Question

the polygon abcd is rotated by 270° anticlockwise about the origin. find the coordinates of the rotated polygon. a (2, -1), b (8, -1), c(8, -6), d (2, -6) a (1, 2), b (1, 8), c(6, 8), d (6, 2) a (1, -2), b (1, -8), c(6, -8), d (6, -2) a (-1, -2), b (-1, -8), c(-6, -8), d (-6, -2)

Explanation:

Step1: Find original coordinates

From the graph, original coordinates: \( A(2,1) \), \( B(8,1) \), \( C(8,6) \), \( D(2,6) \).

Step2: Apply 270° anticlockwise rotation rule

The rule for rotating a point \((x,y)\) 270° anticlockwise about the origin is \((x,y) \to (y, -x)\).

Step3: Rotate each point

• For \( A(2,1) \): \( (1, -2) \)? Wait, no, wait: Wait, 270° anticlockwise is same as 90° clockwise. Wait, correct rule: 270° anticlockwise rotation: \((x,y) \mapsto (y, -x)\)? Wait, no, let's recall:
• 90° anticlockwise: \((x,y) \to (-y, x)\)
• 180° anticlockwise: \((x,y) \to (-x, -y)\)
• 270° anticlockwise: \((x,y) \to (y, -x)\)? Wait, no, actually, 270° anticlockwise is equivalent to 90° clockwise. The rule for 90° clockwise (which is 270° anticlockwise) is \((x,y) \to (y, -x)\)? Wait, no, let's check with a point. Let's take \( A(2,1) \).

Wait, maybe I mixed up. Let's derive: A 270° anticlockwise rotation is three 90° anticlockwise rotations.

First 90° anticlockwise: \((2,1) \to (-1, 2)\)

Second 90° (total 180°): \((-1,2) \to (-2, -1)\)

Third 90° (total 270°): \((-2, -1) \to (-1, 2)\)? No, that's not right. Wait, no, maybe the correct rule for 270° anticlockwise is \((x,y) \to (y, -x)\). Wait, let's test with (0,1). Rotating 270° anticlockwise: it should go to (1,0). Using (y, -x): (1, 0), which is correct. Another test: (1,0) rotated 270° anticlockwise: should go to (0, -1). Using (y, -x): (0, -1), correct. So rule is \((x,y) \to (y, -x)\).

Wait, so for \( A(2,1) \): \( (1, -2) \)? Wait, no, \( x=2, y=1 \), so \( (y, -x) = (1, -2) \)? But let's check the options. Wait, the options have \( A'(1, -2) \)? Wait, no, the third option is \( A'(1, -2) \), \( B'(1, -8) \), \( C'(6, -8) \), \( D'(6, -2) \). Wait, let's check original points:

Wait, original coordinates: Let's re-examine the graph. The grid: A is at (2,1)? Wait, the x-axis: the grid lines. Let's see, A is at (2,1)? Wait, the x-coordinate: the first grid after 1 is 2, so A is (2,1), B is (8,1), C is (8,6), D is (2,6).

Now, applying 270° anticlockwise rotation (rule: (x,y) → (y, -x)):

• A(2,1): (1, -2)
• B(8,1): (1, -8)
• C(8,6): (6, -8)
• D(2,6): (6, -2)

Which matches the third option: \( A'(1, -2) \), \( B'(1, -8) \), \( C'(6, -8) \), \( D'(6, -2) \).

Wait, but let's confirm again. Let's take point A(2,1). Rotate 270° anticlockwise. The rotation matrix for 270° anticlockwise is \(

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

\). So multiplying \(

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

$$\begin{pmatrix} 2 \\ 1 \end{pmatrix}$$

=

$$\begin{pmatrix} 1 \\ -2 \end{pmatrix}$$

\), which is (1, -2). Correct. So A becomes (1, -2), B(8,1) becomes (1, -8), C(8,6) becomes (6, -8), D(2,6) becomes (6, -2). Which is the third option.

Answer:

A' (1, -2), B' (1, -8), C'(6, -8), D' (6, -2) (the third option)