Question

1. lines u and v intersect to form a 25° angle. if the triangle is reflected over u and then reflected over v about a, describe the resulting transformation.
2. in the picture below, u || v. if the parallelogram shown reflects over u then over v, describe the resulting transformation.

Explanation:

Problem 12

Step1: Recall Reflection Properties

When a figure is reflected over two intersecting lines, the composition of the two reflections is a rotation about the point of intersection. The angle of rotation is twice the angle between the two lines.

Step2: Determine Rotation Angle

The angle between lines \( u \) and \( v \) is \( 25^\circ \). So the rotation angle is \( 2 \times 25^\circ = 50^\circ \). The center of rotation is the intersection point \( A \) of lines \( u \) and \( v \).

Step1: Recall Reflection Over Parallel Lines

When a figure is reflected over two parallel lines, the composition of the two reflections is a translation. The distance of the translation is twice the distance between the two parallel lines.

Step2: Calculate Translation Distance

The distance between lines \( u \) and \( v \) is \( \frac{1}{2} \) inch. So the translation distance is \( 2 \times \frac{1}{2} = 1 \) inch. The direction of the translation is parallel to the lines (since the lines are vertical, the translation is horizontal? Wait, no—wait, the lines \( u \) and \( v \) are vertical (from the diagram, arrows are vertical). Wait, the parallelogram is to the left of line \( u \). Reflecting over \( u \) then over \( v \): the distance between \( u \) and \( v \) is \( \frac{1}{2} \) in. So the total translation is \( 2 \times \frac{1}{2} = 1 \) inch in the direction perpendicular to the lines? Wait, no—when reflecting over two parallel lines, the translation is in the direction perpendicular to the lines, and the distance is twice the distance between the lines. Wait, the lines \( u \) and \( v \) are parallel and vertical (from the diagram, arrows are vertical). So the distance between them is \( \frac{1}{2} \) inch (horizontal distance? Wait, no, the diagram shows \( \frac{1}{2} \) in. between \( u \) and \( v \) vertically? Wait, no, the arrows are vertical, so the lines are vertical, and the distance between them is horizontal? Wait, maybe the diagram has \( u \) and \( v \) as parallel vertical lines, with \( \frac{1}{2} \) inch between them horizontally. Then reflecting over \( u \) (a vertical line) and then over \( v \) (another vertical line) would result in a horizontal translation? Wait, no—reflection over a vertical line changes the horizontal position. Let's think: if a point is at distance \( d \) from line \( u \) (to the left), reflecting over \( u \) moves it to \( d \) to the right of \( u \). Then reflecting over \( v \) (which is \( \frac{1}{2} \) in. to the right of \( u \)), the distance from the reflected point (after first reflection) to \( v \) is \( \frac{1}{2} - d \) (if \( d < \frac{1}{2} \))? Wait, maybe better to use the rule: reflection over two parallel lines \( l_1 \) and \( l_2 \) (distance \( h \) apart) results in a translation of \( 2h \) in the direction perpendicular to the lines. Since the lines are vertical, the direction perpendicular is horizontal. Wait, the diagram shows \( \frac{1}{2} \) in. between \( u \) and \( v \) (probably horizontal distance, since the lines are vertical). So the translation distance is \( 2 \times \frac{1}{2} = 1 \) inch, and the direction is parallel to the direction perpendicular to the lines (i.e., horizontal, to the right, since we reflect over \( u \) then \( v \)).

Step3: Describe the Translation

The resulting transformation is a translation. The distance of the translation is \( 1 \) inch (twice the \( \frac{1}{2} \) inch distance between \( u \) and \( v \)), and the direction is parallel to the direction perpendicular to the parallel lines (since \( u \) and \( v \) are vertical, the translation is horizontal, to the right, by \( 1 \) inch).

Answer:

The resulting transformation is a rotation about point \( A \) (the intersection of \( u \) and \( v \)) by \( 50^\circ \) (twice the \( 25^\circ \) angle between \( u \) and \( v \)).

Problem 13