The distance between two points given by their coordinates. Distance between two points on a plane

The distance between two points on a plane.
Coordinate systems

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 vector length is equal to the sum 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 \u003d \ / (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 x"0y" 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 new system coordinates x"0y" it will have other coordinates (x", y").

As an example, consider the point M, located on the axis 0x" 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 x"0y" 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. But the distance between these points does not depend on how the coordinate system is specified. We will make essential use of this important circumstance in the next section.

Exercises

I. Find distances between points of the plane with coordinates:

1) (3.5) and (3.4); 3) (0.5) and (5, 0); 5) (-3.4) and (9, -17);

2) (2, 1) and (- 5, 1); 4) (0.7) and (3.3); 6) (8, 21) and (1, -3).

II. Find the perimeter of a triangle whose sides are given by the equations:

x + y - 1 = 0, 2x - y - 2 = 0 and y = 1.

III. In the x0y coordinate system, points M and N have coordinates (1, 0) and (0,1), respectively. Find the coordinates of these points in the new coordinate system, which is also obtained by rotating the old axes around the starting point by an angle of 30 ° counterclockwise.

IV. In the x0y coordinate system, points M and N have coordinates (2, 0) and (\ / 3/2, - 1/2) respectively. Find the coordinates of these points in the new coordinate system, which is obtained by rotating the old axes around the starting point by an angle of 30° clockwise.

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).

Solution.

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).

Solution.

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.

Solution.

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).

Solution.

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 V.

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).

Solution.

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.

Solution.

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.

Yandex.RTB R-A-339285-1 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 as a coordinate rational number, 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 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.

Solution

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
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.

Solution

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.

Solution

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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THEORETICAL QUESTIONS

ANALYTICAL GEOMETRY ON THE PLANE

1. Coordinate method: number line, coordinates on the line; rectangular (Cartesian) coordinate system on the plane; polar coordinates.

Let's take a look at a straight line. Let's choose a direction on it (then it will become an axis) and some point 0 (the origin). A straight line with a chosen direction and origin is called coordinate line(in this case, we assume that the scale unit is selected).

Let M is an arbitrary point on the coordinate line. Let's put in accordance with the point M real number x, equal to the value OM segment : x=OM. Number x called the coordinate of the point M.

Thus, each point of the coordinate line corresponds to a certain real number - its coordinate. The converse is also true, each real number x corresponds to some point on the coordinate line, namely such a point M, whose coordinate is x. This correspondence is called mutually unambiguous.

So, real numbers can be represented by points of the coordinate line, i.e. the coordinate line serves as an image of the set of all real numbers. Therefore, the set of all real numbers is called number line, and any number is a point of this line. Near a point on a number line, a number is often indicated - its coordinate.

Rectangular (or Cartesian) coordinate system on a plane.

Two mutually perpendicular axes About x and About y having common beginning O and the same scale unit, form rectangular (or Cartesian) coordinate system on the plane.

Axis OH called the x-axis, the axis OY- the y-axis. Dot O the intersection of the axes is called the origin. The plane in which the axes are located OH and OY, is called the coordinate plane and is denoted Oh xy.

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 4 parts, they are called quarters, square or coordinate angles.

Polar coordinates.

The polar coordinate system consists of some point O called pole, and the beam emanating from it OE called polar axis. In addition, the scale unit for measuring the lengths of segments is set. Let a polar coordinate system be given and let M is an arbitrary point of the plane. Denote by R– point distance M from the point O, and through φ - the angle by which the beam is rotated counterclockwise the polar axis to coincide with the beam OM.

polar coordinates points M call the numbers R and φ . Number R considered as the first coordinate and called polar radius, number φ - the second coordinate is called polar angle.

Dot M with polar coordinates R and φ are designated as follows: М( ;φ). Let's establish a connection between the polar coordinates of a point and its rectangular coordinates.
In this case, we will assume that the origin of the rectangular coordinate system is at the pole, and the positive semi-axis of the abscissa coincides with the polar axis.

Let the point M have rectangular coordinates X and Y and polar coordinates R and φ .

(1)

Proof.

Drop from the dots M 1 and M 2 perpendiculars M 1 V and M 1 A,. because (x 2 ; y 2). By theory, if M 1 (x 1) and M 2 (x 2) are any two points and α is the distance between them, then α = ‌‌‌‍‌‌|x 2 - x 1 | .

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.

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).

Solution.

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).

Solution.

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.

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.

Solution.

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).

Solution.

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).

Solution.

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.

Solution.

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:

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