Question

reasoning
in this section, you will be exploring why the rotation of a line by 180° about a point not on the line produces an image parallel to the original. in the diagram shown, a point a, on line m, has rotated about point c by 180°.

1. what is true about points a, c, and a?
2. explain why point a could never lie on the original line m.

(hint: consider where c would need to lie for a to lie on line m.)

1. since no image point could lie on the original line m (the result from #10), then how many points could the original line and image line share in common? what does this imply about the two lines?

Explanation:

9.

When a point \(A\) is rotated \(180^{\circ}\) about a point \(C\), the definition of a \(180^{\circ}\) rotation implies that \(C\) is the mid - point of the segment \(AA'\). By the mid - point formula, if \(A=(x_1,y_1)\) and \(A'=(x_2,y_2)\), then \(C = (\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). So, points \(A\), \(C\), and \(A'\) are collinear and \(C\) is the mid - point of segment \(AA'\).

Suppose \(A'\) lies on line \(m\). If \(A'\) lies on line \(m\) and \(C\) is the mid - point of \(AA'\) (from the property of \(180^{\circ}\) rotation), then by the mid - point formula, if \(A\) is on line \(m\) and \(A'\) is on line \(m\), the mid - point \(C\) of \(AA'\) would also lie on line \(m\) (since the set of mid - points of segments with endpoints on a line is also on that line, if the line is straight). But the problem states that \(C\) is not on line \(m\).

Since no image point (including \(A'\) and all other points obtained by \(180^{\circ}\) rotation of points on line \(m\) about \(C\)) lies on the original line \(m\), the number of common points between the original line \(m\) and its image line (the line containing \(A'\) and other rotated points) is \(0\). In Euclidean geometry, two lines in a plane with no common points are parallel.

Answer:

Points \(A\), \(C\), and \(A'\) are collinear and \(C\) is the mid - point of segment \(AA'\).

10.