Question

reasoning

1. line ( m ) is shown below with an arbitrary point ( a ) marked on the line. point ( c ), not on line ( m ) (obviously), is marked. if point ( a ) is rotated ( 180^{circ} ) about point ( c ) to produce image point ( a ), answer the following:

(a) using a straightedge with a ruler, locate point ( a ) as best as you can.
(b) what must be true about points ( a, c ), and ( a )?
(c) could point ( a ) also be on line ( m )? why or why not?

Explanation:

Part (a)

Step1: Connect \(A\) and \(C\)

Use a straight - edge to draw the line segment \(AC\).

Step2: Extend \(AC\)

Extend the line segment \(AC\) beyond \(C\) such that \(CA'=CA\). The point \(A'\) is the required image of \(A\) after a \(180^{\circ}\) rotation about \(C\).

Part (b)

Step1: Recall rotation property

A \(180^{\circ}\) rotation about a point \(C\) of a point \(A\) to \(A'\) implies that \(C\) is the mid - point of the line segment \(AA'\). By the definition of a mid - point, \(AC = CA'\) and \(A\), \(C\), \(A'\) are collinear. So, the points \(A\), \(C\), and \(A'\) are collinear and \(C\) is the mid - point of \(AA'\).

Part (c)

Step1: Use the property of collinearity and non - collinearity (in general)

If \(A'\) were on line \(m\), then since \(A\) is on line \(m\) and \(A\), \(C\), \(A'\) are collinear (from part (b)), the line \(AA'\) (which is the same as the line \(CA'\) or \(CA\)) would be the line \(m\). But \(C\) is not on line \(m\).
Let's assume, for the sake of contradiction, that \(A'\) is on \(m\). Since \(A\) is on \(m\) and \(A\), \(C\), \(A'\) are collinear (by the \(180^{\circ}\) rotation property), then the line passing through \(A\) and \(A'\) (which is also the line passing through \(C\) and \(A\) and \(C\) and \(A'\)) is line \(m\). But this would mean \(C\) is on line \(m\), which contradicts the given condition that \(C\) is not on line \(m\). So, point \(A'\) cannot be on line \(m\).

Answer:

(a) Locate \(A'\) by extending \(AC\) such that \(CA' = CA\).
(b) Points \(A\), \(C\), and \(A'\) are collinear and \(C\) is the mid - point of \(AA'\).
(c) No. If \(A'\) were on \(m\), then since \(A\) is on \(m\) and \(A\), \(C\), \(A'\) are collinear (from the rotation property), \(C\) would be on \(m\) (because the line through \(A\) and \(A'\) is the same as the line through \(A\), \(C\), and \(A'\)), which contradicts the fact that \(C\) is not on \(m\).