Question

1. which of these sequences of transformations would not return a shape to its original position? a. translate 3 units up, then 3 units down. b. reflect over line p, then reflect over line p again. c. translate 1 unit to the right, then 4 units to the left, then 3 units to the right. d. rotate 120° counterclockwise around center c; then rotate 240° counterclockwise around c again.

Explanation:

Step1: Analyze Option A

Translating 3 units up and then 3 units down.
Let the original \(y -\) coordinate of a point be \(y\). After translating 3 units up, the \(y -\) coordinate becomes \(y + 3\). Then after translating 3 units down, the \(y -\) coordinate is \((y+3)-3=y\). But if we consider the entire shape (assuming it has non - zero \(x\) coordinates in general), a translation up and then down does not affect the \(x\) - coordinates. However, this is a non - identity transformation (unless the shape is only on the \(y\) - axis). In general, a translation up and then down is equivalent to \(T(x,y)=(x,y + 3-3)=(x,y)\) only when we consider the \(y\) - component. But if we think of the transformation as a composition \(T_2\circ T_1\) where \(T_1(x,y)=(x,y + 3)\) and \(T_2(x,y)=(x,y-3)\), the composition \(T_2\circ T_1(x,y)=(x,y)\). But this is a very simple case of translation. In fact, \(T_2\circ T_1\) is the identity transformation for all points \((x,y)\) in the plane.

Step2: Analyze Option B

Reflecting over line \(p\) and then reflecting over line \(p\) again.
Let \(R_p\) be the reflection transformation over line \(p\). If we have a point \(P(x,y)\), and we first apply \(R_p\) to get \(P_1 = R_p(P)\), and then apply \(R_p\) again to \(P_1\), we get \(R_p(P_1)=R_p(R_p(P))\). By the property of reflection, if \(R_p\) is a reflection over a line \(p\), then \(R_p\circ R_p\) is the identity transformation. That is, if \(p\) is the line of reflection, for any point \(P\) in the plane, two reflections over the same line \(p\) will map \(P\) back to its original position.

Step3: Analyze Option C

Translating 1 unit to the right, then 4 units to the left, then 3 units to the right.
Let \(T_{x}\) be the translation in the \(x\) - direction. If \(T_1(x,y)=(x + 1,y)\), \(T_2(x,y)=(x-4,y)\) and \(T_3(x,y)=(x + 3,y)\). Then the composition \(T_3\circ T_2\circ T_1(x,y)=T_3(T_2(T_1(x,y)))=T_3(T_2(x + 1,y))=T_3((x + 1-4,y))=T_3((x-3,y))=(x-3 + 3,y)=(x,y)\)

Step4: Analyze Option D

Rotating \(120^{\circ}\) counter - clockwise around center \(C\) and then rotating \(240^{\circ}\) counter - clockwise around center \(C\).
Let \(R_{C,\theta}\) be the rotation transformation about center \(C\) by an angle \(\theta\). If \(\theta_1 = 120^{\circ}\) and \(\theta_2=240^{\circ}\), then the composition \(R_{C,\theta_2}\circ R_{C,\theta_1}\) of rotations about the same center \(C\) is a rotation by \(\theta=\theta_1+\theta_2\) (by the property of rotation composition \(R_{C,\alpha}\circ R_{C,\beta}=R_{C,\alpha + \beta}\) when the center of rotation is the same). Since \(120^{\circ}+240^{\circ}=360^{\circ}\), and a \(360^{\circ}\) rotation about a center \(C\) is the identity transformation.

Answer:

A. Translate 3 units up, then 3 units down