Math 11 Chapter 1 Lesson 6: The concept of displacement and two equal shapes
1. Summary of theory
1.1. The concept of displacement
a) Definition: Displacement is a transformation that preserves the distance between any two points.
Symbol: F
– If F(M) = M’ and F(N) = N’, then MN = M’N’
b) Comments
– The operations of identification, translation, axial symmetry, center symmetry, rotation are all displacement operations.
– The transformation obtained by performing two consecutive displacements is also a displacement.
1.2. Properties of the displacement

Turn three collinear points into three collinear points and preserve the order between the points.

Turn a line into a line, turn a ray into a ray, turn a line segment into a straight line with it.

Turn triangle into triangle by it, angle into angle by it.

Turn the circle into a circle with the same radius.
1.3. The concept of two congruent figures
Two shapes are said to be equal if there is a transformation that transforms one into the other.
2. Illustrated exercise
Question 1:
a) Given a square ABCD with center O. Find the image of points A, B, O through the displacement obtained by performing two consecutive operations \({Q_{\left( {O{{,90}^0 ). }} \right)}}\) and BD.
b) Observe the figure and say that \(\Delta ABC\) turns into \(\Delta A”B”C”\) through which displacement?
Solution guide
a) We have:
\(\left\{ \begin{array}{l}{Q_{\left( {O{{,90}^0}} \right)}}\left( O \right) = O\\{Q_{ \left( {O{{,90}^0}} \right)}}\left( A \right) = B\\{Q_{\left( {O;{{90}^0}} \right) }}\left( B \right) = C\end{array} \right.\) and DBD(O)=O; BD(B)=B; DBD(C)=A.
So the image of O is O, A is B and B is A.
b) We have:
\({Q_{\left( {C{{,90}^0}} \right)}}\left( {ABC} \right) = A’B’C\)
\({T_{\overrightarrow {AA”} }}\left( {A’B’C} \right) = A”B”C”.\)
So the transformation to look for is a transformation that performs two operations in succession\({Q_{\left( {C{{,90}^0}} \right)}}\) and \({T_{\overrightarrow {AA”} }}.\)
Verse 2: Given a regular hexagon ABCDEF with center O. Determine the image of \(\Delta OAB\) through the displacement by performing successive rotations with center O, rotation angle 60^{0} and vector translation \(\overrightarrow {OE} .\)
Solution guide
We have:
\(\left\{ \begin{array}{l}{Q_{\left( {O{{,60}^0}} \right)}}\left( A \right) = B\\{Q_{ \left( {O{{,60}^0}} \right)}}\left( B \right) = C\end{array} \right.\)\( \Rightarrow {Q_{\left( {O {{,60}^0}} \right)}}\left( {OAB} \right) = OBC\)
\(\left\{ \begin{array}{l}{T_{\overrightarrow {OE} }}\left( O \right) = E\\{T_{\overrightarrow {OE} }}\left( B \ right) = O\\{T_{\overrightarrow {OE} }}\left( C \right) = D\end{array} \right \Rightarrow {T_{\overrightarrow {OE} }}\left( {OBC } \right) = EOD\)
So the image of \(\Delta OAB\) through the given displacement is \(\Delta EOD\).
Question 3: Let ABCD be a rectangle with center O. Let E and F be the midpoints of AD and BC, respectively. Prove that trapezoid AEOB and trapezoid CFOD are congruent.
Solution guide
We have:
Measure(O)=O; MEASURE(A)=C; MEASURE(E)=F; Measure(B)=D.
Inference: MEASURE(AEOB)=CFOD.
So there is a displacement which is center O symmetry that turns trapezoid AEOB into trapezoid CFOD. So these two trapezoids are congruent.
3. Practice
3.1. Essay exercises
Question 1: Given a rectangle ABCD with center O. Find the image of points A, B, O through the displacement obtained by performing two operations consecutively. \({Q_{\left( {O,{{90}^0}} \right)}}\) and the RED spell_{BD}.
Verse 2: Given a regular hexagon ABCDEF with center O. Determine the image of \(\Delta OCD\) through displacement by performing successive rotations of center O, rotation angle of 60^{0} and vector translation \(\overrightarrow {OA} .\)
Question 3: Let ABCD be a rectangle with center O. Let E and F be the midpoints of AB and CD, respectively. Prove that trapezoid ABFD and trapezoid CDEB are congruent.
3.2. Multiple choice exercises
Question 1: Given the rectangle ABCD as shown, the transformation that transforms (1) into shape (3) is to perform the following two displacements consecutively.
A. Center symmetry I and axis symmetry IB.
B. CenterI symmetry and 90degree rotation Icenter.
C. EI axis symmetry and translation by \(\overrightarrow {DI} \).
D. Translation according to \(\overrightarrow {AI} \) and center symmetry I.
Verse 2: Given square ABCD as shown, triangle BIG is the image of triangle DIH through:
A. Center symmetry I
B. Rotation of the center I at an angle of 900
C. Translation by \(\overrightarrow {DI} \)
D. Rotation of the center A at an angle of 900
Question 3: In the Oxy plane, performing consecutive Ocenter symmetry and Ocenter rotation 900 turns the line y = x + 1 into a straight line
A. xy1=0
B. x+y1=0
C. x+y+1=0
D. x+y1=0
Question 4: Given two distinct points A and B. Call DA, NE are symmetries through A, B. For any point M, call M1 = DA(M), M2 = NE(M1). Let F be the transformation that turns M into M2. Choose the correct statement.
A. F is rotation.
B. F is axial symmetry.
C. F is the center symmetry.
D. F is the translation.
Question 5: Given rectangle ABCD. Let E, F, H, K, O, I, J be the midpoints of the sides AB, BC, CD, DA, KF, HC, KO, respectively. Which of the following assertion true?
A. The rotation of the center O angle \({90^0}\) turns the triangle OKA into triangle OCF.
B. There is a displacement by successively performing axial symmetry and translation that transforms the AEJK trapezoid into a FOIC trapezoid.
C. The displacement includes the vector translation \(\overrightarrow {HD} \) and the KF axis symmetry that turns the FIOC trapezoid into an AEJK trapezoid.
D. There is no rotation with center O that turns triangle OKA into triangle OCF.
4. Conclusion
Through this lesson, you should know the following:
 See the common points, the relationship of translations, center symmetry, axis symmetry, rotation
 Understand problem solving methods.
.
=============