Formula for the distance between parallel lines. Distance from a point to a line. Distance between parallel lines

This article focuses on finding the distance between crossing lines using the coordinate method. First, the definition of the distance between intersecting lines is given. Next, an algorithm is obtained that allows one to find the distance between crossing lines. In conclusion, the solution to the example is analyzed in detail.

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Distance between crossing lines - definition.

Before giving the definition of the distance between skew lines, let us recall the definition of skew lines and prove a theorem related to skew lines.

Definition.

- this is the distance between one of the intersecting lines and a plane parallel to it passing through the other line.

In turn, the distance between a straight line and a plane parallel to it is the distance from some point on the straight line to the plane. Then the following formulation of the definition of the distance between crossing lines is valid.

Definition.

Distance between crossing lines is the distance from a certain point of one of the intersecting lines to a plane passing through another line parallel to the first line.

Consider the crossing lines a and b. Let's mark a certain point M 1 on line a, draw a plane parallel to line a through line b, and from point M 1 lower a perpendicular M 1 H 1 to the plane. The length of the perpendicular M 1 H 1 is the distance between the crossing lines a and b.

Finding the distance between crossing lines - theory, examples, solutions.

When finding the distance between crossing lines, the main difficulty is often to see or construct a segment whose length is equal to the desired distance. If such a segment is constructed, then, depending on the conditions of the problem, its length can be found using the Pythagorean theorem, signs of equality or similarity of triangles, etc. This is what we do when finding the distance between intersecting lines in geometry lessons in grades 10-11.

If Oxyz is introduced in three-dimensional space and intersecting lines a and b are given in it, then the coordinate method allows us to cope with the task of calculating the distance between given intersecting lines. Let's look at it in detail.

Let be a plane passing through line b, parallel to line a. Then the required distance between the crossing lines a and b is, by definition, equal to the distance from some point M 1 lying on the line a to the plane. Thus, if we determine the coordinates of a certain point M 1 lying on a straight line a, and obtain the normal equation of the plane in the form, then we can calculate the distance from the point to the plane using the formula (this formula was obtained in the article finding the distance from a point to a plane). And this distance is equal to the required distance between the crossing lines.

Now in detail.

The problem comes down to obtaining the coordinates of the point M 1 lying on the line a and finding the normal equation of the plane.

There are no difficulties in determining the coordinates of point M 1 if you know well the basic types of equations of a straight line in space. But it is worth dwelling in more detail on obtaining the equation of the plane.

If we determine the coordinates of a certain point M 2 through which the plane passes, and also obtain the normal vector of the plane in the form , then we can write the general equation of the plane as .

As a point M 2, you can take any point lying on the line b, since the plane passes through the line b. Thus, the coordinates of point M 2 can be considered found.

It remains to obtain the coordinates of the normal vector of the plane. Let's do this.

The plane passes through line b and is parallel to line a. Consequently, the normal vector of the plane is perpendicular to both the direction vector of the line a (denote it by ), and the direction vector of the line b (denote it by ). Then we can take and as a vector, that is, . Having determined the coordinates and direction vectors of straight lines a and b and calculated , we will find the coordinates of the normal vector of the plane.

So, we have the general equation of the plane: .

All that remains is to bring the general equation of the plane to normal form and calculate the required distance between the crossing lines a and b using the formula.

Thus, to find the distance between crossing lines a and b you need:

Let's look at the solution to the example.

Example.

In three-dimensional space in the rectangular coordinate system Oxyz, two intersecting straight lines a and b are given. The straight line a is determined

A parallelogram is a quadrilateral whose opposite sides are parallel, that is, they lie on parallel lines (Fig. 1).

Theorem 1. On the properties of the sides and angles of a parallelogram. In a parallelogram, opposite sides are equal, opposite angles are equal, and the sum of the angles adjacent to one side of the parallelogram is 180°.

Proof. In this parallelogram ABCD we draw a diagonal AC and get two triangles ABC and ADC (Fig. 2).

These triangles are equal, since ∠ 1 = ∠ 4, ∠ 2 = ∠ 3 (crosswise angles for parallel lines), and side AC is common. From the equality Δ ABC = Δ ADC it follows that AB = CD, BC = AD, ∠ B = ∠ D. The sum of angles adjacent to one side, for example angles A and D, is equal to 180° as one-sided for parallel lines. The theorem has been proven.

Comment. The equality of opposite sides of a parallelogram means that the segments of parallels cut off by parallel ones are equal.

Corollary 1. If two lines are parallel, then all points on one line are at the same distance from the other line.

Proof. Indeed, let a || b (Fig. 3).

Let us draw perpendiculars BA and CD to straight line a from some two points B and C of line b. Since AB || CD, then figure ABCD is a parallelogram, and therefore AB = CD.

The distance between two parallel lines is the distance from an arbitrary point on one of the lines to the other line.

According to what has been proven, it is equal to the length of the perpendicular drawn from some point of one of the parallel lines to the other line.

Example 1. The perimeter of the parallelogram is 122 cm. One of its sides is 25 cm larger than the other. Find the sides of the parallelogram.

Solution. By Theorem 1, opposite sides of a parallelogram are equal. Let's denote one side of the parallelogram by x and the other by y. Then, by condition $$\left\(\begin(matrix) 2x + 2y = 122 \\x - y = 25 \end(matrix)\right.$$ Solving this system, we obtain x = 43, y = 18. Thus Thus, the sides of the parallelogram are 18, 43, 18 and 43 cm.

Example 2.

Solution. Let Figure 4 meet the conditions of the problem.

Let us denote AB by x, and BC by y. According to the condition, the perimeter of the parallelogram is 10 cm, i.e. 2(x + y) = 10, or x + y = 5. The perimeter of triangle ABD is 8 cm. And since AB + AD = x + y = 5 then BD = 8 - 5 = 3. So BD = 3 cm.

Example 3. Find the angles of the parallelogram, knowing that one of them is 50° greater than the other.

Solution. Let Figure 5 meet the conditions of the problem.

Let us denote the degree measure of angle A by x. Then the degree measure of angle D is x + 50°.

Angles BAD and ADC are one-sided interior angles with parallel lines AB and DC and secant AD. Then the sum of these named angles will be 180°, i.e.
x + x + 50° = 180°, or x = 65°. Thus, ∠ A = ∠ C = 65°, a ∠ B = ∠ D = 115°.

Example 4. The sides of the parallelogram are 4.5 dm and 1.2 dm. A bisector is drawn from the vertex of an acute angle. What parts does it divide the larger side of the parallelogram into?

Solution. Let Figure 6 meet the conditions of the problem.

AE is the bisector of an acute angle of a parallelogram. Therefore, ∠ 1 = ∠ 2.

Distance

from point to line

Distance between parallel lines

Geometry, 7th grade

To the textbook by L.S. Atanasyan

mathematics teacher of the highest category

Municipal educational institution "Upshinskaya basic secondary school"

Orsha district of the Republic of Mari El

Perpendicular length drawn from a point to a line, called distance from this point to direct.

AN ⏊ A

M є a, M is different from N

Perpendicular , drawn from a point to a line, less any inclined , drawn from the same point to this line.

AM – inclined, drawn from point A to line a

AN AM

AN - inclined

AN AN

AN AK

AK - inclined

Distance from point to line

M

The distance from point M to straight line c is...

N

The distance from point N to line c is...

With

The distance from point K to straight line c is...

K

The distance from point F to straight line c is...

F

Distance from point to line

AN ⏊ A

AN= 5.2 cm

VK ⏊ A

VK= 2.8 cm

Theorem.

All points of each of two parallel lines are equidistant from the other line

Given: a ǁ b

A є a, B є a,

Prove: the distances from points A and B to line a are equal.

AN ⏊ b,BK ⏊ b,

Prove: AH = BK

Δ ANK = ΔVKA(Why?)

From the equality of triangles it follows AN = BK

The distance from an arbitrary point of one of the parallel lines to another line is called the distance between these lines.

Converse theorem.

All points of the plane located on one side of a given line and equidistant from it lie on a line parallel to the given one.

AN ⏊ b,BK ⏊ b,

АH = BK

Prove: AB ǁ b

Δ ANK = ΔKVA(Why?)

From the equality of triangles it follows … , but these are internal crosswise angles formed … , means AB ǁ NK

What is the distance between lines b and c, if the distance between the lines A and b is equal to 4, and between the lines A and c equals 5?

A ǁ b ǁ c

What is the distance between lines b and a, if the distance between lines b and c is 7, and between the lines A and c equals 2?

What is the distance between lines A and c, if the distance between lines b and c is 10, and between lines b And a equals 6?

What is the set of all points in a plane that are equidistant from two given parallel lines?

A ǁ b

Answer: A line parallel to these lines and located at equal distances from them.

What is the set of all points on a plane located at a given distance from a given line?

Answer: Two lines parallel to a given line and located at a given distance on opposite sides of it.

Proof.

Let's take a point , which lies on the straight line a, then the coordinates of the point M1 satisfy the equation, that is, the equality is true, from where we have .

If font-size:12.0pt;line-height:115%;font-family:Verdana"> b looks likefont-size:12.0pt;line-height:115%;font-family:Verdana">, and if, then the normal equation of the line b looks likefont-size:12.0pt;line-height:115%;font-family:Verdana">.

Then at font-size:12.0pt;line-height:115%;font-family:Verdana">distance from pointto a straight line b calculated by the formula, and when - according to the formula

That is, for any value C2 distance from point to a straight line b can be calculated using the formula. And if we take into account equality, which was obtained above, then the last formula will take the formfont-size:12.0pt;line-height:115%;font-family:Verdana">. The theorem is proven.

2. Solving problems on finding the distance between parallel lines

Example No. 1.

Find the distance between parallel lines And Solution.

Let us obtain general equations for given parallel lines.

For straight font-size:12.0pt; line-height:115%;font-family:Verdana">corresponds to the general equation of a straight line. Let us move from parametric equations of the straight line of the formfont-size:12.0pt;line-height:115%;font-family:Verdana">to the general equation of this line:

font-size:12.0pt; line-height:115%;font-family:Verdana">Coefficients for variables x And y in received general equations parallel lines are equal, so we can immediately apply the formula to calculate the distance between parallel lines on a plane:.

Answer: font-size:12.0pt; line-height:115%;font-family:Verdana">Example No. 2.

A rectangular coordinate system is introduced on the plane Oxy and given the equations of two parallel lines And . Find the distance between the indicated parallel lines.

Solution:

First solution.

Canonical equations of a straight line on a plane of the formfont-size:12.0pt; line-height:115%;font-family:Verdana">allow you to immediately record the coordinates of a point M1 lying on this line:font-size:12.0pt; line-height:115%;font-family:Verdana">. Distance from this point to the straight lineequal to the required distance between parallel lines. Equationis a normal equation of a line, therefore, we can immediately calculate the distance from a point to a straight line font-size:12.0pt;line-height:115%;font-family:Verdana">:.

Second solution.

The general equation of one of the given parallel lines has already been given to usfont-size:12.0pt;line-height:115%;font-family:Verdana">. Let us present the canonical equation of the lineto the general equation of the line:. Coefficients of the variable x in general equations the given parallel lines are equal (with variable y the coefficients are also equal - they are zero), so you can use a formula that allows you to calculate the distance between given parallel lines:.

Answer: 8

3. Homework

Self-test tasks

1. Find the distance between two parallel lines

4.CONCLUSION

All set goals and objectives have been fully achieved. Two lessons have been developed from the section “Relative arrangement of objects on a plane” on the topic “Distance from a point to a line. Distance between parallel lines” using the coordinate method. The material is selected at a level accessible to students, which will allow them to solve geometry problems using simpler and more beautiful methods.

5. REFERENCES

1) , Yudina. Grades 7 – 9: textbook for general education institutions.

2) , Poznyak. Textbook for 10-11 grades of secondary school.

3) , Nikolsky mathematics. Volume one: elements of linear algebra and analytical geometry.

4) , Poznyak geometry.

6.APPLICATIONS

Reference material

General equation of a straight line:

Ah + Wu + C = 0 ,

Where A And IN are not equal to zero at the same time.

Odds A And IN are the coordinates normal vector straight line (i.e., a vector perpendicular to the line). At A = 0 straight line parallel to axis OH, at B = 0 straight line parallel to axis ABOUT Y .

At IN0 we get equation of a line with slope :

Equation of a line passing through a point ( X 0 , at 0) and not parallel to the axisOY, has the form:

at – at 0 = m (x – X 0) ,

Where m – slope , equal to the tangent of the angle formed by the given straight line and the positive direction of the axis OH .

At A font-size:12.0pt;font-family:Verdana;color:black">

Where a = – C / A , b = – C / B . This line passes through the points (a, 0) and (0, b), i.e., it cuts off segments of length on the coordinate axesa And b .

Equation of a line passing through two different points (X 1, at 1) and ( X 2, at 2):

Parametric equation of a line passing through the point ( X 0 , at 0) and parallel direction vector straight line (a, b) :

Condition for parallel lines:

1) for straight lines Ah+ Wu+ C = 0 andDx+Ey+F = 0: A.E. – BD = 0 ,

2) for straight lines at = m x+ k And at= p x+ q : m = p .

This video lesson will be useful for those who want to independently study the topic “Distance from a point to a line. Distance between parallel lines." During the lesson you will learn how to calculate the distance from a point to a line. Then the teacher will give the definition of the distance between parallel lines.

In this lesson we will get acquainted with the concept "distance" generally. We also specify this concept in case of calculation distances between two points, a point and a line, parallel lines

Let's look at Figure 1. It shows 2 points A and B. The distance between two points A and B is a segment having ends at given points, that is, segment AB

Rice. 1. AB - distance between points

It is noteworthy that distance cannot be considered a curve or broken line connecting two points. Distance- This shortest path from one point to another. It is the segment AB that is the smallest of all possible lines connecting points A and B

Consider Figure 2, which shows the straight line A, and point A, which does not belong to this line. Distance from point A to a straight line will be the length of the perpendicular AN.

Rice. 2. AN - distance between a point and a line

It is important to note that AN is the shortest distance, since in the triangle AMN this segment is a leg, and an arbitrary other segment connecting point A and the line A(in this case it is AM) will be the hypotenuse. As you know, the leg is always less than the hypotenuse

Distance designation:

Let's consider parallel lines a and b shown in Figure 3

Rice. 3. Parallel lines a and b

Let's fix two points on a straight line a and drop perpendiculars from them onto a line parallel to it b. Let us prove that if ,

Let us draw segment AM for convenience of proof. Let us consider the resulting triangles ABM and ANM. Since , and , then . Likewise, . These right triangles () have a common side AM. It is the hypotenuse in both triangles. Angles AMN and AMB are internal cross angles with parallel straight lines AB and NM and secant AM. By known property, .

From all of the above it follows that . From the equality of triangles it follows that AN = BM

So, we have proven that in Figure 3 the segments AN and BM are equal. This means that distance between parallel lines is the length of their common perpendicular, and the choice of perpendicular can be arbitrary. Thus,

The converse is also true: a set of points that are at the same distance from a certain line form a line parallel to the given one.

Let's consolidate our knowledge and solve several problems

Example 1: Problem 272 from the textbook “Geometry 7-9”. Author - Atanasyan L.S.

In an equilateral triangle ABC, the bisector AD is drawn. The distance from point D to straight line AC is 6 cm. Find the distance from point A to straight line BC

Rice. 4. Drawing for example 1

Solution:

An equilateral triangle is a triangle with three equal sides (and therefore three equal angles, that is, 60 0 each). Equilateral triangle is a special case of an isosceles triangle, therefore all the properties inherent in an isosceles triangle also apply to an equilateral triangle. Therefore, AD is not only a bisector, but also a height, therefore AD ⊥BC

Since the distance from point D to line AC is the length of the perpendicular drawn from point D to line AC, then DH is this distance. Consider the triangle AND. In it, the angle H = 90 0, since DH is perpendicular to AC (by definition of the distance from a point to a straight line). In addition, in this triangle the leg DH lies opposite the angle, so AD = (cm) (By property)

The distance from point A to straight line BC is the length of the perpendicular dropped onto straight line BC. According to the proven AD ⊥BC, it means .

Answer: 12 cm.

Example 2: Problem 277 from the textbook “Geometry 7-9”. Author - Atanasyan L.S.

The distance between parallel lines a and b is 3 cm, and the distance between parallel lines a and c is 5 cm. Find the distance between parallel lines b and c

Solution:

Rice. 5. Drawing for example 2 (first case)

Since , then = 5 - 3 = 2 (cm).

However, this answer is incomplete. There is another option for locating straight lines on a plane:

Rice. 6. Drawing for example 2 (second case)

In this case.

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1. No. 280, 283. Atanasyan L. S., Butuzov V. F., Kadomtsev S. B., Poznyak E. G., Yudina I. I. edited by Tikhonov A. N. Geometry grades 7-9. M.: Enlightenment. 2010
2. Sum of hypotenuse SE and leg SC right triangle SKE is equal to 31 cm, and their difference is 3 cm. Find the distance from vertex C to straight line KE
3. Based on AB of the isosceles triangle ABC, point M is taken, equidistant from the lateral sides. Prove that CM is the height of triangle ABC
4. Prove that all points of the plane located on one side of a given line and equidistant from it lie on a line parallel to the given one