Solving math problems for students is often accompanied by many difficulties. Helping a student to cope with these difficulties, as well as teach him to apply his existing theoretical knowledge in solving specific problems in all sections of the course of the subject "Mathematics" is the main purpose of our site.
When starting to solve problems on a topic, students should be able to build a point on a plane by its coordinates, as well as find the coordinates of a given point.
The calculation of the distance between two points A (x A; y A) and B (x B; y B) taken on the plane is performed by the formula d = √ ((x A - x B) 2 + (y A - y B) 2), where d is the length of the line 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 according to 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.
In the problem statement it is given: 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 statement of the problem it follows that О 1 А = О 1 В = О 1 С. Let the required point О 1 have coordinates (a; b). By the formula d = √ ((x A - x B) 2 + (y A - y B) 2) we find:
О 1 А = √ ((a - 7) 2 + (b + 1) 2);
О 1 В = √ ((a + 2) 2 + (b - 2) 2);
О 1 С = √ ((a + 1) 2 + (b + 5) 2).
Let's 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 = (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 О 1 (2; -1) is equidistant from three points specified 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 located at a given distance from this 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 statement of the problem it follows that the ordinate of point A is equal to zero and AB = 10.
Denoting the abscissa of point A through a, we write A (a; 0).
AB = √ ((a + 5) 2 + (0 - 6) 2) = √ ((a + 5) 2 + 36).
We obtain 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; a 2 = 3.
We get two points А 1 (-13; 0) and А 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 fit according to the problem statement (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 on the Oy axis a point that is at the same distance from points A (6; 12) and B (-8; 10).
Solution.
Let the coordinates of the point lying on the Oy axis, required by the problem statement, 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 = O 1 B.
By the formula d = √ ((x A - x B) 2 + (y A - y B) 2) we find:
О 1 А = √ ((0 - 6) 2 + (b - 12) 2) = √ (36 + (b - 12) 2);
О 1 В = √ ((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.
The point O 1 (0; 4) required by the problem statement (fig. 4).
5. Calculation of 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 angle, 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.
From the condition of the problem it follows that MA = MP 1 = MP 2, MP 1 = a; MP 2 = | -a |,
those. | -a | = a.
By the formula d = √ ((x A - x B) 2 + (y A - y B) 2) we find:
MA = √ ((- a + 2) 2 + (a - 1) 2).
Let's make the 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; a 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 a given point
Example 6.
Find a point M such that its distance from the ordinate axis and from point A (8; 6) will be equal to 5.
Solution.
It follows from the problem statement that MA = 5 and the abscissa of the point M is 5. Let the ordinate of the point M be equal to b, then M (5; b) (fig. 6).
By 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 the 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 condition of the problem: M 1 (5; 2) and M 2 (5; 10).
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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 omit from points M 1 and M 2 perpendiculars M 1 B and M 2 A, respectively
on the axis Oy and Ox and denote by K the point of intersection of 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 = ôy 2 - y 1 ô. Because ∆М 1 КМ 2 rectangular, then by the Pythagorean theorem d = М 1 М 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 = ôy 2 - at 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 the segment in this respect.
Let an arbitrary segment M 1 M 2 be given on the plane and let M ─ any point of this
a segment different from the point M 2 (Fig. 1.6). The number l, defined by the equality l = 
is called attitude, in which 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 with respect 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 in a similar way. There are two possible cases.
x = x 1 = 
= 
= .
2) Straight line М 1 М 2 is not perpendicular to the Ox axis (Fig. 1.6). Let us drop the perpendiculars from points M 1, M, M 2 to the Ox axis and denote the points of their intersection with the Ox axis, respectively, Р 1, Р, Р 2. By the proportional line segment theorem 
= 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 (for x 1< х 2 они положительны, а при х 1 >x 2 are negative), then
l = = 
,
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 a 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 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), not lying on one
straight line, the area S of the 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 areas of the trapeziums:
S ADEC = 
,
S BCEF = 
S ABFD = 
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 can turn out to be with a “-” sign. Therefore, the modulus sign is put in formula (6).
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. Angle between straight lines, 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 the equation of the line L(in a given coordinate system) if this equation is satisfied by the coordinates of any point lying on the line L, and the coordinates of any point not lying on this line do not.
Examples of equations for lines on a plane.
1) Consider a straight line parallel to the Oy axis of a rectangular coordinate system (Fig. 2.1). Let us denote by the letter A the point of intersection of this straight line with the Ox axis, (a; o) ─ its or-
dinates. 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 straight line and the coordinates of any point not lying on the straight line do not satisfy. If a = 0, then the 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) The equation x 2 + y 2 = 25 ─ the equation of a circle of radius 5 centered at the origin.
Using coordinates, they determine the location of an object on the globe. Coordinates are indicated by latitude and longitude. Latitudes are measured from the equator line on both sides. Latitudes are positive in the Northern Hemisphere and negative in the Southern Hemisphere. Longitude is measured from the initial meridian either to the east or to the west, respectively, either east longitude or west longitude is obtained.According to the generally accepted position, the meridian, which passes through the old Greenwich Observatory in Greenwich, is taken as the initial one. The geographic coordinates of the location can be obtained using a GPS navigator. This device receives signals from the satellite positioning system in the WGS-84 coordinate system, which is the same for the whole world.
Navigator models differ in manufacturers, functionality and interface. Currently, some models of cell phones have built-in GPS-navigators. But any model can record and store the coordinates of the point.
Distance between GPS coordinates
To solve practical and theoretical problems in some industries, it is necessary to be able to determine the distances between points by their coordinates. There are several ways to do this. The canonical form of representation of geographic coordinates: degrees, minutes, seconds.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 distance 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, the result was 3,800,619 m.
The next method is more laborious, but also more visual. You must use any available mapping or navigation software. The programs in which you can create points by coordinates and measure the distances between them include the following applications: BaseCamp (a modern analogue of MapSource), 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 placemarks with 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 above, the Google Earth program returned the result - the length of the distance between point # 1 and point # 2 is 3,817,353 m.
Why there is an error in determining the distance
All distance calculations between coordinates are based on arc length calculations. The radius of the Earth is involved in calculating the arc length. But since the shape of the Earth is close to an oblate ellipsoid, the radius of the Earth at certain points is different. To calculate the distance between the coordinates, the average value of the Earth's radius is taken, which gives an error in the measurement. The greater the measured distance, the greater the error.Point to point distance is the length of the line segment connecting these points, at a given scale. Thus, when it comes to measuring distance, you need to know the scale (unit of length) in which the measurements will be carried out. Therefore, the problem of finding the distance from a point to a point is usually considered either on a coordinate line or in a rectangular Cartesian coordinate system on a plane or in three-dimensional space. In other words, most often it is necessary to calculate the distance between points by their coordinates.
In this article, we, first, recall how the distance from a point to a point on a coordinate line is determined. Next, we will obtain formulas for calculating the distance between two points of a plane or space at given coordinates. In conclusion, let us consider in detail the solutions of typical examples and problems.
Page navigation.
Distance between two points on a coordinate line.
Let's first define the designations. The distance from point A to point B will be denoted as.
From this we can conclude that the distance from point A with coordinate to point B with coordinate is equal to the modulus of the coordinate difference, that is, 
at any location of points on the coordinate line.
Distance from point to point on a plane, formula.
Let's get a formula for calculating the distance between points and, given in a rectangular Cartesian coordinate system on the plane.
Depending on the location of points A and B, the following options are 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 abscissa axis, then the points and coincide, and the distance is equal to the distance. In the previous paragraph, we found out that the distance between two points on the coordinate line is equal to the modulus of the difference in their coordinates, therefore, 
... Hence, .
Similarly, if points A and B lie on a straight line perpendicular to the ordinate, then the distance from point A to point B is found as.
In this case, triangle ABC is rectangular in construction, and 
and . By Pythagorean theorem we can write equality, whence.
Let's summarize all the results obtained: the distance from a point to a point on the plane is found 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 straight line perpendicular to one of the coordinate axes. Indeed, if A and B coincide, then. If points A and B lie on a straight line perpendicular to the Ox axis, then. If A and B lie on a straight line perpendicular to the Oy axis, then.
Distance between points in space, formula.
Let us introduce a rectangular coordinate system Oxyz in space. Let's get the formula for finding the distance from the point 
to the point 
.
In general, points A and B do not lie in a plane parallel to one of the coordinate planes. Let us draw through points A and B planes perpendicular to the coordinate axes Ox, Oy and Oz. The points of intersection of these planes with the coordinate axes will give us the projection of points A and B on these axes. We denote the projections 
.
The desired distance between points A and B is the diagonal of the rectangular parallelepiped shown in the figure. By construction, the dimensions of this parallelepiped are equal 
and . In a high school geometry course, it was proved that the square of the diagonal of a rectangular parallelepiped is equal to the sum of the squares of its three dimensions, therefore,. Based on the information in the first section of this article, we can write the following equalities, therefore,
whence we get formula for finding the distance between points in space .
This formula is also valid if points A and B
- match;
- belong to one of the coordinate axes or a straight line parallel to one of the coordinate axes;
- belong to one of the coordinate planes or a plane parallel to one of the coordinate planes.
Finding the distance from point to point, examples and solutions.
So, we got formulas for finding the distance between two points of the coordinate line, plane and three-dimensional space. It's time to consider solutions to typical examples.
The number of problems in the solution of which the final step is to find the distance between two points by their coordinates is truly enormous. A complete overview of such examples is beyond the scope of this article. Here we will restrict ourselves to examples in which the coordinates of two points are known and it is required to calculate the distance between them.
The distance between two points of the 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 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 has, obviously, 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, which 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 of the plane, if only the coordinates of these points are known
Each time, speaking about the coordinates of a particular point on the plane, we mean a completely definite 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 as a result of 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 "0y" coordinate system it will have different coordinates (x ", y").
As an example, consider the point M located on the 0x "axis and spaced from 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 x coordinate system "0y" 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. But the distance between these points does not depend on how the coordinate system is specified. This important circumstance will be essentially used by us in the next section.
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.