Property of two direct parallel third evidence. Signs of parallelism of two straight lines. Properties of parallel straight lines. Plimming direct - signs and conditions of parallelism

The parallelism of the two direct can be proved on the basis of the theorem, according to which two perpendicular carried out relative to one straight line will be parallel. There are certain signs of parallelism of direct - there are three of all three, and all of them we will look more specifically.

The first sign of parallelism

Direct parallel, if with the intersection of their third straight, formed internal angles lying will be equal.

Suppose, when crossing the straight lines AV and CD, the direct line of EF, angles were formed / 1 and / 2. They are equal, since the straight line EF passes under one bias relative to the other remaining direct. In places intersection of lines, we set the points of L - we turned out the segment of the EF section. I find it the middle and put the point O (Damn 189).

For direct av omit perpendicular from the point O. We call it Ohm. We continue the perpendicular until it intersects with a straight CD. As a result, the initial straight AV is strictly perpendicular to the MN, which means that the CD_ | _MN, but this statement requires proof. As a result of the perpendicular and the crossing line, we have formed two triangles. One of them is mine, the second - Nok. Consider them in more detail. Signs of parallelism of direct grade 7

These triangles are equal, since, in accordance with the conditions of the theorem, / 1 \u003d / 2, and in accordance with the construction of triangles, the side OK \u003d side of the OL. The angle is ML \u003d / NOK, since these are vertical angles. It follows from this that the side and two angle, adjacent to it of one of the triangles, respectively, are equal to the side and two angles adjacent to it, another of the triangles. Thus, the triangle is MOL \u003d triangles, and hence the angle lm \u003d the corner of Kno, but we know that / lm directly, it means that the corresponding corner of Kno is also direct. That is, we managed to prove that to a direct Mn, both direct av and straight CD perpendicular. That is, AB and CD in relation to each other are parallel. It was necessary for us to prove. Consider the remaining signs of the parallelism of the straight lines (grade 7), which differ from the first characteristic of the proof method.

The second sign of parallelism

According to the second sign of the parallelism of direct, we need to prove that the angles obtained in the process of intersection of parallel direct AV and CD direct EF will be equal. Thus, the signs of the parallelism of two direct, both the first and the second, is based on the equity of the angles obtained by the intersection of their third line. We admit that / 3 \u003d / 2, and the angle is 1 \u003d / 3, since it is vertical to it. Thus, and / 2 will be equal to the corner1, but it should be borne in mind that both an angle of 1 and an angle 2 are internal, and inhibit lying angles. Consequently, we have to apply our knowledge, namely, that two segments will be parallel if they are directly directly educated when they are intersected, the underlying angles will be equal. Thus, we found out that Av || Cd.

We managed to prove that under the condition of the parallelity of two perpendicular to one straight line, according to the corresponding theorem, the sign of the parallelism of direct is obvious.

Third sign parallelism

There is also a third sign of parallelism, which is proved by the amount of one-sided internal angles. Such proof of a sign of parallelism of direct allows us to conclude that two straight lines will be parallel, if the third direct, the sum of the obtained single-sided angles will be 2d. See Figure 192.

Parallelism - very useful property in geometry. In real life, parallel sides allow you to create beautiful, symmetrical things, pleasant to any eye, so geometry has always needed ways to check this parallel. On the signs of parallel straight lines we will talk in this article.

Definition for parallelism

We highlight the definitions that you need to know to prove the signs of the parallelism of the two straight lines.

Direct is called parallel if they do not have intersection points. In addition, in solutions, usually parallel straight lines go in conjunction with a securing line.

The secure direct is called straight, which crosses both parallel straight. In this case, the liar, respective and unilateral angles are formed. There will be pairs of angles 1 and 4 who lie under bed; 2 and 3; 8 and 6; 7 and 5. will be appropriate 7 and 2; 1 and 6; 8 and 4; 3 and 5.

One-sided 1 and 2; 7 and 6; 8 and 5; 3 and 4.

With proper decoration, it is written: "Low the underlying angles with two parallel straight lines A and B and the sequential C", because for two parallel straight lines there can be an infinite set of sequers, so it is necessary to specify what kind of section you mean, you mean.

Also, for the proof, you will need the external triangle theorem, which states that the external angle of the triangle is equal to the sum of the two corners of the triangle of non-negative with it.

Signs

All signs of parallel straight lines are tied to the knowledge of the properties of the corners and the theorem about the external angle of the triangle.

Sign 1.

Two straight parallel if the underlying corners are equal.

Consider two straight a and b from the secant. Low the underlying angles 1 and 4 are equal. Suppose that the straight lines are not parallel. Means the direct intersection point M. Then the triangle of AVM with an outer angle is formed 1. The outer angle should be equal to the amount of the angles 4 and AVM as non-negative with it by the external angle theorem in the triangle. But then it turns out that angle 1 is more angle 4, and this contradicts the condition of the problem, which means that the points M does not exist, the straight lines do not intersect, that is, parallel.

Fig. 1. Drawing to the proof.

Sign 2.

Two straight parallel, if corresponding angles at the unit are equal.

Consider two straight a and b from the secant. The corresponding angles 7 and 2 are equal. Pay attention to the angle 3. It is vertical for angle 7. So, the angles 7 and 3 are equal. It means that the angles 3 and 2 are also equal because<7=<2 и <7=<3. А угол 3 и угол 2 являются накрест лежащими. Следовательно, прямые параллельны, что и требовалось доказать.

Fig. 2. Drawing to the proof.

Sign 3.

Two straight parallels if the sum of one-sided corners is 180 degrees.

Fig. 3. Drawing to the proof.

Consider two straight a and b from the secant. The sum of one-sided angles 1 and 2 is 180 degrees. Pay attention to the angles 1 and 7. They are adjacent. I.e:

$$<1+<7=180$$

$$<1+<2=180$$

Submount from the first expression second:

$$(<1+<7)-(<1+<2)=180-180$$

$$(<1+<7)-(<1+<2)=0$$

$$<1+<7-<1-<2=0$$

$$<7-<2=0$$

$<7=<2$ - а они являются соответственными. Значит, прямые параллельны.

What did we know?

We disassembled in details, what angles are obtained in the dissection of parallel direct third line, allocated and detail the proof of three signs of the parallelism of the straight lines.

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Definition 1.

Direct $ with $ called sale For direct $ a $ and $ b $, if it crosses them in two points.

Consider two straight $ a $ and $ b $ and the requested direct $ with $.

When they are intersection there are angles that designate numbers from $ 1 $ to $ 8 $.

Each of these corners have a name that often has to be used in mathematics:

couples Corners $ 3 $ and $ 5 $, $ 4 $ and $ 6 $ called living Lying;
couples corners $ 1 $ and $ 5 $, $ 4 $ and $ 8 $, $ 2 $ and $ 6 $, $ 3 $ and $ 7 $ called respective;
couples corners $ 4 $ and $ 5 $, $ 5 $ and $ 6 $ called one-sided.

Signs of parallelism direct

Theorem 1.

Equality of a pair of passage of lying angles for direct $ a $ and $ b $ and the section $ C $ and $ says that direct $ a $ and $ b $ - parallel:

Evidence.

Let the underlying angles for direct $ a $ and $ b $ and the section $ C $ are equal to: $ ∠1 \u003d ∠2 $.

We show that $ A \\ Parallel B $.

Provided that the corners of $ 1 $ and $ 2 will be straight, we obtain that direct $ a $ and $ b $ will be perpendicular to the direct $ AV $, and therefore parallel.

Provided that the corners of $ 1 $ and $ 2 $ are not direct, carry out from the point $ about $ - the middle of the segment $ Av $, perpendicular $ it $ to direct $ a $.

On a direct $ b $ to postpone the segment $ BH_1 \u003d ah $ and spend the segment $ oh_1 $. We receive two equal triangles $ one and $ oh_1 in two sides and the corner between them ($ ∠1 \u003d ∠2 $, $ \u003d ao $, $ bh_1 \u003d ah $), therefore $ ∠3 \u003d ∠4 $ and $ ∠5 \u003d ∠6 $. Because $ ∠3 \u003d ∠4 $, then the $ H_1 $ point lies on a $-$ beam, thus the $ H $, $ and $ H_1 $ point belongs to one straight line. Because $ ∠5 \u003d ∠6 $, then $ ∠6 \u003d 90 ^ (\\ CIRC) $. Thus, straight $ a $ and $ b $ are perpendicular to direct direct $ HH_1 $ are parallel. Theorem is proved.

Theorem 2.

Equality of the pair of respective angles for direct $ a $ and $ b $ and the section $ with $ indicates that direct $ a $ and $ b $ - parallel:

if $ ∠1 \u003d ∠2 $, then $ A \\ Parallel B $.

Evidence.

Let the corresponding angles for direct $ a $ and $ b $ and the section $ C $ are equal to: $ ∠1 \u003d ∠2 $. Corners $ 2 $ and $ 3 $ are vertical, therefore $ ∠2 \u003d ∠3 $. So $ ∠1 \u003d ∠3 $. Because The corners of $ 1 $ and $ 3 $ - the liar, then direct $ a $ and $ b $ are parallel. Theorem is proved.

Theorem 3.

If the sum of two one-sided corners for direct $ a $ and $ b $ and the section $ C $ is $ 180 ^ (\\ CIRC) C $, then direct $ a $ and $ b $ - parallel:

if $ ∠1 + ∠4 \u003d 180 ^ (\\ CIRC) $, then $ A \\ Parallel B $.

Evidence.

Let one-sided corners for direct $ a $ and $ b $ and the section $ C $ in the amount give $ 180 ^ (\\ CIRC) $, for example

$ ∠1 + ∠4 \u003d 180 ^ (\\ CIRC) $.

COALS $ 3 $ and $ 4 $ are adjacent, so

$ ∠3 + ∠4 \u003d 180 ^ (\\ CIRC) $.

It can be seen from the equalities obtained that the underlying corners of $ ∠1 \u003d ∠3 $, from which it follows that direct $ a $ and $ b $ are parallel.

Theorem is proved.

From the considered features, the parallelism of direct.

Examples of solving problems

Example 1.

The intersection point divides the segment $ AB $ and $ CD $ in half. Prove that $ AC \\ Parallel BD $.

Dano: $ AO \u003d OB $, $ CO \u003d OD $.

Prove: $ AC \\ Parallel BD $.

Evidence.

From the condition of the problem $ AO \u003d OB $, $ Co \u003d OD $ and the equality of vertical angles $ ∠1 \u003d ∠2 $ according to the i-MU, the sign of the equality of triangles follows that $ \\ BigtriangleUP COA \u003d \\ Bigtriangle Dob $. Thus, $ ∠3 \u003d ∠4 $.

COGNES $ 3 $ and $ 4 $ - the lies in two direct $ AC $ and $ BD $ and the $ AB $. Then according to the i-th, a sign of direct $ AC \\ Parallel BD $. The statement is proved.

Example 2.

Dan an angle $ ∠2 \u003d 45 ^ (\\ CIRC) $, and $ ∠ $ $ $ 3 $ 3 times more than this angle. Prove that $ A \\ Parallel B $.

Dano: $ ∠2 \u003d 45 ^ (\\ CIRC) $, $ ∠7 \u003d 3∠2 $.

Prove: $ A \\ Parallel B $.

Evidence:

Find an angle value of $ 7 $:

$ ∠7 \u003d 3 \\ CDOT 45 ^ (\\ CIRC) \u003d 135 ^ (\\ CIRC) $.

Vertical angles $ ∠5 \u003d ∠7 \u003d 135 ^ (\\ CIRC) $, $ ∠2 \u003d ∠4 \u003d 45 ^ (\\ CIRC) $.
We find the amount of inner angles $ ∠5 + ∠4 \u003d 135 ^ (\\ CIRC) +45 ^ (\\ CIRC) \u003d 180 ^ (\\ CIRC) $.

According to the III, a sign of directly direct $ A \\ Parallel B $. The statement is proved.

Example 3.

Dano: $ \\ BigTRIANGEUP ABC \u003d \\ BIGTRIANGEUP ADB $.

Prove: $ AC \\ Parallel BD $, $ AD \\ Parallel BC $.

Evidence:

The pictures under consideration have the side $ AB $ - common.

Because $ ABC $ and $ adb $ are equal, then $ ad \u003d Cb $, $ AC \u003d BD $, as well as the corresponding angles are equal to $ ∠1 \u003d ∠2 $, $ ∠3 \u003d ∠4 $, $ ∠5 \u003d ∠6 $.

The pair of corners of $ 3 $ and $ 4 $ - the lies lying for direct $ speakers $ and $ BD $ and the corresponding section of $ AU $, therefore, according to the i-MU, a sign of direct $ AC \\ Parallel BD $.

A corners of the corners of $ 5 and $ 6 $ - the crosslies lying for direct $ ad $ and $ BC $ and the corresponding section of $ AB $, therefore, according to the i-th, a sign of direct $ AD \\ Parallel BC $.

This article is about parallel direct and on direct parallelism. At first, the definition of parallel direct on the plane and in space, the notation was introduced, examples and graphic illustrations of parallel straight lines are given. Next disassemble the signs and conditions of the parallelism of direct. The conclusion shows the solutions of the characteristic tasks on the proof of the parallelism of direct, which are given by some equations of the straight line in the rectangular coordinate system on the plane and in three-dimensional space.

Navigating page.

Parallel direct - basic information.

Definition.

Two straight planes are called parallelIf they do not have common points.

Definition.

Two straight in three-dimensional space are called parallelIf they lie in the same plane and do not have common points.

Please note that the reservation "If they lie in the same plane" in the definition of parallel direct in space is very important. Let us explain this moment: two straight in three-dimensional space that do not have common points and do not lie in the same plane are not parallel, but are crossing.

Here are some examples of parallel straight lines. The opposite edges of the notebook leaf lie on parallel straight lines. Direct, by which the plane of the wall of the house crosses the plane of the ceiling and floor, are parallel. Railway rails on a flat terrain can also be viewed as parallel straight.

To indicate parallel direct use the symbol "". That is, if direct a and b are parallel, then you can briefly record a b.

Please note: if direct a and b are parallel, then we can say that direct A is parallel to the straight line B, and also that straight b is parallel to direct a.

Let's voice the statement that plays an important role in the study of parallel straight lines on the plane: through a point that does not lie on this direct, the only straight line, parallel to this. This statement is adopted as a fact (it cannot be proven on the basis of the planimetry by the axis), and it is called axiom of parallel straight lines.

For the case in space, the theorem is valid: through any point of space that does not lie on a given straight line, the only straight line, parallel to this. This theorem is easily proved with the help of the above axiom parallel direct (its proof you can find in the geometry textbook 10-11, which is specified at the end of the article on the literature).

Parallelism of direct - signs and conditions of parallelism.

A sign of direct parallelism It is a sufficient condition for parallelism of direct, that is, such a condition, the execution of which guarantees the parallelity of direct. In other words, the implementation of this condition is enough to state the fact of parallelism of direct.

There are also necessary and sufficient conditions for parallelism of direct on the plane and in three-dimensional space.

Let us explain the meaning of the phrase "necessary and sufficient condition of parallelism of direct".

With a sufficient condition of parallelism, we have already figured out. And what is the "necessary condition of parallelism of direct"? By the name "necessary" it is clear that the execution of this condition is necessary for the parallelism of direct. In other words, if the required condition of direct parallelism is not fulfilled, then the straight lines are not parallel. In this way, required and sufficient condition parallelism - This condition, the execution of which is both necessary and enough for the parallelism of the straight lines. That is, on the one hand, this is a sign of direct parallelism, and on the other hand, this is a property that has parallel straight.

Before formulating the necessary and sufficient condition for parallelism of direct, it is advisable to remind several auxiliary definitions.

Singing straight - This is a straight line that crosses each of the two defined uncompanying straight lines.

When crossing two direct secant, eight non-verminated. In the wording of the necessary and sufficient condition of the parallelism of direct participate so-called led, respectively and one-sided corners. Show them in the drawing.

Theorem.

If two direct on the plane are crossed by the unit, then for their parallel, it is necessary and enough so that the underlying angles are equal to, or the corresponding angles were equal, or the sum of one-sided corners was 180 degrees.

We show the graphic illustration of this necessary and sufficient condition of parallelism directly on the plane.

The evidence of these conditions of parallelism direct can be found in geometry textbooks for 7 -9 classes.

Note that these conditions can also be used in three-dimensional space - the main thing is that two straight and secant lay in the same plane.

We give a few theorems that are often used in the proof of parallelism of direct.

Theorem.

If two straight line on the plane are parallel to the third straight, then they are parallel. The proof of this feature follows from the axiom of parallel direct.

There is a similar condition for parallelism of direct in three-dimensional space.

Theorem.

If the two straight in the space are parallel to the third straight, then they are parallel. The proof of this feature is considered in the lessons of geometry in the 10th grade.

We illustrate the voiced theorems.

We give another theorem that allows you to prove the parallelism of direct on the plane.

Theorem.

If two straight ones are perpendicular to the third straight, then they are parallel.

There is a similar theorem for direct in space.

Theorem.

If two direct in three-dimensional space are perpendicular to one plane, then they are parallel.

I will depict the drawings corresponding to these theorems.

All the theorems formulated above, signs and necessary and sufficient conditions are perfectly suitable for evidence of parallelism of direct geometry methods. That is, to prove the parallelism of the two specified directs to show that they are parallel to the third straight, or to show the equality of the passage of lying angles, etc. Many such tasks are solved in the lessons of geometry in high school. However, it should be noted that in many cases it is convenient to use the coordinate method to proof the parallelism of direct on the plane or in three-dimensional space. We formulate the necessary and sufficient conditions for the parallelism of direct, which are specified in the rectangular coordinate system.

Parallelism of direct in the rectangular coordinate system.

In this paragraph of the article we will formulate required and sufficient conditions of direct parallelism In a rectangular coordinate system, depending on the type of equations that determine these direct, and also give detailed solutions to the characteristic tasks.

Let's start with the condition of the parallelism of two direct on the plane in the rectangular Oxy coordinate system. The basis of its proof is the definition of the guide vector direct and the definition of the normal vectors on the plane.

Theorem.

For the parallelity of the two inconsistent straight lines on the plane, it is necessary and enough that the guide vectors of these lines were collinear, or the normal vectors of these straight lines were collinear, or the director of one straight was perpendicular to the normal vector of the second direct.

Obviously, the condition of parallelism of two straight lines on the plane is reduced to (guide vectors of direct or normal vectors of straight lines) or K (guide vector of one straight and normal vector second line). Thus, if both - direct vectors of direct a and b, and and - Normal vectors of straight lines a and b, respectively, then the necessary and sufficient condition of parallelism of direct a and b will be recorded as , or , or, where T is some valid number. In turn, the coordinates of the guide and (or) normal vectors of straight lines A and B are located according to the well-known equations of direct.

In particular, if direct A in the rectangular system of Oxy coordinates on the plane sets the general equation of direct type , and straight b - , The normal vectors of these directs have coordinates and, accordingly,, and the condition of parallelism of direct a and b will be recorded as.

If the direct A corresponds to the equation of a straight line with an angular coefficient of the species, and direct b -, then the normal vectors of these directs have coordinates and, and the condition of the parallelism of these direct will take the form . Therefore, if direct on the plane in the rectangular coordinate system is parallel and can be set by equations of direct with angular coefficients, the corner coefficients will be equal. And back: if the coordinate-straight lines on the plane in the rectangular coordinate system can be given by the equations of direct with equal angular coefficients, then such direct are parallel.

If direct a and straight b in a rectangular coordinate system define canonical equations direct on the plane of the species and , or parametric equations direct on the plane of the species and Accordingly, the guide vectors of these directs have coordinates and, and the condition of parallelism of direct a and b is recorded as.

We will analyze the solutions of several examples.

Example.

Whether straight lines are parallel and?

Decision.

I rewrite the equation is straight in segments in the form of a general direct equation: . Now it can be seen that - normal vector straight , and - normal vector straight. These vectors are not collinear, since there is no such valid number T for which the equality is true ( ). Therefore, the necessary and sufficient condition of parallelism of direct on the plane is not performed, therefore, the specified straight lines are not parallel.

Answer:

No, straight is not parallel.

Example.

Are the straight and parallel?

Decision.

We give the canonical equation direct to the equation direct with the angular coefficient :. It is obvious that the equations of direct and not the same (in this case, the specified straight lines would be coinciding) and the angular coefficients of the direct are equal, therefore, the initial straight parallels.

The second way to solve.

First, we show that the source straight lines do not match: take any point direct, for example, (0, 1), the coordinates of this point do not satisfy the equation direct, therefore, the straight lines do not coincide. Now check the fulfillment of the condition of the parallelism of these direct. Normal vector line eating vector, and direct vector direct eating vector. Calculate and: . Consequently, vectors and perpendicular, it means that the necessary and sufficient condition for the parallelism of the specified direct are performed. Thus, straight parallel.

Answer:

The specified straight parallels.

To prove the parallelism of direct in the rectangular coordinate system in three-dimensional space, use the following necessary and sufficient condition.

Theorem.

For the parallelism of the incomprehensive straight lines in the three-dimensional space, it is necessary and enough for their guide vectors to be collinear.

Thus, if the equations of direct in the rectangular coordinate system in three-dimensional space are known and you need to answer the question parallel to these direct or not, then you need to find the coordinates of the guide vectors of these direct and verify the condition of the collinearity of the guide vectors. In other words, if and - Direct vectors a specified directs have coordinates and. As then. Thus, a necessary and sufficient condition for the parallelism of two direct in space is performed. This proven parallelism of direct and .

Bibliography.

Atanasyan L.S., Butuzov V.F., Kadomtsev S.B., Poznyak E.G., Yudina I.I. Geometry. 7 - 9 Classes: Textbook for general education institutions.
Atanasyan L.S., Buduzov V.F., Kadomtsev S.B., Kiseleva L.S., Poznyak E.G. Geometry. Tutorial for the 10-11 high school classes.
Pogorelov A.V., Geometry. Tutorial for 7-11 classes of general education institutions.
Bugrov Ya.S., Nikolsky S.M. Higher mathematics. Volume One: elements of linear algebra and analytical geometry.
Ilyin V.A., Poznyak E.G. Analytic geometry.

In this article we will tell about parallel direct, we will give the definitions, we denote the signs and conditions of parallelism. For clarity of theoretical material, we will use illustrations and solutions of typical examples.

Definition 1.

Parallel straight on the plane - Two straight ones on planes that do not have common points.

Definition 2.

Parallel straight in three-dimensional space - Two straight in three-dimensional space lying in the same plane and do not have common points.

It is necessary to note that to determine the parallel direct in space, the clarification "lying in the same plane" is extremely important: two straight lines in three-dimensional space that do not have common points and not lying in the same plane are not parallel, but crossing.

To designate parallelism of direct, generally accepted using the symbol ∥. Those., If the specified straight lines a and b are parallel, briefly write this condition like this: a ‖ b. Wonderfully parallelism of direct is indicated as follows: straight a and b are parallel, or straight and parallel to direct b, or straight b parallel to direct a.

We formulate an approval that plays an important role in the topic studied.

Axiom

Through a point that does not belong to the specified direct, the only straight line, parallel to the specified one. This statement is impossible to prove on the basis of famous planimeters.

In the case when it comes to space, the theorem is true:

Theorem 1.

Through any point of space that does not belong to the specified direct direct, the only straight line, parallel specified.

This theorem is simply proved on the basis of the above axiom (program of geometry 10 - 11 classes).

A sign of parallelism is a sufficient condition, when executing which the parallelism of the straight lines is guaranteed. In other words, the fulfillment of this condition is enough to confirm the fact of parallelism.

Including the necessary and sufficient conditions for the parallelism of direct on the plane and in space. Let us explain: necessary - it means that the condition, the execution of which is necessary for the parallelity of the straight lines; If it is not executed - direct are not parallel.

Summarizing, the necessary and sufficient condition for the parallelism of direct - such a condition, the observance of which is necessary and enough for the straight lines are parallel to each other. On the one hand, this is a sign of parallelism, on the other, the property inherent in parallel direct.

Before you give an accurate wording of the necessary and sufficient condition, we recall some more additional concepts.

Definition 3.

Singing straight - Direct, intersecting each of the two specified uncompanying straight lines.

Crossing two straight, the sequer forms eight uniform angles. In order to formulate the necessary and sufficient condition, we will use such types of corners as the liar, respective and unilateral. We will demonstrate them in the illustration:

Theorem 2.

If the two direct on the plane intersects the unit, then for the parallelism of the specified direct, it is necessary and sufficient so that the underlying angles can be equal or equal to the corresponding angles, or the sum of one-sided corners was 180 degrees.

We illustrate the graphically necessary and sufficient condition for parallelism directly on the plane:

The proof of these conditions is present in the geometry program for 7 - 9 classes.

In general, these conditions are applicable for three-dimensional space, despite the fact that the two straight and secant belong to the same plane.

We indicate a few more theorems often used in the proof of the fact of parallelism of direct.

Theorem 3.

On the plane two straight, parallel to the third, parallel between themselves. This feature is proved on the basis of the axiom of the parallelism indicated above.

Theorem 4.

In three-dimensional space, two straight, parallel to the third, parallel between themselves.

The proof of the attribute is studied in the Grade 10 geometry program.

Let's take an illustration of the following theorems:

We indicate another pair of theorems that are proof of the parallelism of direct.

Theorem 5.

On the plane, two straight, perpendicular to the third, parallel between themselves.

We formulate a similar one for three-dimensional space.

Theorem 6.

In three-dimensional space, two straight, perpendicular to the third, parallel between themselves.

We illustrate:

All the above theorems, features and conditions allow us to conveniently prove the parallelism of direct geometry methods. Those, to bring the proof of the parallelism of direct, it can be shown that the corresponding angles are equal, or demonstrate the fact that the two specified direct perpendicular to the third, etc. But we note that it is often for the evidence of parallelism of direct on the plane or in three-dimensional space it is more convenient to use the coordinate method.

Parallelism of direct in the rectangular coordinate system

In a given rectangular coordinate system, the straight line is determined by the direct equation on the plane of one of the possible species. So the straight line specified in the rectangular coordinate system in three-dimensional space corresponds to some equations direct in space.

We write the necessary and sufficient conditions for the parallelism of direct in the rectangular coordinate system depending on the type of equation describing the specified direct.

Let's start with the condition of parallelism of direct on the plane. It is based on the definitions of the guide vector of the straight and normal vector line on the plane.

Theorem 7.

In order for two inconsistent plane, the straight lines were parallel, it is necessary and enough that the vectors of the specified direct vectors were collinear, or were the collinear normal vectors of the specified direct, or the director of one straight line was perpendicular to the normal vector of another direct.

It becomes obvious that the condition of parallelism of direct on the plane is based on the condition of the collinearity of vectors or the condition of the perpendicularity of two vectors. Those., If A → \u003d (a x, a y) and b → \u003d (b x, b y) are guide vectors of direct a and b;

and NB → \u003d (NBX, NBY) are normal vectors of direct a and b, the above-mentioned needed and sufficient condition will write this: a → \u003d t · b → ⇔ ax \u003d t · bxay \u003d t · by or na → \u003d t · NB → ⇔ NAX \u003d T · NBXnay \u003d T · NBY or A →, NB → \u003d 0 ⇔ AX · NBX + AY · NBY \u003d 0, where T is some valid number. The coordinates of the guide or direct vectors are determined by the specified direct equations. Consider the main examples.

Direct A in the rectangular coordinate system is determined by the general equation direct: a 1 x + b 1 y + C 1 \u003d 0; Direct B - A 2 X + B 2 Y + C 2 \u003d 0. Then the normal vectors of the specified direct will have coordinates (A 1, in 1) and (A 2, B 2), respectively. The condition of parallelism will write down this:

A 1 \u003d t · a 2 b 1 \u003d t · b 2

Direct A is described by the straight equation with an angular coefficient of the form y \u003d k 1 x + b 1. Straight b - y \u003d k 2 x + b 2. Then the normal vectors of the specified direct will have coordinates (K 1, - 1) and (k 2, - 1), respectively, and the condition of parallelism will write down this:

k 1 \u003d t · k 2 - 1 \u003d t · (- 1) ⇔ k 1 \u003d t · k 2 t \u003d 1 ⇔ k 1 \u003d k 2

Thus, if parallel straight lines on the plane in a rectangular coordinate system are set by equations with angular coefficients, the angular coefficients of the specified direct will be equal. And the reverse statement is true: if the coordinate-straight lines on the plane in the rectangular coordinate system are determined by the direct equations with the same angular coefficients, then these specified direct are parallel.

Straight A and B in a rectangular coordinate system are set by canonical equations direct on the plane: x - x 1 ax \u003d y - y 1 ay and x - x 2 bx \u003d y - y 2 by or parametric equations direct on the plane: x \u003d x 1 + λ · axy \u003d y 1 + λ · ay and x \u003d x 2 + λ · bxy \u003d y 2 + λ · by.

Then the guide vectors of the specified direct will be: a x, a y and b x, b y, respectively, and the condition of parallelism will write this:

a x \u003d t · b x a y \u003d t · b y

We will analyze examples.

Example 1.

Two straight lines are given: 2 x - 3 y + 1 \u003d 0 and x 1 2 + y 5 \u003d 1. It is necessary to determine whether they are parallel.

Decision

We write the equation direct in segments in the form of a general equation:

x 1 2 + y 5 \u003d 1 ⇔ 2 x + 1 5 y - 1 \u003d 0

We see that n a → \u003d (2, - 3) is the normal vector of the straight line 2 x - 3 y + 1 \u003d 0, and n b → \u003d 2, 1 5 - normal vector straight x 1 2 + y 5 \u003d 1.

The resulting vectors are not collinear, because There is no such value t, in which equality will be true:

2 \u003d T · 2 - 3 \u003d T · 1 5 ⇔ T \u003d 1 - 3 \u003d T · 1 5 ⇔ T \u003d 1 - 3 \u003d 1 5

Thus, the necessary and sufficient condition of parallelism of direct on the plane is not performed, which means that the specified straight lines are not parallel.

Answer: The specified straight lines are not parallel.

Example 2.

The straight lines are given y \u003d 2 x + 1 and x 1 \u003d y - 4 2. Are they parallel?

Decision

We transform the canonical equation direct x 1 \u003d y - 4 2 to the equation direct with the angular coefficient:

x 1 \u003d y - 4 2 ⇔ 1 · (y - 4) \u003d 2 x ⇔ y \u003d 2 x + 4

We see that the equations of direct y \u003d 2 x + 1 and y \u003d 2 x + 4 are not the same (if it were different, the straight lines would be coinciding) and the angular coefficients of the lines are equal, which means the specified straight lines are parallel.

Let's try to solve the task otherwise. First check whether the specified straight lines coincide. We use any point direct y \u003d 2 x + 1, for example, (0, 1), the coordinates of this point do not correspond to the equation direct x 1 \u003d y - 4 2, which means that the direct does not match.

The next step is to determine the fulfillment of the condition of the parallelism of the specified direct.

Normal vector straight y \u003d 2 x + 1 is a vector n a → \u003d (2, - 1), and the vector guide of the second specified direct is B → \u003d (1, 2). The scalar product of these vectors is zero:

n A →, B → \u003d 2 · 1 + (- 1) · 2 \u003d 0

Thus, vectors are perpendicular: it demonstrates us to perform the necessary and sufficient condition for the parallelism of the source direct. Those. The specified straight parallels.

Answer: Direct data are parallel.

To prove the parallel nature of direct in the rectangular coordinate system of three-dimensional space, the following necessary and sufficient condition is used.

Theorem 8.

In order for two inconsistent straight lines in the three-dimensional space, it is necessary and enough for the director vectors of these directs to be collinear.

Those. For given equations of direct in three-dimensional space, the answer to the question: they are parallel or not, it is located with the help of determining the coordinates of the guide vectors of the specified direct, as well as testing the conditions of their collinearity. In other words, if A → \u003d (AX, AY, AZ) and B → \u003d (BX, BZ) are guide vectors of direct a and b, respectively, then so that they are parallel, the existence of such a valid number T, To make equality:

a → \u003d t · b → ⇔ a x \u003d t · b x a y \u003d t · b y a z \u003d t · b z

Example 3.

Are the straight lines x 1 \u003d y - 2 0 \u003d z + 1 - 3 and x \u003d 2 + 2 λ y \u003d 1 z \u003d - 3 - 6 λ. It is necessary to prove the parallelism of these direct.

Decision

The conditions of the problem are set by canonical equations of one direct in space and parametric equations of another direct in space. Guide vectors A → I. B → the specified directs have coordinates: (1, 0, - 3) and (2, 0, - 6).

1 \u003d T · 2 0 \u003d T · 0 - 3 \u003d T · - 6 ⇔ T \u003d 1 2, then A → \u003d 1 2 · B →.

Consequently, the necessary and sufficient condition of the parallelism of direct in space is made.

Answer: The parallelism of the specified direct is proved.

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