The distance between two points by geographic coordinates. Determining the distance between two points using longlat coordinates only

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 the perpendiculars M 1 B and M 2 A from points M 1 and M 2, respectively

on the Oy and Ox axis 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, 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 = ôу 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 = M 1 M 2 = M 1 K = ôx 2 – x 1 ô= =

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

M 1 M 2 = KM 2 = ôу 2 - y 1 ô= = .

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

d = M 1 M 2 = O = .

Division of a segment in this respect.

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

segment different from point M 2 (Fig. 1.6). The number l, defined by the equality l = , called attitude, at which point M divides the segment M 1 M 2.

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

x = , y = , (4)

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

Proof. Let us prove the first of formulas (4). The second formula is proven in a similar way. There are two possible cases.

x = x 1 = = = .

2) Straight line M 1 M 2 is not perpendicular to the Ox axis (Fig. 1.6). Let us lower the perpendiculars from points M 1, M, M 2 to the Ox axis and designate the points of their intersection with the Ox axis as P 1, P, P 2, respectively. By the theorem of proportional segments = l.

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

x – x 1 = l(x 2 – x), x + lx = 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 point M(x;y) is the middle of the segment M 1 M 2, then

x = , y = (5)

Proof. Since M 1 M = M 2 M, then l = 1 and using 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 = ô(x 2 – x 1)(y 3 – y 1) – (x 3 – x 1)(y 2 – y 1)ô (6)

Proof. Area ∆ ABC shown in Fig. 1.7, we calculate as follows

S ABC = S ADEC + S BCEF – S ABFD .

We calculate the area of ​​trapezoids:

S ADEC =
,

S BCEF =

Now we have

S ABC = ((x 3 – x 1)(y 3 + y 1) + (x 3 – x 2)(y 3 + y 2) - (x 2 – -x 1)(y 1 + y 2)) = (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) = (x 3 y 1 – x 3 y 2 + x 1 y 2 – x 2 y 1 + x 2 y 3 –

X 1 y 3) = (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)) = (x 1 (y 2 – y 1) – x 3 (y 2 – y 1) + +y 1 (x 1 – x 2) – y 3 (x 1 – x 2)) = ((x 1 – x 3)( y 2 – y 1) + (x 1 – x 2)(y 1 – y 3)) = ((x 2 – x 1)(y 3 – y 1) –

- (x 3 – x 1)(y 2 – y 1)).

For another location ∆ ABC, formula (6) is proved in a similar way, but it may turn out with a “-” sign. Therefore, in formula (6) they put the modulus sign.

Lecture 2.

Equation of a straight line on a plane: equation of a straight line with a principal coefficient, general equation of a straight line, equation of a straight line in segments, equation of a straight line passing through two points. The angle between straight lines, the conditions of parallelism and perpendicularity of straight 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, connecting 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 Oy axis of the rectangular coordinate system (Fig. 2.1). Let us denote by the letter A the point of intersection of this line with the Ox axis, (a;o) ─ its or-

Dinata. 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 is not satisfied by the coordinates of any point not lying on the line. If a = 0, then the straight line coincides with the Oy axis, which has the equation x = 0.

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

3) The equation x 2 - y 2 = 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) Equation x 2 + y 2 = 25 ─ equation of a circle of radius 5 with center at the origin.

Mathematics

§2. Coordinates of a point on the plane

3. Distance between two points.

You and I can now 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 the 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 relationships between numbers. Now let's try to consistently translate other geometric concepts and facts into the language of numbers.

We will start with a simple and common task.

Find the distance between two points on the plane.

Solution:
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 a drawing, but the rule itself should not contain any references to the drawing, but should only show what actions and in what order must be performed on the given numbers - the coordinates of the points - in order to obtain the desired number - the distance between dots.

Perhaps some readers will find this approach to solving the problem strange and far-fetched. What is simpler, they will say, the points are given, even by 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 doesn’t have a ruler, and she doesn’t draw, but she can count so quickly that it’s not a problem for her at all. Note that our problem is formulated so that the rule for calculating the distance between two points consists of commands that can be executed by a machine.

It is better to first solve the problem posed for the special case when one of these points lies at the origin of coordinates. Start with a few numerical examples: find the distance from the origin of the points; And .

Note: Use the Pythagorean theorem.

Now write a general formula to calculate 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 conditions stated above. In particular, it can be used in calculations on machines that can multiply numbers, add them, and extract square roots.

Now let's solve the general problem

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

Solution:
Let us denote by , , , the projections of points and on the coordinate axes.

Let us denote the point of intersection of the lines with the letter . From a right triangle using the Pythagorean theorem we obtain:

But the length of the segment is equal to the length of the segment. The 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 equal to .

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

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

Definition 1

Distance between points is the length of the segment connecting them, on the existing scale. It is necessary to set a 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 on 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 length of a certain segment is assessed in comparison with a segment taken as a unit of length on a given scale.

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

For example, point A corresponds to the number 3 - to get to it from point O, you will need to lay off three unit segments. If point A has coordinate - 4, unit segments are laid out in a similar way, but in a different, negative direction. Thus, in the first case, the distance O A is equal to 3; in the second case O A = 4.

If point A has a rational number as a coordinate, then from the origin (point O) we plot 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 plot the fraction 4 111 on the coordinate line.

Using the above method, it is completely impossible to plot 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 = x A (the number is taken as the 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.

To summarize: the distance from the origin to the point that corresponds to a real number on the coordinate line is equal to:

• 0 if the point coincides with 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 point O to point A with the coordinate x A: O A = x A

The following 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 for any location and having corresponding 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 us draw perpendiculars through points A and B to the coordinate axes O x and O y and obtain as a result the projection points: A x, A y, B x, B y. Based on the location of points A and B, the following options are then 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 coincide, and | A B | = | A 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 = 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 (ordinate 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 will find the distance between them by deriving the calculation formula:

We see that triangle A B C is rectangular in construction. In this case, A C = A x B x and B C = A y B y. Using the Pythagorean theorem, we create 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 draw 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 = (x B - x A) 2 + (y B - y A) 2

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

For a situation where 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 ordinate 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: a 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.

Let's consider the general case when points A and B do not lie in a plane parallel to one of the coordinate planes. Let us draw planes perpendicular to the coordinate axes through points A and B and 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 measurements of this parallelepiped: A x B x , A y B y and A z B z

From the geometry course we know 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 = 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 when:

The points coincide;

They lie on one 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: 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 origin point O to point A and between points A and B.

Solution

1. The distance from the reference point to the point is equal to the modulus of the coordinate of this point, respectively O A = 1 - 2 = 2 - 1
2. We define the distance between points A and B 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: a rectangular coordinate system and two points lying on it A (1, - 1) and B (λ + 1, 3) are given. λ is some real number. It is necessary to find all values ​​of this number at 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 coordinate values, we get: A B = (λ + 1 - 1) 2 + (3 - (- 1)) 2 = λ 2 + 16

We also 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 = 5 if λ = ± 3.

Example 3

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

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 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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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 0A coming out from point 0 - the origin of coordinates.

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 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 = (x 2 - x 1) 2 + (y 2 - y 1) 2.

d = \/ (x 2 - x 1) 2 + (y 2 - y 1) 2

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

Every time we talk about the coordinates of a particular point on the plane, we mean a well-defined coordinate system x0y. In general, the coordinate system on a plane can be chosen in different ways. So, instead of the x0y coordinate system, you 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 a certain point of the plane in the coordinate system x0y had coordinates (x, y), then in the new coordinate system x"0y" it will have different coordinates (x, y").

As an example, consider point M, located on the 0x-axis and separated from point 0 at a distance of 1.

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

The coordinates of any two points on the plane A and B depend on how the coordinate system is specified in this plane. But the distance between these points does not depend on the method of specifying the coordinate system. We will make significant use of this important circumstance in the next paragraph.

Exercises

I. Find the 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 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 is often accompanied by many difficulties for students. Helping the student cope with these difficulties, as well as teach them to apply their existing theoretical knowledge when solving specific problems in all sections of the course in the subject “Mathematics” is the main purpose of our site.

When starting to solve problems on the topic, students should be able to construct a point on a plane using its coordinates, as well as find the coordinates of a given point.

Calculation of the distance between two points A(x A; y A) and B(x B; y B) taken on a plane is performed using the formula d = √((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 of coordinates, 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. Calculation of the distance between two points based on the given coordinates of these points

Example 1.

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

Solution.

The problem statement states: x A = 2; x B = -4; y A = -5 and y B = 3. Find d.

Applying the formula d = √((x A – x B) 2 + (y A – y B) 2), we get:

d = AB = √((2 – (-4)) 2 + (-5 – 3) 2) = 10.

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

Example 2.

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

Solution.

From the formulation of the problem conditions it follows that O 1 A = O 1 B = O 1 C. Let the desired point O 1 have coordinates (a; b). Using the formula d = √((x A – x B) 2 + (y A – y B) 2) we find:

O 1 A = √((a – 7) 2 + (b + 1) 2);

O 1 B = √((a + 2) 2 + (b – 2) 2);

O 1 C = √((a + 1) 2 + (b + 5) 2).

Let's create 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 = (a + 2) 2 + (b – 2) 2,
((a – 7) 2 + (b + 1) 2 = (a + 1) 2 + (b + 5) 2.

Simplifying, let's 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 specified in the condition that do not lie on the same 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 a given point

Example 3.

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

Solution.

From the formulation of the problem conditions it follows that the ordinate of point A is equal to zero and AB = 10.

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

AB = √((a + 5) 2 + (0 – 6) 2) = √((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 are a 1 = -13; and 2 = 3.

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

Examination:

A 1 B = √((-13 + 5) 2 + (0 – 6) 2) = 10.

A 2 B = √((3 + 5) 2 + (0 – 6) 2) = 10.

Both obtained points are suitable according to the conditions 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 conditions of the problem, lying on the Oy axis, be O 1 (0; b) (at the point lying on the Oy axis, the abscissa is zero). It follows from the condition that O 1 A = O 1 B.

Using the formula d = √((x A – x B) 2 + (y A – y B) 2) we find:

O 1 A = √((0 – 6) 2 + (b – 12) 2) = √(36 + (b – 12) 2);

O 1 B = √((a + 8) 2 + (b – 10) 2) = √(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.

Point O 1 (0; 4) required by the conditions of the problem (Fig. 4).

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

From the conditions of the problem it follows that MA = MR 1 = MR 2, MR 1 = a; MP 2 = |-a|,

those. |-a| = a.

Using the formula d = √((x A – x B) 2 + (y A – y B) 2) we find:

MA = √((-а + 2) 2 + (а – 1) 2).

Let's make an equation:

√((-а + 2) 2 + (а – 1) 2) = а.

After squaring and simplification we have: a 2 – 6a + 5 = 0. Solve the equation, find a 1 = 1; and 2 = 5.

We obtain two points M 1 (-1; 1) and M 2 (-5; 5) that satisfy the conditions of the problem.

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

Example 6.

Find a point M such that its distance from the ordinate axis and from point A(8; 6) is equal to 5.

Solution.

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

According to the formula d = √((x A – x B) 2 + (y A – y B) 2) we have:

MA = √((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 = 10. Consequently, there are two points that satisfy the conditions of the problem: M 1 (5; 2) and M 2 (5; 10).

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