Mapping of a Plane with Motion Graphing

Geometric mapping of planes

The axial symmetry and the central symmetry are examples of the geometric mapping of the plane.

 

When a plane is mapped onto itself:

1) every point of the plane is put in correspondence with the unique point of the plane.

2) each point of the plane is placed in correspondence with some point of the plane.

 

Definition:

The motion is a map of the plane onto itself such that the distance between the points is kept.

 

Examples: the point A lies on the line L,

the point B does not lie on the line L

a) draw the point A relative to the line L:

A → A (see Figure a)

Draw the point B relative to the line L:

B → B₁

 

1) the segment BB₁ is perpendicular to the line L

2) the segments OB = OB₁ are equal (see Figure a)

b) draw the points A, B and the segment AB relative to the line L (see Figure b):

A → A₁

B → B₁

AB → A₁B₁

c) draw the segment AB relative to the line L (see Figure c):

the segments are parallel: AB || A₁B₁

d) draw the segment AB relative to the point O (see Figure d):

A → A₁ : AO = A₁O

B → B₁ : BO = B₁O

***

 

 

Problem 139.

Given:

L – is the axis of the symmetry,

the straight line a is parallel to the line L

a || L

a₁ – is the mapping of a

a → a₁

 

Prove:

the straight line a₁ is parallel to the line L

a₁ || L

Proof:

Let us take any two points A and B on the line a. We obtain the segment AB a, lying on the line a. Find the mapping of the segment AB. To get that, we draw perpendiculars to the axis L from the points B and A.

Having measured the distance from the points A or B to the axis L, we postpone the same distance by a compass, i.e.

A → A₁

B → B₁

We connect the point A₁ with the point B₁ and obtain the segment A₁B₁ a₁, lying on the line a₁.

It follows, by the construction:

AO = OA₁, BO₁ = O₁B₁

Since AB || L , we see that AO = BO₁ = O₁B₁ = OA₁

Then OA₁ = OB₁, i.e. the points A₁ and B₁ are at an equal distance from the axis L.

Therefore, the line a₁ is parallel to the line L

a₁ || L

***

 

Problem 140.

Given:

a is the axis of the symmetry, the line

ABCD is the quadrilateral

 

Build:

a figure F is a map of the given quadrilateral

Construction:

1) draw perpendiculars from points A, B, C, D to the axis a.

2) measure the distance with the help of a compass from the point O to the points A and D and postpone them so that the distance between the points A and O, D and O is equal to the distance between the symmetrical distances D₁O and A₁O.

3) Measure the distance with the help of the compass from the point O₁ to the points B and C and postpone them so that the distance between the points B and O₁, C and O₁ is equal to the distance between the symmetrical distances B₁O₁ and C₁O₁.

4) Connect in series the points A₁, B₁, C₁, D₁.

By construction, the figure F is the quadrilateral.

***

 

Problem 141.

Given:

L is the axis of the symmetry

the straight line a is perpendicular to the line L

a L

 

Prove:

the straight line a is mapped onto itself relative to the axis L

a a

Proof:

1) any point A a, lying on the line a, is placed in correspondence with the point A₁ a, also lying on the line a (by the definition of the symmetry relative to the line L; a L ).

2) any point A₁ a, lying on the line a, is placed in correspondence with the point A a, lying on the line a.

Therefore, the line a is mapped onto itself relative to the axis L

a a

***

 

Motion

Theorem:

When moving, the segment is mapped to the segment.

 

Given:

MN – is the segment

M M₁

N N₁

Prove:

the segment MN is mapped to the segment M₁N₁

MN M₁N₁

Proof:

1) We take a point P MN, lying on the segment MN.

Then MP + PN = MN (*)

By moving the point P is mapped to the point P₁

P P₁

Since the motion keeps the distance between the points, then

MN = M₁N₁; MP = M₁P₁, PN = P₁N₁ (1)

Then it follows from equalities (*) and (1) that

M₁P₁ + P₁N₁ = M₁N₁ , therefore the point P₁ lies on the segment M₁N₁

P₁ M₁N₁

Then the segment MN is mapped to the segment M₁N₁

MN M₁N₁

 

2) We verify the second condition.

Any point P₁ M₁N₁, lying on the segment M₁N₁, gives the equality M₁P₁ + P₁N₁ = M₁N₁ (**)

By moving the point P₁ is translated to the point P

P₁ P

It follows from (**) and (1) that MP + PN = MN

Therefore, the point P MN lies on the segment MN.

So, by movement, the segment MN is translated into the segment M₁N₁:

MN M₁N₁

***

 

Consequence 1:

The triangle is translated to a triangle equal to it.

 

Given:

the triangle ΔABC

A A₁

B B₁

C C₁

 

Prove:

the triangle ΔABC is translated into the equal triangle ΔA₁B₁C₁

ΔABC ΔA₁B₁C₁: ΔABC = ΔA₁B₁C₁

Proof:

By the proved theorem, the segment is translated into the equal segment.

Then

1) by the motion, the segment AB is translated into its equal segment A₁B₁ , i.e. AB A₁B₁ , so that AB = A₁B₁

2) by the motion, the segment AC is translated into its equal segment A₁C₁ , i.e. AC A₁C₁ , so that AC = A₁C₁

3) by the motion, the segment BC is translated into its equal segment B₁C₁ , i.e. BC B₁C₁ , so that BC = B₁C₁

 

Therefore, by the motion, the triangle ΔABC is translated into an equal triangle ΔA₁B₁C₁, i.e.

ΔABC ΔA₁B₁C₁:

ΔABC = ΔA₁B₁C₁ (according to three sides)

***

 

 

Consequence 2:

The straight line is translated into a straight line, the ray is translated into a ray, the angle is translated into an equal angle.

 

Problem 142.

Given:

two straight lines are parallel a || b

a a₁

b b₁

 

Prove:

parallelism of the lines, i.e. two straight lines are parallel a₁ || b₁

Proof (by contradiction):

Suppose that two lines are not parallel a₁ b₁, then these lines intersect a₁ ∩ b₁

Each point of the lines a and b corresponds to a unique point of the lines a₁ and b₁.

Each point of the lines a₁ and b₁ corresponds to the point of the straight lines a and b.

Then the point of the intersection of the lines a₁ and b₁ corresponds to the intersection point of the lines a and b, but this claim contradicts the condition of the problem.

Therefore, the two lines are parallel a₁ || b₁

***

 

Problem 143.

Given:

O is the point of the central symmetry

b is the straight line

 

Build:

the straight line b₁ is a map of the line b relative to the point O

b b₁

 

Construction:

1) Take any two points, for example, A and B, on the line b.

2) Draw the ray BO.

3) Draw the ray AO.

4) Let us measure the segment AO with the help of the compass and on the continuation of the ray AO let us postpone from the point O the distance equal to the distance between the points A and O.

5) Let us measure the segment BO with the help of the compass and, on the continuation of the ray BO, postpone the distance equal to the distance between the points B and O from the point O.

6) We obtain two equal segments AB = A₁B₁

7) We draw the line b₁ through A₁B₁, then we obtain a straight line b₁ is a map of the line b relative to the point O

b b₁

***

 

Problem 144.

Given:

ABCD is the rectangle

A₁B₁C₁D₁ is the rectangle

a) AB = A₁B₁, BC = B₁C₁

 

b) A₁B₁ = AB

AC = A₁C₁ are the diagonals

 

Prove: ABCD = A₁B₁C₁D₁

Proof:

a) Let us prove it by contradiction.

Suppose that ABCD ≠ A₁B₁C₁D₁

Then AB ≠ A₁B₁ and BC ≠ B₁C₁, but this claim contradicts this condition.

Therefore, the rectangles are equal ABCD = A₁B₁C₁D₁

***

b) We prove by contradiction.

Suppose that ABCD ≠ A₁B₁C₁D₁

Then AB ≠ A₁B₁ and BC ≠ B₁C₁

Since the rectangles are not equal, we see that the diagonals will not be equal, but this claim contradicts the condition of the problem.

Therefore, the rectangles are equal ABCD = A₁B₁C₁D₁

***