Question

quadrilateral ( abcd ) is the image of quadrilateral ( abcd ) under a rotation about the origin, ( (0,0) ).
determine the angles of rotation.
choose all answers that apply:
a ( 90^{circ} ) clockwise
b ( 90^{circ} ) counterclockwise
c ( 180^{circ} )
d ( 270^{circ} ) clockwise
e ( 270^{circ} ) counterclockwise

Explanation:

Step1: Recall rotation rules

- Rotation of \(90^{\circ}\) clockwise: \((x,y)\to(y, -x)\)
- Rotation of \(90^{\circ}\) counter - clockwise: \((x,y)\to(-y,x)\)
- Rotation of \(180^{\circ}\): \((x,y)\to(-x,-y)\)
- Rotation of \(270^{\circ}\) clockwise: \((x,y)\to(-y,x)\) (same as \(90^{\circ}\) counter - clockwise)
- Rotation of \(270^{\circ}\) counter - clockwise: \((x,y)\to(y, -x)\) (same as \(90^{\circ}\) clockwise)

Let's take a point, say \(B(1,5)\). Its image \(B'(- 5,-1)\)

If we use the rotation formula:
For \(90^{\circ}\) clockwise rotation \((x,y)\to(y, -x)\). If \(x = 1,y = 5\), then \((y,-x)=(5,-1)\) (not correct)
For \(90^{\circ}\) counter - clockwise rotation \((x,y)\to(-y,x)\). If \(x = 1,y = 5\), then \((-y,x)=(-5,1)\) (not correct)
For \(180^{\circ}\) rotation \((x,y)\to(-x,-y)\). If \(x = 1,y = 5\), then \((-x,-y)=(-1,-5)\) (not correct)
For \(270^{\circ}\) clockwise rotation \((x,y)\to(-y,x)\). If \(x = 1,y = 5\), then \((-y,x)=(-5,1)\) (not correct)
For \(270^{\circ}\) counter - clockwise rotation \((x,y)\to(y, -x)\). If \(x = 1,y = 5\), then \((y,-x)=(5,-1)\) (not correct)

Let's take another approach. The general rule for rotation about the origin:
We know that a rotation of \(270^{\circ}\) counter - clockwise is equivalent to a rotation of \(90^{\circ}\) clockwise.
If we consider the transformation of the quadrilateral.
The rotation of a point \((x,y)\) by \(270^{\circ}\) counter - clockwise:
Let’s take point \(A(2,4)\). After \(270^{\circ}\) counter - clockwise rotation \((x,y)\to(y, -x)\), we get \((4,-2)\) (not matching).
Let’s take point \(D(2,3)\). After \(270^{\circ}\) counter - clockwise rotation \((2,3)\to(3,-2)\) (not matching)
Let’s take point \(C(-5,5)\). After \(270^{\circ}\) counter - clockwise rotation \((-5,5)\to(5,5)\) (not matching)

Let’s use the property of rotation. The angle between the pre - image and image vectors.
The rotation of \(270^{\circ}\) clockwise is equivalent to \(90^{\circ}\) counter - clockwise.
Let’s check the orientation.
If we rotate a figure \(270^{\circ}\) clockwise (or \(90^{\circ}\) counter - clockwise) about the origin.
The transformation of a point \((x,y)\) to \((-y,x)\) (for \(90^{\circ}\) counter - clockwise or \(270^{\circ}\) clockwise)
Take point \(B(1,5)\). If we rotate \(270^{\circ}\) clockwise (using \((x,y)\to(-y,x)\)), we get \((-5,1)\) (not correct)
Take point \(A(2,4)\). If we rotate \(270^{\circ}\) counter - clockwise (using \((x,y)\to(y, -x)\)), we get \((4,-2)\) (not correct)

Let’s use the fact that rotation is a rigid transformation.
The rotation of \(270^{\circ}\) counter - clockwise:
Let’s assume a rotation of \(270^{\circ}\) counter - clockwise.
The rotation matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\)
For a point \((x,y)\), the new coordinates \((x',y')=(y,-x)\)
Take point \(C(-5,5)\). After rotation \((5,5)\) (not correct)

Let’s consider the rotation of \(270^{\circ}\) clockwise.
The rotation matrix is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\), and \((x,y)\to(-y,x)\)
Take point \(B(1,5)\to(-5,1)\) (not correct)

Let’s consider the rotation of \(90^{\circ}\) clockwise.
The rotation matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\) (wait no, \(90^{\circ}\) clockwise: \((x,y)\to(y,-x)\)
Take point \(C(-5,5)\to(5,5)\) (not correct)

Let’s consider the rotation of \(90^{\circ}\) counter - clockwise: \((x,y)\to(-y,x)\)
Take point \(B(1,5)\to(-5,1)\) (not correct)

Let’s use the property of the entire figure.
If we rotate a figure \(270^{\circ}\) counter - clockwise (equivalent to \(90^{\circ}\) clockwise)
The orientation of the quadrilateral:
We know that a \(270^{\circ}\) counter - clockwise rotation (or \(90^{\circ}\) clockwise) changes the position of the quadrilateral.
Let’s check the direction.
If we consider the "hand - like" movement of the quadrilateral.
A \(270^{\circ}\) counter - clockwise rotation (or \(90^{\circ}\) clockwise)
Let’s take two adjacent points, say \(B(1,5)\) and \(A(2,4)\)
After rotation:
If it is \(270^{\circ}\) counter - clockwise (\((x,y)\to(y, -x)\)):
\(B(1,5)\to(5,-1)\), \(A(2,4)\to(4,-2)\) (not matching)
If it is \(270^{\circ}\) clockwise (\((x,y)\to(-y,x)\)):
\(B(1,5)\to(-5,1)\), \(A(2,4)\to(-4,2)\) (not matching)

Let’s use the fact that \(360^{\circ}-\theta\) rotation in one direction is equivalent to \(\theta\) rotation in the opposite direction.
A \(270^{\circ}\) counter - clockwise rotation is equivalent to a \(90^{\circ}\) clockwise rotation.
Let’s check the coordinates:
Take point \(C(-5,5)\)
For \(90^{\circ}\) clockwise rotation \((x,y)\to(y, -x)\): \((5,5)\) (not correct)
For \(270^{\circ}\) counter - clockwise rotation \((x,y)\to(y, -x)\): \((5,5)\) (not correct)

Let’s take another approach.
The rotation of a figure. If we assume the rotation is \(270^{\circ}\) counter - clockwise (or \(90^{\circ}\) clockwise)
We know that the rotation of a point \((x,y)\) about the origin:
If we consider the transformation of the quadrilateral \(ABCD\) to \(A'B'C'D'\)
Let’s use the slope of the sides.
The slope of \(AB\): \(m_{AB}=\frac{4 - 5}{2-1}=-1\)
The slope of \(A'B'\): Let \(A'(-1,-4)\), \(B'(-1,-5)\), \(m_{A'B'}=\frac{-5+4}{-1 + 1}\) (undefined, not helpful)

Let’s use the distance formula. Since rotation is a rigid transformation (distance is preserved)
\(AB=\sqrt{(2 - 1)^{2}+(4 - 5)^{2}}=\sqrt{1 + 1}=\sqrt{2}\)
\(A'B'=\sqrt{(-1+1)^{2}+(-4 + 5)^{2}} = 1\) (This approach is wrong, let's get back)

Let’s use the standard rotation rules:
For a point \(P(x,y)\)
- \(90^{\circ}\) clockwise: \(P(x,y)\to P'(y,-x)\)
- \(270^{\circ}\) counter - clockwise: \(P(x,y)\to P'(y,-x)\)

Take \(B(1,5)\), if rotated \(270^{\circ}\) counter - clockwise (or \(90^{\circ}\) clockwise) \((1,5)\to(5,-1)\) (not \(B'(-1,-5)\))
Take \(C(-5,5)\), if rotated \(270^{\circ}\) counter - clockwise (or \(90^{\circ}\) clockwise) \((-5,5)\to(5,5)\) (not \(C'(5,-5)\))

Let’s use the correct rotation formula:
The rotation of \(270^{\circ}\) clockwise (\((x,y)\to(-y,x)\))
Take \(B(1,5)\to(-5,1)\) (not \(B'(-1,-5)\))
Take \(C(-5,5)\to(-5,-5)\) (not \(C'(5,-5)\))

Let’s use the rotation of \(180^{\circ}\): \((x,y)\to(-x,-y)\)
Take \(B(1,5)\to(-1,-5)\), \(A(2,4)\to(-2,-4)\), \(C(-5,5)\to(5,-5)\), \(D(2,3)\to(-2,-3)\)

Answer:

C. \(180^{\circ}\)