Question

from unit 1, lesson 14
three line segments form the letter n.
rotate the letter n clockwise around the midpoint of segment bc by 180°.
describe the result.
learning targets
+ i can prove alternate interior angles are congruent.
+ i can prove corresponding angles are congruent.

Explanation:

Step1: Recall the property of a \(180^{\circ}\) rotation

A \(180^{\circ}\) rotation about a point \(O\) (in this case, the mid - point of \(BC\)) maps a point \(P(x,y)\) to a point \(P'(-x + 2a,-y+2b)\) if \(O=(a,b)\). Geometrically, a \(180^{\circ}\) rotation about a point is equivalent to a point - reflection. When we rotate a line segment \(AB\) about the mid - point of \(BC\) by \(180^{\circ}\), the image of segment \(AB\) will be parallel and equal in length to the original segment. Similarly, the image of segment \(CD\) will be parallel and equal in length to the original segment.

Step2: Analyze the transformation of the letter \(N\)

The letter \(N\) is composed of three line segments \(AB\), \(BC\), and \(CD\). When we rotate the letter \(N\) \(180^{\circ}\) clockwise around the mid - point of \(BC\):

• The segment \(AB\) will be rotated to a position parallel to its original position. Let the mid - point of \(BC\) be \(M\). If we consider the vector from \(M\) to \(A\), after a \(180^{\circ}\) rotation, the new point \(A'\) (the image of \(A\)) will be such that \(M\) is the mid - point of \(AA'\).
• The segment \(BC\) is rotated about its mid - point. A \(180^{\circ}\) rotation of a line segment about its mid - point maps the segment onto itself (since if \(M\) is the mid - point of \(BC\), and we rotate \(B\) \(180^{\circ}\) about \(M\), we get \(C\) and vice - versa).
• The segment \(CD\) will be rotated to a position parallel to its original position.

The result of rotating the letter \(N\) \(180^{\circ}\) clockwise around the mid - point of \(BC\) is the letter \(N\) itself.

Answer:

The result of the rotation is the letter \(N\) itself.