Distance

from point to line

Distance between parallel lines

Geometry, 7th grade

To the textbook by L.S. Atanasyan

mathematics teacher of the highest category

Municipal educational institution "Upshinskaya basic secondary school"

Orsha district of the Republic of Mari El

Perpendicular length drawn from a point to a line, called distance from this point to straight.

AN ⏊ A

M є a, M is different from N

Perpendicular , drawn from a point to a line, less any inclined , drawn from the same point to this line.

AM – inclined, drawn from point A to line a

AN AM

AN - inclined

AN AN

AN AK

AK - inclined

Distance from point to line

M

The distance from point M to straight line c is...

N

The distance from point N to line c is...

With

The distance from point K to straight line c is...

K

The distance from point F to straight line c is...

F

Distance from point to line

AN ⏊ A

AN= 5.2 cm

VC ⏊ A

VC= 2.8 cm

Theorem.

All points of each of two parallel lines are equidistant from the other line

Given: a ǁ b

A є a, B є a,

Prove: the distances from points A and B to line a are equal.

AN ⏊ b,BK ⏊ b,

Prove: AH = BK

Δ ANK = ΔVKA(Why?)

From the equality of triangles it follows AN = BK

The distance from an arbitrary point of one of the parallel lines to another line is called the distance between these lines.

Converse theorem.

All points of the plane located on one side of a given line and equidistant from it lie on a line parallel to the given one.

AN ⏊ b,BK ⏊ b,

АH = BK

Prove: AB ǁ b

Δ ANK = ΔKVA(Why?)

From the equality of triangles it follows … , but these are internal crosswise angles formed … , means AB ǁ NK

What is the distance between lines b and c, if the distance between the lines A and b is equal to 4, and between the lines A and c equals 5?

A ǁ b ǁ c

What is the distance between lines b and a, if the distance between lines b and c is 7, and between the lines A and c equals 2?

What is the distance between lines A and c, if the distance between lines b and c is 10, and between lines b And a equals 6?

What is the set of all points in a plane that are equidistant from two given parallel lines?

A ǁ b

Answer: A line parallel to these lines and located at equal distances from them.

What is the set of all points on a plane located at a given distance from a given line?

Answer: Two lines parallel to a given line and located at a given distance on opposite sides of it.

In this article, using the example of solving problem C2 from the Unified State Examination, the method of finding using the coordinate method is analyzed. Recall that straight lines are skew if they do not lie in the same plane. In particular, if one line lies in a plane, and the second line intersects this plane at a point that does not lie on the first line, then such lines are intersecting (see figure).

To find distances between intersecting lines necessary:

1. Draw a plane through one of the intersecting lines that is parallel to the other intersecting line.
2. Drop a perpendicular from any point of the second line onto the resulting plane. The length of this perpendicular will be the required distance between the lines.

Let us analyze this algorithm in more detail using the example of solving problem C2 from the Unified State Examination in mathematics.

Distance between lines in space

Task. In a unit cube ABCDA 1 B 1 C 1 D 1 find the distance between the lines B.A. 1 and D.B. 1 .

Rice. 1. Drawing for the task

Solution. Through the middle of the diagonal of the cube D.B. 1 (point O) draw a line parallel to the line A 1 B. Points of intersection of this line with edges B.C. And A 1 D 1 is denoted accordingly N And M. Straight MN lies in a plane MNB 1 and parallel to the line A 1 B, which does not lie in this plane. This means that the straight line A 1 B parallel to the plane MNB 1 based on the parallelism of a straight line and a plane (Fig. 2).

Rice. 2. The required distance between crossing lines is equal to the distance from any point of the selected line to the depicted plane

Now we are looking for the distance from some point on the line A 1 B to plane MNB 1 . This distance, by definition, will be the required distance between the crossing lines.

To find this distance we will use the coordinate method. Let us introduce a rectangular Cartesian coordinate system so that its origin coincides with point B, axis X was directed along the edge B.A., axis Y- along the edge B.C., axis Z- along the edge BB 1 (Fig. 3).

Rice. 3. We choose a rectangular Cartesian coordinate system as shown in the figure

Finding the equation of the plane MNB 1 in this coordinate system. To do this, we first determine the coordinates of the points M, N And B 1: We substitute the resulting coordinates into the general equation of the straight line and obtain the following system of equations:

From the second equation of the system we obtain from the third we obtain after which from the first we obtain Substitute the obtained values ​​into the general equation of the straight line:

We note that otherwise the plane MNB 1 would pass through the origin. Divide both sides of this equation by and we get:

The distance from a point to a plane is determined by the formula.

Using this online calculator you can find the distance between lines in space. A detailed solution with explanations is given. To calculate the distance between lines in space, set the type of equation of lines ("canonical" or "parametric"), enter the coefficients of the equations of lines in the cells and click on the "Solve" button.

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Data entry instructions. Numbers are entered as integers (examples: 487, 5, -7623, etc.), decimals (ex. 67., 102.54, etc.) or fractions. The fraction must be entered in the form a/b, where a and b (b>0) are integers or decimals. Examples 45/5, 6.6/76.4, -7/6.7, etc.

Distance between lines in space - theory, examples and solutions

Let a Cartesian rectangular coordinate system be given Oxyz L 1 and L 2:

,

(2)

Where M 1 (x 1 , y 1 , z 1) and M 2 (x 2 , y 2 , z 2) − points lying on straight lines L 1 and L 2, a q 1 ={m 1 , p 1 , l 1) and q 2 ={m 2 , p 2 , l 2 ) – direction vectors of straight lines L 1 and L 2, respectively.

Lines (1) and (2) in space can coincide, be parallel, intersect, or be intersecting. If lines in space intersect or coincide, then the distance between them is zero. We will consider two cases. The first is that the lines are parallel, and the second is that the lines intersect. The rest are common cases. If, when calculating the distance between parallel lines, we get the distance equal to zero, then this means that these lines coincide. If the distance between intersecting lines is zero, then these lines intersect.

1. Distance between parallel lines in space

Let's look at two methods for calculating the distance between lines.

Method 1. From a point M 1 straight L 1 draw a plane α , perpendicular to the line L 2. Finding a point M 3 (x 3 , y 3 , y 3) plane intersections α and straight L 3. Essentially we find the projection of the point M 1 straight L 2. How to find the projection of a point onto a line, look. Next we calculate the distance between the points M 1 (x 1 , y 1 , z 1) and M 3 (x 3 , y 3 , z 3):

Straight L 2 passes through the point M 2 (x 2 , y 2 , z 2)=M

Substituting values m 2 , p 2 , l 2 , x 1 , y 1 , z 1 in (5) we get:

Let's find the point of intersection of the line L 2 and plane α , for this we construct a parametric equation of the straight line L 2 .

To find the intersection point of a line L 2 and plane α , substitute the values ​​of the variables x, y, z from (7) to (6):

Substituting the resulting value t in (7), we obtain the intersection point of the straight line L 2 and plane α :

It remains to find the distance between the points M 1 and M 3:

L 1 and L 2 equals d=7.2506.

Method 2. Find the distance between lines L 1 and L 2 (equations (1) and (2)). First, we check the parallelism of the lines L 1 and L 2. If the direction vectors of straight lines L 1 and L 2 are collinear, i.e. if there is a number λ such that the equality q 1 =λ q 2, then straight L 1 and L 2 are parallel.

This method of calculating the distance between parallel vectors is based on the concept of the vector product of vectors. It is known that the norm of the vector product of vectors and q 1 gives the area of ​​the parallelogram formed by these vectors (Fig. 2). Once you know the area of ​​a parallelogram, you can find the vertex of the parallelogram d, dividing the area by the base q 1 parallelogram.

q 1:

.

Distance between lines L 1 and L 2 equals:

,

Example 2. Let's solve example 1 using method 2. Find the distance between lines

Straight L 2 passes through the point M 2 (x 2 , y 2 , z 2)=M 2 (8, 4, 1) and has a direction vector

Vectors q 1 and q 2 are collinear. Therefore straight L 1 and L 2 are parallel. To calculate the distance between parallel lines, we use the vector product of vectors.

Let's build a vector =( x 2 −x 1 , y 2 −y 1 , z 2 −z 1 }={7, 2, 0}.

Let's calculate the vector product of vectors and q 1 . To do this, we create a 3×3 matrix, the first row of which is the basis vectors i, j, k, and the remaining lines are filled with elements of vectors and q 1:

Thus, the result of the vector product of vectors and q 1 will be a vector:

Answer: Distance between lines L 1 and L 2 equals d=7.25061.

2. Distance between crossing lines in space

Let a Cartesian rectangular coordinate system be given Oxyz and let straight lines be given in this coordinate system L 1 and L 2 (equations (1) and (2)).

Let straight L 1 and L 2 are not parallel (we discussed parallel lines in the previous paragraph). To find the distance between lines L 1 and L 2 you need to build parallel planes α 1 and α 2 so that it is straight L 1 lay on a plane α 1 a straight L 2 - on the plane α 2. Then the distance between the lines L 1 and L 2 is equal to the distance between the planes L 1 and L 2 (Fig. 3).

Where n 1 ={A 1 , B 1 , C 1 ) − normal vector of the plane α 1 . In order for the plane α 1 passed through a straight line L 1, normal vector n 1 must be orthogonal to the direction vector q 1 straight L 1, i.e. the scalar product of these vectors must be equal to zero:

Solving the system of linear equations (27)−(29), with three equations and four unknowns A 1 , B 1 , C 1 , D 1, and substituting into the equation

Planes α 1 and α 2 are parallel, therefore the resulting normal vectors n 1 ={A 1 , B 1 , C 1) and n 2 ={A 2 , B 2 , C 2) these planes are collinear. If these vectors are not equal, then we can multiply (31) by a certain number so that the resulting normal vector n 2 coincided with the normal vector of equation (30).

Then the distance between parallel planes is calculated by the formula:

(33)

Solution. Straight L 1 passes through the point M 1 (x 1 , y 1 , z 1)=M 1 (2, 1, 4) and has a direction vector q 1 ={m 1 , p 1 , l 1 }={1, 3, −2}.

Straight L 2 passes through the point M 2 (x 2 , y 2 , z 2)=M 2 (6, −1, 2) and has a direction vector q 2 ={m 2 , p 2 , l 2 }={2, −3, 7}.

Let's build a plane α 1 passing through the line L 1, parallel to the straight line L 2 .

Since the plane α 1 passes through the line L 1, then it also passes through the point M 1 (x 1 , y 1 , z 1)=M 1 (2, 1, 4) and normal vector n 1 ={m 1 , p 1 , l 1) plane α 1 perpendicular to the direction vector q 1 straight L 1 . Then the equation of the plane must satisfy the condition:

Since the plane α 1 must be parallel to the line L 2, then the following condition must be met:

Let's present these equations in matrix form:

(40)

Let us solve the system of linear equations (40) with respect to A 1 , B 1 , C 1 , D 1.

This video lesson will be useful for those who want to independently study the topic “Distance from a point to a line. Distance between parallel lines." During the lesson you will learn how to calculate the distance from a point to a line. Then the teacher will give the definition of the distance between parallel lines.

In this lesson we will get acquainted with the concept "distance" generally. We also specify this concept in the case of calculating distances between two points, a point and a line, parallel lines

Let's look at Figure 1. It shows 2 points A and B. The distance between two points A and B is a segment having ends at given points, that is, segment AB

Rice. 1. AB - distance between points

It is noteworthy that distance cannot be considered a curve or broken line connecting two points. Distance- this is the shortest path from one point to another. It is the segment AB that is the smallest of all possible lines connecting points A and B

Consider Figure 2, which shows the straight line A, and point A, which does not belong to this line. Distance from point A to a straight line will be the length of the perpendicular AN.

Rice. 2. AN - distance between a point and a line

It is important to note that AN is the shortest distance, since in the triangle AMN this segment is a leg, and an arbitrary other segment connecting point A and the line A(in this case it is AM) will be the hypotenuse. As you know, the leg is always less than the hypotenuse

Distance designation:

Let's consider parallel lines a and b shown in Figure 3

Rice. 3. Parallel lines a and b

Let's fix two points on a straight line a and drop perpendiculars from them onto a line parallel to it b. Let us prove that if ,

Let us draw segment AM for convenience of proof. Let us consider the resulting triangles ABM and ANM. Since , and , then . Likewise, . These right triangles () have a common side AM. It is the hypotenuse in both triangles. Angles AMN and AMB are internal cross angles with parallel straight lines AB and NM and secant AM. According to the well-known property, .

From all of the above it follows that . From the equality of triangles it follows that AN = BM

So, we have proven that in Figure 3 the segments AN and BM are equal. It means that distance between parallel lines is the length of their common perpendicular, and the choice of perpendicular can be arbitrary. Thus,

The converse is also true: a set of points that are at the same distance from a certain line form a line parallel to the given one.

Let's consolidate our knowledge and solve several problems

Example 1: Problem 272 from the textbook “Geometry 7-9”. Author - Atanasyan L.S.

In an equilateral triangle ABC, the bisector AD is drawn. The distance from point D to straight line AC is 6 cm. Find the distance from point A to straight line BC

Rice. 4. Drawing for example 1

Solution:

An equilateral triangle is a triangle with three equal sides (and therefore three equal angles, that is, 60 0 each). An equilateral triangle is a special case of an isosceles triangle, therefore all the properties inherent in an isosceles triangle also apply to an equilateral triangle. Therefore, AD is not only a bisector, but also a height, therefore AD ⊥BC

Since the distance from point D to line AC is the length of the perpendicular drawn from point D to line AC, then DH is this distance. Consider the triangle AND. In it, the angle H = 90 0, since DH is perpendicular to AC (by definition of the distance from a point to a straight line). In addition, in this triangle the leg DH lies opposite the angle, so AD = (cm) (By property)

The distance from point A to straight line BC is the length of the perpendicular dropped onto straight line BC. According to the proven AD ⊥BC, it means .

Answer: 12 cm.

Example 2: Problem 277 from the textbook “Geometry 7-9”. Author - Atanasyan L.S.

The distance between parallel lines a and b is 3 cm, and the distance between parallel lines a and c is 5 cm. Find the distance between parallel lines b and c

Solution:

Rice. 5. Drawing for example 2 (first case)

Since , then = 5 - 3 = 2 (cm).

However, this answer is incomplete. There is another option for locating straight lines on a plane:

Rice. 6. Drawing for example 2 (second case)

In this case .

1. Unified collection of digital educational resources ().
2. Math tutor ().

1. No. 280, 283. Atanasyan L. S., Butuzov V. F., Kadomtsev S. B., Poznyak E. G., Yudina I. I. edited by Tikhonov A. N. Geometry grades 7-9. M.: Enlightenment. 2010
2. The sum of the hypotenuse CE and leg CK of the right triangle SKE is 31 cm, and their difference is 3 cm. Find the distance from vertex C to straight line KE
3. Based on AB of the isosceles triangle ABC, point M is taken, equidistant from the lateral sides. Prove that CM is the height of triangle ABC
4. Prove that all points of the plane located on one side of a given line and equidistant from it lie on a line parallel to the given one

Lesson outline

Triangle Angle Sum Theorem

1. Full name: Sayfetdinova Gulnara Vasilevna

2. Place of work: Municipal budgetary educational institution "Knyazevskaya secondary school" of the Tukaevsky district of the Republic of Tatarstan

3. Job title: mathematic teacher

4. Item: geometry

5. Class: 7th grade

6. Lesson topic: Distance from a point to a line. Distance between parallel lines.

7. Basic tutorial: Geometry.7-9 grades: textbook for educational institutions / author. L.S. Atanasyan, V.F. Butuzov,

S.B. Kadomtsev et al., 2010

8.Goals:

Activity goal: create conditions for independent formulation and proof of the properties of oblique and perpedicular dropped from a point to a line, the theorem on the equidistance of points on parallel lines; organize the activities of students to perceive, comprehend and initially consolidate new knowledge and methods of activity.

Educational goal:

Subject:

apply the concepts of distance from a point to a line, distance between lines when solving problems

Metasubject:

Regulatory UUD:

Cognitive UUD:

Communication UUD:

Personal UUD:

10. Teaching methods: problematic, research.
11.Forms of organizing educational activities: frontal, group, pair, individual, training structures.

12.Equipment, technical conditions:

Computer, projector, screen, Internet, software: Microsoft Power Point, class seating - 4 people per table.

13.Lesson duration: 45 minutes

14.Lesson plan

I . Organizing time.

II . Updating knowledge.

III . Setting a lesson goal . Introduction of new material.

VI. Summarizing. Reflection.

I . Organizing time.

Target: preparing students for work, activating attention for quick inclusion in activities.

Teacher : Hello guys? How are you feeling? Let's raise him up and start the lesson with a smile! Let's smile at our partner's face! Let's smile at our partner's shoulder!

II . Updating knowledge.

Teacher : You’ve been studying a new subject of geometry for six months now and you probably know what a theorem is. What methods of proof do you know?

Possible student answers: Method by contradiction, constructive method, method of proof based on axioms and previously proven theorems (slide No. 2).

Teacher: Guys, what associations do you have with the word distance?

Possible student answers: The distance between cities, the distance between pillars, the distance from something to something (slide number 3).

Teacher: What is the distance between two points?

Possible student answers: Section length (slide number 4).

Teacher: Make an entry in the technological map in paragraph 1

Teacher: Please note that in geometry, distance refers to the shortest distance. Make an entry in the technological map in paragraph 2

Teacher: What can be said about the relative position of straight line AN and straight line a?

Teacher: What are these lines called?

Teacher: A What is the name of segment AN?

Teacher: Remember: A perpendicular is a segment. Make an entry in the technological map in paragraph 3.

III. Setting the lesson goal.Introduction of new material.

Teacher: Practical task:

We are in a field; a road runs through the field. Draw a mathematical model of the situation. We need to get on the road. Draw the trajectory (slide No. 6).

Teacher: How can this trajectory be defined in mathematical language? Possible student answers: Perpendicular

Teacher: Why not? –

Try giving it a name (slide number 7).

Possible student answers: Inclined.

Teacher: How many inclined lines can be drawn from this point?

Possible student answers: A bunch of.

(slide number 7).

Teacher: So you think the shortest path is a perpendicular? Prove it.

Teacher: Now prove that any inclined line is greater than a perpendicular line.

What do we see in the picture?

Possible student answers: right triangle (slide No. 8).

Teacher: What are the names of perpendicular and oblique in this triangle? Possible student answers: leg and hypotenuse.

Teacher: Why is the hypotenuse larger than the leg?

Possible student answers: Opposite the larger angle is the larger side. The largest angle in a right triangle is a right angle. Opposite it lies the hypotenuse.

Teacher. What else can you call segment AC? What if we return to the content of the task?

Possible student answers: Distance from point to line .

Teacher: Formulate the definition: “The distance from a point to a line is... (the length of the perpendicular drawn from this point to the line)” (slide No. 9). Make an entry in the technological map in paragraph 4.

Teacher: Practical task.

Find the distance from point B to straight lines A D AndDC using a drawing triangle and a ruler (slide No. 10). technological map point 6

Teacher: Practical task. Construct two parallel lines a and b. On line a, mark point A. Drop a perpendicular from point A to line b. Place point B at the base of the perpendicular.

What can you say about segment AB? (slide number 11).

It is perpendicular to both line a and line b.

Teacher: Therefore, it is called the common perpendicular (slide No. 13). Make an entry in the technological map in paragraph 5

Teacher: Make an entry in the technological map in paragraph 6

Teacher: Task. It is required to lay linoleum on the floor in a long corridor. It is known that two opposite walls are parallel. A common perpendicular was drawn at one end of the corridor, and its length turned out to be 4 m. Is it worth rechecking the lengths of the common perpendiculars in other places in the corridor? (slide number 14).

Possible student answers: No need, their lengths will also be equal to 4.

Teacher: Prove it. But first, draw a mathematical model of this situation. To prove, highlight what is known and what needs to be proven.

How is the equality of segments and angles usually proved in geometry?

Possible student answers: Through the equality of triangles containing these segments and angles. Come up with a construction that would allow us to prove the equality of these triangles.

Structure SingleRoundRobin:

2. Four students in a team answer once.

Teacher: Prove equality segments AB and CD through the equality of triangles . On the sign board, write down the three conditions for the triangle equality test.

1.The teacher asks a question and gives time to think

Students perform additional constructions, prove the equality of triangles, draw a conclusion about the equality of segments AB and CD (slide No. 15-17).

Teacher: Segments AB and CD are equal. What can be said about points A and C relative to straight line BD?

Possible student answers: They are at equal distance. They are equidistant (slide number 18).

Teacher: Does this property hold for any points?

Possible student answers: Yes

Teacher: Let's try to formulate this property. What does a property statement consist of?

Possible student answers: From the condition and conclusion (slide No. 19,20).

Possible student answers: If the points lie on one of the parallel lines, then they are equidistant from the second line.

Teacher: Edit this property without conjunctions: if, then (slide number 21).

Possible student answers: Points lying on one of the parallel lines are equidistant from the second line.

Think-Write-Round Robin structure:

1.The teacher asks a question and gives time to think

2. Students think and write down the answer on their piece of paper

3. Students take turns reading their answer from a piece of paper.

Teacher: Which statement is called the converse?

Possible student answers: If the condition and conclusion are swapped.

Teacher: Formulate the opposite statement (slide number 22).

Possible student answers: If points lying on one of two lines are equidistant from the second line, then the lines are parallel.

Teacher: Make an entry in the technological map in paragraphs 7,8.

Teacher: Is it possible to define such a concept as the distance between parallel lines?

Possible student answers: Yes

Teacher: What can be called the distance between parallel lines

Possible student answers: The length of the common perpendicular. Make an entry in the technological map in paragraph 5.

IV. Application of the theorem, executionpractical work.

Teacher: Practical work. Find the width of the strip.

What mathematical concept is the width of a strip?

Teacher: Where else are these theorems used in practical life?

VI. Summarizing. Reflection.

Teacher: What new concepts did you become familiar with?

What did you learn in the lesson?

Where in life will we apply this?

(slide No. 26-28)

Teacher: Make an entry in the technological map in paragraph 9

Homework No. 276.279 – proof of the converse theorem.

Self-analysis of the lesson

Goals:

Activity goal: create conditions for independently formulating and proving the properties of inclined and perpedicular dropped from a point to a straight line, create conditions for proving the theorem on the equidistance of points on parallel lines; organize the activities of students to perceive, comprehend and initially consolidate new knowledge and methods of activity.

Educational goal: develop the knowledge that a perpendicular is less than any inclined, drawn from one point to a straight line, all points of each of two parallel lines are equidistant from the other straight line.

Subject: the student will have the opportunity to learn:

apply the theorem to solve practical problems

analyze, compare, generalize, draw conclusions to solve practical problems.

Metasubject:

Regulatory UUD:

the ability to independently set goals, select and create algorithms for solving educational mathematical problems;

ability to plan and carry out activities aimed at solving research problems.

Cognitive UUD:

• • the ability to establish cause-and-effect relationships, build logical reasoning, inferences, conclusions;

  • the ability to put forward hypotheses when solving educational problems and understand the need to test them; the ability to use inductive and deductive methods of reasoning, to see different strategies for solving problems;

  • develop initial ideas about the ideas and methods of mathematics as a universal language of science, a means of modeling phenomena and processes;

  • ability to understand and use drawings and drawings for illustration, interpretation, argumentation.

Communication UUD:

• the ability to organize educational cooperation and joint activities with the teacher and students, determine goals, distribute the functions and roles of participants, general ways of working;

• ability to work in a group: find a common solution and resolve conflicts based on coordinating positions and taking into account interests, listening to a partner, formulating, arguing and defending one’s opinion.

Personal UUD:

• formation of communicative competence in communication and cooperation in joint educational and research activities;

  development of the ability to clearly, accurately, competently express one’s thoughts in oral and written speech, understand the meaning of the task, build an argument, give examples and counterexamples;

  development of critical thinking, the ability to recognize logically incorrect statements, distinguish a hypothesis from a fact;

  develop creative thinking, initiative, resourcefulness, and activity in solving geometric problems.

The structure of the lesson fragment corresponded to the type - a lesson on discovering new knowledge. In accordance with the goals and content of the material, the lesson was structured according to the following stages:

I . Organizing time.

II . Updating knowledge.

III . Setting a lesson goal . Introduction of new material.

IV. Application of the theorem, implementation of practical work.

VI. Summarizing.

All structural elements of the lesson were followed. The organization of the educational process is based on the activity method.

The purpose of the first stageIt was easy to quickly integrate students into the business rhythm.

At the second stage the knowledge necessary to work on new material was updated.

At the third stageIn order to define the concepts of distance from a point to a line, the concept of an inclined line attracted children to practical activities with search elements. First, on an intuitive level, students put forward a hypothesis, then independently proved the property of a perpendicular and an oblique drawn from one point to a straight line.

In general, I used practical tasks throughout the entire lesson, including during the initial consolidation. They help to attract students to independent cognitive activity, and solve the problems of a competency-based approach to learning.

To formulate and prove the theorem on the equidistance of points on parallel lines, I used a problematic task, which contributed to the formulation of a hypothesis about the properties of the objects under consideration and the subsequent search for proof of the validity of the assumption put forward.

By organizing work on formulating the theorem, and then the inverse theorem, I achieved my goaldevelopment of initial ideas about the ideas and methods of mathematics as a universal language of science, a means of modeling phenomena and processes.

Educational and cognitive activities were organized through frontal work, individual and group work. This organization made it possible to include each student in active activities to achieve the goal. Students cooperated with each other, providing mutual assistance.

Time, I believe, was distributed rationally. In a short period of time, it was possible to introduce the concepts of distance from a point to a straight line, an inclined line, the distance between parallel straight lines, formulate and prove two theorems, and consider the application of the theorem in practice.

For clarity, I used a presentation during the lesson. I used a special program for a demonstration to compare the length of an oblique and a perpendicular, in which geometric shapes come to life. During the lesson, I used student work on the signal board, which solves the problems of equal student participation in the lesson, control over the learning of the material, and, of course, activates the student in the lesson.

The students were active during the lesson, I managed to involve them in research activities, creative activities, with a constructive method of proving the theorem, formulating the theorem

At the end of the lesson, the students formulated the topic themselves.

Reflection