Let a rectangular coordinate system be given.

Theorem 1.1. For any two points M 1 (x 1; y 1) and M 2 (x 2; y 2) of the plane, the distance d between them is expressed by the formula

Proof. Let us drop from the points M 1 and M 2 the perpendiculars M 1 B and M 2 A, respectively

on the Oy and Ox axes and denote by K the point of intersection of the lines M 1 B and M 2 A (Fig. 1.4). The following cases are possible:

1) Points M 1, M 2 and K are different. Obviously, the point K has coordinates (x 2; y 1). It is easy to see that M 1 K = ôx 2 – x 1 ô, M 2 K = ôy 2 – y 1 ô. Because ∆M 1 KM 2 is rectangular, then by the Pythagorean theorem d = M 1 M 2 = = .

2) Point K coincides with point M 2, but is different from point M 1 (Fig. 1.5). In this case y 2 = y 1

and d \u003d M 1 M 2 \u003d M 1 K \u003d ôx 2 - x 1 ô \u003d =

3) The point K coincides with the point M 1, but is different from the point M 2. In this case x 2 = x 1 and d =

M 1 M 2 \u003d KM 2 \u003d ôy 2 - y 1 ô \u003d = .

4) Point M 2 coincides with point M 1. Then x 1 \u003d x 2, y 1 \u003d y 2 and

d \u003d M 1 M 2 \u003d O \u003d.

The division of the segment in this respect.

Let an arbitrary segment M 1 M 2 be given on the plane and let M be any point of this

segment other than the point M 2 (Fig. 1.6). The number l defined by the equality l = , is called attitude, in which the point M divides the segment M 1 M 2.

Theorem 1.2. If the point M (x; y) divides the segment M 1 M 2 in relation to l, then the coordinates of this are determined by the formulas

x = , y = , (4)

where (x 1; y 1) are the coordinates of the point M 1, (x 2; y 2) are the coordinates of the point M 2.

Proof. Let us prove the first of formulas (4). The second formula is proved similarly. Two cases are possible.

x = x 1 = = = .

2) The straight line M 1 M 2 is not perpendicular to the Ox axis (Fig. 1.6). Let's drop the perpendiculars from the points M 1 , M, M 2 to the axis Ox and denote the points of their intersection with the axis Ox respectively P 1 , P, P 2 . According to the proportional segments theorem =l.

Because P 1 P \u003d ôx - x 1 ô, PP 2 \u003d ôx 2 - xô and the numbers (x - x 1) and (x 2 - x) have the same sign (for x 1< х 2 они положительны, а при х 1 >x 2 are negative), then

l == ,

x - x 1 \u003d l (x 2 - x), x + lx \u003d x 1 + lx 2,

x = .

Corollary 1.2.1. If M 1 (x 1; y 1) and M 2 (x 2; y 2) are two arbitrary points and the point M (x; y) is the midpoint of the segment M 1 M 2, then

x = , y = (5)

Proof. Since M 1 M = M 2 M, then l = 1 and by formulas (4) we obtain formulas (5).

Area of ​​a triangle.

Theorem 1.3. For any points A (x 1; y 1), B (x 2; y 2) and C (x 3; y 3) that do not lie on the same

straight line, the area S of triangle ABC is expressed by the formula

S \u003d ô (x 2 - x 1) (y 3 - y 1) - (x 3 - x 1) (y 2 - y 1) ô (6)

Proof. The area ∆ ABC shown in fig. 1.7, we calculate as follows

S ABC \u003d S ADEC + S BCEF - S ABFD.

Calculate the area of ​​the trapezoid:

S-ADEC=
,

SBCEF=

S ABFD =

Now we have

S ABC \u003d ((x 3 - x 1) (y 3 + y 1) + (x 3 - x 2) (y 3 + y 2) - (x 2 - -x 1) (y 1 + y 2)) \u003d (x 3 y 3 - x 1 y 3 + x 3 y 1 - x 1 y 1 + + x 2 y 3 - -x 3 y 3 + x 2 y 2 - x 3 y 2 - x 2 y 1 + x 1 y 1 - x 2 y 2 + x 1 y 2) \u003d (x 3 y 1 - x 3 y 2 + x 1 y 2 - x 2 y 1 + x 2 y 3 -

X 1 y 3) \u003d (x 3 (y 1 - y 2) + x 1 y 2 - x 1 y 1 + x 1 y 1 - x 2 y 1 + y 3 (x 2 - x 1)) \u003d (x 1 (y 2 - y 1) - x 3 (y 2 - y 1) + + y 1 (x 1 - x 2) - y 3 (x 1 - x 2)) \u003d ((x 1 - x 3) ( y 2 - y 1) + (x 1 - x 2) (y 1 - y 3)) \u003d ((x 2 - x 1) (y 3 - y 1) -

- (x 3 - x 1) (y 2 - y 1)).

For another location ∆ ABC, formula (6) is proved similarly, but it can be obtained with the “-” sign. Therefore, in the formula (6) put the sign of the modulus.

Lecture 2

The equation of a straight line on a plane: the equation of a straight line with the main coefficient, the general equation of a straight line, the equation of a straight line in segments, the equation of a straight line passing through two points. Angle between lines, conditions of parallelism and perpendicularity of lines on a plane.

2.1. Let a rectangular coordinate system and some line L be given on the plane.

Definition 2.1. An equation of the form F(x;y) = 0 relating the variables x and y is called line equation L(in a given coordinate system) if this equation is satisfied by the coordinates of any point lying on the line L, and not by the coordinates of any point not lying on this line.

Examples of equations of lines on a plane.

1) Consider a straight line parallel to the axis Oy of a rectangular coordinate system (Fig. 2.1). Let us denote by the letter A the point of intersection of this line with the axis Ox, (a; o) ─ its or-

dinats. The equation x = a is the equation of the given line. Indeed, this equation is satisfied by the coordinates of any point M(a;y) of this line and the coordinates of any point not lying on the line are satisfied. If a = 0, then the line coincides with the Oy axis, which has the equation x = 0.

2) The equation x - y \u003d 0 defines the set of points in the plane that make up the bisectors of I and III coordinate angles.

3) The equation x 2 - y 2 \u003d 0 is the equation of two bisectors of coordinate angles.

4) The equation x 2 + y 2 = 0 defines a single point O(0;0) on the plane.

5) The equation x 2 + y 2 \u003d 25 is the equation of a circle of radius 5 centered at the origin.

Each point A of the plane is characterized by its coordinates (x, y). They coincide with the coordinates of the vector 0А , coming out of the point 0 - the origin.

Let A and B be arbitrary points of the plane with coordinates (x 1 y 1) and (x 2, y 2), respectively.

Then the vector AB obviously has the coordinates (x 2 - x 1, y 2 - y 1). It is known that the square of the length of a vector is equal to the sum of the squares of its coordinates. Therefore, the distance d between points A and B, or, what is the same, the length of the vector AB, is determined from the condition

d 2 \u003d (x 2 - x 1) 2 + (y 2 - y 1) 2.

$$ d = \sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2) $$

The resulting formula allows you to find the distance between any two points of the plane, if only the coordinates of these points are known

Each time, speaking about the coordinates of one or another point of the plane, we have in mind a well-defined coordinate system x0y. In general, the coordinate system on the plane can be chosen in different ways. So, instead of the x0y coordinate system, we can consider the xִy’ coordinate system, which is obtained by rotating the old coordinate axes around the starting point 0 counter-clockwise arrows on the corner α .

If some point of the plane in the x0y coordinate system had coordinates (x, y), then in the new x-y’ coordinate system it will have other coordinates (x’, y’).

As an example, consider a point M located on the 0x' axis and spaced from the point 0 at a distance equal to 1.

Obviously, in the x0y coordinate system, this point has coordinates (cos α , sin α ), and in the coordinate system хִу’ the coordinates are (1,0).

The coordinates of any two points of the plane A and B depend on how the coordinate system is set in this plane. And here the distance between these points does not depend on how the coordinate system is specified .

Other materials

Mathematics

§2. Point coordinates on the plane

3. Distance between two points.

We now know how to talk about points in the language of numbers. For example, we no longer need to explain: take a point that is three units to the right of the axis and five units below the axis. Suffice it to say simply: take a point.

We have already said that this creates certain advantages. So, we can transmit a drawing made up of dots by telegraph, communicate it to a computer, which does not understand drawings at all, but understands numbers well.

In the previous paragraph, we defined some sets of points on the plane using relations between numbers. Now let's try to consistently translate other geometric concepts and facts into the language of numbers.

We'll start with a simple and common task.

Find the distance between two points on the plane.

Decision:
As always, we assume that the points are given by their coordinates, and then our task is to find a rule by which we can calculate the distance between points, knowing their coordinates. When deriving this rule, of course, it is allowed to resort to the drawing, but the rule itself should not contain any references to the drawing, but should only show what actions and in what order should be performed on the given numbers - the coordinates of the points in order to get the desired number - the distance between dots.

Perhaps, some of the readers will find this approach to solving the problem strange and far-fetched. What is simpler, they will say, the points are given, even if they are coordinates. Draw these points, take a ruler and measure the distance between them.

This method is sometimes not so bad. However, imagine again that you are dealing with a computer. She does not have a ruler, and she does not draw, but she can count so quickly that this is not a problem for her at all. Note that our task is set up so that the rule for calculating the distance between two points consists of commands that the machine can execute.

It is better to solve the problem at first for the particular case when one of the given points lies at the origin. Start with a few numerical examples: find the distance from the origin of the points ; and .

Instruction. Use the Pythagorean theorem.

Now write a general formula for calculating the distance of a point from the origin.

The distance of a point from the origin is determined by the formula:

Obviously, the rule expressed by this formula satisfies the above conditions. In particular, it can be used in computing on machines that can multiply numbers, add them, and take square roots.

Now let's solve the general problem

Given two points on a plane and find the distance between them.

Decision:
Denote by , , , the projections of the points and on the coordinate axes.

The point of intersection of the lines and will be denoted by the letter . From a right triangle, according to the Pythagorean theorem, we get:

But the length of the segment is equal to the length of the segment. Points and , lie on the axis and have coordinates and , respectively. According to the formula obtained in paragraph 3 of paragraph 2, the distance between them is .

Arguing similarly, we get that the length of the segment is equal to . Substituting the found values ​​and into the formula we get.

Solving problems in mathematics for students is often accompanied by many difficulties. To help the student cope with these difficulties, as well as to teach him how to apply his theoretical knowledge in solving specific problems in all sections of the course of the subject "Mathematics" is the main purpose of our site.

Starting to solve problems on the topic, students should be able to build a point on a plane according to its coordinates, as well as find the coordinates of a given point.

The calculation of the distance between two points taken on the plane A (x A; y A) and B (x B; y B) is performed by the formula d \u003d √ ((x A - x B) 2 + (y A - y B) 2), where d is the length of the segment that connects these points on the plane.

If one of the ends of the segment coincides with the origin, and the other has coordinates M (x M; y M), then the formula for calculating d will take the form OM = √ (x M 2 + y M 2).

1. Calculating the distance between two points given the coordinates of these points

Example 1.

Find the length of the segment that connects the points A(2; -5) and B(-4; 3) on the coordinate plane (Fig. 1).

Decision.

The condition of the problem is given: x A = 2; x B \u003d -4; y A = -5 and y B = 3. Find d.

Applying the formula d \u003d √ ((x A - x B) 2 + (y A - y B) 2), we get:

d \u003d AB \u003d √ ((2 - (-4)) 2 + (-5 - 3) 2) \u003d 10.

2. Calculating the coordinates of a point that is equidistant from three given points

Example 2

Find the coordinates of the point O 1, which is equidistant from the three points A(7; -1) and B(-2; 2) and C(-1; -5).

Decision.

From the formulation of the condition of the problem it follows that O 1 A \u003d O 1 B \u003d O 1 C. Let the desired point O 1 have coordinates (a; b). According to the formula d \u003d √ ((x A - x B) 2 + (y A - y B) 2) we find:

O 1 A \u003d √ ((a - 7) 2 + (b + 1) 2);

O 1 V \u003d √ ((a + 2) 2 + (b - 2) 2);

O 1 C \u003d √ ((a + 1) 2 + (b + 5) 2).

We compose a system of two equations:

(√((a - 7) 2 + (b + 1) 2) = √((a + 2) 2 + (b - 2) 2),
(√((a - 7) 2 + (b + 1) 2) = √((a + 1) 2 + (b + 5) 2).

After squaring the left and right sides of the equations, we write:

((a - 7) 2 + (b + 1) 2 \u003d (a + 2) 2 + (b - 2) 2,
((a - 7) 2 + (b + 1) 2 = (a + 1) 2 + (b + 5) 2 .

Simplifying, we write

(-3a + b + 7 = 0,
(-2a - b + 3 = 0.

Having solved the system, we get: a = 2; b = -1.

Point O 1 (2; -1) is equidistant from the three points given in the condition that do not lie on one straight line. This point is the center of a circle passing through three given points. (Fig. 2).

3. Calculation of the abscissa (ordinate) of a point that lies on the abscissa (ordinate) axis and is at a given distance from this point

Example 3

The distance from point B(-5; 6) to point A lying on the x-axis is 10. Find point A.

Decision.

It follows from the formulation of the condition of the problem that the ordinate of point A is zero and AB = 10.

Denoting the abscissa of the point A through a, we write A(a; 0).

AB \u003d √ ((a + 5) 2 + (0 - 6) 2) \u003d √ ((a + 5) 2 + 36).

We get the equation √((a + 5) 2 + 36) = 10. Simplifying it, we have

a 2 + 10a - 39 = 0.

The roots of this equation a 1 = -13; and 2 = 3.

We get two points A 1 (-13; 0) and A 2 (3; 0).

Examination:

A 1 B \u003d √ ((-13 + 5) 2 + (0 - 6) 2) \u003d 10.

A 2 B \u003d √ ((3 + 5) 2 + (0 - 6) 2) \u003d 10.

Both obtained points fit the condition of the problem (Fig. 3).

4. Calculation of the abscissa (ordinate) of a point that lies on the abscissa (ordinate) axis and is at the same distance from two given points

Example 4

Find a point on the Oy axis that is at the same distance from points A (6; 12) and B (-8; 10).

Decision.

Let the coordinates of the point required by the condition of the problem, lying on the Oy axis, be O 1 (0; b) (at the point lying on the Oy axis, the abscissa is equal to zero). It follows from the condition that O 1 A \u003d O 1 B.

According to the formula d \u003d √ ((x A - x B) 2 + (y A - y B) 2) we find:

O 1 A \u003d √ ((0 - 6) 2 + (b - 12) 2) \u003d √ (36 + (b - 12) 2);

O 1 V \u003d √ ((a + 8) 2 + (b - 10) 2) \u003d √ (64 + (b - 10) 2).

We have the equation √(36 + (b - 12) 2) = √(64 + (b - 10) 2) or 36 + (b - 12) 2 = 64 + (b - 10) 2 .

After simplification, we get: b - 4 = 0, b = 4.

Required by the condition of the problem point O 1 (0; 4) (Fig. 4).

5. Calculating the coordinates of a point that is at the same distance from the coordinate axes and some given point

Example 5

Find point M located on the coordinate plane at the same distance from the coordinate axes and from point A (-2; 1).

Decision.

The required point M, like point A (-2; 1), is located in the second coordinate corner, since it is equidistant from points A, P 1 and P 2 (Fig. 5). The distances of the point M from the coordinate axes are the same, therefore, its coordinates will be (-a; a), where a > 0.

It follows from the conditions of the problem that MA = MP 1 = MP 2, MP 1 = a; MP 2 = |-a|,

those. |-a| = a.

According to the formula d \u003d √ ((x A - x B) 2 + (y A - y B) 2) we find:

MA \u003d √ ((-a + 2) 2 + (a - 1) 2).

Let's make an equation:

√ ((-a + 2) 2 + (a - 1) 2) = a.

After squaring and simplifying, we have: a 2 - 6a + 5 = 0. We solve the equation, we find a 1 = 1; and 2 = 5.

We get two points M 1 (-1; 1) and M 2 (-5; 5), satisfying the condition of the problem.

6. Calculation of the coordinates of a point that is at the same specified distance from the abscissa (ordinate) axis and from this point

Example 6

Find a point M such that its distance from the y-axis and from the point A (8; 6) will be equal to 5.

Decision.

It follows from the condition of the problem that MA = 5 and the abscissa of the point M is equal to 5. Let the ordinate of the point M be equal to b, then M(5; b) (Fig. 6).

According to the formula d \u003d √ ((x A - x B) 2 + (y A - y B) 2) we have:

MA \u003d √ ((5 - 8) 2 + (b - 6) 2).

Let's make an equation:

√((5 - 8) 2 + (b - 6) 2) = 5. Simplifying it, we get: b 2 - 12b + 20 = 0. The roots of this equation are b 1 = 2; b 2 \u003d 10. Therefore, there are two points that satisfy the condition of the problem: M 1 (5; 2) and M 2 (5; 10).

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In this article, we will consider ways to determine the distance from a point to a point theoretically and on the example of specific tasks. Let's start with some definitions.

Definition 1

Distance between points- this is the length of the segment connecting them, in the existing 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 the coordinate line, in the coordinate plane or three-dimensional space.

Initial data: the coordinate line O x and an arbitrary point A lying on it. One real number is inherent in any point of the line: let this be a certain number for point A xA, it is 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 on a given scale.

If point A corresponds to an integer real number, having set aside successively from point O to a point along a straight line O A segments - units of length, we can determine the length of segment O A by the total number of pending single segments.

For example, point A corresponds to the number 3 - in order to get to it from point O, it will be necessary to set aside three unit segments. If point A has a coordinate of - 4, single segments are plotted in a similar way, but in a different, negative direction. Thus, in the first case, the distance O A is 3; in the second case, O A \u003d 4.

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

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 a distance); if the coordinate is less than zero, then O A = - x A . In general, these statements are true for any real number x A .

Summarizing: the distance from the origin to the point, which corresponds to a real number on the coordinate line, is equal to:

• 0 if the point is the same as the origin;

• x A if x A > 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 the distance from the point O to the point A with the coordinate x A: O A = x A

The correct statement would be: the distance from one point to another will be equal to the modulus of the difference in coordinates. Those. for points A and B lying on the same coordinate line at any location and having, respectively, the coordinates x A and x B: A B = 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 the points and coincide, and | A B | = | A y B y | . Since the distance between the points is equal to the modulus of the difference between their coordinates, then A y B y = y B - y A , and, therefore, A B = A y B y = y B - y A .

If points A and B lie on a straight line perpendicular to the O y axis (y-axis) - by analogy with the previous paragraph: A B = A x B x = 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 by deriving the calculation formula:

We see that the triangle A B C is right-angled by construction. In this case, A C = A x B x and B C = A y B y . Using the Pythagorean theorem, we compose the equality: A B 2 = A C 2 + B C 2 ⇔ A B 2 = A x B x 2 + A y B y 2 , and then transform it: A B = A x B x 2 + A y B y 2 = x B - x A 2 + y B - y A 2 = (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 the calculation by the formula using the coordinates of these points

A B = (x B - x A) 2 + (y B - y A) 2

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

For the situation when points A and B lie on a straight line perpendicular to the x-axis:

A B = (x B - x A) 2 + (y B - y A) 2 = 0 2 + (y B - y A) 2 = y B - y A

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

A B = (x B - x A) 2 + (y B - y A) 2 = (x B - x A) 2 + 0 2 = 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. Draw through points A and B planes perpendicular to the coordinate axes, and get 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 box. According to the construction of the measurement of this box: A x B x , A y B y and A z B z

From the course of geometry it is known that the square of the diagonal of a parallelepiped is equal to the sum of the squares of its dimensions. 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 = x B - x A , A y B y = y B - y A , A z B z = z B - z A

Let's transform the expression:

A B 2 = A x B x 2 + A y B y 2 + A z B z 2 = x B - x A 2 + y B - y A 2 + z B - z A 2 = = (x B - x A) 2 + (y B - y A) 2 + z B - z A 2

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

A B = x B - x A 2 + y B - y A 2 + (z B - z A) 2

The resulting formula is also valid for cases where:

The dots match;

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

Examples of solving problems for finding the distance between points

Example 1

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

Decision

1. The distance from the reference point to the point is equal to the module 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 = 11 + 2 - (1 - 2) = 10 + 2 2

Answer: O A = 2 - 1, A B = 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 values ​​of this number for which the distance A B will be equal to 5.

Decision

To find the distance between points A and B, you must use the formula A B = (x B - x A) 2 + y B - y A 2

Substituting the real values ​​of the coordinates, we get: A B = (λ + 1 - 1) 2 + (3 - (- 1)) 2 = λ 2 + 16

And also we use the existing condition that A B = 5 and then the equality will be true:

λ 2 + 16 = 5 λ 2 + 16 = 25 λ = ± 3

Answer: A B \u003d 5 if λ \u003d ± 3.

Example 3

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

Decision

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

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

Answer: | A B | = 9

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