( 90^{\circ} \) counter - clockwise. The rotation matrix for \( 90^{\circ} \) counter - clockwise is \(

$$\begin{pmatrix}0&- 1\\1&0\end{pmatrix}$$

\). So if we have a point \( (x,y) \), the new point \( (x',y')=

$$\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$

$$\begin{pmatrix}x\\y\end{pmatrix}$$

=

$$\begin{pmatrix}-y\\x\end{pmatrix}$$

\)

Step 2: Apply the rule to \( B(7,7) \)

For \( x = 7 \) and \( y = 7 \), \( x'=-7 \), \( y' = 7 \). So \( B'=(-7,7) \)

Problem 3: Find \( C' \) after \( C(7,1) \) is rotated \( 180^{\circ} \) clockwise.

Step 1: Recall the rotation rule for \( 180^{\circ} \) rotation.

The rule for rotating a point \( (x,y) \) \( 180^{\circ} \) (clockwise or counter - clockwise) is \( (x,y)\to(-x,-y) \)

Step 2: Apply the rule to \( C(7,1) \)

For \( x = 7 \) and \( y = 1 \), \( x'=-7 \), \( y'=-1 \). So \( C'=(-7,-1) \)

Problem 4: Find \( D' \) after \( D(2,1) \) is rotated \( 90^{\circ} \) clockwise.

Step 1: Recall the \( 90^{\circ} \) clockwise rotation rule \( (x,y)\to(y, - x) \)

Step 2: Apply the rule to \( D(2,1) \)

For \( x = 2 \) and \( y = 1 \), \( x' = 1 \), \( y'=-2 \). So \( D'=(1,-2) \)

Problem 5: Find \( A' \) after \( A(2,7) \) is rotated \( 180^{\circ} \) counter - clockwise.

Step 1: Recall the \( 180^{\circ} \) rotation rule \( (x,y)\to(-x,-y) \)

Step 2: Apply the rule to \( A(2,7) \)

For \( x = 2 \) and \( y = 7 \), \( x'=-2 \), \( y'=-7 \). So \( A'=(-2,-7) \)

Problem 6: Find \( B' \) after \( B(7,7) \) is rotated \( 270^{\circ} \) counter - clockwise.

Step 1: Recall the rotation rule for \( 270^{\circ} \) counter - clockwise. The rotation matrix for \( 270^{\circ} \) counter - clockwise is \(

$$\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

\), so the rule is \( (x,y)\to(y,x) \)? Wait, no. Let's think in terms of direction. A \( 270^{\circ} \) counter - clockwise rotation is the same as a \( 90^{\circ} \) clockwise rotation. Wait, no: \( 90^{\circ} \) counter - clockwise: \( (x,y)\to(-y,x) \), \( 180^{\circ} \) counter - clockwise: \( (x,y)\to(-x,-y) \), \( 270^{\circ} \) counter - clockwise: \( (x,y)\to(y,-x) \)

Let's take a point \( (1,0) \). A \( 270^{\circ} \) counter - clockwise rotation of \( (1,0) \) should land at \( (0,1) \)? Wait, no. Let's use the unit circle. A point \( (x,y) \) on the unit circle at angle \( \theta \) has coordinates \( (\cos\theta,\sin\theta) \). A \( 270^{\circ} \) counter - clockwise rotation is adding \( 270^{\circ} \) to \( \theta \). So \( \cos(\theta + 270^{\circ})=\sin\theta \), \( \sin(\theta + 270^{\circ})=-\cos\theta \). So the rule is \( (x,y)\to(y,-x) \)

Step 2: Apply the rule to \( B(7,7) \)

For \( x = 7 \) and \( y = 7 \), \( x' = 7 \), \( y'=-7 \). So \( B'=(7,-7) \)

Problem 7: Find \( C' \) after \( C(7,1) \) is rotated \( 90^{\circ} \) counter - clockwise.

Step 1: Recall the \( 90^{\circ} \) counter - clockwise rotation rule \( (x,y)\to(-y,x) \)

Step 2: Apply the rule to \( C(7,1) \)

For \( x = 7 \) and \( y = 1 \), \( x'=-1 \), \( y' = 7 \). So \( C'=(-1,7) \)

Problem 8: Find \( D' \) after \( D(2,1) \) is rotated \( 180^{\circ} \) counter - clockwise.

Step 1: Recall the \( 180^{\circ} \) rotation rule \( (x,y)\to(-x,-y) \)

Step 2: Apply the rule to \( D(2,1) \)

For \( x = 2 \) and \( y = 1 \), \( x'=-2 \), \( y'=-1 \). So \( D'=(-2,-1) \)

Problem 9: Find \( A' \) after \( A(2,7) \) is rotated \( 270^{\circ} \) clockwise.

Step 1: Recall the \( 270^{\circ} \) clockwise rotation rule \( (x,y)\to(-y,x) \) (since \( 270^{\circ} \) clockwise is equivalent to \( 90^{\circ} \) counter - clockwise)

Step 2: Apply the rule to \( A(2,7) \)

For \( x = 2 \) and \( y = 7 \), \( x'=-7 \), \( y' = 2 \). So \( A'=(-7,2) \)

Problem 10: Find \( B' \) after \( B(7,7) \) is rotated \( 180^{\circ} \) clockwise.

Step 1: Recall the \( 180^{\circ} \) rotation rule \( (x,y)\to(-x,-y) \)

Step 2: Apply the rule to \( B(7,7) \)

For \( x = 7 \) and \( y = 7 \), \( x'=-7 \), \( y'=-7 \). So \( B'=(-7,-7) \)

Problem 11: Find \( C' \) after \( C(7,1) \) is rotated \( 90^{\circ} \) clockwise.

Step 1: Recall the \( 90^{\circ} \) clockwise rotation rule \( (x,y)\to(y, - x) \)

Step 2: Apply the rule to \( C(7,1) \)

For \( x = 7 \) and \( y = 1 \), \( x' = 1 \), \( y'=-7 \). So \( C'=(1,-7) \)

Problem 12: Find \( D' \) after \( D(2,1) \) is rotated \( 90^{\circ} \) counter - clockwise.

Step 1: Recall the \( 90^{\circ} \) counter - clockwise rotation rule \( (x,y)\to(-y,x) \)

Step 2: Apply the rule to \( D(2,1) \)

For \( x = 2 \) and \( y = 1 \), \( x'=-1 \), \( y' = 2 \). So \( D'=(-1,2) \)

Now, we match the answers with the given table:

1. \( A'=(7,-2) \) which is \( A \)
2. \( B'=(-7,7) \) which is \( N \)
3. \( C'=(-7,-1) \) which is \( O \)
4. \( D'=(1,-2) \) which is \( I \) (Wait, no, in the table \( I=(-7,2) \), maybe my initial coordinate assumption is wrong. Let's re - check the grid. Let's assume the coordinates correctly:

Looking at the grid, let's find the coordinates of \( A \), \( B \), \( C \), \( D \):

• \( D \) is at \( (2,1) \) (since it's 2 units right on x - axis, 1 unit up on y - axis)
• \( A \) is directly above \( D \), so \( A=(2,7) \) (since from \( y = 1 \) to \( y = 7 \) is 6 units up)
• \( B \) is to the right of \( A \), same y - coordinate. The distance from \( A \) to \( B \): from \( x = 2 \) to \( x = 7 \), so \( B=(7,7) \)
• \( C \) is directly below \( B \), same x - coordinate, so \( C=(7,1) \)

Now, let's re - check the table:

• \( A=(7,-2) \) (matches our first problem's answer)
• \( N=(-7,7) \) (matches problem 2's answer)
• \( O=(-7,-1) \) (matches problem 3's answer)
• \( I=(1,-2) \) (matches problem 4's answer)
• \( C=(-2,-7) \) (matches problem 5's answer: \( A'=(-2,-7) \))
• \( S=(7,-7) \) (matches problem 6's answer: \( B'=(7,-7) \))
• \( P=(-1,7) \) (matches problem 7's answer: \( C'=(-1,7) \))
• \( U=(-2,-1) \) (matches problem 8's answer: \( D'=(-2,-1) \))
• \( I=(-7,2) \) (matches problem 9's answer: \( A'=(-7,2) \))
• \( B=(-7,-7) \) (matches problem 10's answer: \( B'=(-7,-7) \))
• \( L=(1,-7) \) (matches problem 11's answer: \( C'=(1,-7) \))
• \( E=(-1,2) \) (matches problem 12's answer: \( D'=(-1,2) \))

Now, let's list the answers for each problem (1 - 12) with their corresponding letters:

1. \( A'=(7,-2) \) → \( A \)
2. \( B'=(-7,7) \) → \( N \)
3. \( C'=(-7,-1) \) → \( O \)
4. \( D'=(1,-2) \) → \( I \)
5. \( A'=(-2,-7) \) → \( C \)
6. \( B'=(7,-7) \) → \( S \)
7. \( C'=(-1,7) \) → \( P \)
8. \( D'=(-2,-1) \) → \( U \)
9. \( A'=(-7,2) \) → \( I \) (Wait, no, \( I=(-7,2) \) in the table, yes)
10. \( B'=(-7,-7) \) → \( B \)
11. \( C'=(1,-7) \) → \( L \)
12. \( D'=(-1,2) \) → \( E \)

Now, let's arrange the letters according to the blanks:

The blanks are: \( \overline{10}\ \overline{12}\ \overline{5}\ \overline{1}\ \overline{8}\ \overline{6}\ \overline{12}\ \ \overline{9}\ \overline{4}\ \overline{6}\ \ \overline{7}\ \overline{3}\ \overline{9}\ \overline{2}\ \overline{4}\ \overline{11}\ \overline{12}\ \overline{6}\ \overline{6} \)

Let's map each number to the letter:

• 1: \( A \)
• 2: \( N \)
• 3: \( O \)
• 4: \( I \)
• 5: \( C \)
• 6: \( S \)
• 7: \( P \)
• 8: \( U \)
• 9: \( I \)
• 10: \( B \)
• 11: \( L \)
• 12: \( E \)

Now, let's fill in the blanks:

• \( \overline{10} \): \( B \)
• \( \overline{12} \): \( E \)
• \( \overline{5} \): \( C \)
• \( \overline{1} \): \( A \)
• \( \overline{8} \): \( U \)
• \( \overline{6} \): \( S \)
• \( \overline{12} \): \( E \)
• \( \overline{9} \): \( I \)
• \( \overline{4} \): \( I \)
• \( \overline{6} \): \( S \)
• \( \overline{7} \): \( P \)
• \( \overline{3} \): \( O \)
• \( \overline{9} \): \( I \)
• \( \overline{2} \): \( N \)
• \( \overline{4} \): \( I \)
• \( \overline{11} \): \( L \)
• \( \overline{12} \): \( E \)
• \( \overline{6} \): \( S \)
• \( \overline{6} \): \( S \)

Putting it all together: \( B\ E\ C\ A\ U\ S\ E\ I\ I\ S\ P\ O\ I\ N\ I\ L\ E\ S\ S \)

Wait, but the riddle is "WHY DOES NOBODY TALK TO CIRCLES?". The answer should be related to the fact that circles have no "angles" (or "they're too round", but in the context of rotations and