Question

rotating a point (2, 7) 270 degrees clockwise around the origin will move it to what coordinates?
○ a. (-7,2)
○ b. (-2,7)
○ c. (7,-2)
○ d. (2,-7)

Explanation:

Step1: Recall rotation rule

The rule for rotating a point \((x, y)\) 270° clockwise (or 90° counterclockwise) around the origin is \((x, y) \to (-y, x)\).

Step2: Apply the rule to \((2, 7)\)

Here, \(x = 2\) and \(y = 7\). Substitute into the rule: \(-y=-7\) and \(x = 2\)? Wait, no, wait. Wait, correction: The correct rule for 270° clockwise rotation is \((x,y)\to(-y,x)\)? Wait, no, let's recheck. Actually, the rule for 270° clockwise rotation (or 90° counterclockwise) is \((x, y) \mapsto (-y, x)\)? Wait, no, let's derive it. A 90° counterclockwise rotation is \((x,y)\to(-y,x)\), and 270° clockwise is the same as 90° counterclockwise. Wait, no: 270° clockwise is equivalent to 90° counterclockwise. Wait, let's take a point \((x,y)\). Rotating 270° clockwise: first, 90° clockwise is \((y, -x)\), 180° clockwise is \((-x, -y)\), 270° clockwise is \((-y, x)\)? Wait, no, let's use the standard rotation matrices. The rotation matrix for \(\theta\) degrees 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 \(

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

\). So multiplying by \(

$$\begin{pmatrix}x\\y\end{pmatrix}$$

\) 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, first row: \(0x + (-1)y=-y\), second row: \(1*x + 0*y = x\). So the new point is \((-y, x)\). Wait, but let's test with a point. Take (1,0). Rotating 270° clockwise: it should go to (0,1)? Wait, no. Wait, (1,0) rotated 90° clockwise is (0,-1), 180° is (-1,0), 270° is (0,1). 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 (0,1). So the rule is \((x,y)\to(-y,x)\)? Wait, (1,0) becomes (0,1)? Wait, no, (x,y)=(1,0), -y=0, x=1, so (0,1). Correct. Now take (2,7). Then -y = -7, x=2? Wait, no, wait the matrix gave (-y, x)? Wait, no, the matrix multiplication gave \(

$$\begin{pmatrix}-y\\x\end{pmatrix}$$

\), so the coordinates are \((-y, x)\)? Wait, no, the first element is -y, the second is x. So (x,y)=(2,7) becomes (-7, 2)? Wait, but that's option a. Wait, but let's check another way. Rotating 270° clockwise is the same as rotating 90° counterclockwise. The rule for 90° counterclockwise is \((x,y)\to(-y,x)\). So (2,7) rotated 90° counterclockwise (270° clockwise) is (-7, 2). Let's verify with a graph. The point (2,7) is in the first quadrant. Rotating 270° clockwise: that's three 90° clockwise turns. First 90° clockwise: (7, -2) (since 90° clockwise rule is (x,y)→(y, -x)). Second 90° clockwise: (-2, -7) (applying (y, -x) to (7, -2) gives (-2, -7)). Third 90° clockwise: (-7, 2) (applying (y, -x) to (-2, -7) gives (-7, 2)). Yes, so after three 90° clockwise rotations (270° total), (2,7) becomes (-7,2).

Answer:

a. \((-7, 2)\)