Question

quadrilateral pqrs is rotated 90° clockwise about the origin to form quadrilateral pqrs. what is the y-coordinate of point p?

Explanation:

Step1: Find coordinates of P

From the graph, point \( P \) has coordinates \( (-1, 2) \).

Step2: Apply 90° clockwise rotation rule

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

For point \( P(-1, 2) \), applying the rule:
\( x = -1 \), \( y = 2 \)
New \( x \)-coordinate: \( 2 \)
New \( y \)-coordinate: \( -(-1) = 1 \)? Wait, no, wait. Wait, the rule is \( (x, y) \) rotated 90° clockwise becomes \( (y, -x) \). Wait, let's recheck. Wait, 90° clockwise rotation: the formula is \( (x, y) \mapsto (y, -x) \)? Wait, no, actually, the correct rule for 90° clockwise rotation about the origin is \( (x, y) \to (y, -x) \)? Wait, no, let's recall:

Wait, standard rotation: 90° clockwise: \( (x, y) \) becomes \( (y, -x) \)? Wait, no, maybe I mixed up. Wait, let's take a point (1,0). Rotate 90° clockwise: it should go to (0, -1)? No, wait, (1,0) rotated 90° clockwise around origin: the rotation matrix for 90° clockwise is \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \). So applying that to \( \begin{pmatrix} x \\ y \end{pmatrix} \) 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, but (1,0) would become (0, -1)? But actually, (1,0) rotated 90° clockwise is (0, -1)? Wait, no, (1,0) is on the positive x-axis. Rotating 90° clockwise would point it down the negative y-axis? Wait, no, 90° clockwise from positive x-axis is positive y-axis? Wait, no, maybe I have the rotation direction wrong. Wait, 90° clockwise: imagine the point (1,0) (right on x-axis). Rotating 90° clockwise (turning right) would move it to (0, -1)? Wait, no, that's 270° counterclockwise. Wait, maybe I got the rule reversed. Let's check with a point (0,1) (top of y-axis). Rotating 90° clockwise: it should go to (1, 0) (right on x-axis). Let's apply the matrix: \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \), which is correct. So (0,1) becomes (1,0). So the rule is \( (x, y) \to (y, -x) \)? Wait, (0,1) has x=0, y=1. Then (y, -x) is (1, 0), which is correct. Another example: (1,1). Rotating 90° clockwise: should be (1, -1)? Wait, no, (1,1) rotated 90° clockwise: using the matrix, \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \). Let's visualize: (1,1) is in the first quadrant. Rotating 90° clockwise would move it to the fourth quadrant, with x=1, y=-1? Wait, no, maybe I'm confused. Wait, maybe the correct rule is 90° clockwise: \( (x, y) \to (y, -x) \), and 90° counterclockwise is \( (x, y) \to (-y, x) \). Let's confirm with (2,3). 90° clockwise: (3, -2). 90° counterclockwise: (-3, 2). Let's see on a graph: (2,3) is in Q1. 90° clockwise: turn right 90°, so the x becomes the original y, and y becomes negative original x. So (3, -2) is in Q4. 90° counterclockwise: turn left 90°, x becomes -original y, y becomes original x: (-3, 2) is in Q2. That makes sense.

So back to point \( P(-1, 2) \). Applying 90° clockwise rotation: \( (x, y) \to (y, -x) \). So \( x = -1 \), \( y = 2 \). So new x is \( 2 \), new y is \( -(-1) = 1 \)? Wait, no: \( -x \) when x is -1 is \( -(-1) = 1 \). Wait, but let's check the coordinates. Wait, point \( P \) is at (-1, 2). So x = -1, y = 2. Applying the rule \( (x, y) \to (y, -x) \), so new coordinates are (2, -(-1)) = (2, 1). Wait, but let's visualize. Rotating (-1, 2) 90° clockwise around origin. Let's plot (-1,2): it's in Q2 (x negative, y positive). Rotating 90° clockwise: it should move to Q1 (x positive, y positive)? Wait, no, Q2 to Q1? Wait, 90° clockwise from Q2: the original point is (-1,2). The vector from origin is (-1,2). Rotating 90° clockwise: the new vector should be (2, 1), because the rotation matrix for 90° clockwise is \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \), so multiplying by \( \begin{pmatrix} -1 \\ 2 \end{pmatrix} \) gives \( \begin{pmatrix} 0*(-1) + 1*2 \\ -1*(-1) + 0*2 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \). So the new point \( P' \) is (2, 1). Therefore, the y-coordinate of \( P' \) is 1.

Wait, but let me check again. Maybe I made a mistake in the rotation rule. Let's use another method. The formula for 90° clockwise rotation is \( (x, y) ightarrow (y, -x) \). So for \( P(-1, 2) \), x = -1, y = 2. So new x is y = 2, new y is -x = -(-1) = 1. So yes, the y-coordinate is 1.

Answer:

1