The distance between the points of the coordinate line. Lesson on the topic the distance between the points of the coordinate line. Distance between points in space, formula

Lesson plan.

Distance between two points on a straight line.

Rectangular (Cartesian) coordinate system.

Distance between two points on a straight line.

Theorem 3.If A (x) and B (y) are any two points, then d - the distance between them is calculated by the formula: d \u003d lу - хl.

Evidence. According to Theorem 2, we have AB \u003d y - x. But the distance between points A and B is equal to the length of the segment AB, those. the length of the vector AB. Therefore, d \u003d lАВl \u003d lу-хl.

Since the numbers y-x and x-y are taken modulo, we can write d \u003d lx-yl. So, to find the distance between points on the coordinate line, you need to find the modulus of the difference between their coordinates.

Example 4... Given points A (2) and B (-6), find the distance between them.

Decision. Let's substitute in the formula instead of x \u003d 2 and y \u003d -6. We get AB \u003d lу-хl \u003d l-6-2l \u003d l-8l \u003d 8.

Example 5. Construct a point symmetrical to point M (4) relative to the origin.

Decision. Because from point M to point O 4 unit segments, set aside on the right, then in order to build a point symmetric to it, we postpone 4 unit segments to the left from point O, we get point M "(-4).

Example 6. Construct point C (x) symmetric to point A (-4) relative to point B (2).

Decision. Let's mark the points A (-4) and B (2) on the number line. Find the distance between the points according to Theorem 3, we get 6. Then the distance between the points B and C should also be 6. We put off 6 unit segments from point B to the right, we get point C (8).

Exercises. 1) Find the distance between points A and B: a) A (3) and B (11), b) A (5) and B (2), c) A (-1) and B (3), d) A (-5) and B (-3), e) A (-1) and B (3), (Answer: a) 8, b) 3, c) 4, d) 2, e) 2).

2) Construct a point C (x) symmetric to point A (-5) relative to point B (-1). (Answer: C (3)).

Rectangular (Cartesian) coordinate system.

Two mutually perpendicular axes Ox and Oy, having a common origin O and the same scale unit, form rectangular (or cartesian) plane coordinate system.

Axis Oh is called abscissa, and the Oy axis is the ordinate... The point O of the intersection of the axes is called origin... The plane in which the axes Ox and Oy are located is called the coordinate plane and is denoted by Ox.

Let M be an arbitrary point of the plane. Let us omit from it the perpendiculars MA and MB, respectively, on the Ox and Oy axes. The points of intersection of A and B eith of perpendiculars with the axes are called projections points M on the coordinate axis.

Points A and B correspond to certain numbers x and y - their coordinates on the axes Ox and Oy. The number x is called abscissa point M, the number y - her ordinate.

The fact that the point M has coordinates x and y is symbolically denoted as follows: M (x, y). In this case, the first in parentheses indicate the abscissa, and the second - the ordinate. The origin has coordinates (0,0).

Thus, for the chosen coordinate system, each point M of the plane corresponds to a pair of numbers (x, y) - its rectangular coordinates and, conversely, to each pair of numbers (x, y) there corresponds, and moreover, one point M on the plane Oxy such that its the abscissa is x and the ordinate is y.

So, a rectangular coordinate system on a plane establishes a one-to-one correspondence between the set of all points of the plane and the set of pairs of numbers, which makes it possible to apply algebraic methods when solving geometric problems.

The coordinate axes divide the plane into four parts, they are called quarters, quadrants or coordinate angles and are numbered with Roman numerals I, II, III, IV as shown in the figure (hyperlink).

The figure also shows the signs of the coordinates of the points, depending on their location. (for example, in the first quarter, both coordinates are positive).

Example 7. Construct points: A (3; 5), B (-3; 2), C (2; -4), D (-5; -1).

Decision. Let's construct point A (3; 5). First of all, we introduce a rectangular coordinate system. Then, along the abscissa axis, we postpone 3 scale units to the right, and along the ordinate axis - 5 scale units upwards and through the final division points we draw straight lines parallel to the coordinate axes. The intersection point of these lines is the desired point A (3; 5). The rest of the points are constructed in the same way (see the picture-hyperlink).

Exercises.

Without drawing point A (2; -4), find out which quarter it belongs to.

In what quarters can a point be located if its ordinate is positive?

On the Oy axis, a point with a coordinate of -5 is taken. What are its coordinates on the plane? (Answer: since the point lies on the Oy axis, then its abscissa is 0, the ordinate is given by condition, so the coordinates of the point are (0; -5)).

Points are given: a) A (2; 3), b) B (-3; 2), c) C (-1; -1), d) D (x; y). Find the coordinates of the points symmetrical to them about the Ox axis. Plot all these points. (answer: a) (2; -3), b) (-3; -2), c) (-1; 1), d) (x; -y)).

Points are given: a) A (-1; 2), b) B (3; -1), c) C (-2; -2), d) D (x; y). Find the coordinates of the points symmetrical to them about the Oy axis. Plot all these points. (answer: a) (1; 2), b) (-3; -1), c) (2; -2), d) (-x; y)).

Points are given: a) A (3; 3), b) B (2; -4), c) C (-2; 1), d) D (x; y). Find the coordinates of the points symmetrical to them about the origin. Plot all these points. (answer: a) (-3; -3), b) (-2; 4), c) (2; -1), d) (-x; -y)).

Point M (3; -1) is given. Find the coordinates of the points symmetrical to it about the Ox axis, the Oy axis and the origin. Plot all the points. (Answer: (3; 1), (-3; -1), (-3; 1)).

Determine in which quarters the point M (x; y) can be located if: a) xy\u003e 0, b) xy< 0, в) х-у=0, г) х+у=0. (ответ: а) в первой и третьей, б)во второй и четвертой, в) в первой и третьей, г) во второй и четвертой).

Determine the coordinates of the vertices of an equilateral triangle with side equal to 10, lying in the first quarter, if one of its vertices coincides with the origin of coordinates O, and the base of the triangle is located on the Ox axis. Draw a drawing. (Answer: (0; 0), (10; 0), (5; 5v3)).

Using the coordinate method, determine the coordinates of all vertices of the regular hexagon ABCDEF. (Answer: A (0; 0), B (1; 0), C (1.5; v3 / 2), D (1; v3), E (0; v3), F (-0.5; v3 / 2). Note: take point A as the origin of coordinates, direct the abscissa axis from A to B, take the length of the side AB as the unit of scale. It is convenient to draw large diagonals of the hexagon.)

In mathematics, both algebra and geometry pose problems of finding the distance to a point or straight line from a given object. It is found in completely different ways, the choice of which depends on the initial data. Let's consider how to find the distance between the given objects in different conditions.

At the initial stage of mastering mathematical science, they teach how to use elementary tools (such as a ruler, protractor, compasses, triangle, and others). Finding the distance between points or straight lines with their help is not difficult at all. It is enough to attach the scale of divisions and write down the answer. One has only to know that the distance will be equal to the length of the straight line that can be drawn between the points, and in the case of parallel lines, the perpendicular between them.

Use of theorems and axioms of geometry

In learning to measure distance without the help of special devices or This requires numerous theorems, axioms and their proofs. Often the tasks of how to find the distance come down to education and the search for its sides. To solve such problems, it is enough to know the Pythagorean theorem, the properties of triangles and how to transform them.

If there are two points and their position on the coordinate axis is given, then how to find the distance from one to the other? The solution will include several stages:

1. We connect the points with a straight line, the length of which will be the distance between them.
2. We find the difference in the values \u200b\u200bof the coordinates of the points (k; p) of each axis: | k 1 - k 2 | \u003d q 1 and | p 1 - p 2 | \u003d q 2 (we take the values \u200b\u200bmodulo, since the distance cannot be negative) ...
3. After that, we square the resulting numbers and find their sum: q 1 2 + q 2 2
4. The final step will be to extract from the resulting number. This will be the distance between the points: q \u003d V (q 1 2 + q 2 2).

As a result, the whole solution is carried out according to one formula, where the distance is equal to the square root of the sum of the squares of the difference of coordinates:

q \u003d V (| k 1 - k 2 | 2 + | p 1 - p 2 | 2)

If the question arises of how to find the distance from one point to another, then the search for an answer to it will not be very different from the above. The decision will be made according to the following formula:

q \u003d V (| k 1 - k 2 | 2 + | p 1 - p 2 | 2 + | f 1 - f 2 | 2)

A perpendicular drawn from any point lying on one straight line to a parallel will be the distance. When solving problems in a plane, it is necessary to find the coordinates of any point of one of the straight lines. And then calculate the distance from it to the second straight line. To do this, we bring them to the general form Ax + Vy + C \u003d 0. It is known from the properties of parallel lines that their coefficients A and B will be equal. In this case, you can find it by the formula:

q \u003d | C 1 - C 2 | / V (A 2 + B 2)

Thus, when answering the question of how to find the distance from a given object, it is necessary to be guided by the condition of the problem and the tools provided for its solution. They can be both measuring devices and theorems and formulas.

In this article, we will consider ways to determine the distance from point to point theoretically and using the example of specific tasks. And to begin with, let's introduce some definitions.

Definition 1

Distance between points Is the length of the segment connecting them, on the available scale. It is necessary to set the scale in order to have a unit of length for measurement. Therefore, basically the problem of finding the distance between points is solved by using their coordinates on a coordinate line, in a coordinate plane or three-dimensional space.

Initial data: coordinate line O x and an arbitrary point A lying on it. Any point of the line has one real number: let it be a certain number for point A x A,it is also the coordinate of point A.

In general, we can say that the estimation of the length of a certain segment occurs in comparison with the segment taken as a unit of length in a given scale.

If point A corresponds to an integer real number, sequentially postponing from point O to point along the straight line OA segments - units of length, we can determine the length of the segment O A by the total number of pending unit segments.

For example, point A corresponds to the number 3 - to get to it from point O, you will need to postpone three unit segments. If point A has a coordinate - 4 - unit segments are plotted in the same way, but in a different, negative direction. Thus, in the first case, the distance O And is equal to 3; in the second case О А \u003d 4.

If point A has a rational number as a coordinate, then from the origin (point O) we postpone an integer number of unit segments, and then its necessary part. But it is not always geometrically possible to make a measurement. For example, it seems difficult to postpone the fraction 4 111 on the coordinate straight line.

In the above way, it is completely impossible to postpone an irrational number on a straight line. For example, when the coordinate of point A is 11. In this case, it is possible to turn to abstraction: if the given coordinate of point A is greater than zero, then O A \u003d x A (the number is taken as the distance); if the coordinate is less than zero, then O A \u003d - x A. In general, these statements are true for any real number x A.

To summarize: the distance from the origin to the point corresponding to the real number on the coordinate line is:

• 0 if the point coincides with the origin;

• x A if x A\u003e 0;

• - x A if x A< 0 .

In this case, it is obvious that the length of the segment itself cannot be negative, therefore, using the modulus sign, we write down the distance from point O to point A with the coordinate x A: O A \u003d x A

The statement will be true: the distance from one point to another will be equal to the modulus of the coordinate difference.Those. for points A and B lying on the same coordinate line at any of their locations and having coordinates, respectively x A and x B: A B \u003d x B - x A.

Initial data: points A and B lying on a plane in a rectangular coordinate system O x y with given coordinates: A (x A, y A) and B (x B, y B).

Let's draw perpendiculars to the coordinate axes O x and O y through points A and B and get the projection points as a result: A x, A y, B x, B y. Based on the location of points A and B, the following options are further possible:

If points A and B coincide, then the distance between them is zero;

If points A and B lie on a straight line perpendicular to the O x axis (abscissa axis), then points and coincide, and | A B | \u003d | А y B y | ... Since the distance between the points is equal to the modulus of the difference of their coordinates, then A y B y \u003d y B - y A, and therefore A B \u003d A y B y \u003d y B - y A.

If points A and B lie on a straight line perpendicular to the O y axis (ordinate axis) - by analogy with the previous paragraph: A B \u003d A x B x \u003d x B - x A

If points A and B do not lie on a straight line perpendicular to one of the coordinate axes, we find the distance between them, deriving the calculation formula:

We see that triangle ABC is rectangular in construction. Moreover, A C \u003d A x B x and B C \u003d A y B y. Using the Pythagorean theorem, we compose the equality: AB 2 \u003d AC 2 + BC 2 ⇔ AB 2 \u003d A x B x 2 + A y B y 2, and then we transform it: AB \u003d A x B x 2 + A y B y 2 \u003d x B - x A 2 + y B - y A 2 \u003d (x B - x A) 2 + (y B - y A) 2

Let's form a conclusion from the result obtained: the distance from point A to point B on the plane is determined by calculation using the formula using the coordinates of these points

A B \u003d (x B - x A) 2 + (y B - y A) 2

The resulting formula also confirms the previously formed statements for cases of coincidence of points or situations when the points lie on straight lines perpendicular to the axes. So, for the case of coincidence of points A and B, the equality will be true: A B \u003d (x B - x A) 2 + (y B - y A) 2 \u003d 0 2 + 0 2 \u003d 0

For a situation where points A and B lie on a straight line perpendicular to the abscissa axis:

A B \u003d (x B - x A) 2 + (y B - y A) 2 \u003d 0 2 + (y B - y A) 2 \u003d y B - y A

For the case when points A and B lie on a straight line perpendicular to the ordinate axis:

A B \u003d (x B - x A) 2 + (y B - y A) 2 \u003d (x B - x A) 2 + 0 2 \u003d x B - x A

Initial data: rectangular coordinate system O x y z with arbitrary points lying on it with given coordinates A (x A, y A, z A) and B (x B, y B, z B). It is necessary to determine the distance between these points.

Consider the general case when points A and B do not lie in a plane parallel to one of the coordinate planes. We draw through points A and B planes perpendicular to the coordinate axes, and we obtain the corresponding projection points: A x, A y, A z, B x, B y, B z

The distance between points A and B is the diagonal of the resulting parallelepiped. According to the construction of the measurement of this parallelepiped: A x B x, A y B y and A z B z

It is known from the geometry course that the square of the diagonal of a parallelepiped is equal to the sum of the squares of its measurements. Based on this statement, we obtain the equality: A B 2 \u003d A x B x 2 + A y B y 2 + A z B z 2

Using the conclusions obtained earlier, we write the following:

A x B x \u003d x B - x A, A y B y \u003d y B - y A, A z B z \u003d z B - z A

Let's transform the expression:

AB 2 \u003d A x B x 2 + A y B y 2 + A z B z 2 \u003d x B - x A 2 + y B - y A 2 + z B - z A 2 \u003d \u003d (x B - x A) 2 + (y B - y A) 2 + z B - z A 2

The final formula for determining the distance between points in space will look like this:

A B \u003d x B - x A 2 + y B - y A 2 + (z B - z A) 2

The resulting formula is also valid for cases when:

The points match;

They lie on the same coordinate axis or a straight line parallel to one of the coordinate axes.

Examples of solving problems on finding the distance between points

Initial data: given a coordinate line and points lying on it with the given coordinates A (1 - 2) and B (11 + 2). It is necessary to find the distance from the origin point O to point A and between points A and B.

Decision

1. The distance from the origin to the point is equal to the modulus of the coordinate of this point, respectively O A \u003d 1 - 2 \u003d 2 - 1
2. The distance between points A and B is defined as the modulus of the difference between the coordinates of these points: A B \u003d 11 + 2 - (1 - 2) \u003d 10 + 2 2

Answer: O A \u003d 2 - 1, A B \u003d 10 + 2 2

Example 2

Initial data: given a rectangular coordinate system and two points lying on it, A (1, - 1) and B (λ + 1, 3). λ is some real number. It is necessary to find all the values \u200b\u200bof this number at which the distance AB will be equal to 5.

Decision

To find the distance between points A and B, use the formula A B \u003d (x B - x A) 2 + y B - y A 2

Substituting the real values \u200b\u200bof the coordinates, we get: A B \u003d (λ + 1 - 1) 2 + (3 - (- 1)) 2 \u003d λ 2 + 16

And we also use the existing condition that AB \u003d 5 and then the equality will be true:

λ 2 + 16 \u003d 5 λ 2 + 16 \u003d 25 λ \u003d ± 3

Answer: AB \u003d 5 if λ \u003d ± 3.

Example 3

Initial data: given a three-dimensional space in a rectangular coordinate system O x y z and points A (1, 2, 3) and B - 7, - 2, 4 lying in it.

Decision

To solve the problem, we use the formula A B \u003d x B - x A 2 + y B - y A 2 + (z B - z A) 2

Substituting real values, we get: A B \u003d (- 7 - 1) 2 + (- 2 - 2) 2 + (4 - 3) 2 \u003d 81 \u003d 9

Answer: | A B | \u003d 9

If you notice an error in the text, please select it and press Ctrl + Enter

Point to point distance is the length of the line segment connecting these points in a given scale. Thus, when it comes to measuring distance, you need to know the scale (unit of length) in which the measurements will be carried out. Therefore, the problem of finding the distance from point to point is usually considered either on a coordinate line or in a rectangular Cartesian coordinate system on a plane or in three-dimensional space. In other words, most often it is necessary to calculate the distance between points by their coordinates.

In this article, we, firstly, recall how the distance from point to point on the coordinate line is determined. Next, we will obtain formulas for calculating the distance between two points of a plane or space at given coordinates. In conclusion, let us consider in detail the solutions of typical examples and tasks.

Page navigation.

Distance between two points on a coordinate line.

Let's first define the designations. The distance from point A to point B will be denoted as.

Hence we can conclude that the distance from point A with coordinate to point B with coordinate is equal to the modulus of the coordinate difference, i.e, at any location of points on the coordinate line.

Distance from point to point on a plane, formula.

Let's get a formula for calculating the distance between points and, given in a rectangular Cartesian coordinate system on the plane.

Depending on the location of points A and B, the following options are possible.

If points A and B coincide, then the distance between them is zero.

If points A and B lie on a straight line perpendicular to the abscissa axis, then the points and coincide, and the distance is equal to the distance. In the previous paragraph, we found out that the distance between two points on the coordinate line is equal to the modulus of the difference between their coordinates, therefore, ... Consequently, .

Similarly, if points A and B lie on a straight line perpendicular to the ordinate, then the distance from point A to point B is found as.

In this case, triangle ABC is rectangular in construction, and and. By the Pythagorean theorem we can write equality, whence.

Let's summarize all the results obtained: the distance from a point to a point on the plane is found through the coordinates of the points by the formula .

The resulting formula for finding the distance between points can be used when points A and B coincide or lie on a straight line perpendicular to one of the coordinate axes. Indeed, if A and B coincide, then. If points A and B lie on a straight line perpendicular to the Ox axis, then. If A and B lie on a straight line perpendicular to the Oy axis, then.

Distance between points in space, formula.

Let's introduce a rectangular coordinate system Oxyz in space. Let's get the formula for finding the distance from the point to the point .

In general, points A and B do not lie in a plane parallel to one of the coordinate planes. Let us draw through points A and B planes perpendicular to the coordinate axes Ox, Oy and Oz. The points of intersection of these planes with the coordinate axes will give us the projection of points A and B on these axes. We denote projections .

The desired distance between points A and B is the diagonal of the rectangular parallelepiped shown in the figure. By construction, the dimensions of this parallelepiped are and. In a high school geometry course, it was proved that the square of the diagonal of a rectangular parallelepiped is equal to the sum of the squares of its three dimensions, therefore,. Based on the information in the first section of this article, we can write the following equalities, therefore,

whence we get formula for finding the distance between points in space .

This formula is also valid if points A and B

• match;
• belong to one of the coordinate axes or a straight line parallel to one of the coordinate axes;
• belong to one of the coordinate planes or a plane parallel to one of the coordinate planes.

Finding the distance from point to point, examples and solutions.

So, we got formulas for finding the distance between two points of the coordinate line, plane and three-dimensional space. It's time to consider solutions to typical examples.

The number of problems in the solution of which the final step is to find the distance between two points by their coordinates is truly enormous. A complete overview of such examples is beyond the scope of this article. Here we will restrict ourselves to examples in which the coordinates of two points are known and it is required to calculate the distance between them.

Lesson # 3

TOPIC: Distance between points of a coordinate line

The purpose of the teacher: create conditions for mastering the skills to find the distance between points on the coordinate line, calculating the modulus of the difference, the coordinates of the midpoint of the segment.

Planned results of studying the topic:

Personal: show a cognitive interest in the study of the subject.

Subject: know how to find the distance between points on the coordinate line, calculating the modulus of the difference, the coordinates of the midpoint of the segment.

Meta-subject results of studying the topic (universal educational actions):

cognitive: focus on a variety of ways to solve problems; know how to generalize and organize information;

regulatory: take into account the rule in planning and controlling the solution;

communicative: take into account different opinions and strive to coordinate different positions in cooperation.

Lesson script.

I .Org moment.
Hello guys. Today at the guest We welcome them!

Sit down.

Our lesson is not quite ordinary. Knowledge generalization lesson. We must show what we have learned, what we have learned.

What topic are we working on lately? (Comparison, addition of rational numbers)

As the epigraph of the lesson, I took these words : We will go for science today

Let's take a fantasy to help

We won't turn off the straight road

And so that we can reach our goals sooner

We must climb the stairs up!

2. Updating knowledge .

Task "Ladder".

Option work, validation and self-assessment

3 Well done, we continue to move upward for knowledge.Let's check our homework.

1. Find the distance between the points of the coordinate line: Д / З

a) A (-4) and B (-6); b) A (5) and B (-7); c) A (3) and B (-18).

DECISION: a) AB \u003d | -6 - (- 4) | \u003d | -2 | \u003d 2

b) AB \u003d | -7-5 | \u003d 12

c) AB \u003d | -18-3 | \u003d 21

2. Find the coordinates of points remote from the point:

a) A (-8) by 5; b) B (6) by -2.7; c) C (4) by -3.2

Decision: a) -8 + 5 \u003d -3 AND 1 (-3) and -8-5 \u003d -13 AND 2 (-13)

b) 6 + (- 2.7) \u003d 3.3 IN 1 (3,3) and 6 - (- 2.7) \u003d 8.7 IN 2 (8,7)

c) 4 + (- 3.2) \u003d 0.8 FROM 1 (0,8) 4-(-3,2) = 7,2 FROM 2 (7,2)

3) Find the coordinate of point C, the middle of the segment, if:

a) A (-12) B (1) b) A (-7) and B (9) c) A (16) and B (-8)

DECISION:

12 + 1 \u003d -11 B) -7 + 9 \u003d 2 C) 16 + (- 8) \u003d 8

11: 2=-5,5 2:2=1 8:2 =4

M (-5.5) s (1) M (4)

You have a homework standard on your tables. Check and put the mark on the self-assessment sheet.

4 ... Blitz - poll :

1. What is a coordinate line?

2. What rules for comparing rational numbers do you know?

3.What is the modulus of number?

4.How to add two numbers with the same sign?

5.How to add two numbers with different signs?

6. How to determine the distance between the points of the coordinate line?

Well, now let's show how we can apply our knowledge in practice.

5 fix bugs

12+4 =-16 -12+(-18) =6 9-14=5

16 +(-10)=6 30 +(-10) =-20 5 –(-3)=2

6 –(-5) =11 -20 -14 =-34 -2 +7=9

11-28 =-39 -34 -5 =-29 9 -13=22

Perform a self-test.

12+4 =--8 -12+(-18) =30 9-14= -5

16 +(-10)=-26 30 +(-10) =20 5 –(-3)=8

26 –(-5) =-21 -20 -14 =-34 -2 +7=5

11-28 =--17 -34 -5 =-41 9 -13=-4

6. Determine the distance between the points: and find the midpoint of the segment (by options)

(exchange of notebooks and mutual check.)

7. Well, now we will rest. Our eyes must rest

8. Independent work (in a notebook) marking.

Option 1 Option 2

1,5-4,6 0,8 -1,2

-2,8 +3,8 4-9,4

0,45 -1 -4,3 +(-1,2) (Slide 9)

Purpose: test the ability to apply the laws of addition to transform expressions; develop cognitive interest, independence; to cultivate perseverance and perseverance in achieving the goal.




Find the meaning of the expression and, according to the result obtained, in accordance with the table, color the gnome. (the card with the gnome remains with the students as a talisman)

Well done boys!

You have completed the assignments

And they flashed with knowledge.

And the magic key to learning is

Your persistence and patience!