Question

90° clockwise rotation
rule (x, y): (□, □)
s(□, □)
t(□, □)
u(□, □)
v(□, □)

Explanation:

Step1: Recall 90° clockwise rotation rule

The rule for a \(90^\circ\) clockwise rotation about the origin is \((x, y) \to (y, -x)\).

Step2: Find original coordinates

First, identify the original coordinates of points \(S\), \(T\), \(U\), \(V\):

• \(S\): From the graph, \(S\) is at \((2, 3)\)
• \(T\): \(T\) is at \((-1, 0)\)
• \(U\): \(U\) is at \((4, -1)\)
• \(V\): \(V\) is at \((5, 2)\)

Step3: Apply rotation rule to each point

• For \(S(2, 3)\): Apply \((x, y) \to (y, -x)\), so \(S'\) is \((3, -2)\)
• For \(T(-1, 0)\): Apply the rule, \(T'\) is \((0, 1)\) (since \(x=-1\), \(-x = 1\))
• For \(U(4, -1)\): Apply the rule, \(U'\) is \((-1, -4)\) (Wait, no: \(x = 4\), \(y=-1\); so \((y, -x)=(-1, -4)\)? Wait, no, wait: \(90^\circ\) clockwise: \((x,y)\to(y, -x)\). So \(x = 4\), \(y=-1\): \(y=-1\), \(-x=-4\), so \((-1, -4)\)? Wait, no, let's recheck. Wait, original \(U\) is at \((4, -1)\)? Wait, looking at the graph, \(U\) is at (4, -1)? Wait, the grid: x-axis from -5 to 5, y-axis from -5 to 5. Let's re-express:

Wait, maybe I misread the coordinates. Let's re-identify:

• \(S\): x=2, y=3 (correct, as per the graph, 2 units right on x, 3 up on y)
• \(T\): x=-1, y=0 (on the origin? Wait, no, T is at (-1, 0)? Wait, the graph shows T at (-1, 0)? Wait, the line from T to S: T is at (-1, 0), S at (2,3), T to S: slope (3-0)/(2 - (-1))= 3/3=1, which matches the line. Then U: looking at the graph, U is at (4, -1) (x=4, y=-1). V is at (5, 2) (x=5, y=2).

Now apply \(90^\circ\) clockwise rotation: \((x, y) \to (y, -x)\)

• \(S(2, 3)\): \(y = 3\), \(-x = -2\) → \((3, -2)\)
• \(T(-1, 0)\): \(y = 0\), \(-x = 1\) → \((0, 1)\)
• \(U(4, -1)\): \(y = -1\), \(-x = -4\) → \((-1, -4)\)? Wait, no, wait: \(90^\circ\) clockwise rotation: the formula is \((x, y) \to (y, -x)\). So for \(U(4, -1)\): \(x=4\), \(y=-1\), so new x is \(y=-1\), new y is \(-x=-4\), so \((-1, -4)\). Wait, but let's check with another method. A \(90^\circ\) clockwise rotation is equivalent to a \(270^\circ\) counterclockwise rotation. Alternatively, matrix transformation: \(

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

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

=

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

\), which confirms the rule.

• \(V(5, 2)\): \(y = 2\), \(-x = -5\) → \((2, -5)\)

Wait, but let's verify with an example. Take a point (1,0), rotate 90° clockwise: should be (0, -1). Using the rule: (1,0)→(0, -1), correct. Take (0,1), rotate 90° clockwise: (1, 0), correct (since (0,1)→(1, 0) (y=1, -x=0? Wait, no: (0,1)→(1, 0)? Wait, no, wait: (x,y)=(0,1), so (y, -x)=(1, 0). Yes, that's correct. (0,1) rotated 90° clockwise is (1,0), which matches. So the rule is correct.

So rechecking each point:

• \(S(2, 3)\): \((y, -x) = (3, -2)\) → \(S'(3, -2)\)
• \(T(-1, 0)\): \((0, -(-1)) = (0, 1)\) → \(T'(0, 1)\) (since \(x=-1\), \(-x = 1\))
• \(U(4, -1)\): \((-1, -4)\) → \(U'(-1, -4)\)
• \(V(5, 2)\): \((2, -5)\) → \(V'(2, -5)\)

Wait, but let's check the original coordinates again. Maybe I misread U. Let's look at the graph: U is at (4, -1)? The grid lines: x=4, y=-1 (since it's below the x-axis, y=-1). Yes. V is at (5, 2) (x=5, y=2). T is at (-1, 0) (x=-1, y=0). S is at (2, 3) (x=2, y=3).

So applying the rotation rule:

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

So:

• \(S(2, 3)\): \(y=3\), \(-x=-2\) → (3, -2)
• \(T(-1, 0)\): \(y=0\), \(-x=1\) → (0, 1)
• \(U(4, -1)\): \(y=-1\), \(-x=-4\) → (-1, -4)
• \(V(5, 2)\): \(y=2\), \(-x=-5\) → (2, -5)

Answer:

Rule: \((x, y) \to (y, -x)\)

\(S'(3, -2)\)

\(T'(0, 1)\)

\(U'(-1, -4)\)

\(V'(2, -5)\)