Question

for exercises 3-7, use the figure to answer the questions. 3. where will w be if it is rotated 90° clockwise about a?

Explanation:

Step1: Understand Rotation Rules

A \( 90^\circ \) clockwise rotation about a point \( (x,y) \) has the transformation rule: for a point \( (x_1,y_1) \), the new point \( (x_2,y_2) \) is given by \( (y_1 - y + x, -x_1 + x + y) \) (or more simply, for rotation about the origin, \( (x,y) \to (y, -x) \); adjusting for the center \( A \)). First, identify the position of \( W \) relative to \( A \). Let's assume \( A \) is the center (from the figure, \( A \) is the midpoint of \( WY \), and the figure is a square - like grid). Let's set coordinates: let \( A \) be at \( (0,0) \) for simplicity. Then \( W \) is at \( (-2,0) \) (assuming the grid has units, with \( W \) two units left of \( A \), \( Y \) two units right, etc.).

Step2: Apply \( 90^\circ \) Clockwise Rotation

The rule for \( 90^\circ \) clockwise rotation about a point \( (h,k) \) (here \( h = 0,k = 0 \)) is \( (x,y)\to(y, -x) \). For \( W(-2,0) \), applying the rule: \( x=-2,y = 0 \), so new \( x = 0 \), new \( y=-(-2)=2 \)? Wait, maybe better to use the figure's structure. Looking at the figure, \( W \) is on the left side, \( A \) is the center. Rotating \( W \) \( 90^\circ \) clockwise about \( A \): the vector from \( A \) to \( W \) is left along the horizontal. A \( 90^\circ \) clockwise rotation of a horizontal left vector (along negative x - axis) will turn it into a vertical up vector (along positive y - axis). So the new position of \( W \) after rotating \( 90^\circ \) clockwise about \( A \) should be the point \( Q \)? Wait, no, let's look at the figure again. The figure has \( W \), \( A \), \( Y \) on the bottom horizontal line, \( Z \) on top, \( Q \) and \( R \) on the middle horizontal line. Wait, maybe the coordinates: let's assume \( A \) is at \( (0,0) \), \( W \) is at \( (-1, - 1) \), \( A \) at \( (0,0) \), then \( 90^\circ \) clockwise rotation: \( (x,y)\to(y, -x) \), so \( (-1,-1)\to(-1,1) \)? No, maybe the figure is a square with \( A \) as the center. Wait, the key is: when you rotate a point \( 90^\circ \) clockwise about a center, the horizontal distance from the center becomes vertical distance in the positive y - direction (if original was negative x - direction). Looking at the figure, \( W \) is on the left, \( A \) is the center. Rotating \( W \) \( 90^\circ \) clockwise about \( A \) will move it to the position of \( Q \)? Wait, no, maybe to the position of the point directly above \( A \)? Wait, no, let's think of the rotation: a \( 90^\circ \) clockwise rotation about \( A \): the line \( AW \) is horizontal (left - right). Rotating \( AW \) \( 90^\circ \) clockwise will make it vertical (up - down), with \( W \) moving to a point above \( A \)? Wait, no, the correct way: if \( W \) is at \( (x,y) \) relative to \( A \), after \( 90^\circ \) clockwise rotation, it's \( (y, -x) \). Suppose \( A \) is \( (0,0) \), \( W \) is \( (-a,0) \) (a > 0), then after rotation, it's \( (0,a) \), which would be the point \( Q \)? Wait, in the figure, \( Q \) is on the left - top, \( R \) on the right - top, \( Z \) on top - center. Wait, maybe the answer is \( Q \)? No, wait, let's check the figure again. The figure has \( W \) at the bottom - left, \( A \) at the center of the bottom side (midpoint of \( WY \)), \( Z \) at the top - center, \( Q \) is the top - left of the small square, \( R \) top - right. So when we rotate \( W \) \( 90^\circ \) clockwise about \( A \): the vector \( \overrightarrow{AW} \) is from \( A \) to \( W \), which is left along the bottom edge. A \( 90^\circ \) clockwise rotation of this vector will point upwards (since clockwise rotation of left - pointing vector is up - pointing). So the new point will be along the left - top edge, at the same distance from \( A \) as \( W \) is. So that point is \( Q \)? Wait, no, maybe the point is \( Z \)? No, \( Z \) is at the top - center. Wait, maybe the correct position is the point \( Q \) or the point directly above \( A \)? Wait, I think I made a mistake. Let's use the standard rotation: for a point \( (x,y) \) rotated \( 90^\circ \) clockwise about the origin, it's \( (y, -x) \). If \( A \) is the origin, and \( W \) is at \( (-1,0) \) (assuming unit grid), then after rotation, it's \( (0,1) \), which would be the point \( Q \) if \( Q \) is at \( (0,1) \)? Wait, maybe the answer is \( Q \). Wait, no, let's look at the figure again. The figure has \( W \), \( Q \), \( Z \) on the left edge (vertical), \( Y \), \( R \), \( Z \) on the right edge. \( A \) is the center of the bottom horizontal line \( WY \). So \( W \) is bottom - left, \( Y \) bottom - right, \( A \) middle. Rotating \( W \) \( 90^\circ \) clockwise about \( A \): the rotation will move \( W \) to a point such that the angle between \( AW \) and \( A \) - new point is \( 90^\circ \) clockwise. So \( AW \) is horizontal (left), so new vector is vertical (up). So the new point is the point above \( A \) on the left - top, which is \( Q \). Wait, but maybe the correct answer is \( Q \) or the point \( Z \)? No, \( Z \) is top - center. Wait, I think the correct position is \( Q \). Wait, no, let's do it with coordinates. Let \( A=(0,0) \), \( W=(-2,0) \), \( Y=(2,0) \), \( Z=(0,2) \), \( Q=(-1,1) \), \( R=(1,1) \). Wait, no, maybe the grid is 2x2. So \( W=(-1,-1) \), \( A=(0,0) \), \( Y=(1,-1) \), \( Z=(0,1) \), \( Q=(-1,0) \)? No, this is getting confusing. The key is: when you rotate a point \( 90^\circ \) clockwise about a center, the horizontal distance becomes vertical distance in the positive y - direction (if original was negative x - direction). So if \( W \) is to the left of \( A \), after \( 90^\circ \) clockwise rotation, it will be above \( A \), at the same distance. So in the figure, the point above \( A \) on the left side is \( Q \)? Wait, maybe the answer is \( Q \). Wait, no, let's check the standard rotation: for a point \( (x,y) \) rotated \( 90^\circ \) clockwise about \( (h,k) \), the formula is \( (h+(y - k),k-(x - h)) \). Let \( A=(h,k)=(0,0) \), \( W=(x,y)=(-a,0) \). Then new \( x = 0+(0 - 0)=0 \)? No, wait the formula is: \( x'=h+(y - k) \), \( y'=k-(x - h) \). So for \( W(-a,0) \), \( h = 0,k = 0 \), so \( x'=0+(0 - 0)=0 \), \( y'=0-(-a - 0)=a \). So the new point is \( (0,a) \), which is \( Z \)? Wait, \( Z \) is at \( (0,2) \) if \( a = 2 \). Oh! I see, \( Z \) is at the top - center. So if \( W \) is at \( (-2,0) \), \( A \) at \( (0,0) \), then rotating \( 90^\circ \) clockwise gives \( (0,2) \), which is \( Z \)? Wait, no, \( (x,y)\to(y, -x) \) for rotation about origin. So \( (-2,0)\to(0,2) \), which is \( Z \). Yes! Because \( Z \) is at the top - center. So the vector from \( A \) to \( W \) is \( (-2,0) \), rotating \( 90^\circ \) clockwise, the vector becomes \( (0,2) \), so the new point is \( A+(0,2)=(0,2) \), which is \( Z \).

Answer:

The point \( W \) will be at \( Z \) after a \( 90^\circ \) clockwise rotation about \( A \).