According to the generally accepted position, the prime meridian is taken to be the one that passes through the old Greenwich Observatory in Greenwich. Geographic coordinates of the location can be obtained using a GPS navigator. This device receives satellite positioning system signals in the WGS-84 coordinate system, uniform for the whole world.
Navigator models differ in manufacturer, functionality and interface. Currently, built-in GPS navigators are also available in some cell phone models. But any model can record and save the coordinates of a point.
For example, you can determine the distance between the following coordinates: point No. 1 - latitude 55°45′07″ N, longitude 37°36′56″ E; point No. 2 - latitude 58°00′02″ N, longitude 102°39′42″ E.
The easiest way is to use a calculator to calculate the length between two points. In the browser search engine, you must set the following search parameters: online - to calculate the distance between two coordinates. In the online calculator, latitude and longitude values are entered into the query fields for the first and second coordinates. When calculating, the online calculator gave the result - 3,800,619 m.
The next method is more labor-intensive, but also more visual. You must use any available mapping or navigation program. Programs in which you can create points using coordinates and measure distances between them include the following applications: BaseCamp (a modern analogue of the MapSource program), Google Earth, SAS.Planet.
All of the above programs are available to any network user. For example, to calculate the distance between two coordinates in Google Earth, you need to create two labels indicating the coordinates of the first point and the second point. Then, using the “Ruler” tool, you need to connect the first and second marks with a line, the program will automatically display the measurement result and show the path on the satellite image of the Earth.
In the case of the example given above, the Google Earth program returned the result - the length of the distance between point No. 1 and point No. 2 is 3,817,353 m.
In this article we will look at ways to determine the distance from point to point theoretically and using the example of specific problems. 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:
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 xA: 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 xA 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 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.
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
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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Calculating distances between points based on their coordinates on a plane is elementary; on the Earth’s surface it is a little more complicated: we will consider measuring the distance and initial azimuth between points without projection transformations. First, let's understand the terminology.
Through any two points on the surface of a sphere, if they are not directly opposite to each other (that is, they are not antipodes), a unique great circle can be drawn. Two points divide a large circle into two arcs. The length of a short arc is the shortest distance between two points. An infinite number of large circles can be drawn between two antipodal points, but the distance between them will be the same on any circle and equal to half the circumference of the circle, or π*R, where R is the radius of the sphere.
On a plane (in a rectangular coordinate system), large circles and their fragments, as mentioned above, represent arcs in all projections except the gnomonic one, where large circles are straight lines. In practice, this means that airplanes and other air transport always use the route of the minimum distance between points to save fuel, that is, the flight is carried out along a great circle distance, on a plane it looks like an arc.
The shape of the Earth can be described as a sphere, so great circle distance equations are important for calculating the shortest distance between points on the Earth's surface and are often used in navigation. Calculating distance by this method is more efficient and in many cases more accurate than calculating it for projected coordinates (in rectangular coordinate systems), since, firstly, it does not require converting geographic coordinates to a rectangular coordinate system (carry out projection transformations) and, secondly, many projections, if incorrectly selected, can lead to significant length distortions due to the nature of projection distortions. It is known that it is not a sphere, but an ellipsoid that describes the shape of the Earth more accurately, however, this article discusses the calculation of distances on a sphere; for calculations, a sphere with a radius of 6,372,795 meters is used, which can lead to an error in calculating distances of the order of 0.5%.
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Consider a coordinate line on which 2 points are marked: A A A And B B B. To find the distance between these points, you need to find the length of the segment A B AB A B. This is done using the following formula:
Distance between two points on a lineA B = ∣ a − b ∣ AB=|a-b|A B =∣ a −b∣,
Where a , b a, b a, b- coordinates of these points on a straight line (coordinate line).
Due to the fact that the formula contains a modulus, when solving it, it is not important which coordinate to subtract from which (since the absolute value of this difference is taken).
∣ a − b ∣ = ∣ b − a ∣ |a-b|=|b-a|∣ a −b ∣ =∣ b −a∣
Let's look at an example to better understand the solution to such problems.
Example 1Points are marked on the coordinate line A A A, whose coordinate is equal to 9 9 9 and period B B B with coordinate − 1 -1 − 1 . We need to find the distance between these two points.
Solution
Here a = 9 , b = − 1 a=9, b=-1 a =9 , b =− 1
We use the formula and substitute the values:
A B = ∣ a − b ∣ = ∣ 9 − (− 1) ∣ = ∣ 10 ∣ = 10 AB=|a-b|=|9-(-1)|=|10|=10A B =∣ a −b ∣ =∣ 9 − (− 1 ) ∣ = ∣ 1 0 ∣ = 1 0
Answer
Consider two points given on a plane. From each point marked on the plane, you need to lower two perpendiculars: To the axis O X OX O X and on the axle 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 the segment A B AB A B, which is also the distance between points, can be done using the Pythagorean theorem. We have:
A B 2 = A C 2 + B C 2 AB^2=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 to x B − x A x_B-x_A x B − x A , and the length B C BC B C equal to y B − y A y_B-y_A y B − y A , this formula can be rewritten as follows:
Distance between two points on a planeA B = (x B − x A) 2 + (y B − y A) 2 AB=\sqrt((x_B-x_A)^2+(y_B-y_A)^2)A B =(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 points A A A And B B B respectively.
Example 2It is necessary to find the distance between points C C C And F F F, if the coordinates of the first (8 ; − 1) (8;-1) (8 ; − 1 ) , and the second - (4 ; 2) (4;2) (4 ; 2 ) .
Solution
X C = 8 x_C=8 x C
=
8
y C = − 1 y_C=-1 y C
=
−
1
x F = 4 x_F=4 x F
=
4
y F = 2 y_F=2 y F
=
2
C F = (x F − x C) 2 + (y F − y C) 2 = (4 − 8) 2 + (2 − (− 1)) 2 = 16 + 9 = 25 = 5 CF=\sqrt(( x_F-x_C)^2+(y_F-y_C)^2)=\sqrt((4-8)^2+(2-(-1))^2)=\sqrt(16+9)=\sqrt( 25)=5C F =(x F − x C ) 2 + (y F − y C ) 2 = (4 − 8 ) 2 + (2 − (− 1 ) ) 2 = 1 6 + 9 = 2 5 = 5
Answer
Finding the distance between two points in this case is similar to the previous one, except that the coordinates of the point in space are specified by three numbers; accordingly, the coordinate of the applicate axis must also be added to the formula. The formula will look like this:
Distance between two points in spaceA B = (x B − x A) 2 + (y B − y A) 2 + (z B − z A) 2 AB=\sqrt((x_B-x_A)^2+(y_B-y_A)^2+( z_B-z_A)^2)A B =(x B − x A ) 2 + (y B − y A ) 2 + (z B − zA ) 2
Example 3Find the length of the segment FK FKFK in space if the coordinates of the points at its ends are as follows: (− 1 ; − 1 ; 8) (-1;-1;8) (− 1 ; − 1 ; 8 ) And (− 3 ; 6 ; 0) (-3;6;0) (− 3 ; 6 ; 0 ) . Round your answer to the nearest whole number.
Solution
F = (− 1 ; − 1 ; 8) F=(-1;-1;8)F=
(−
1
;
−
1
;
8
)
K = (− 3 ; 6 ; 0) K=(-3;6;0)K=
(−
3
;
6
;
0
)
F K = (x K − x F) 2 + (y K − y F) 2 + (z K − z F) 2 = (− 3 − (− 1)) 2 + (6 − (− 1)) 2 + (0 − 8) 2 = 117 ≈ 10.8 FK=\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)\approx10.8FK= (xK − xF ) 2 + (yK − yF ) 2 + (zK − zF ) 2 = (− 3 − (− 1 ) ) 2 + (6 − (− 1 ) ) 2 + (0 − 8 ) 2 = 1 1 7 ≈ 1 0 . 8
According to the conditions of the problem, we need to round the answer to a whole number.
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