Each point and the plane is characterized by its coordinates (x, y). They coincide with the coordinates of the vector 0a, leaving the point 0 - the beginning of the coordinates.

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

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

$$ d \u003d \\ sqrt ((x_2 - x_1) ^ 2 + (y_2 - y_1) ^ 2) $$

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 a particular point of the plane, we mean a well-defined X0U coordinate system. In general, the coordinate system on the plane can be chosen differently. So, instead of the X0U coordinate system, you can consider the coordinate system x ִ y, which is obtained by turning the old axes of the coordinates around the starting point 0 counter-clockwise Arrows at corner α .

If some plane point in the X0U coordinate system has coordinates (x, y), then in the new coordinate system x ִ, it will have other coordinates (x ', y).

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

Obviously, in the coordinate system X0U, this point has coordinates (COS α sin. α ), and in the coordinate system x ִ in 'coordinates (1.0).

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

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Distance between two points on a straight line

Consider the coordinate direct on which 2 points are marked: A A. A. and B B. B.. To find the distance between these dots, you need to find the length of the segment A B AB A B.. This is done using the following formula:

A B \u003d | A - B | AB \u003d | A-B |A b \u003d.| A -b |,

where A, B A, B a, B. - The coordinates of these points on a straight line (coordinate direct).

Due to the fact that the module is present in the formula, when solving it is not fundamentally, from which coordinate to deduct (since the absolute value of this difference is taken).

| A - b | \u003d | b - a | | a-b | \u003d | b-a || A -b | \u003d| B -a |

We will analyze an example in order to better understand the solution of such tasks.

On the coordinate direct point marked A A. A.whose coordinate is equal 9 9 9 and point B B. B. With coordinate − 1 -1 − 1 . It is necessary to find the distance between these two points.

Decision

Here a \u003d 9, b \u003d - 1 a \u003d 9, b \u003d -1 a \u003d.9, b \u003d− 1

We use the formula and substitute the value:

A b \u003d | a - b | \u003d | 9 - (- 1) | \u003d | 10 | \u003d 10 AB \u003d | A-B | \u003d | 9 - (- 1) | \u003d | 10 | \u003d 10A b \u003d.| A -b | \u003d∣ 9 − (− 1 ) ∣ = ∣ 1 0 ∣ = 1 0

Answer

Distance between two points on the plane

Consider two points specified on the plane. From each point marked on the plane, you need to lower two perpendicular: on the axis O X OX. O X. And on the axis O y oy. O y.. Then the triangle is considered A B C ABC A B C.. Since it is rectangular ( B c bc. B C. Perpendicular A C AC A C.), then find a segment A B AB A B.It is also a distance between the points, you can use the Pythagores theorem. We have:

A B 2 \u003d A C 2 + B C 2 AB ^ 2 \u003d AC ^ 2 + BC ^ 2A. B. 2 = A. C. 2 + B. C. 2

But based on the fact that the length A C AC A C. equal x b - x a x_b-x_a x. B.​ − x. A.​ , and length B c bc. B C. equal y b - y a y_b-y_a y. B.​ − y. A.​ This formula can be rewritten in the following form:

A b \u003d (x b - x a) 2 + (y b - y a) 2 ab \u003d \\ sqrt ((x_b-x_a) ^ 2 + (y_b-y_a) ^ 2)A b \u003d.(x. B.​ − x. A.​ ) 2 + (y. B.​ − y. A.​ ) 2 ​ ,

where x a, y a x_a, y_a x. A.​ , y. A.​ and x b, y b x_b, y_b x. B.​ , y. B.​ - coordinates of the point A A. A. and B B. B. respectively.

It is necessary to find the distance between the points. C C. C. and F F. F.if the coordinates of the first (8 ; − 1) (8;-1) (8 ; − 1 ) , and second - (4 ; 2) (4;2) (4 ; 2 ) .

Decision

X C \u003d 8 x_c \u003d 8 x. C.​ = 8
y C \u003d - 1 y_c \u003d -1 y. C.​ = − 1
x f \u003d 4 x_f \u003d 4 x. F.​ = 4
Y f \u003d 2 y_f \u003d 2 y. F.​ = 2

Cf \u003d (x f - x c) 2 + (y f - y c) 2 \u003d (4 - 8) 2 + (2 - (- 1)) 2 \u003d 16 + 9 \u003d 25 \u003d 5 cf \u003d \\ sqrt (( x_f-x_c) ^ 2 + (y_f-y_c) ^ 2) \u003d \\ sqrt ((4-8) ^ 2 + (2 - (- 1)) ^ 2) \u003d \\ SQRT (16 + 9) \u003d \\ SQRT ( 25) \u003d 5C F \u003d.(x. F.​ − x. C.​ ) 2 + (y. F.​ − y. C.​ ) 2 ​ = (4 − 8 ) 2 + (2 − (− 1 ) ) 2 ​ = 1 6 + 9 ​ = 2 5 ​ = 5

Answer

Distance between two points in space

Finding the distance between the two points in this case occurs similarly to the previous one, except that the coordinates of the point in space are set by three numbers, respectively, in the formula you also need to add the coordinate of the appliquet axis. The formula will take this kind:

AB \u003d (x b - x a) 2 + (y b - y a) 2 + (z b - z a) 2 ab \u003d \\ sqrt ((x_b-x_a) ^ 2 + (y_b-y_a) ^ 2 + ( z_b-z_a) ^ 2)A b \u003d.(x. B.​ − x. A.​ ) 2 + (y. B.​ − y. A.​ ) 2 + (z. B.​ − z.A.​ ) 2 ​

Find the length of the cut F K FK.F.K. In space, if the coordinates of the points of its ends are as follows: (− 1 ; − 1 ; 8) (-1;-1;8) (− 1 ; − 1 ; 8 ) and (− 3 ; 6 ; 0) (-3;6;0) (− 3 ; 6 ; 0 ) . The answer is rounded to an integer.

Decision

$F.=(-1;-1;8)K\; \backslash u003d\; (-\; 3;\; 6;\; 0)\; k\; \backslash u003d\; (-\; 3;\; 6;\; 0)K.=(-3;6;0)$

$F.K.=\sqrt{({x.}^{K.-{x.}^{F.{)}^{2+({y.}^{K.-{y.}^{F.{)}^{2+({z.}^{K.-{z.}^{F.{)}^{2=\sqrt{(-3-(-1){)}^{2+(6-(-1){)}^{2+(0-8{)}^{2=\sqrt{117\approx 10.8}}}}}}}}}}}}}}}$

By the condition of the task, we need to round the answer to an integer.

Distance from point to point - This is the length of the segment connecting these points on a specified scale. Thus, when it comes to measuring the distance, it is required to know the scale (a unit of length) in which measurements will be carried out. Therefore, the task of finding the distance from the point to the point is usually considered either on the coordinate direct, or in a rectangular decartular coordinate system on the plane or in three-dimensional space. In other words, most often you have to calculate the distance between points by their coordinates.

In this article, we, firstly, we recall how the distance from the point to the point is determined to the coordinate direct. Next, we obtain the formula for calculating the distance between the two points of the plane or space according to the specified coordinates. In conclusion, we consider in detail the solutions of characteristic examples and tasks.

Navigating page.

The distance between two points on the coordinate direct.

Let's first define with the designations. The distance from the point A to the point in we denote as.

From here you can conclude that the distance from the point A with the coordinate to the point in the coordinate is equal to the coordinate difference module, i.e, For any location of points on the coordinate direct.

Distance from point to point on the plane, formula.

We obtain a formula for calculating the distance between the dots and specified in the rectangular Cartesian coordinate system on the plane.

Depending on the location of points A and in possible the following options.

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

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

Similarly, if points A and B lie on the direct, perpendicular axis of the ordinate, then the distance from the point A to the point B is like.

In this case, the ABC triangle is rectangular by construction, and and. By pythagora theorem We can record equality, from where.

By summarizing all the results obtained: the distance from the point to the point on the plane is 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 direct perpendicular to one of the coordinate axes. Indeed, if a and in coincide, then. If points A and B lie on a straight line, perpendicular axis oh, then. If A and B lie on a straight line, perpendicular to the AU axis, then.

Distance between points in space, formula.

We introduce a rectangular oxyz coordinate system in space. We get a formula for finding the distance from the point to the point .

In general, points A and B are not lying in the plane parallel to one of the coordinate planes. We carry out through points A and in the plane perpendicular to the coordinate axes OH, OU and OZ. The intersection points of these planes with the coordinate axes will give us projection points A and B for these axes. Denote by projection .

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

Where do you get the formula for finding the distance between points in space .

This formula is also valid if points A and in

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

Finding distance from point to point, examples and solutions.

So, we obtained formulas for finding the distance between the two points of the coordinate direct, plane and three-dimensional space. It's time to consider solving the characteristic examples.

The number of tasks, when solving which the final step is to find the distance between the two points according to their coordinates, is truly huge. A complete review of such examples is beyond the scope of this article. Here we are limited to examples in which two-point coordinates are known and you need to calculate the distance between them.

The solution of tasks in mathematics in students is often accompanied by many difficulties. Help the student to cope with these difficulties, as well as to teach the applied theoretical knowledge of it in solving specific tasks for all sections of the subject "Mathematics" - the main purpose of our site.

Getting Started to solve problems on the topic, students should be able to build a point on the plane on its coordinates, as well as find the coordinates of the specified point.

Calculation of the distance between the plane taken on the plane two points A (x A; U a) and in (x in; y), is performed by the formula d \u003d √ ((x a - x c) 2 + (at a - y y) 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 the coordinates, and the other has the coordinates of M (x m; y), then the formula for calculating D will take the form om \u003d √ (x m 2 + y 2).

1. Calculate the distance between two points according to these coordinates of these points

Example 1..

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

Decision.

The problem of the problem is given: x a \u003d 2; x B \u003d -4; U \u003d -5 and y \u003d 3. Find D.

By applying the formula d \u003d √ ((x a - x c) 2 + (at a - at c) 2), we get:

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

2. Calculation of the coordinate point that is equidistant from three specified points

Example 2.

Find the coordinates of the point O 1, which is equal to three points A (7; -1) and in (-2; 2) and c (-1; -5).

Decision.

From the wording, the conditions of the problem follows that o 1 A \u003d O 1 B \u003d O 1 C. Let the desired point of 1 has coordinates (A; b). According to the formula d \u003d √ ((x a - x c) 2 + (A - at c) 2) we will find:

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

O 1 B \u003d √ ((A + 2) 2 + (B - 2) 2);

O 1 C \u003d √ ((A + 1) 2 + (B + 5) 2).

Make a system of two equations:

(√ ((A - 7) 2 + (B + 1) 2) \u003d √ ((A + 2) 2 + (B - 2) 2),
(√ ((A - 7) 2 + (B + 1) 2) \u003d √ ((A + 1) 2 + (B + 5) 2).

After the construction of the left and right parts of the equations to write:

((A - 7) 2 + (B + 1) 2 \u003d (A + 2) 2 + (B - 2) 2,
((A - 7) 2 + (B + 1) 2 \u003d (A + 1) 2 + (B + 5) 2.

Steering, write

(-3a + b + 7 \u003d 0,
(-2a - b + 3 \u003d 0.

Solving the system, we get: a \u003d 2; B \u003d -1.

The point O 1 (2; -1) is equal to the three points defined in the condition that do not lie on one straight line. This point is a center of a circle passing through three setpoints. (Fig. 2).

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

Example 3.

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

Decision.

From the wording of the conditions of the problem, it follows that the order of the point A is zero and av \u003d 10.

Designating the abscissa of points and through A, write a (a; 0).

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

We obtain the equation √ ((a + 5) 2 + 36) \u003d 10. Similarizing it, we have

a 2 + 10A - 39 \u003d 0.

Roots of this equation a 1 \u003d -13; a 2 \u003d 3.

We obtain two points a 1 (-13; 0) and a 2 (3; 0).

Check:

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 points obtained are suitable under the task condition. (Fig. 3).

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

Example 4.

Find on the OU axis point, which is at the same distance from points A (6; 12) and in (-8; 10).

Decision.

Let the coordinates needed by the condition of the problem of the point lying on the OU axis will be about 1 (0; b) (at the point lying on the OU axis, the abscissa is zero). From the condition it follows that about 1 A \u003d O 1 V.

According to the formula d \u003d √ ((x a - x) 2 + (A - U C) 2) we find:

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

O 1 B \u003d √ ((A + 8) 2 + (B - 10) 2) \u003d √ (64 + (b - 10) 2).

We have an equation √ (36 + (b - 12) 2) \u003d √ (64 + (b - 10) 2) or 36 + (b - 12) 2 \u003d 64 + (b - 10) 2.

After simplification, we obtain: b - 4 \u003d 0, b \u003d 4.

Necessary under the condition of the problem point O 1 (0; 4) (Fig. 4).

5. Calculation of the coordinates of the point that is at the same distance from the axes of the coordinates and some specified point.

Example 5.

Find the 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 the point A (-2; 1), is located in the second coordinate corner, since it is equidistant from the 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\u003e 0.

From the condition of the problem it follows that ma \u003d mp 1 \u003d mp 2, mp 1 \u003d a; MP 2 \u003d |a |,

those. | -a | \u003d a.

According to the formula d \u003d √ ((x a - x) 2 + (A - U C) 2) we find:

Ma \u003d √ ((A + 2) 2 + (A - 1) 2).

Make an equation:

√ ((- a + 2) 2 + (a - 1) 2) \u003d a.

After the construction of the square and simplification, we have: A 2 - 6A + 5 \u003d 0. I solve the equation, we will find a 1 \u003d 1; A 2 \u003d 5.

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

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

Example 6.

Find a point M such that the distance of it from the axis of the ordinate and from point A (8; 6) will be equal to 5.

Decision.

From the condition of the problem it follows that Ma \u003d 5 and the abscissa point M is equal to 5. Let the ordinate point M is equal to B, then m (5; b) (Fig. 6).

According to the formula d \u003d √ ((x a - x c) 2 + (at a - y y) 2) we have:

Ma \u003d √ ((5 - 8) 2 + (B - 6) 2).

Make an equation:

√ ((5 - 8) 2 + (b - 6) 2) \u003d 5. Steering it, we obtain: b 2 - 12b + 20 \u003d 0. The roots of this equation B 1 \u003d 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).

It is known that many students when dealing with tasks need permanent consultations on receptions and methods of their decision. Often, find the way to solve the problem without the help of the teacher is not a student. Necessary advice to solve the tasks of the student and can get on our website.

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Calculation of distances between points according to their coordinates on the plane is element, on the surface of the Earth - a little more complicated: we will look at the measurement of the distance and the initial azimuth between points without projection transformations. To begin with, we will understand the terminology.

Introduction

Through any two points on the surface of the sphere, if they are not directly opposite to each other (that is, they are not antipodes), you can spend a unique large circle. Two points, divide a large circle into two arcs. The length of the short arc is the shortest distance between two points. There can be an infinite number of large circles between two points, but the distance between them will be the same on any circle and equal half the circle circumference, or π * r, where R is the sphere radius.

On the plane (in the rectangular coordinate system), large circles and their fragments, as mentioned above, are arcs in all projections, except for the gnomonic, where large circles are straight lines. In practice, this means that aircraft and other air transport always uses the minimum distance route between points to save fuel, that is, the flight is carried out at a distance of a large circle, it looks like an arc on the plane.

The shape of the Earth can be described as a sphere, therefore the equations for calculating distances on a large circle are important for calculating the shortest distance between points on the ground surface and are often used in navigation. The calculation of the distance with this method is more efficient and in many cases more accurately than calculating it for designed coordinates (in rectangular coordinate systems), since, firstly, it is not necessary to translate geographical coordinates to a rectangular coordinate system (carry out projection transformations) and, Secondly, many projections, if incorrectly selected, can lead to significant distortions of lengths due to the features of projection distortion. It is known that more accurately describes the form of the Earth, but the ellipsoid, but this article discusses the range calculation precisely on the sphere, for calculations, the sphere is used by a radius of 6372795 meters, which can lead to an error of calculating the distance of about 0.5%.

Formulas

My implementation on PNP

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