Question

lesson 9 practice problems 1 a. draw parallel lines ab and cd. b. pick any point e. rotate ab 90° clockwise around e. c. rotate line cd 90° clockwise around e. d. what do you notice?

Explanation:

Part a: Drawing Parallel Lines AB and CD

To draw parallel lines \( AB \) and \( CD \):

1. Draw Line \( AB \): Use a ruler to draw a straight line segment and label its endpoints as \( A \) and \( B \).
2. Draw Line \( CD \) Parallel to \( AB \):

• Choose a point \( C \) not on line \( AB \).
• Use a set square or a ruler with a right angle to ensure that the angle between the new line and a transversal (if drawn) is the same as for \( AB \), or use the property that parallel lines have the same slope (in coordinate geometry) or are equidistant at all points. Draw line \( CD \) such that it never intersects \( AB \) and is oriented the same way as \( AB \).

Part b: Rotating \( AB \) \( 90^\circ \) Clockwise Around \( E \)

1. Choose Point \( E \): Select any point (it can be on \( AB \), off \( AB \), or even on the plane outside the initial lines). Let's say \( E \) is a point in the plane.
2. Rotate \( AB \) Around \( E \) by \( 90^\circ \) Clockwise:

• For a point \( A \) on line \( AB \), the vector from \( E \) to \( A \) is \( \vec{EA} \). To rotate this vector \( 90^\circ \) clockwise, if \( \vec{EA}=(x,y) \), the rotated vector \( \vec{EA'} \) becomes \( (y, -x) \) (using the rotation matrix for \( 90^\circ \) clockwise: \(

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

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

=

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

\)).

• Do the same for point \( B \) to get \( B' \). Then draw the line \( A'B' \) which is the rotated line \( AB \) around \( E \).

Part c: Rotating \( CD \) \( 90^\circ \) Clockwise Around \( E \)

1. Rotate \( CD \) Around \( E \) by \( 90^\circ \) Clockwise:

• For a point \( C \) on line \( CD \), find the vector \( \vec{EC} \). Rotate this vector \( 90^\circ \) clockwise using the same rotation matrix as above. If \( \vec{EC}=(m,n) \), the rotated vector \( \vec{EC'}=(n, -m) \).
• Do the same for point \( D \) to get \( D' \). Then draw the line \( C'D' \) which is the rotated line \( CD \) around \( E \).

Part d: What Do You Notice?

When we rotate two parallel lines (\( AB \) and \( CD \)) by the same angle (\( 90^\circ \) clockwise) around the same point (\( E \)):

• The resulting lines \( A'B' \) and \( C'D' \) are also parallel.
• This is because a rotation is a rigid transformation, and rigid transformations preserve the angle between lines. Since \( AB \parallel CD \), the angle between \( AB \) and any transversal is equal to the angle between \( CD \) and the same transversal. After rotating both lines by \( 90^\circ \), the angle between the rotated lines and the transversal (or between each other) remains such that they are parallel. Also, the distance between the original parallel lines and the distance between the rotated parallel lines have a consistent relationship (related to the rotation and the distance from \( E \) to the lines), but the key observation is that the rotated lines are parallel to each other, just as the original lines were.

Answer:

Part a: Drawing Parallel Lines AB and CD

To draw parallel lines \( AB \) and \( CD \):

1. Draw Line \( AB \): Use a ruler to draw a straight line segment and label its endpoints as \( A \) and \( B \).
2. Draw Line \( CD \) Parallel to \( AB \):

• Choose a point \( C \) not on line \( AB \).
• Use a set square or a ruler with a right angle to ensure that the angle between the new line and a transversal (if drawn) is the same as for \( AB \), or use the property that parallel lines have the same slope (in coordinate geometry) or are equidistant at all points. Draw line \( CD \) such that it never intersects \( AB \) and is oriented the same way as \( AB \).

Part b: Rotating \( AB \) \( 90^\circ \) Clockwise Around \( E \)

1. Choose Point \( E \): Select any point (it can be on \( AB \), off \( AB \), or even on the plane outside the initial lines). Let's say \( E \) is a point in the plane.
2. Rotate \( AB \) Around \( E \) by \( 90^\circ \) Clockwise:

• For a point \( A \) on line \( AB \), the vector from \( E \) to \( A \) is \( \vec{EA} \). To rotate this vector \( 90^\circ \) clockwise, if \( \vec{EA}=(x,y) \), the rotated vector \( \vec{EA'} \) becomes \( (y, -x) \) (using the rotation matrix for \( 90^\circ \) clockwise: \(

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

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

=

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

\)).

• Do the same for point \( B \) to get \( B' \). Then draw the line \( A'B' \) which is the rotated line \( AB \) around \( E \).

Part c: Rotating \( CD \) \( 90^\circ \) Clockwise Around \( E \)

1. Rotate \( CD \) Around \( E \) by \( 90^\circ \) Clockwise:

• For a point \( C \) on line \( CD \), find the vector \( \vec{EC} \). Rotate this vector \( 90^\circ \) clockwise using the same rotation matrix as above. If \( \vec{EC}=(m,n) \), the rotated vector \( \vec{EC'}=(n, -m) \).
• Do the same for point \( D \) to get \( D' \). Then draw the line \( C'D' \) which is the rotated line \( CD \) around \( E \).

Part d: What Do You Notice?

When we rotate two parallel lines (\( AB \) and \( CD \)) by the same angle (\( 90^\circ \) clockwise) around the same point (\( E \)):

• The resulting lines \( A'B' \) and \( C'D' \) are also parallel.
• This is because a rotation is a rigid transformation, and rigid transformations preserve the angle between lines. Since \( AB \parallel CD \), the angle between \( AB \) and any transversal is equal to the angle between \( CD \) and the same transversal. After rotating both lines by \( 90^\circ \), the angle between the rotated lines and the transversal (or between each other) remains such that they are parallel. Also, the distance between the original parallel lines and the distance between the rotated parallel lines have a consistent relationship (related to the rotation and the distance from \( E \) to the lines), but the key observation is that the rotated lines are parallel to each other, just as the original lines were.