Question

the coordinate grid below shows rectangle wxyz. rectangle wxyz is rotated 270° counter - clockwise about the origin to create rectangle wxyz. which rule describes this transformation? a ((x,y)\to(-x,y)) b ((x,y)\to(x,-y)) c ((x,y)\to(-y,x)) d ((x,y)\to(y,-x))

Explanation:

Step1: Recall rotation rules

For a \(270^\circ\) counter - clockwise rotation about the origin, the rule for a point \((x,y)\) is \((x,y)\to(-y,x)\). Wait, let's re - derive it. A \(90^\circ\) counter - clockwise rotation is \((x,y)\to(-y,x)\), a \(180^\circ\) counter - clockwise rotation is \((x,y)\to(-x,-y)\), and a \(270^\circ\) counter - clockwise rotation is equivalent to a \(90^\circ\) clockwise rotation. The rule for a \(90^\circ\) clockwise rotation (which is the same as \(270^\circ\) counter - clockwise) is \((x,y)\to(y,-x)\)? Wait, no, let's use the standard rotation matrices. The rotation matrix for an angle \(\theta\) counter - clockwise is \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\). For \(\theta = 270^\circ\), \(\cos(270^\circ)=0\), \(\sin(270^\circ)= - 1\). So the matrix is \(\begin{pmatrix}0&1\\- 1&0\end{pmatrix}\). Multiplying this matrix by the vector \(\begin{pmatrix}x\\y\end{pmatrix}\) gives \(\begin{pmatrix}y\\-x\end{pmatrix}\)? Wait, no, \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\-x\end{pmatrix}\)? Wait, no, let's do the multiplication: \(0\times x+1\times y = y\) and \(- 1\times x+0\times y=-x\). So the new point is \((y,-x)\). Wait, but let's take an example. Suppose we have a point \((1,0)\). A \(270^\circ\) counter - clockwise rotation about the origin should take \((1,0)\) to \((0, - 1)\)? Wait, no, a \(90^\circ\) counter - clockwise rotation takes \((1,0)\) to \((0,1)\), a \(180^\circ\) takes it to \((-1,0)\), a \(270^\circ\) counter - clockwise takes it to \((0,-1)\). Using the matrix \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\) on \((1,0)\) gives \((0,-1)\), which is correct. Another example: \((0,1)\) rotated \(270^\circ\) counter - clockwise should be \((-1,0)\). The matrix gives \((1,0)\)? Wait, no, I must have messed up the matrix. The correct rotation matrix for counter - clockwise rotation by \(\theta\) is \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\). For \(\theta = 270^\circ\), \(\cos(270^\circ)=0\), \(\sin(270^\circ)=-1\). So the matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\)? Wait, no, \(-\sin(270^\circ)=-(-1) = 1\), \(\sin(270^\circ)=-1\), \(\cos(270^\circ)=0\). So the matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). Wait, when we apply it to \((0,1)\): \(0\times0 + 1\times1=1\), \(-1\times0+0\times1 = 0\). So we get \((1,0)\), but a \(270^\circ\) counter - clockwise rotation of \((0,1)\) should be \((-1,0)\)? Wait, no, I think I have the direction wrong. A \(270^\circ\) counter - clockwise rotation is the same as a \(90^\circ\) clockwise rotation. Let's use clockwise rotation. The rotation matrix for clockwise rotation by \(\theta\) is \(\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}\). For \(\theta = 90^\circ\) (clockwise), which is the same as \(270^\circ\) counter - clockwise, \(\cos(90^\circ)=0\), \(\sin(90^\circ)=1\). So the matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). Wait, applying this to \((0,1)\): \(0\times0+1\times1 = 1\), \(-1\times0 + 0\times1=0\), so \((1,0)\). But a \(90^\circ\) clockwise rotation of \((0,1)\) is \((1,0)\), which is correct. A \(90^\circ\) clockwise rotation of \((1,0)\) is \((0,-1)\), which is correct. So the rule for \(270^\circ\) counter - clockwise (or \(90^\circ\) clockwise) rotation about the origin is \((x,y)\to(y,-x)\)? Wait, no, in the options, we have:

A. \((x,y)\to(-x,y)\) (this is \(90^\circ\) counter - clockwise? No, \(90^\circ\) counter - clockwise is \((x,y)\to(-y,x)\))

B. \((x,y)\to(x,-y)\) (this is reflection over x - axis)

C. \((x,y)\to(-y,x)\) (this is \(90^\circ\) counter - clockwise)

D. \((x,y)\to(y,-x)\) (this is \(270^\circ\) counter - clockwise or \(90^\circ\) clockwise)

Wait, maybe I made a mistake in the matrix. Let's take a point from the rectangle. Suppose one of the vertices of the rectangle WXYZ is, say, W. Let's assume the coordinates. From the graph, let's say W is at \((2,-3)\), X is at \((4,-3)\), Y is at \((4,-1)\), Z is at \((2,-1)\) (just estimating from the grid). Now, rotate \(270^\circ\) counter - clockwise about the origin. Using the rule \((x,y)\to(y,-x)\):

For W \((2,-3)\): new point is \((-3,-2)\)? Wait, no, \((x,y)=(2,-3)\), so \(y=-3\), \(-x=-2\), so \((-3,-2)\)? Wait, that doesn't seem right. Wait, maybe the correct rule for \(270^\circ\) counter - clockwise is \((x,y)\to(-y,x)\). Let's test with \((1,0)\): \(-y = 0\), \(x = 1\), so \((0,1)\)? No, that's \(90^\circ\) counter - clockwise. Wait, I'm confused. Let's recall the standard rotation rules:

- \(90^\circ\) counter - clockwise: \((x,y)\to(-y,x)\)

- \(180^\circ\) counter - clockwise: \((x,y)\to(-x,-y)\)

- \(270^\circ\) counter - clockwise: \((x,y)\to(y,-x)\)

- \(90^\circ\) clockwise: \((x,y)\to(y,-x)\) (same as \(270^\circ\) counter - clockwise)

- \(180^\circ\) clockwise: same as \(180^\circ\) counter - clockwise

- \(270^\circ\) clockwise: same as \(90^\circ\) counter - clockwise

So if we have a \(270^\circ\) counter - clockwise rotation, the rule is \((x,y)\to(y,-x)\), which is option D? Wait, no, in the options, option D is \((x,y)\to(y,-x)\)? Wait, the options are:

A. \((x,y)\to(-x,y)\)

B. \((x,y)\to(x,-y)\)

C. \((x,y)\to(-y,x)\)

D. \((x,y)\to(y,-x)\)

Wait, maybe the rectangle has vertices, let's say, let's take a simple rectangle. Suppose the rectangle has vertices at (1,1), (3,1), (3,3), (1,3). Rotating \(270^\circ\) counter - clockwise about the origin. Using the rule \((x,y)\to(y,-x)\):

(1,1) becomes (1, - 1)? No, that can't be. Wait, I think I have the rotation direction wrong. Let's use the right - hand rule. For counter - clockwise rotation, a \(90^\circ\) turn: (x,y) -> (-y,x). \(180^\circ\): (-x,-y). \(270^\circ\): (y,-x). Let's take (1,0): \(270^\circ\) counter - clockwise: (0,-1). Using (y,-x): y = 0, -x=-1, so (0,-1). Correct. (0,1): \(270^\circ\) counter - clockwise: (-1,0). Using (y,-x): y = 1, -x = 0, so (1,0)? No, that's wrong. Wait, no, (0,1): x = 0, y = 1. So (y,-x)=(1,0). But a \(270^\circ\) counter - clockwise rotation of (0,1) should be (-1,0). So there's a mistake in my formula. The correct rotation matrix for \(270^\circ\) counter - clockwise is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\). Let's check: \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-y\\x\end{pmatrix}\). Ah! Here's the mistake. The rotation matrix for \(\theta\) counter - clockwise is \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\). For \(\theta = 270^\circ\), \(\cos(270^\circ)=0\), \(\sin(270^\circ)=-1\). So the matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\)? No, \(-\sin\theta=-\sin(270^\circ)=-(-1) = 1\), \(\sin\theta=-1\), \(\cos\theta = 0\). So the matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). Wait, when we multiply by \((x,y)\), we get \((y,-x)\). But when we take (0,1), we get (1,0), which is a \(90^\circ\) clockwise rotation, which is the same as \(270^\circ\) counter - clockwise. So the correct rule for \(270^\circ\) counter - clockwise rotation about the origin is \((x,y)\to(y,-x)\). But in the options, option D is \((x,y)\to(y,-x)\)? Wait, the options are:

A. \((x,y)\to(-x,y)\)

B. \((x,y)\to(x,-y)\)

C. \((x,y)\to(-y,x)\)

D. \((x,y)\to(y,-x)\)

Wait, maybe I misread the options. Let's check again. The user's options:

A. \((x,y)\to(-x,y)\)

B. \((x,y)\to(x,-y)\)

C. \((x,y)\to(-y,x)\)

D. \((x,y)\to(y,-x)\)

Wait, no, maybe the correct rule for \(270^\circ\) counter - clockwise is \((x,y)\to(y,-x)\), which is option D? Wait, no, let's take a point (2,3). Rotate \(270^\circ\) counter - clockwise. Using the matrix \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\), we get (3,-2). So (x,y)=(2,3) becomes (y,-x)=(3,-2). Correct. So the rule is \((x,y)\to(y,-x)\), which is option D? Wait, but in some sources, the rule for \(270^\circ\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\). Wait, I think I have the angle direction wrong. A \(270^\circ\) counter - clockwise rotation is the same as a \(90^\circ\) clockwise rotation. The rule for \(90^\circ\) clockwise rotation is \((x,y)\to(y,-x)\), which is what we derived. So the correct option should be D? Wait, no, let's check the options again. Wait, the options are:

A. \((x,y)\to(-x,y)\) (this is a \(90^\circ\) counter - clockwise? No, \(90^\circ\) counter - clockwise is \((x,y)\to(-y,x)\))

B. \((x,y)\to(x,-y)\) (reflection over x - axis)

C. \((x,y)\to(-y,x)\) ( \(90^\circ\) counter - clockwise)

D. \((x,y)\to(y,-x)\) ( \(270^\circ\) counter - clockwise or \(90^\circ\) clockwise)

So if the rectangle is rotated \(270^\circ\) counter - clockwise about the origin, the transformation rule is \((x,y)\to(y,-x)\), which is option D.

Step2: Confirm with rotation rules

The rotation of a point \((x,y)\) by \(270^\circ\) counter - clockwise about the origin follows the rule \((x,y)\to(y,-x)\). This is derived from the rotation matrix or by understanding that a \(270^\circ\) counter - clockwise rotation is equivalent to a \(90^\circ\) clockwise rotation, and the rule for \(90^\circ\) clockwise rotation is \((x,y)\to(y,-x)\). By testing with sample points and recalling the rotation rules, we can confirm that this is the correct transformation rule.

Answer:

D. \((x,y)\to(y,-x)\)