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? 4. where will r be if it is rotated 180° about a?

Explanation:

Problem 3:

Step1: Understand Rotation Direction

A \( 90^\circ \) clockwise rotation about a point \( A \) means we rotate the point \( W \) around \( A \) such that the angle of rotation is \( 90^\circ \) in the clockwise direction (right - turning direction). In a coordinate - like system (assuming \( A \) is the center), for a point \( (x,y) \) relative to the center \( A \), a \( 90^\circ \) clockwise rotation transforms the coordinates \((x,y)\) to \((y, - x)\) (in a standard coordinate system with \( A \) at the origin). Looking at the figure, point \( W \) is to the left of \( A \). When we rotate \( W \) \( 90^\circ \) clockwise about \( A \), we can visualize the rotation. The horizontal and vertical distances from \( W \) to \( A \) will swap and the sign of the horizontal distance (in terms of direction) will change. From the figure, after a \( 90^\circ \) clockwise rotation about \( A \), the position of \( W \) will coincide with the position of \( T \). Wait, no, let's re - examine. Let's consider the square - like structure. The figure is a square with center \( A \). The vertices around \( A \): \( W \) is on the left, \( Z \) on top, \( Y \) on the right, \( X \) at the bottom. The mid - points: \( Q \), \( R \), \( S \), \( T \). When we rotate \( W \) \( 90^\circ \) clockwise about \( A \), the vector from \( A \) to \( W \) is horizontal (left - ward). After a \( 90^\circ \) clockwise rotation, this vector becomes vertical (down - ward). So the new position of \( W \) after \( 90^\circ \) clockwise rotation about \( A \) is \( T \)? Wait, no, let's think again. Let's assume the coordinates: Let \( A \) be \((0,0)\), \( W \) be \((- a,0)\) (for some \( a>0 \)). A \( 90^\circ \) clockwise rotation of a point \((x,y)\) about the origin is given by the transformation \((x,y)\to(y, - x)\). So for \( W = (-a,0)\), the new point is \((0,a)\)? No, that's not right. Wait, the standard rotation matrix for \( 90^\circ \) clockwise is \(

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

\). So if \( W \) has coordinates \((x,y)\) relative to \( A \) (the center), then after rotation, the new coordinates \((x',y')\) are given by \( x'=y \) and \( y'=-x \). If \( W \) is at \((-k,0)\) (left of \( A \)), then \( x=-k \), \( y = 0 \). Then \( x'=0 \), \( y'=k \). But in the figure, the top - midpoint is \( Q \)? Wait, maybe my coordinate system is wrong. Let's look at the figure: The square has vertices \( W \), \( Z \), \( Y \), \( X \) (in order, left, top, right, bottom). The mid - sides: \( Q \) (mid of \( WZ \)), \( R \) (mid of \( ZY \)), \( S \) (mid of \( YX \)), \( T \) (mid of \( XW \)). Center \( A \). So \( W \) is on the left side, \( T \) is the mid - point of the bottom - left side? No, \( T \) is the mid - point of \( XW \), \( S \) mid - point of \( YX \), \( R \) mid - point of \( ZY \), \( Q \) mid - point of \( WZ \). So when we rotate \( W \) \( 90^\circ \) clockwise about \( A \), the direction from \( A \) to \( W \) is left. After a \( 90^\circ \) clockwise rotation, the direction becomes down. So the point \( W \) will move to the mid - point of the bottom side? No, \( T \) is the mid - point of \( XW \). Wait, maybe a better way: Rotating a point \( 90^\circ \) clockwise about the center \( A \) in a square - like figure. The left vertex \( W \), when rotated \( 90^\circ \) clockwise, will end up at the bottom - midpoint? No, let's take specific points. Let's assume the square has side length \( 2 \), \( A \) at \((0,0)\), \( W=(- 1,0)\), \( Z=(0,1)\), \( Y=(1,0)\), \( X=(0, - 1)\), \( Q=(-0.5,0.5)\), \( R=(0.5,0.5)\), \(…

Step1: Understand \( 180^\circ \) Rotation

A \( 180^\circ \) rotation about a point \( A \) means that the point \( R \) is rotated such that the angle between the original position vector \( \vec{AR} \) and the new position vector \( \vec{AR'} \) is \( 180^\circ \). In other words, the new point \( R' \) is such that \( A \) is the mid - point of the segment \( RR' \). For a point \((x,y)\) relative to \( A \) (the center), a \( 180^\circ \) rotation transforms the coordinates \((x,y)\) to \((-x, - y)\). Looking at the figure, point \( R \) is on the top - right mid - side. The mid - point of the bottom - left mid - side is \( T \)? No, let's consider the coordinates. Let \( A=(0,0) \), \( R=(a,b) \). After a \( 180^\circ \) rotation, the new coordinates are \((-a, - b)\). In the figure, \( R \) is the mid - point of \( ZY \) (top - right side), and \( T \) is the mid - point of \( XW \) (bottom - left side). The center \( A \) is the mid - point between \( R \) and \( T \) because a \( 180^\circ \) rotation about \( A \) will map \( R \) to a point such that \( A \) is the mid - point. So when we rotate \( R \) \( 180^\circ \) about \( A \), the new position of \( R \) is \( T \)? Wait, no, let's take the earlier coordinate example where \( R=(0.5,0.5) \), \( A=(0,0) \). A \( 180^\circ \) rotation of \( (0.5,0.5) \) about \( (0,0) \) is \( (-0.5, - 0.5) \), which is \( T \).

Step2: Confirm the Rotation

By the property of \( 180^\circ \) rotation about the center \( A \), the point \( R \) (on the top - right mid - side) when rotated \( 180^\circ \) about \( A \) will be at the position of \( T \) (on the bottom - left mid - side) since \( A \) is the mid - point of the segment joining \( R \) and \( T \).

Answer:

\( T \)

Problem 4: