Question

ABCD rotation 270 counterclockwise around the origin

Answer 1

The trapezoid ABCD, after a 270-degree counterclockwise rotation around the origin, has vertices A'(9, 9), B'(9, 5), C'(10, 3), D'(10, 10).

To find the coordinates of the trapezoid ABCD after a 270-degree counterclockwise rotation around the origin, we can use the rotation matrix formula:

`\[ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \]`

where
`\(\theta\)` is the angle of rotation. For a 270-degree counterclockwise rotation,
`\(\theta = (3\pi)/(2)\)`

Let's apply this rotation to each vertex of the trapezoid:

1. For point A(-9, 9):

`\[ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -9 \\ 9 \end{bmatrix} = \begin{bmatrix} 9 \\ 9 \end{bmatrix} \]`

2. For point B(-5, 9):

`\[ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -5 \\ 9 \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \end{bmatrix} \]`

3. For point C(-3, 10):

`\[ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -3 \\ 10 \end{bmatrix} = \begin{bmatrix} 10 \\ 3 \end{bmatrix} \]`

4. For point D(-10, 10):

`\[ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 10 \\ 10 \end{bmatrix} \]`

So, after a 270-degree counterclockwise rotation, the new coordinates of the trapezoid ABCD are:

A'(9, 9), B'(9, 5), C'(10, 3), D'(10, 10).

The question probable may be:

What would be the coordinates of the trapezoid ABCD after a rotation fo 270 degrees counterclockwise around the origin.

The original coordinates are

A( -9 , 9 ) B(-5 , 9) C(-3 , 10) D(-10 , 10 )