How to find the distance between parallel lines. Mutual arrangement of lines in space. Problems with a straight line in space
A parallelogram is a quadrilateral whose opposite sides are parallel, that is, they lie on parallel lines (Fig. 1).
Theorem 1. On the properties of sides and angles of a parallelogram. In a parallelogram, opposite sides are equal, opposite angles are equal, and the sum of the angles adjacent to one side of the parallelogram is 180°.
Proof. In this parallelogram ABCD, draw a diagonal AC and get two triangles ABC and ADC (Fig. 2).
These triangles are equal, since ∠ 1 = ∠ 4, ∠ 2 = ∠ 3 (cross-lying angles at parallel lines), and side AC is common. From the equality Δ ABC = Δ ADC it follows that AB \u003d CD, BC \u003d AD, ∠ B \u003d ∠ D. The sum of the angles adjacent to one side, for example, angles A and D, is equal to 180 ° as one-sided with parallel lines. The theorem has been proven.
Comment. The equality of the opposite sides of a parallelogram means that the segments of the parallel ones cut off by the parallel ones are equal.
Corollary 1. If two lines are parallel, then all points of one line are at the same distance from the other line.
Proof. Indeed, let a || b (Fig. 3).
Let us draw from some two points B and C of the line b the perpendiculars BA and CD to the line a. Since AB || CD, then the figure ABCD is a parallelogram, and therefore AB = CD.
The distance between two parallel lines is the distance from an arbitrary point on one of the lines to the other line.
By what has been proved, it is equal to the length of the perpendicular drawn from some point of one of the parallel lines to the other line.
Example 1 The perimeter of the parallelogram is 122 cm. One of its sides is 25 cm longer than the other. Find the sides of the parallelogram.
Decision. By Theorem 1, opposite sides of a parallelogram are equal. Let's denote one side of the parallelogram as x, the other as y. Then by condition $$\left\(\begin(matrix) 2x + 2y = 122 \\x - y = 25 \end(matrix)\right.$$ Solving this system, we get x = 43, y = 18. Thus Thus, the sides of the parallelogram are 18, 43, 18 and 43 cm.
Example 2
Decision. Let figure 4 correspond to the condition of the problem.
Denote AB by x and BC by y. By condition, the perimeter of the parallelogram is 10 cm, i.e. 2(x + y) = 10, or x + y = 5. The perimeter of the triangle ABD is 8 cm. And since AB + AD = x + y = 5, then BD = 8 - 5 = 3 . So BD = 3 cm.
Example 3 Find the angles of the parallelogram, knowing that one of them is 50° greater than the other.
Decision. Let figure 5 correspond to the condition of the problem.
Let us denote the degree measure of angle A as x. Then the degree measure of the angle D is x + 50°.
Angles BAD and ADC are internal one-sided with parallel lines AB and DC and secant AD. Then the sum of these named angles will be 180°, i.e.
x + x + 50° = 180°, or x = 65°. Thus, ∠ A = ∠ C = 65°, a ∠ B = ∠ D = 115°.
Example 4 The sides of the parallelogram are 4.5 dm and 1.2 dm. A bisector is drawn from the vertex of an acute angle. What parts does it divide the long side of the parallelogram into?
Decision. Let figure 6 correspond to the condition of the problem.
AE is the bisector of the acute angle of the parallelogram. Therefore, ∠ 1 = ∠ 2.
In this article, attention is focused on finding the distance between skew lines using the coordinate method. First, the definition of the distance between skew lines is given. Next, an algorithm is obtained that allows you to find the distance between skew lines. In conclusion, the solution of the example is analyzed in detail.
Page navigation.
The distance between skew lines is a definition.
Before giving a definition of the distance between skew lines, we recall the definition of skew lines and prove a theorem related to skew lines.
Definition.
is the distance between one of the intersecting lines and a plane parallel to it passing through the other line.
In turn, the distance between a line and a plane parallel to it is the distance from some point on the line to the plane. Then the following formulation of the definition of the distance between skew lines is valid.
Definition.
Distance between intersecting lines is the distance from some point of one of the skew lines to a plane passing through the other line parallel to the first line.
Consider intersecting lines a and b. We mark a certain point M 1 on the line a, through the line b we draw a plane parallel to the line a, and from the point M 1 we drop the perpendicular M 1 H 1 onto the plane. The length of the perpendicular M 1 H 1 is the distance between the intersecting lines a and b.
Finding the distance between crossing lines - theory, examples, solutions.
When finding the distance between intersecting lines, the main difficulty often lies in seeing or constructing a segment whose length is equal to the required distance. If such a segment is constructed, then, depending on the conditions of the problem, its length can be found using the Pythagorean theorem, signs of equality or similarity of triangles, etc. This is what we do when finding the distance between intersecting lines in geometry lessons in grades 10-11.
If Oxyz is introduced in three-dimensional space and skew lines a and b are given in it, then the coordinate method allows to cope with the task of calculating the distance between the given skew lines. Let's analyze it in detail.
Let be a plane passing through the line b, parallel to the line a. Then the desired distance between the intersecting lines a and b, by definition, is equal to the distance from some point M 1 lying on the line a to the plane. Thus, if we determine the coordinates of some point M 1 lying on the line a, and get the normal equation of the plane in the form, then we can calculate the distance from the point 
to the plane by the formula (this formula was obtained in the article finding the distance from a point to a plane). And this distance is equal to the desired distance between the skew lines.
Now in detail.
The task is reduced to obtaining the coordinates of the point M 1 lying on the line a, and to finding the normal equation of the plane.
There are no difficulties with determining the coordinates of the point M 1 if you know well the main types of straight line equations in space. But it is worth dwelling on obtaining the equation of the plane in more detail.
If we determine the coordinates of some point M 2 through which the plane passes, and also get the normal vector of the plane in the form 
, then we can write the general equation of the plane as .
As a point M 2, you can take any point lying on the line b, since the plane passes through the line b. Thus, the coordinates of the point M 2 can be considered found.
It remains to obtain the coordinates of the normal vector of the plane . Let's do it.
The plane passes through line b and is parallel to line a. Therefore, the normal vector of the plane is perpendicular to both the directing vector of the straight line a (let's denote it ) and the directing vector of the straight line b (let's denote it ). Then we can take and as a vector, that is, . Having determined the coordinates and direction vectors of lines a and b and calculating 
, we will find the coordinates of the normal vector of the plane .
So, we have the general equation of the plane: .
It remains only to bring the general equation of the plane to normal form and calculate the desired distance between the intersecting lines a and b using the formula.
Thus, to find the distance between intersecting lines a and b you need:
Let's take a look at an example solution.
Example.
In three-dimensional space in a rectangular coordinate system Oxyz two intersecting straight lines a and b are given. The line a is defined
Distance
point to line
Distance between parallel lines
Geometry, 7th grade
To the textbook by L.S. Atanasyan
mathematics teacher of the highest category
MOU "Upshinsky basic comprehensive 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 H
Perpendicular drawn from a point to a line, smaller any oblique drawn from the same point to this line.
AM – oblique, drawn from point A to line a
AN AM
AN - oblique
AN AN
AN AK
AK - oblique
Distance from point to line
M
The distance from point M to line c is...
N
The distance from point N to line c is ...
with
The distance from point K to line c is ...
K
The distance from point F to line c is ...
F
Distance from point to line
AN ⏊ a
AN= 5.2 cm
VK ⏊ a
VK= 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 ANC = ΔVKA(why?)
From the equality of triangles follows AN = VK
The distance from an arbitrary point of one of the parallel lines to another line is called the distance between these lines.
Inverse theorem.
All points of a plane that are on the same side of a given line and are equidistant from it lie on a line parallel to the given line.
AN ⏊ b, BK ⏊ b,
AH = BK
Prove: AB ǁ b
Δ ANC = ΔKVA(why?)
From the equality of triangles it follows … , but these are internal cross-lying angles formed by … , so AB ǁ NK
What is the distance between lines b and c if the distance between lines a and b is 4, and between the lines a and c is 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 lines a and c is 2 ?
What is the distance between the lines a and c, if the distance between lines b and c is 10, and between lines b and a equal to 6?
What is the set of all points in a plane equidistant from two given parallel lines?
a ǁ b
Answer: A line parallel to the given lines and at equal distances from them.
What is the set of all points in a plane 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.
Oh-oh-oh-oh-oh ... well, it's tinny, as if you read the sentence to yourself =) However, then relaxation will help, especially since I bought suitable accessories today. Therefore, let's proceed to the first section, I hope, by the end of the article I will keep a cheerful mood.
Mutual arrangement of two straight lines
The case when the hall sings along in chorus. Two lines can:
1) match;
2) be parallel: ;
3) or intersect at a single point: .
Help for dummies : please remember the mathematical sign of the intersection , it will occur very often. The entry means that the line intersects with the line at the point.
How to determine the relative position of two lines?
Let's start with the first case:
Two lines coincide if and only if their respective coefficients are proportional, that is, there is such a number "lambda" that the equalities
Let's consider straight lines and compose three equations from the corresponding coefficients: . From each equation it follows that, therefore, these lines coincide.
Indeed, if all the coefficients of the equation 
multiply by -1 (change signs), and all the coefficients of the equation 
reduce by 2, you get the same equation: .
The second case when the lines are parallel:
Two lines are parallel if and only if their coefficients at the variables are proportional: 
, but.
As an example, consider two straight lines. We check the proportionality of the corresponding coefficients for the variables : 
However, it is clear that .
And the third case, when the lines intersect:
Two lines intersect if and only if their coefficients of the variables are NOT proportional, that is, there is NOT such a value of "lambda" that the equalities are fulfilled 
So, for straight lines we will compose a system: 
From the first equation it follows that , and from the second equation: , hence, the system is inconsistent(no solutions). Thus, the coefficients at the variables are not proportional.
Conclusion: lines intersect
In practical problems, the solution scheme just considered can be used. By the way, it is very similar to the algorithm for checking vectors for collinearity, which we considered in the lesson. The concept of linear (non) dependence of vectors. Vector basis. But there is a more civilized package:
Example 1
Find out the relative position of the lines: 
Decision based on the study of directing vectors of straight lines:
a) From the equations we find the direction vectors of the lines: 
.
, so the vectors are not collinear and the lines intersect.
Just in case, I will put a stone with pointers at the crossroads:
The rest jump over the stone and follow on, straight to Kashchei the Deathless =)
b) Find the direction vectors of the lines: 
The lines have the same direction vector, which means they are either parallel or the same. Here the determinant is not necessary.
Obviously, the coefficients of the unknowns are proportional, while .
Let's find out if the equality is true: 
Thus,
c) Find the direction vectors of the lines: 
Let's calculate the determinant, composed of the coordinates of these vectors: 
, therefore, the direction vectors are collinear. The lines are either parallel or coincide.
The proportionality factor "lambda" is easy to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: 
.
Now let's find out if the equality is true. Both free terms are zero, so:
The resulting value satisfies this equation (any number generally satisfies it).
Thus, the lines coincide.
Answer:
Very soon you will learn (or even have already learned) to solve the considered problem verbally literally in a matter of seconds. In this regard, I see no reason to offer something for an independent solution, it is better to lay one more important brick in the geometric foundation:
How to draw a line parallel to a given one?
For ignorance of this simplest task, the Nightingale the Robber severely punishes.
Example 2
The straight line is given by the equation . Write an equation for a parallel line that passes through the point.
Decision: Denote the unknown line by the letter . What does the condition say about it? The line passes through the point. And if the lines are parallel, then it is obvious that the directing vector of the line "ce" is also suitable for constructing the line "de".
We take out the direction vector from the equation:
Answer:
The geometry of the example looks simple: 
Analytical verification consists of the following steps:
1) We check that the lines have the same direction vector (if the equation of the line is not properly simplified, then the vectors will be collinear).
2) Check if the point satisfies the resulting equation.
Analytical verification in most cases is easy to perform verbally. Look at the two equations and many of you will quickly figure out how the lines are parallel without any drawing.
Examples for self-solving today will be creative. Because you still have to compete with Baba Yaga, and she, you know, is a lover of all kinds of riddles.
Example 3
Write an equation for a line passing through a point parallel to the line if
There is a rational and not very rational way to solve. The shortest way is at the end of the lesson.
We did a little work with parallel lines and will return to them later. The case of coinciding lines is of little interest, so let's consider a problem that is well known to you from the school curriculum:
How to find the point of intersection of two lines?
If straight 
intersect at the point , then its coordinates are the solution systems of linear equations 
How to find the point of intersection of lines? Solve the system.
Here's to you geometric meaning of a system of two linear equations with two unknowns are two intersecting (most often) straight lines on a plane.
Example 4
Find the point of intersection of lines
Decision: There are two ways to solve - graphical and analytical.
The graphical way is to simply draw the given lines and find out the point of intersection directly from the drawing: 
Here is our point: . To check, you should substitute its coordinates into each equation of a straight line, they should fit both there and there. In other words, the coordinates of a point are the solution of the system . In fact, we considered a graphical way to solve systems of linear equations with two equations, two unknowns.
The graphical method, of course, is not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide this way, the point is that it will take time to make a correct and EXACT drawing. In addition, some lines are not so easy to construct, and the intersection point itself may be somewhere in the thirtieth kingdom outside the notebook sheet.
Therefore, it is more expedient to search for the intersection point by the analytical method. Let's solve the system: 
To solve the system, the method of termwise addition of equations was used. To develop the relevant skills, visit the lesson How to solve a system of equations?
Answer:
The verification is trivial - the coordinates of the intersection point must satisfy each equation of the system.
Example 5
Find the point of intersection of the lines if they intersect.
This is a do-it-yourself example. The task can be conveniently divided into several stages. Analysis of the condition suggests that it is necessary:
1) Write the equation of a straight line.
2) Write the equation of a straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.
The development of an action algorithm is typical for many geometric problems, and I will repeatedly focus on this.
Full solution and answer at the end of the tutorial:
A pair of shoes has not yet been worn out, as we got to the second section of the lesson:
Perpendicular lines. The distance from a point to a line.
Angle between lines
Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to the given one, and now the hut on chicken legs will turn 90 degrees:
How to draw a line perpendicular to a given one?
Example 6
The straight line is given by the equation . Write an equation for a perpendicular line passing through a point.
Decision: It is known by assumption that . It would be nice to find the direction vector of the straight line. Since the lines are perpendicular, the trick is simple:
From the equation we “remove” the normal vector: , which will be the directing vector of the straight line.
We compose the equation of a straight line by a point and a directing vector:
Answer: 
Let's unfold the geometric sketch:
Hmmm... Orange sky, orange sea, orange camel.
Analytical verification of the solution:
1) Extract the direction vectors from the equations 
and with the help dot product of vectors we conclude that the lines are indeed perpendicular: .
By the way, you can use normal vectors, it's even easier.
2) Check if the point satisfies the resulting equation 
.
Verification, again, is easy to perform verbally.
Example 7
Find the point of intersection of perpendicular lines, if the equation is known 
and dot.
This is a do-it-yourself example. There are several actions in the task, so it is convenient to arrange the solution point by point.
Our exciting journey continues:
Distance from point to line
Before us is a straight strip of the river and our task is to reach it in the shortest way. There are no obstacles, and the most optimal route will be movement along the perpendicular. That is, the distance from a point to a line is the length of the perpendicular segment.
The distance in geometry is traditionally denoted by the Greek letter "ro", for example: - the distance from the point "em" to the straight line "de".
Distance from point to line 
is expressed by the formula
Example 8
Find the distance from a point to a line 
Decision: all you need is to carefully substitute the numbers into the formula and do the calculations:
Answer: 
Let's execute the drawing: 
The distance found from the point to the line is exactly the length of the red segment. If you make a drawing on checkered paper on a scale of 1 unit. \u003d 1 cm (2 cells), then the distance can be measured with an ordinary ruler.
Consider another task according to the same drawing:
The task is to find the coordinates of the point , which is symmetrical to the point with respect to the line 
. I propose to perform the actions on your own, however, I will outline the solution algorithm with intermediate results:
1) Find a line that is perpendicular to a line.
2) Find the point of intersection of the lines: 
.
Both actions are discussed in detail in this lesson.
3) The point is the midpoint of the segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the middle of the segment find .
It will not be superfluous to check that the distance is also equal to 2.2 units.
Difficulties here may arise in calculations, but in the tower a microcalculator helps out a lot, allowing you to count ordinary fractions. Have advised many times and will recommend again.
How to find the distance between two parallel lines?
Example 9
Find the distance between two parallel lines
This is another example for an independent solution. A little hint: there are infinitely many ways to solve. Debriefing at the end of the lesson, but better try to guess for yourself, I think you managed to disperse your ingenuity well.
Angle between two lines
Whatever the corner, then the jamb: 
In geometry, the angle between two straight lines is taken as the SMALLER angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered to be the angle between intersecting lines. And its “green” neighbor or oppositely oriented crimson corner.
If the lines are perpendicular, then any of the 4 angles can be taken as the angle between them.
How are the angles different? Orientation. First, the direction of "scrolling" the corner is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example, if .
Why did I say this? It seems that you can get by with the usual concept of an angle. The fact is that in the formulas by which we will find the angles, a negative result can easily be obtained, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing for a negative angle, it is imperative to indicate its orientation (clockwise) with an arrow.
How to find the angle between two lines? There are two working formulas:
Example 10
Find the angle between lines
Decision and Method one
Consider two straight lines given by equations in general form: 
If straight not perpendicular, then oriented the angle between them can be calculated using the formula: 
Let's pay close attention to the denominator - this is exactly scalar product direction vectors of straight lines:
If , then the denominator of the formula vanishes, and the vectors will be orthogonal and the lines will be perpendicular. That is why a reservation was made about the non-perpendicularity of the lines in the formulation.
Based on the foregoing, the solution is conveniently formalized in two steps:
1) Calculate the scalar product of directing vectors of straight lines:
so the lines are not perpendicular.
2) We find the angle between the lines by the formula:
Using the inverse function, it is easy to find the angle itself. In this case, we use the oddness of the arc tangent (see Fig. Graphs and properties of elementary functions):
Answer: 
In the answer, we indicate the exact value, as well as the approximate value (preferably both in degrees and in radians), calculated using a calculator.
Well, minus, so minus, it's okay. Here is a geometric illustration: 
It is not surprising that the angle turned out to be of a negative orientation, because in the condition of the problem the first number is a straight line and the “twisting” of the angle began precisely from it.
If you really want to get a positive angle, you need to swap the straight lines, that is, take the coefficients from the second equation 
, and take the coefficients from the first equation . In short, you need to start with a direct 
.
Lesson outline
Triangle sum of angles theorem
1. Full name: Sayfetdinova Gulnara Vasilevna
2. Place of work: Municipal budgetary educational institution "Knyazevskaya secondary school" Tukaevsky district of the Republic of Tatarstan
3. Position: mathematic teacher
4. Thing: geometry
5. Class: 7th grade
6. Lesson topic: The distance from a point to a line. Distance between parallel lines.
7. Basic Tutorial: Geometry. 7-9 grades: a textbook for educational institutions / ed. L.S. Atanasyan, V.F. Butuzov,
S.B. Kadomtsev et al., 2010
8.Purposes:
Activity goal: create conditions for independent formulation and proof of the properties of oblique and perpedicular omitted from a point to a line, the theorem on the equidistance of points on parallel lines; organize the activities of students in the perception, comprehension and primary consolidation of new knowledge and methods of activity.
Educational Purpose:
Subject:
apply the concepts of distance from a point to a line, distance between lines when solving problems
Metasubject:
Regulatory UUD:
Cognitive UUD:
Communicative UUD:
Personal UUD:
10.
Teaching methods: problem, research.
11. Forms of organization of educational activities: frontal, group, pair, individual, learning structures.
12. Equipment, specifications:
Computer, projector, screen, internet, software: Microsoft Power Point, seating in the classroom - 4 people at the table.
13. Duration of the lesson: 45 min
14. Lesson plan
I . Organizing time.
II . Knowledge update.
III . Setting the goal of the lesson . 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? And let's pick it up and start the lesson with a smile! Let's smile at our partner! Let's smile at our shoulder partner!
II . Knowledge update.
Teacher : You have been studying a new subject of geometry for half a year already and probably know what a theorem is. What proof methods do you know?
Possible student responses: Contradiction method, constructive method, method of proof based on axioms and previously proven theorems (slide No. 2).
Teacher: Guys, what are your associations with the word - distance?
Possible student responses: Distance between cities, distance between poles, distance from something to something (slide number 3).
Teacher: What is the distance between two points called?
Possible student responses: Cut length (slide number 4).
Teacher: Make an entry in the technological map in step 1
Teacher: Note that in geometry, distance refers to the shortest distance. Make an entry in the technological map in step 2
Teacher: What can be said about the relative position of the line AH and the line a?
Teacher: What are these lines called?
Teacher: BUT What is the name of segment AN?
Teacher: Remember: Perpendicular is a line segment. Make an entry in the technological map in step 3.
III. Setting the goal of the lesson.Introduction of new material.
Teacher: Practical task:
We are on the field, the road passes through the field. Draw a mathematical model of the situation. We need to get on the road. Draw a trajectory (slide number 6).
Teacher: And how can this trajectory be defined in mathematical language? Possible student responses: Perpendicular
Teacher: Why not? -
Try to give it a name (slide number 7).
Possible student responses: Inclined.
Teacher: How many slopes can be drawn from this point?
Possible student responses: A bunch of.
(slide number 7).
Teacher: So you think that the shortest path is a perpendicular? Prove it.
Teacher: Now prove that any oblique is greater than a perpendicular.
What do we see in the picture?
Possible student responses: right triangle (slide number 8).
Teacher: What is the name of the perpendicular and oblique in this triangle? Possible student responses: leg and hypotenuse.
Teacher: Why is the hypotenuse longer than the leg?
Possible student responses: Opposite the larger angle lies the larger side. The largest angle in a right triangle is a right angle. Opposite it lies the hypotenuse.
Teacher. What is another name for segment AC? And if we return to the content of the task?
Possible student responses: Distance from point to line .
Teacher: Formulate a definition: “The distance from a point to a line is ... (the length of the perpendicular dropped from this point to the line)” (slide No. 9). Make an entry in the technological map in step 4.
Teacher: Practical task.
Find the distance from point B to lines A D andDC using a drawing triangle and ruler (slide No. 10). technological map p. 6
Teacher: Practical task. Construct two parallel lines a and b . On the line a, mark point A. Drop the 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 in a long corridor on the floor. 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 double-checking the lengths of the common perpendiculars in other places of the corridor? (slide number 14).
Possible student responses: 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, what needs to be proved.
How is the equality of segments and angles usually proved in geometry?
Possible student responses: 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 signal board, write down the three conditions for the sign of equality of triangles.
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 the point A and C relative to the line BD?
Possible student responses: They are at equal distance. They are equidistant (slide number 18).
Teacher: Does this property hold for any points?
Possible student responses: Yes
Teacher: Let's try to formulate this property. What is a property assertion?
Possible student responses: From condition and conclusion (slide No. 19,20).
Possible student responses: 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 responses: 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. Pupils think and write down the answer on their piece of paper
3. Pupils take turns reading their answer from a piece of paper.
Teacher: What is the reverse statement?
Possible student responses: If the condition and conclusion are interchanged.
Teacher: Formulate the inverse statement (slide number 22).
Possible student responses: If the 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 paragraph 7.8.
Teacher: Is it possible to define such a concept as the distance between parallel lines?
Possible student responses: Yes
Teacher: What is the distance between parallel lines
Possible student responses: The length of the common perpendicular. Make an entry in the technological map in paragraph 5.
IV. Application of the theorem, fulfillment of npractical work.
Teacher: Practical work. Find the width of the strip.
What is the mathematical concept - the width of the strip?
Teacher: Where else are these theorems applied in practical life?
VI. Summarizing. Reflection.
Teacher: What new concepts did you get?
What did you learn in the lesson?
Where in life will we use it?
(slide №№26-28)
Teacher: Make an entry in the technological map in paragraph 9
Homework No. 276,279 - proof of the inverse theorem.
Introspection of the lesson
Goals:
Activity goal: create conditions for self-formulation and proof of the properties of oblique and perpedicular dropped from a point to a line, create conditions for proving the theorem on the equidistance of points on parallel lines; organize the activities of students in the perception, comprehension and primary consolidation of new knowledge and methods of activity.
Educational Purpose: develop the knowledge that the perpendicular is less than any oblique drawn from one point to a straight line, all points of each of the two parallel lines are equidistant from the other straight line.
Subject: The student will have the opportunity to learn:
apply the theorem in solving practical problems
analyze, compare, generalize, draw conclusions to solve practical problems.
Metasubject:
Regulatory UUD:
the ability to independently set goals, choose 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, inference, conclusions;
the ability to put forward hypotheses when solving educational problems and understand the need to test them; the ability to apply 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;
the ability to understand and use drawings and drawings to illustrate, interpret, argue.
Communicative 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;
the ability to work in a group: to find a common solution and resolve conflicts based on the coordination of positions and taking into account interests, listen to a partner, formulate, argue and defend 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 arguments, give examples and counterexamples;
development of critical thinking, the ability to recognize logically incorrect statements, to distinguish a hypothesis from a fact;
develop creative thinking, initiative, resourcefulness, activity in solving geometric problems.
The structure of the lesson fragment corresponded to the type - the lesson of discovering new knowledge. In accordance with the goals and content of the material, the lesson was built according to the following stages:
I . Organizing time.
II . Knowledge update.
III . Setting the goal of the lesson . Introduction of new material.
IV. Application of the theorem, implementation of practical work.
VI. Summarizing.
All structural elements of the lesson were sustained. The organization of the educational process is built by the activity method.
The aim of the first stageit was quick to turn the students into a business rhythm.
At the second stage the knowledge necessary for working on new material was updated.
At the third stageIn order to define the concepts of distance from a point to a straight line, the concept of an oblique 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 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 putting forward a hypothesis about the properties of the objects under consideration and, with the subsequent search for evidence of the validity of the proposed assumption.
By organizing the work on the formulation of the theorem, and then the inverse theorem, I achieved the 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 activity was organized through frontal work, individual, group work. Such an organization allowed each student to be actively involved in achieving the goal. Students collaborated with each other to help each other.
Time, I believe, was distributed rationally. For a short period, it was possible to introduce the concepts of distance from a point to a straight line, an inclined one, the distance between parallel lines, formulate two theorems and prove, consider the application of the theorem in practice.
For clarity, a presentation was used during the lesson. Used a special demo program to compare the length of an oblique and a perpendicular, in which geometric shapes come to life. During the lesson, I used the work of students on the signal board, which solves the problems of equal participation of students in the lesson, control over the assimilation 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 a theorem, formulating a theorem
At the end of the lesson, the students themselves formulated the topic.
Reflection