Question

if p = (2, -1), find the image of p under the following rotation. 270° counterclockwise about the origin (?, ) enter the number that belongs in the green box

Explanation:

Step1: Recall rotation rule

The rule for a \(270^\circ\) counterclockwise rotation about the origin is \((x,y)\to(y, -x)\).

Step2: Apply the rule to point \(P=(2,-1)\)

Here, \(x = 2\) and \(y=-1\). Using the rule, the new \(x\)-coordinate is \(y=-1\) and the new \(y\)-coordinate is \(-x=-2\)? Wait, no, wait. Wait, the correct rule for \(270^\circ\) counterclockwise (or \(90^\circ\) clockwise) is \((x,y)\to(y, -x)\)? Wait, no, let's correct. The standard rotation rules:
- \(90^\circ\) counterclockwise: \((x,y)\to(-y,x)\)
- \(180^\circ\) counterclockwise: \((x,y)\to(-x,-y)\)
- \(270^\circ\) counterclockwise: \((x,y)\to(y, -x)\)? Wait, no, actually, \(270^\circ\) counterclockwise is equivalent to \(90^\circ\) clockwise. The formula for \(90^\circ\) clockwise (which is same as \(270^\circ\) counterclockwise) is \((x,y)\to(y, -x)\)? Wait, no, let's derive it.

A \(270^\circ\) counterclockwise rotation about the origin. Let's use the rotation matrix. The rotation matrix for \(\theta\) counterclockwise is \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\). For \(\theta = 270^\circ\), \(\cos270^\circ=0\), \(\sin270^\circ=-1\). So the matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\) (wait, no: \(\cos270 = 0\), \(-\sin270= -(-1)=1\); \(\sin270=-1\), \(\cos270 = 0\). So the matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). So when we multiply this matrix with the vector \(\begin{pmatrix}x\\y\end{pmatrix}\), we get \(\begin{pmatrix}0\times x + 1\times y\\-1\times x+0\times y\end{pmatrix}=\begin{pmatrix}y\\-x\end{pmatrix}\). Wait, no, wait: \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\cdot x + 1\cdot y\\-1\cdot x + 0\cdot y\end{pmatrix}=\begin{pmatrix}y\\-x\end{pmatrix}\). Wait, but let's test with a point. Let's take \((1,0)\). Rotating \(270^\circ\) counterclockwise: it should go to \((0,1)\)? Wait, no. Wait, \((1,0)\) rotated \(90^\circ\) counterclockwise is \((0,1)\), \(180^\circ\) is \((-1,0)\), \(270^\circ\) is \((0,-1)\). Wait, using the matrix: \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}0\\-1\end{pmatrix}\), which is correct. So the rule is \((x,y)\to(y, -x)\)? Wait, for \((1,0)\), \(y = 0\), \(-x=-1\)? No, wait, the result was \((0,-1)\). Wait, I messed up the matrix. Let's recalculate the rotation matrix. The correct rotation matrix for \(\theta\) counterclockwise is \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\). For \(\theta = 270^\circ\), \(\cos270^\circ = 0\), \(\sin270^\circ=-1\). So:

\(\cos\theta = 0\), \(-\sin\theta = -(-1)=1\)

\(\sin\theta=-1\), \(\cos\theta = 0\)

So the matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). So multiplying by \(\begin{pmatrix}x\\y\end{pmatrix}\) gives \(\begin{pmatrix}0\cdot x + 1\cdot y\\-1\cdot x + 0\cdot y\end{pmatrix}=\begin{pmatrix}y\\-x\end{pmatrix}\). Wait, for \((1,0)\), that gives \((0,-1)\), which is correct (270 degrees counterclockwise from (1,0) is (0,-1)). For \((0,1)\), it gives \((1,0)\), which is correct (270 degrees counterclockwise from (0,1) is (1,0)). For \((2,-1)\), applying the rule: \(x = 2\), \(y=-1\). So new \(x\) is \(y=-1\), new \(y\) is \(-x=-2\)? Wait, no, wait the matrix gave \(\begin{pmatrix}y\\-x\end{pmatrix}\), so the new point is \((y, -x)\). So for \(P=(2,-1)\), \(y=-1\), \(-x=-2\)? Wait, but let's check with another method. Rotating 270 degrees counterclockwise is the same as rotating 90 degrees clockwise. The rule for 90 degrees clockwise is \((x,y)\to(y, -x)\)? Wait, no, the rule for 90 degrees clockwise is \((x,y)\to(y, -x)\)? Wait, no, let's use a different approach. Let's consider the quadrants. The point \(P=(2,-1)\) is in the fourth quadrant. Rotating 270 degrees counterclockwise: 90 degrees is first quadrant, 180 degrees is second, 270 degrees is third? Wait, no. Wait, 270 degrees counterclockwise from fourth quadrant: let's think of the rotation direction. Counterclockwise 270 degrees is the same as clockwise 90 degrees. So a point \((x,y)\) rotated 90 degrees clockwise: the x and y swap, and the new x is the old y, the new y is the negative of the old x. Wait, let's take (2,-1). Rotate 90 degrees clockwise: the image should be (-1, -2)? No, wait, no. Wait, let's use the rotation formula correctly. The correct formula for a \(270^\circ\) counterclockwise rotation about the origin is \((x, y) \to (y, -x)\)? Wait, no, I think I had it wrong. Let's check with an example. Let's take point (1, 2). Rotate 270 degrees counterclockwise. The rotation matrix:

\(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}2\\-1\end{pmatrix}\). So (1,2) becomes (2,-1). Now, rotate (2,-1) 270 degrees counterclockwise: \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}-1\\-2\end{pmatrix}\). Wait, but let's see the angle. Alternatively, the rule for \(270^\circ\) counterclockwise is the same as \(90^\circ\) clockwise, and the formula for \(90^\circ\) clockwise is \((x, y) \to (y, -x)\)? Wait, no, in the example above, (1,2) rotated 90 degrees clockwise should be (2, -1), which matches the matrix result. So (x,y) → (y, -x) for 90 degrees clockwise (270 degrees counterclockwise). So for point \(P=(2,-1)\), applying the rule: \(x = 2\), \(y = -1\). So the new x-coordinate is \(y = -1\), and the new y-coordinate is \(-x = -2\)? Wait, but in the matrix example, (1,2) became (2,-1), which is (y, -x) because y=2, -x=-1? No, wait (1,2): y=2, -x=-1? No, (1,2) → (2, -1), which is (y, -x) where y=2, -x=-1? Wait, no, 1's negative is -1, so (y, -x) is (2, -1), which is correct. So for (2,-1), y=-1, -x=-2, so the image is (-1, -2)? Wait, no, wait the matrix multiplication for (2,-1) is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}0*2 + 1*(-1)\\-1*2 + 0*(-1)\end{pmatrix}=\begin{pmatrix}-1\\-2\end{pmatrix}\). So the image is (-1, -2). Wait, but the problem says "the number that belongs in the green box" – maybe the first coordinate? Wait, the green box is the first [?], so the x-coordinate of the image. Wait, let's re-express the rotation rule correctly. Wait, maybe I mixed up the rotation direction. Let's check the standard rotation rules:

- \(90^\circ\) counterclockwise: \((x, y) \to (-y, x)\)
- \(180^\circ\) counterclockwise: \((x, y) \to (-x, -y)\)
- \(270^\circ\) counterclockwise: \((x, y) \to (y, -x)\) Wait, no, that can't be. Wait, let's use a different source. The correct rule for \(270^\circ\) counterclockwise rotation about the origin is \((x, y) \to (y, -x)\)? No, actually, the correct formula is: when rotating a point \((x, y)\) \(270^\circ\) counterclockwise about the origin, the new coordinates are \((y, -x)\). Wait, no, let's take (3,4). Rotate 270 degrees counterclockwise. Using the rotation matrix:

\(\cos270 = 0\), \(\sin270 = -1\)

Rotation matrix: \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\)

So \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}3\\4\end{pmatrix}=\begin{pmatrix}4\\-3\end{pmatrix}\). So (3,4) becomes (4, -3), which is (y, -x) where y=4, -x=-3. So that's correct. So for (2, -1), applying the rule: y = -1, -x = -2. So the image is (-1, -2)? Wait, no, (2, -1): x=2, y=-1. So (y, -x) is (-1, -2). So the first coordinate (the green box) is -1? Wait, but let's check with another method. Let's consider the rotation as three 90-degree counterclockwise rotations. A 90-degree counterclockwise rotation of (2, -1) is (-(-1), 2) = (1, 2) (using the 90-degree rule (x,y)→(-y,x)). Then a second 90-degree (180 total) is (-1, -2). Then a third 90-degree (270 total) is (2, 1)? Wait, no, that's conflicting. Wait, no, 90-degree counterclockwise: (x,y)→(-y,x). So (2,-1) rotated 90 degrees counterclockwise: -y = 1, x=2 → (1, 2). Rotated another 90 (180 total): -y = -2, x=1 → (-2, 1). Rotated another 90 (270 total): -y = -1, x=-2 → (-1, -2). Ah, there we go. So after 270 degrees counterclockwise, (2,-1) becomes (-1, -2). So the first coordinate (the green box) is -1.

Wait, but let's confirm with the rotation matrix again. For (2, -1):

\(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}0*2 + 1*(-1)\\-1*2 + 0*(-1)\end{pmatrix}=\begin{pmatrix}-1\\-2\end{pmatrix}\). Yes, so the image is (-1, -2). So the number in the green box (the first coordinate) is -1.

Answer:

-1