Lesson on the topic of the distance between the points of the coordinate line. Video lesson “Distance between points of a coordinate line Distance between two points on a coordinate line

Lesson plan.

The distance between two points on a straight line.

Rectangular (Cartesian) coordinate system.

The distance between two points on a straight line.

Theorem 3. If A(x) and B(y) are any two points, then d - the distance between them is calculated by the formula: d = lу - xl.

Proof. According to Theorem 2, we have AB = y - x. But the distance between points A and B is equal to the length of the segment AB, those. the length of the vector AB . Therefore, d \u003d lABl \u003d lu-xl.

Since the numbers y-x and x-y are taken modulo, we can write d =lx-ul. So, to find the distance between points on the coordinate line, you need to find the modulus of the difference between their coordinates.

Example 4. Given points A(2) and B(-6), find the distance between them.

Solution. Substitute in the formula instead of x=2 and y=-6. We get, AB=lу-хl=l-6-2l=l-8l=8.

Example 5 Construct a point symmetrical to the point M(4) with respect to the origin.

Solution. Because from the point M to the point O 4 single segments, set aside on the right, then, in order to build a point symmetrical to it, we postpone 4 single segments from the point O to the left, we get the point M "(-4).

Example 6 Construct a point C(x) symmetrical to point A(-4) with respect to point B(2).

Solution. Note the points A(-4) and B(2) on the number line. Let's find the distance between the points according to Theorem 3, we get 6. Then the distance between points B and C should also be equal to 6. We set aside 6 unit segments from point B to the right, we get point C (8).

Exercises. 1) Find the distance between points A and B: a) A(3) and B(11), b) A(5) and B(2), c) A(-1) and B(3), d) A (-5) and B (-3), e) A (-1) and B (3), (Answer: a) 8, b) 3, c) 4, d) 2, e) 2).

2) Construct a point C(x) symmetrical to point A(-5) with respect to point B(-1). (Answer: C(3)).

Rectangular (Cartesian) coordinate system.

Two mutually perpendicular axes Ox and Oy, having a common origin O and the same scale unit, form rectangular(or Cartesian) coordinate system on the plane.

The Ox axis is called x-axis, and the y-axis y-axis. The point O of intersection of the axes is called origin. The plane in which the axes Ox and Oy are located is called the coordinate plane and is denoted Oxy.

Let M be an arbitrary point of the plane. Let us drop from it the perpendiculars MA and MB, respectively, on the axes Ox and Oy. The intersection points A and B of the eith perpendiculars with the axes are called projections points M on the coordinate axis.

Points A and B correspond to certain numbers x and y - their coordinates on the axes Ox and Oy. The number x is called abscissa points M, number y - her ordinate.

The fact that the point M has coordinates x and y is symbolically denoted as follows: M (x, y). In this case, the first in brackets indicate the abscissa, and the second - the ordinate. The origin has coordinates (0,0).

Thus, with the chosen coordinate system, each point M of the plane corresponds to a pair of numbers (x, y) - its rectangular coordinates and, conversely, to each pair of numbers (x, y) corresponds, and moreover, one point M on the Oxy plane such that its the abscissa is x and the ordinate is y.

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 apply algebraic methods when solving geometric problems.

The coordinate axes divide the plane into four parts, they are called quarters, quadrants or coordinate angles and numbered with Roman numerals I, II, III, IV as shown in the figure (hyperlink).

The figure also shows the signs of the coordinates of the points depending on their location. (for example, in the first quarter, both coordinates are positive).

Example 7 Build points: A(3;5), B(-3;2), C(2;-4), D (-5;-1).

Solution. Let's construct the point A(3;5). First of all, we introduce a rectangular coordinate system. Then, along the abscissa axis, we set aside 3 scale units to the right, and along the ordinate axis, 5 scale units up, and through the final division points we draw straight lines parallel to the coordinate axes. The point of intersection of these lines is the required point A(3;5). The rest of the points are constructed in the same way (see the hyperlink figure).

Exercises.

    Without drawing point A(2;-4), find out which quarter it belongs to.

    What quarters can a point be in if its ordinate is positive?

    A point with coordinate -5 is taken on the Oy axis. What are its coordinates on the plane? (answer: since the point lies on the Oy axis, then its abscissa is 0, the ordinate is given by condition, so the coordinates of the point are (0; -5)).

    Points are given: a) A(2;3), b) B(-3;2), c) C(-1;-1), d) D(x;y). Find the coordinates of the points that are symmetrical to them about the x-axis. Plot all these points. (answer: a) (2; -3), b) (-3; -2), c) (-1; 1), d) (x; -y)).

    Points are given: a) A(-1;2), b) B(3;-1), c) C(-2;-2), d) D(x;y). Find the coordinates of the points that are symmetrical to them about the y-axis. Plot all these points. (answer: a) (1; 2), b) (-3; -1), c) (2; -2), d) (-x; y)).

    Points are given: a) A(3;3), b) B(2;-4), c) C(-2;1), d) D(x;y). Find the coordinates of points that are symmetrical to them about the origin. Plot all these points. (answer: a) (-3; -3), b) (-2; 4), c) (2; -1), d) (-x;-y)).

    Given a point M(3;-1). Find the coordinates of points that are symmetrical to it about the Ox axis, the Oy axis and the origin. Plot all points. (answer: (3;1), (-3;-1), (-3;1)).

    Determine in which quarters the point M (x; y) can be located if: a) xy> 0, b) xy< 0, в) х-у=0, г) х+у=0. (ответ: а) в первой и третьей, б)во второй и четвертой, в) в первой и третьей, г) во второй и четвертой).

    Determine the coordinates of the vertices of an equilateral triangle with a side equal to 10, lying in the first quarter, if one of its vertices coincides with the origin O, and the base of the triangle is located on the Ox axis. Make a drawing. (answer: (0;0), (10;0), (5;5v3)).

    Using the coordinate method, determine the coordinates of all the vertices of the regular hexagon ABCDEF. (answer: A (0;0), B (1;0), C (1.5;v3/2) , D (1;v3), E (0;v3 ), F (-0.5;v3 /2).Indication: take point A as the origin of coordinates, direct the abscissa axis from A to B, take the length of side AB as the scale unit.It is convenient to draw large diagonals of the hexagon.)

In this article, we will consider ways to determine the distance from a point to a point theoretically and using an example specific tasks. Let's start with some definitions.

Definition 1

Distance between points- this is the length of the segment connecting them, in the existing scale. It is necessary to set the 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 the coordinate line, in the coordinate plane or three-dimensional space.

Initial data: the coordinate line O x and an arbitrary point A lying on it. One real number is inherent in any point of the line: let this be a certain number for point A xA, it is the coordinate of point A.

In general, we can say that the estimation of the length of a certain segment occurs in comparison with the segment taken as a unit of length on a given scale.

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

For example, point A corresponds to the number 3 - in order to get to it from point O, it will be necessary to set aside three unit segments. If point A has a coordinate of - 4, single segments are plotted in a similar way, but in a different, negative direction. Thus, in the first case, the distance O A is 3; in the second case, O A \u003d 4.

If point A has a rational number as a coordinate, 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 put aside the coordinate direct fraction 4 111 .

In the above way, it is completely impossible to postpone 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 \u003d x A (the number is taken as a 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 .

Summarizing: the distance from the origin to the point, which corresponds to a real number on the coordinate line, is equal to:

  • 0 if the point is the same as 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 the point O to the point A with the coordinate x A: O A = x A

The correct statement would be: the distance from one point to another will be equal to the modulus of the difference in coordinates. Those. for points A and B lying on the same coordinate line at any location and having, respectively, the 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's draw perpendiculars to the coordinate axes O x and O y through points A and B and get the projection points as a result: A x , A y , B x , B y . Based on the location of points A and B, the following options are further 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 and coincide, and | A B | = | A y B y | . Since the distance between the points is equal to the modulus of the difference between 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 (y-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 line perpendicular to one of the coordinate axes, find the distance between them, deriving the calculation formula:

We see that the triangle A B C is right-angled by construction. In this case, A C = A x B x and B C = A y B y . Using the Pythagorean theorem, we compose 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 form a conclusion from the result obtained: the distance from point A to point B on the plane is determined by the 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 the previously formed statements for the cases of coincidence of points or situations when the points lie on straight lines perpendicular to the axes. So, for the case of the coincidence of points A and B, the equality will be true: A B = (x B - x A) 2 + (y B - y A) 2 = 0 2 + 0 2 = 0

For the situation when 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 y-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: 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.

Consider the general case when points A and B do not lie in a plane parallel to one of coordinate planes. Draw through points A and B planes perpendicular to the coordinate axes, and get 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 box. According to the construction of the measurement of this box: A x B x , A y B y and A z B z

From the course of geometry it is known 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 \u003d 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 where:

The dots match;

They lie on the same coordinate axis or on a straight line parallel to one of the coordinate axes.

Examples of solving problems for 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 reference 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 module of the coordinate of this point, respectively O A \u003d 1 - 2 \u003d 2 - 1
  2. The distance between points A and B is defined 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: given a rectangular coordinate system and two points lying on it A (1 , - 1) and B (λ + 1 , 3) ​​. λ is some real number. It is necessary to find all values ​​of this number for 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 values ​​of the coordinates, we get: A B = (λ + 1 - 1) 2 + (3 - (- 1)) 2 = λ 2 + 16

And also we 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 \u003d 5 if λ \u003d ± 3.

Example 3

Initial data: given 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 the 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 points on the coordinate line - 6 class.

The formula for finding the distance between points on a coordinate line

Algorithm for finding the coordinates of a point - the middle of a segment

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Slides captions:

Distance between points on the coordinate line x 0 1 A B AB \u003d ρ (A, B)

Distance between points on a coordinate line The purpose of the lesson: - Find a way (formula, rule) to find the distance between points on a coordinate line. - Learn to find the distance between points on a coordinate line using the found rule.

1. Oral counting 15 -22 +8 -31 +43 -27 -14

2. Orally solve the task using the coordinate line: how many integers are enclosed between the numbers: a) - 8.9 and 2 b) - 10.4 and - 3.7 c) - 1.2 and 4.6? a) 10 b) 8 c) 6

0 1 2 7 positive numbers -1 -5 negative numbers Distance from home to stadium 6 Distance from home to school 6 Coordinate line

0 1 2 7 -1 -5 Distance from stadium to home 6 Distance from school to home 6 Finding the distance between points on the coordinate line ρ (-5 ; 1)=6 ρ (7 ; 1)=6 The distance between points will be denoted by the letter ρ (rho)

0 1 2 7 -1 -5 Distance from stadium to home 6 Distance from school to home 6 Finding the distance between points on the coordinate line ρ (-5 ; 1)=6 ρ (7 ; 1)=6 ρ (a; b) = ? | a-b |

The distance between points a and b is equal to the modulus of the difference between the coordinates of these points. ρ (a; b)= | a-b | Distance between points on a coordinate line

Geometric meaning of the modulus of a real number a b a a=b b x x x Distance between two points

0 1 2 7 -1 -5 Find the distances between points on the coordinate line - 2 - 3 - 4 3 4 5 6 -6 ρ (-6 ; 2)= ρ (6 ; 3)= ρ (0 ; 7)= ρ (1 ; -4) = 8 3 7 5

0 1 2 7 -1 -5 Find the distances between points on the coordinate line - 2 - 3 - 4 3 4 5 6 -6 ρ (2 ; -6)= ρ (3 ; 6)= ρ (7 ; 0)= ρ (-4 ; 1) = 8 3 7 5

Output: expression values ​​| a-b | and | b-a | are equal for any values ​​of a and b =

–16 –2 0 –3 +8 0 +4 +17 0 ρ(–3; 8) = 11; |(–3) – (+8)| = 11; |(+8) – (–3)| = 11. ρ(–16; –2) = 14; |(–16) – (–2)| = 14; |(–2) – (–16)| = 14. ρ(4; 17) = 13; |(+4) – (+17)| = 13; |(+17) – (+4)| = 13. Distance between points of the coordinate line

Find ρ(x; y)if: 1) x = -14, y = -23; ρ(x; y)=| x – y |=|–14–(– 23)|=|–14+23|=| 9 |=9 2) x = 5.9, y = -6.8; ρ(x; y)=|5, 9 –(– 6.8)|=|5.9+6.8|=| 12.7 |=12.7

Continue the sentence 1. A coordinate line is a line with ... 2. The distance between two points is ... 3. Opposite numbers are numbers, ... 4. The modulus of the number X is called ... 5. - Compare the values ​​​​of expressions a - b V b – a conclude … - Compare the values ​​of expressions | a-b | v | b-a | c conclude...

Vintik and Shpuntik are walking along coordinate beam. The screw is at point B(236), Shpuntik is at point W(193) How far are Screw and Shpuntik from each other? ρ(B, W) = 43

Find the distance between points A (0), B (1) A (2), B (5) A (0), B (- 3) A (- 10), B (1) AB \u003d 1 AB \u003d 3 AB \u003d 3 AB = 11

Find the distance between points A (- 3.5), B (1.4) K (1.8), B (4.3) A (- 10), C (3)

Check AB = KV = AC =

C (- 5) C (- 3) Find the coordinate of the point - the middle of the segment BA

Points A (–3.25) and B (2.65) are marked on the coordinate line. Find the coordinate of the point O - the midpoint of the segment AB. Solution: 1) ρ(А;В)= |–3.25 – 2.65| = |–5.9| \u003d 5.9 2) 5.9: 2 \u003d 2.95 3) -3.25 + 2.95 \u003d - 0.3 or 2.65 - 2.95 \u003d - 0.3 Answer: O (-0, 3)

Points С(–5.17) and D(2.33) are marked on the coordinate line. Find the coordinate of point A - the midpoint of the segment CD. Solution: 1) ρ(С; D)= |– 5 , 17 – 2, 33 | = |– 7 , 5 | \u003d 7, 5 2) 7, 5: 2 \u003d 3, 7 5 3) - 5, 17 + 3, 7 5 \u003d - 1, 42 or 2, 33 - 3, 7 5 \u003d - 1, 42 Answer: A ( - 1, 42)

Conclusion: Algorithm for finding the coordinates of the point - the middle this segment: 1. Find the distance between the points - the ends of this segment = 2. Divide the result-1 by 2 (half the value) = c 3. Add the result-2 to the a coordinate or subtract the result-2 from the a + c or - c coordinate 4. Result-3 is the coordinate of the point - the midpoint of the given segment

Work with the textbook: §19, p.112, A. No. 573, 575 V. No. 578, 580 Homework: §19, p.112, A. no. 574, 576, B. no. 579, 581 rational numbers. Distance between points on a coordinate line "

Today I learned… It was interesting… I realized that… Now I can… I learned… I succeeded… I will try… I was surprised… I wanted to…

§ 1 Rule for finding the distance between points of a coordinate line

In this lesson, we will derive a rule for finding the distance between points of a coordinate line, and also learn how to find the length of a segment using this rule.

Let's do the task:

Compare Expressions

1. a = 9, b = 5;

2. a = 9, b = -5;

3. a = -9, b = 5;

4. a = -9, b = -5.

Substitute the values ​​in the expressions and find the result:

The modulus of the difference of 9 and 5 is modulo 4, the modulus of 4 is 4. The modulus of the difference of 5 and 9 is modulo minus 4, the modulus of -4 is 4.

The module of the difference between 9 and -5 is equal to the module 14, the module 14 is equal to 14. The module of the difference minus 5 and 9 is equal to the module -14, the module is -14=14.

The modulus of the difference minus 9 and 5 is equal to the modulus of minus 14, the modulus of minus 14 is 14. The modulus of the difference of 5 and minus 9 is modulo 14, the modulus of 14 is 14

The module of the difference minus 9 and minus 5 is equal to the module minus 4, the module -4 is 4. The module of the difference minus 5 and minus 9 is equal to the module 4, the module 4 is (l-9 - (-5)l \u003d l-4l \u003d 4; l -5 - (-9)l = l4l = 4)

In each case, equal results were obtained, therefore, we can conclude:

The values ​​of the expressions modulus of difference a and b and modulus of difference b and a are equal for any values ​​of a and b.

One more task:

Find the distance between the points of the coordinate line

1.A(9) and B(5)

2.A(9) and B(-5)

On the coordinate line, mark the points A(9) and B(5).

Let's count the number of unit segments between these points. There are 4 of them, which means the distance between points A and B is 4. Similarly, we find the distance between two other points. We mark points A (9) and B (-5) on the coordinate line, determine the distance between these points along the coordinate line, the distance is 14.

Compare the results with previous tasks.

The modulus of the difference between 9 and 5 is 4, and the distance between the points with coordinates 9 and 5 is also 4. The modulus of the difference between 9 and minus 5 is 14, the distance between the points with coordinates 9 and minus 5 is 14.

It begs the conclusion:

The distance between points A(a) and B(b) of the coordinate line is equal to the modulus of the difference between the coordinates of these points l a - b l.

Moreover, the distance can also be found as the modulus of the difference between b and a, since the number of unit segments will not change from the point from which we count them.

§ 2 The rule for finding the length of a segment from the coordinates of two points

Find the length of the segment CD, if on the coordinate line С(16), D(8).

We know that the length of a segment is equal to the distance from one end of the segment to the other, i.e. from point C to point D on the coordinate line.

Let's use the rule:

and find the modulus of the difference of coordinates c and d

So, the length of segment CD is 8.

Consider another case:

Let's find the length of the segment MN, the coordinates of which have different signs M (20), N (-23).

Substitute the values

we know that -(-23) = +23

so the modulus of the difference of 20 and minus 23 is equal to the modulus of the sum of 20 and 23

Let's find the sum of modules of coordinates of the given segment:

The value of the modulus of the difference of coordinates and the sum of the modules of coordinates in this case turned out to be the same.

We can conclude:

If the coordinates of two points have different signs, then the distance between the points is equal to the sum of the modules of the coordinates.

In the lesson, we got acquainted with the rule for finding the distance between two points of a coordinate line and learned how to find the length of a segment using this rule.

List of used literature:

  1. Mathematics. Grade 6: lesson plans for the textbook by I.I. Zubareva, A.G. Mordkovich // Compiled by L.A. Topilin. – M.: Mnemosyne 2009.
  2. Mathematics. Grade 6: student textbook educational institutions. I.I. Zubareva, A.G. Mordkovich. - M.: Mnemosyne, 2013.
  3. Mathematics. Grade 6: textbook for students of educational institutions./N.Ya. Vilenkin, V.I. Zhokhov, A.S. Chesnokov, S.I. Schwarzburd. - M.: Mnemosyne, 2013.
  4. Mathematics Handbook - http://lyudmilanik.com.ua
  5. Handbook for students in high school http://shkolo.ru

In mathematics, both algebra and geometry pose problems of finding the distance to a point or line from a given object. It is located completely different ways, the choice of which depends on the initial data. Consider how to find the distance between given objects in different conditions.

Using Measuring Tools

At the initial stage of development mathematical science learn how to use elementary tools (such as a ruler, protractor, compass, triangle, and others). Finding the distance between points or lines with their help is not difficult at all. It is enough to attach the scale of divisions and write down the answer. One has only to know that the distance will be equal to the length of the straight line that can be drawn between the points, and in the case of parallel lines, the perpendicular between them.

Using theorems and axioms of geometry

In learning to measure distance without the help of special devices or For this, numerous theorems, axioms and their proofs are needed. Often the tasks of how to find the distance come down to the formation and search for its sides. To solve such problems, it is enough to know the Pythagorean theorem, the properties of triangles and how to transform them.

Points on the coordinate plane

If there are two points and their position on the coordinate axis is given, then how to find the distance from one to the other? The solution will include several steps:

  1. We connect the points with a straight line, the length of which will be the distance between them.
  2. Find the difference between the values ​​of the coordinates of the points (k; p) of each axis: |k 1 - k 2 |= d 1 and | p 1 - p 2 |= d 2 (we take the values ​​modulo, because the distance cannot be negative) .
  3. After that, we square the resulting numbers and find their sum: d 1 2 + d 2 2
  4. The final step is to extract from the resulting number. This will be the distance between the points: d \u003d V (d 1 2 + d 2 2).

As a result, the whole solution is carried out according to one formula, where the distance is equal to square root from the sum of squares of the difference of coordinates:

d \u003d V (| k 1 - k 2 | 2 + | p 1 - p 2 | 2)

If the question arises of how to find the distance from one point to another, then the search for an answer to it will not differ much from the above. The decision will be made according to the following formula:

d \u003d V ( | k 1 - k 2 | 2 + | p 1 - p 2 | 2 + | e 1 - e 2 | 2)

Parallel lines

The perpendicular drawn from any point lying on one straight line to the parallel will be the distance. When solving problems in a plane, it is necessary to find the coordinates of any point of one of the lines. And then calculate the distance from it to the second straight line. To do this, we bring them to the general form Ax + Vy + C = 0. It is known from the properties of parallel lines that their coefficients A and B will be equal. In this case, you can find by the formula:

d \u003d | C 1 - C 2 | / V (A 2 + B 2)

Thus, when answering the question of how to find the distance from a given object, it is necessary to be guided by the condition of the problem and the tools provided for its solution. They can be both measuring devices, and theorems and formulas.

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