Question

3.05: swbat write and perform a series of rigid motions on the coordinate plane
directions: complete the following questions by showing all work and annotations. keep work organized and box any final answer. all work must be shown in order to receive full credit.
**#1.)
a. write the sequence of rigid motions that maps abcd onto pqrs using proper notation.
b. use the properties of rigid motions to explain why abcd ≅ pqrs.
#2.) graph △abc, the image of △abc, after undergoing the series t_{-5,2}∘t_{y - axis}.
1^st transformation:
2^nd transformation:

Explanation:

Step1: Analyze the translation

First, find the horizontal and vertical distances between corresponding points of \(ABCD\) and \(PQRS\). For example, if we consider point \(A(-1,0)\) and \(P(5,0)\), the horizontal translation \(h\) can be found by \(x - \) coordinate difference: \(5-(-1)=6\). But since we also need to consider rotation. Let's use a better approach.

We know that a translation \(T_{(x,y)}\) moves a point \((a,b)\) to \((a + x,b + y)\) and a rotation \(R_{O,\theta}\) rotates a point around the origin \(O\) by an angle \(\theta\).

Looking at the orientation of \(ABCD\) and \(PQRS\), we first translate \(ABCD\) to the right. The \(x\) - coordinate of \(A(-1,0)\) and \(P(5,0)\): we can first translate \(ABCD\) by \(T_{(6,0)}\) (move 6 units to the right). But then we need to rotate.

Alternatively, we can first rotate \(ABCD\) by \(R_{O,90^{\circ}}\) (clock - wise rotation around the origin). The rule for a clock - wise rotation of \(90^{\circ}\) around the origin is \((x,y)\to(y, - x)\). Then we translate.

Let's use another method. If we consider the vectors.

The sequence of rigid motions: First, translate \(ABCD\) by \(T_{(6,0)}\) (move 6 units to the right). Then rotate the translated figure by \(R_{O,90^{\circ}}\) (clock - wise rotation around the origin).

In proper notation: \(R_{O,90^{\circ}}\circ T_{(6,0)}\)

Step2: Use the properties of rigid motions

Rigid motions (translations \(T\), rotations \(R\), and reflections \(F\)) preserve side lengths and angle measures.

A translation \(T_{(a,b)}\) moves every point of a figure by the same distance in the same direction. A rotation \(R_{O,\theta}\) rotates every point of a figure around a fixed point \(O\) by an angle \(\theta\).

Since \(ABCD\) is mapped to \(PQRS\) using a translation (which preserves distances and angles: \(d(A,B)=d(T(A),T(B))\), \(\angle ABC=\angle T(A)T(B)T(C)\)) and a rotation (which also preserves distances \(d(A,B)=d(R(A),R(B))\) and angles \(\angle ABC=\angle R(A)R(B)R(C)\)), the two quadrilaterals \(ABCD\) and \(PQRS\) have all corresponding side lengths equal (\(AB = PQ\), \(BC=QR\), \(CD = RS\), \(DA=SP\)) and all corresponding angle measures equal (\(\angle A=\angle P\), \(\angle B=\angle Q\), \(\angle C=\angle R\), \(\angle D=\angle S\))

Answer:

a. The sequence of rigid motions is \(R_{O,90^{\circ}}\circ T_{(6,0)}\) (or \(T_{(6,0)}\circ R_{O,90^{\circ}}\), since translation and rotation are commutative in some cases when the translation is parallel - like in this coordinate - based situation)
b. Rigid motions (translation and rotation) preserve side lengths and angle measures. A translation moves every point of \(ABCD\) by the same vector, so it preserves distances between points (e.g., \(AB\) and its image after translation) and angle measures (e.g., \(\angle ABC\) and its image after translation). A rotation rotates every point of the translated figure around a fixed point (the origin in our case) by \(90^{\circ}\), and it also preserves distances (e.g., if \(X\) is a point on the translated figure, \(d(X,Y)=d(R(X),R(Y))\)) and angle measures (e.g., \(\angle XYZ=\angle R(X)R(Y)R(Z)\)). So, \(ABCD\cong PQRS\) because all corresponding side lengths and angle measures are equal.