Calculate the distance between points. Distance between two points on a plane

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 about 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

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 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, area S triangle ABC 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 =

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 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 line, equation of a line in segments, equation of a 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, relating variables x and y are called line equation L(V given system coordinates), 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 the straight line, parallel to the axis Oy 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-

dinats. 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 I and III coordinate corners

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.

Hello,

PHP used:

Sincerely, Alexander.

Hello,

I've been struggling with a problem for quite some time now: I'm trying to calculate the distance between two arbitrary points, which are located at a distance from 30 to 1500 meters from each other.

PHP used:

$cx=31.319738; //x coordinate of the first point
$cy=60.901638; //y coordinate of the first point
$x=31.333312; //x coordinate of the second point
$y=60.933981; //y coordinate of the second point
$mx=abs($cx-$x); //calculate the difference in X (first leg right triangle), function abs(x) - returns the modulus of the number x x
$my=abs($cy-$y); //calculate the difference between the players (the second leg of the right triangle)
$dist=sqrt(pow($mx,2)+pow($my,2)); //Get the distance to the metro (the length of the hypotenuse according to the rule, the hypotenuse is equal to the root of the sum of the squares of the legs)

If it’s not clear, let me explain: I imagine that the distance between two points is the hypotenuse of a right triangle. Then the difference between the X's of each of the two points will be one of the legs, and the other leg will be the difference of the Y's of the same two points. Then, by calculating the differences between the X's and Y's, you can use the formula to calculate the length of the hypotenuse (i.e., the distance between two points).

I know that this rule works well for the Cartesian coordinate system, however, it should more or less work through longlat coordinates, because the measured distance between two points is negligible (from 30 to 1500 meters).

However, the distance according to this algorithm is calculated incorrectly (for example, distance 1 calculated by this algorithm exceeds distance 2 by only 13%, while in reality distance 1 is equal to 1450 meters, and distance 2 is equal to 970 meters, that is, in fact the difference reaches almost 50% ).

If anyone can help, I would be very grateful.

Sincerely, Alexander.

","contentType":"text/html"),"proposedBody":("source":"

Hello,

I’ve been struggling with a problem for quite some time now: I’m trying to calculate the distance between two arbitrary points that are located at a distance of 30 to 1500 meters from each other.

PHP used:

$cx=31.319738; //x coordinate of the first point
$cy=60.901638; //y coordinate of the first point
$x=31.333312; //x coordinate of the second point
$y=60.933981; //y coordinate of the second point
$mx=abs($cx-$x); //calculate the difference in x (the first leg of a right triangle), function abs(x) - returns the modulus of the number x x
$my=abs($cy-$y); //calculate the difference between the players (the second leg of the right triangle)
$dist=sqrt(pow($mx,2)+pow($my,2)); //Get the distance to the metro (the length of the hypotenuse according to the rule, the hypotenuse is equal to the root of the sum of the squares of the legs)

If anyone can help, I would be very grateful.

Sincerely, Alexander.

Hello,

I’ve been struggling with a problem for quite some time now: I’m trying to calculate the distance between two arbitrary points that are located at a distance of 30 to 1500 meters from each other.

PHP used:

$cx=31.319738; //x coordinate of the first point
$cy=60.901638; //y coordinate of the first point
$x=31.333312; //x coordinate of the second point
$y=60.933981; //y coordinate of the second point
$mx=abs($cx-$x); //calculate the difference in x (the first leg of a right triangle), function abs(x) - returns the modulus of the number x x
$my=abs($cy-$y); //calculate the difference between the players (the second leg of the right triangle)
$dist=sqrt(pow($mx,2)+pow($my,2)); //Get the distance to the metro (the length of the hypotenuse according to the rule, the hypotenuse is equal to the root of the sum of the squares of the legs)

If anyone can help, I would be very grateful.

Sincerely, Alexander.

","contentType":"text/html"),"authorId":"108613929","slug":"15001","canEdit":false,"canComment":false,"isBanned":false,"canPublish" :false,"viewType":"old","isDraft":false,"isOnModeration":false,"isSubscriber":false,"commentsCount":14,"modificationDate":"Wed Jun 27 2012 20:07:00 GMT +0000 (UTC)","showPreview":true,"approvedPreview":("source":"

Hello,

I’ve been struggling with a problem for quite some time now: I’m trying to calculate the distance between two arbitrary points that are located at a distance of 30 to 1500 meters from each other.

PHP used:

$cx=31.319738; //x coordinate of the first point
$cy=60.901638; //y coordinate of the first point
$x=31.333312; //x coordinate of the second point
$y=60.933981; //y coordinate of the second point
$mx=abs($cx-$x); //calculate the difference in x (the first leg of a right triangle), function abs(x) - returns the modulus of the number x x
$my=abs($cy-$y); //calculate the difference between the players (the second leg of the right triangle)
$dist=sqrt(pow($mx,2)+pow($my,2)); //Get the distance to the metro (the length of the hypotenuse according to the rule, the hypotenuse is equal to the root of the sum of the squares of the legs)

If anyone can help, I would be very grateful.

Sincerely, Alexander.

","html":"Hello,","contentType":"text/html"),"proposedPreview":("source":"

Hello,

I’ve been struggling with a problem for quite some time now: I’m trying to calculate the distance between two arbitrary points that are located at a distance of 30 to 1500 meters from each other.

PHP used:

$cx=31.319738; //x coordinate of the first point
$cy=60.901638; //y coordinate of the first point
$x=31.333312; //x coordinate of the second point
$y=60.933981; //y coordinate of the second point
$mx=abs($cx-$x); //calculate the difference in x (the first leg of a right triangle), function abs(x) - returns the modulus of the number x x
$my=abs($cy-$y); //calculate the difference between the players (the second leg of the right triangle)
$dist=sqrt(pow($mx,2)+pow($my,2)); //Get the distance to the metro (the length of the hypotenuse according to the rule, the hypotenuse is equal to the root of the sum of the squares of the legs)

If anyone can help, I would be very grateful.

Sincerely, Alexander.

","html":"Hello,","contentType":"text/html"),"titleImage":null,"tags":[("displayName":"distance measurement","slug":"izmerenie- rasstoyaniy","categoryId":"10615601","url":"/blog/mapsapi??tag=izmerenie-rasstoyaniy"),("displayName":"API 1.x","slug":"api-1 -x","categoryId":"150000131","url":"/blog/mapsapi??tag=api-1-x")],"isModerator":false,"commentsEnabled":true,"url": "/blog/mapsapi/15001","urlTemplate":"/blog/mapsapi/%slug%","fullBlogUrl":"https://yandex.ru/blog/mapsapi","addCommentUrl":"/blog/ createComment/mapsapi/15001","updateCommentUrl":"/blog/updateComment/mapsapi/15001","addCommentWithCaptcha":"/blog/createWithCaptcha/mapsapi/15001","changeCaptchaUrl":"/blog/api/captcha/new ","putImageUrl":"/blog/image/put","urlBlog":"/blog/mapsapi","urlEditPost":"/blog/56a98d48b15b79e31e0d54c8/edit","urlSlug":"/blog/post/generateSlug ","urlPublishPost":"/blog/56a98d48b15b79e31e0d54c8/publish","urlUnpublishPost":"/blog/56a98d48b15b79e31e0d54c8/unpublish","urlRemovePost":"/blog/56a98d48b15b79e31e0d5 4c8/removePost","urlDraft":"/blog/mapsapi /15001/draft","urlDraftTemplate":"/blog/mapsapi/%slug%/draft","urlRemoveDraft":"/blog/56a98d48b15b79e31e0d54c8/removeDraft","urlTagSuggest":"/blog/api/suggest/mapsapi" ,"urlAfterDelete":"/blog/mapsapi","isAuthor":false,"subscribeUrl":"/blog/api/subscribe/56a98d48b15b79e31e0d54c8","unsubscribeUrl":"/blog/api/unsubscribe/56a98d48b15b79e31e0d54c8","urlEditPost Page ":"/blog/mapsapi/56a98d48b15b79e31e0d54c8/edit","urlForTranslate":"/blog/post/translate","urlRelateIssue":"/blog/post/updateIssue","urlUpdateTranslate":"/blog/post/updateTranslate ","urlLoadTranslate":"/blog/post/loadTranslate","urlTranslationStatus":"/blog/mapsapi/15001/translationInfo","urlRelatedArticles":"/blog/api/relatedArticles/mapsapi/15001","author" :("id":"108613929","uid":("value":"108613929","lite":false,"hosted":false),"aliases":(),"login":"mrdds" ,"display_name":("name":"mrdds","avatar":("default":"0/0-0","empty":true)),,"address":" [email protected]","defaultAvatar":"0/0-0","imageSrc":"https://avatars.mds.yandex.net/get-yapic/0/0-0/islands-middle","isYandexStaff": false),"originalModificationDate":"2012-06-27T16:07:49.000Z","socialImage":("orig":("fullPath":"https://avatars.mds.yandex.net/get-yablogs /47421/file_1456488726678/orig")))))">

Determining the distance between two points ONLY using longlat coordinates.

$my=abs($cy-$y); //calculate the difference between the players (the second leg of the right triangle)

$dist=sqrt(pow($mx,2)+pow($my,2)); //Get the distance to the metro (the length of the hypotenuse according to the rule, the hypotenuse is equal to the root of the sum of the squares of the legs)

If anyone can help, I would be very grateful.

Sincerely, Alexander.

Solving problems in mathematics is often accompanied by many difficulties for students. Help the student cope with these difficulties, as well as teach him to use what he has theoretical knowledge when deciding specific tasks in all sections of the course of the subject “Mathematics” - 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 desired point O 1 has 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 located on 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 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 = √((-a + 2) 2 + (a – 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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THEORETICAL ISSUES

ANALYTICAL GEOMETRY ON THE PLANE

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

Let's consider some straight line. Let's choose a direction on it (then it will become an axis) and some point 0 (the origin of coordinates). A straight line with a chosen direction and origin is called coordinate line(we assume that the scale unit is selected).

Let M– an arbitrary point on the coordinate line. Let's put it 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 on the coordinate line corresponds to a certain real number - its coordinate. The converse is also true: each real number x corresponds to a certain point on the coordinate line, namely such a point M, whose coordinate is x. This correspondence is called one-to-one.

So, real numbers can be represented by points of a 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 on 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 general beginning ABOUT and the same unit of scale, form rectangular (or Cartesian) coordinate system on a plane.

Axis OH called the abscissa axis, the axis OY– ordinate axis. Dot ABOUT 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 About 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 solve geometric problems apply algebraic methods. The coordinate axes divide the plane into 4 parts, they are called in quarters, square or coordinate angles.

Polar coordinates.

The polar coordinate system consists of a certain point ABOUT, called pole, and the ray emanating from it OE, called polar axis. In addition, the scale unit for measuring the lengths of segments is set. Let it be given polar system coordinates and let M– arbitrary point of the plane. Let us denote by R– point distance M from point ABOUT, and through φ – the angle by which the beam is rotated counterclockwise to align the polar axis with the beam OM.

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

Dot M with polar coordinates R And φ are designated as follows: M( ;φ). Let us 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-abscissa axis coincides with the polar axis.

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

(1)

Proof.

Drop from dots M 1 And M 2 perpendiculars M 1 V And M 1 A,. because (x 2 ; y 2). By theorem, 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 | .

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.

Yandex.RTB R-A-339285-1 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 characteristic real number: let for point A this be a certain number 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 as its 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 plot the fraction 4 111 on the coordinate line.

Using the above method, put it on a straight line irrational number and completely impossible. 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 the plane in rectangular system coordinates O x y s 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 axis O x (abscissa axes), 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 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 general case, when points A and B do not lie in a plane parallel to one of coordinate planes. Let us draw planes perpendicular to the coordinate axes through points A and B and obtain corresponding points projections: 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 it is known that the square of the diagonal of a parallelepiped equal to the sum squares of its measurements. 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;

Lying 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

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 origin 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 modulus of the coordinate of this point, respectively O A = 1 - 2 = 2 - 1
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 it will be true equality:

λ 2 + 16 = 5 λ 2 + 16 = 25 λ = ± 3

Answer: A B = 5 if λ = ± 3.

Example 3

Initial data: specified 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.

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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