The distance between parallel lines is equal. Mutual arrangement of two straight lines

Proof.

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

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

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

That is, for any value C2 distance from the point to straight b can be calculated using the formula. And given the 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 proved.

2. Solving problems on finding the distance between parallel lines

Example #1.

Find the distance between parallel lines and Decision.

We obtain the general equations of given parallel lines.

For straight font-size:12.0pt line-height:115%;font-family:Verdana">corresponds to the general equation of a line. Let us pass from the parametric equations of the direct 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">Variable Coefficients x and y in the obtained general equations, parallel lines are equal, so we can immediately apply the formula for calculating the distance between parallel lines in a plane:.

Answer: font-size:12.0pt line-height:115%;font-family:Verdana">Example #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 given parallel lines.

Decision:

The 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 the point M1 lying on this line:font-size:12.0pt line-height:115%;font-family:Verdana">. Distance from this point to the lineequal to the desired distance between parallel lines. The equationis a normal equation of a straight line, therefore, we can immediately calculate the distance from the point to straight font-size:12.0pt;line-height:115%;font-family:Verdana">:.

The 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">. Here is the canonical equation of the lineto the general equation of a straight line:. Variable coefficients x in general equations, the given parallel lines are equal (with a variable y the coefficients are also equal - they are equal to zero), so you can use a formula that allows you to calculate the distance between given parallel lines:.

Answer: 8

3. Homework

Tasks for self-test

1. Find the distance between two parallel lines

4. CONCLUSION

All goals and objectives set have been fully achieved. Two lessons have been developed from the section “Mutual 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 an accessible level for students, which will allow solving problems in geometry with simpler and more beautiful methods.

5. LIST OF LITERATURE

1) , Yudina. Grades 7 - 9: a textbook for educational institutions.

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

3) , Nikolsky Mathematics. Volume One: Elements of Linear Algebra and Analytic Geometry.

4) , Poznyak geometry.

6.APPS

Reference material

General equation of a straight line:

Ah + Wu + C = 0 ,

where BUT and AT not equal to zero at the same time.

Odds BUT and AT are coordinates normal vector straight line (i.e., a vector perpendicular to the straight line). At A = 0 straight line parallel to the axis OH, at B = 0 straight line parallel to the axis O Y .

At AT0 get slope equation :

The equation of a straight line passing through a point ( X 0 , at 0) and not parallel to the axisOY, looks like:

at – at 0 = m (x – X 0) ,

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

At BUT 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. cuts off on the coordinate axes segments of lengtha and b .

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

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

Condition of parallel lines:

1) for straight lines Ax + Vy + C = 0 andDx+Ey+F = 0: AE – BD = 0 ,

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

In the material of this article, we will analyze the question of finding the distance between two parallel lines, in particular, using the coordinate method. Analysis of typical examples will help to consolidate the theoretical knowledge gained.

Yandex.RTB R-A-339285-1 Definition 1

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

Here is an illustration for clarity:

The drawing shows two parallel lines. a and b. Point M 1 belongs to the line a, a perpendicular to the line is dropped from it b. The resulting segment M 1 H 1 is the distance between two parallel lines a and b.

The specified definition of the distance between two parallel lines is valid both on the plane and for lines in three-dimensional space. Moreover, this definition is related to the following theorem.

Theorem

When two lines are parallel, all points of one of them are equidistant from the other line.

Proof

Let us be given two parallel lines a and b. Set on a straight line a points M 1 and M 2, we drop perpendiculars from them to the line b, denoting their bases, respectively, as H 1 and H 2. M 1 H 1 is the distance between two parallel lines by definition, and we need to prove that | M 1 H 1 | = | M 2 H 2 | .

Let there also be some secant that intersects two given parallel lines. The condition of parallelism of lines, considered in the corresponding article, gives us the right to assert that in this case, the internal cross-lying angles formed at the intersection of the secant of the given lines are equal: ∠ M 2 M 1 H 2 = ∠ H 1 H 2 M 1. The line M 2 H 2 is perpendicular to the line b by construction, and, of course, perpendicular to the line a. The resulting triangles M 1 H 1 H 2 and M 2 M 1 H 2 are rectangular and equal to each other in terms of the hypotenuse and the acute angle: M 1 H 2 is the common hypotenuse, ∠ M 2 M 1 H 2 = ∠ H 1 H 2 M 1 . Based on the equality of triangles, we can talk about the equality of their sides, ie: | M 1 H 1 | = | M 2 H 2 | . The theorem has been proven.

Note that the distance between two parallel lines is the smallest of the distances from points on one line to points on the other.

Finding the distance between parallel lines

We have already found out that, in fact, in order to find the distance between two parallel lines, it is necessary to determine the length of the perpendicular dropped from a certain point on one line to another. There are several ways to do this. In some problems it is convenient to use the Pythagorean theorem; others involve the use of signs of equality or similarity of triangles, etc. In cases where lines are given in a rectangular coordinate system, it is possible to calculate the distance between two parallel lines using the coordinate method. Let's consider it in more detail.

Let's set the conditions. Suppose a rectangular coordinate system is fixed, in which two parallel lines a and b are given. It is necessary to determine the distance between the given lines.

We will build the solution of the problem on determining the distance between parallel lines: to find the distance between two given parallel lines, it is necessary:

Find the coordinates of some point M 1 belonging to one of the given lines;

Calculate the distance from point M 1 to a given straight line to which this point does not belong.

Based on the skills of working with the equations of a straight line in a plane or in space, it is easy to determine the coordinates of the point M 1. When finding the distance from point M 1 to a straight line, the material of the article on finding the distance from a point to a straight line is useful.

Let's go back to the example. Let the line a be described by the general equation A x + B y + C 1 = 0 , and the line b be described by the equation A x + B y + C 2 = 0 . Then the distance between two given parallel lines can be calculated using the formula:

M 1 H 1 = C 2 - C 1 A 2 + B 2

Let's derive this formula.

We use some point М 1 (x 1 , y 1) belonging to the line a . In this case, the coordinates of the point M 1 will satisfy the equation A x 1 + B y 1 + C 1 = 0. Thus, the equality is fair: A x 1 + B y 1 + C 1 = 0; from it we get: A x 1 + B y 1 = - C 1 .

When C 2< 0 , нормальное уравнение прямой b будет иметь вид:

A A 2 + B 2 x + B A 2 + B 2 y + C 2 A 2 + B 2 = 0

With C 2 ≥ 0, the normal equation of the straight line b will look like this:

A A 2 + B 2 x + B A 2 + B 2 y - C 2 A 2 + B 2 = 0

And then for cases when C 2< 0 , применима формула: M 1 H 1 = A A 2 + B 2 x 1 + B A 2 + B 2 y 1 + C 2 A 2 + B 2 .

And for C 2 ≥ 0, the desired distance is determined by the formula M 1 H 1 = - A A 2 + B 2 x 1 - B A 2 + B 2 y 1 - C 2 A 2 + B 2 = = A A 2 + B 2 x 1 + B A 2 + B 2 y 1 + C 2 A 2 + B 2

Thus, for any value of the number C 2, the length of the segment | M 1 H 1 | (from point M 1 to line b) is calculated by the formula: M 1 H 1 \u003d A A 2 + B 2 x 1 + B A 2 + B 2 y 1 + C 2 A 2 + B 2

Above we got: A x 1 + B y 1 \u003d - C 1, then we can transform the formula: M 1 H 1 \u003d - C 1 A 2 + B 2 + C 2 A 2 + B 2 \u003d C 2 - C 1 A 2 +B2. So we, in fact, received the formula specified in the algorithm of the coordinate method.

Let's analyze the theory with examples.

Example 1

Given two parallel lines y = 2 3 x - 1 and x = 4 + 3 · λ y = - 5 + 2 · λ . It is necessary to determine the distance between them.

Decision

The initial parametric equations make it possible to set the coordinates of the point through which the straight line passes, described by the parametric equations. Thus, we get the point M 1 (4, - 5) . The required distance is the distance between the point M 1 (4, - 5) to the straight line y = 2 3 x - 1, let's calculate it.

The given equation of a straight line with slope y = 2 3 x - 1 is converted into a normal equation of a straight line. To this end, we first make the transition to the general equation of a straight line:

y = 2 3 x - 1 ⇔ 2 3 x - y - 1 = 0 ⇔ 2 x - 3 y - 3 = 0

Let's calculate the normalizing factor: 1 2 2 + (- 3) 2 = 1 13 . We multiply both parts of the last equation by it and, finally, we get the opportunity to write the normal equation of the straight line: 1 13 2 x - 3 y - 3 = 1 13 0 ⇔ 2 13 x - 3 13 y - 3 13 = 0.

For x = 4, and y = - 5, we calculate the desired distance as the modulus of the value of the extreme equality:

2 13 4 - 3 13 - 5 - 3 13 = 20 13

Answer: 20 13 .

Example 2

In a fixed rectangular coordinate system O x y, two parallel lines are given, defined by the equations x - 3 = 0 and x + 5 0 = y - 1 1 . It is necessary to find the distance between the given parallel lines.

Decision

The conditions of the problem define one general equation, given by one of the original lines: x-3=0. Let's transform the original canonical equation into a general one: x + 5 0 = y - 1 1 ⇔ x + 5 = 0 . For the variable x, the coefficients in both equations are equal (also equal for y - zero), and therefore we have the opportunity to apply the formula for finding the distance between parallel lines:

M 1 H 1 \u003d C 2 - C 1 A 2 + B 2 \u003d 5 - (- 3) 1 2 + 0 2 \u003d 8

Answer: 8 .

Finally, consider the problem of finding the distance between two parallel lines in three-dimensional space.

Example 3

In a rectangular coordinate system O x y z, two parallel lines are given, described by the canonical equations of a straight line in space: x - 3 1 = y - 1 = z + 2 4 and x + 5 1 = y - 1 - 1 = z - 2 4 . Find the distance between these lines.

Decision

From the equation x - 3 1 \u003d y - 1 \u003d z + 2 4, the coordinates of the point through which the straight line passes, described by this equation, can be easily determined: M 1 (3, 0, - 2) . Let's calculate the distance | M 1 H 1 | from point M 1 to line x + 5 1 = y - 1 - 1 = z - 2 4 .

The straight line x + 5 1 \u003d y - 1 - 1 \u003d z - 2 4 passes through the point M 2 (- 5, 1, 2). We write the direction vector of the straight line x + 5 1 = y - 1 - 1 = z - 2 4 as b → with coordinates (1 , - 1 , 4) . Let us determine the coordinates of the vector M 2 M → :

M 2 M 1 → = 3 - (- 5 , 0 - 1 , - 2 - 2) ⇔ M 2 M 1 → = 8 , - 1 , - 4

Let's calculate the cross product of vectors:

b → × M 2 M 1 → = i → j → k → 1 - 1 4 8 - 1 - 4 = 8 i → + 36 j → + 7 k → ⇒ b → × M 2 M 1 → = ( 8, 36, 7)

Let's apply the formula for calculating the distance from a point to a straight line in space:

M 1 H 1 = b → × M 2 M 1 → b → = 8 2 + 36 2 + 7 2 1 2 + (- 1) 2 + 4 2 = 1409 3 2

Answer: 1409 3 2 .

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

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 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, draw a diagonal AC and get two triangles ABC and ADC (Fig. 2).

These triangles are equal, since ∠ 1 = ∠ 4, ∠ 2 = ∠ 3 (cross-lying angles at parallel lines), and side AC is common. From the equality Δ ABC = Δ ADC it follows that AB \u003d CD, BC \u003d AD, ∠ B \u003d ∠ D. The sum of the angles adjacent to one side, for example, angles A and D, is equal to 180 ° as one-sided with parallel lines. The theorem has been proven.

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

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

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

Let us draw from some two points B and C of the line b the perpendiculars BA and CD to the line a. Since AB || CD, then the 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.

By what has been proved, 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 longer than the other. Find the sides of the parallelogram.

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

Example 2

Decision. Let figure 4 correspond to the condition of the problem.

Denote AB by x and BC by y. By condition, the perimeter of the parallelogram is 10 cm, i.e. 2(x + y) = 10, or x + y = 5. The perimeter of the 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.

Decision. Let figure 5 correspond to the condition of the problem.

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

Angles BAD and ADC are internal one-sided 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 long side of the parallelogram into?

Decision. Let figure 6 correspond to the condition of the problem.

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

In this article, using the example of solving problem C2 from the Unified State Examination, the method of finding coordinates using the method is analyzed. Recall that lines are skew if they do not lie in the same plane. In particular, if one line lies in a plane, and the second line intersects this plane at a point that does not lie on the first line, then such lines are skew (see figure).

For finding distances between intersecting lines necessary:

1. Draw a plane through one of the skew lines that is parallel to the other skew line.
2. Drop a perpendicular from any point of the second straight line to the resulting plane. The length of this perpendicular will be the desired distance between the lines.

Let us analyze this algorithm in more detail using the example of solving problem C2 from the Unified State Examination in mathematics.

Distance between lines in space

Task. in a single cube ABCDA 1 B 1 C 1 D 1 find the distance between the lines BA 1 and D.B. 1 .

Rice. 1. Drawing for the task

Decision. Through the midpoint of the diagonal of the cube D.B. 1 (dot O) draw a line parallel to the line A 1 B. Points of intersection of a given line with edges BC and A 1 D 1 denote respectively N and M. Straight MN lies in the plane MNB 1 and parallel to the line A 1 B, which does not lie in this plane. This means that the direct A 1 B parallel to the plane MNB 1 on the basis of parallelism of a straight line and a plane (Fig. 2).

Rice. 2. The desired distance between the crossing lines is equal to the distance from any point of the selected line to the depicted plane

We are now looking for the distance from some point on the straight line A 1 B up to the plane MNB one . This distance, by definition, will be the desired distance between the skew lines.

To find this distance, we use the coordinate method. We introduce a rectangular Cartesian coordinate system so that its origin coincides with point B, the axis X was directed along the edge BA, axis Y- along the rib BC, axis Z- along the rib BB 1 (Fig. 3).

Rice. 3. We choose a rectangular Cartesian coordinate system as shown in the figure

We find the equation of the plane MNB 1 in this coordinate system. To do this, we first determine the coordinates of the points M, N and B 1: We substitute the obtained coordinates into the general equation of a straight line and obtain the following system of equations:

From the second equation of the system, we obtain from the third one, and then from the first we obtain. We substitute the obtained values ​​into the general equation of the straight line:

Note that otherwise the plane MNB 1 would pass through the origin. We divide both sides of this equation by and we get:

The distance from a point to a plane is determined by the formula.

Oh-oh-oh-oh-oh ... well, it's tinny, as if you read the sentence to yourself =) However, then relaxation will help, especially since today I bought suitable accessories. Therefore, let's proceed to the first section, I hope, by the end of the article I will keep a cheerful mood.

Mutual arrangement of two straight lines

The case when the hall sings along in chorus. Two lines can:

1) match;

2) be parallel: ;

3) or intersect at a single point: .

Help for dummies : please remember the mathematical sign of the intersection , it will occur very often. The entry means that the line intersects with the line at the point.

How to determine the relative position of two lines?

Let's start with the first case:

Two lines coincide if and only if their respective coefficients are proportional, that is, there is such a number "lambda" that the equalities

Let's consider straight lines and compose three equations from the corresponding coefficients: . From each equation it follows that, therefore, these lines coincide.

Indeed, if all the coefficients of the equation multiply by -1 (change signs), and all the coefficients of the equation reduce by 2, you get the same equation: .

The second case when the lines are parallel:

Two lines are parallel if and only if their coefficients at the variables are proportional: , but.

As an example, consider two straight lines. We check the proportionality of the corresponding coefficients for the variables :

However, it is clear that .

And the third case, when the lines intersect:

Two lines intersect if and only if their coefficients of the variables are NOT proportional, that is, there is NOT such a value of "lambda" that the equalities are fulfilled

So, for straight lines we will compose a system:

From the first equation it follows that , and from the second equation: , hence, the system is inconsistent(no solutions). Thus, the coefficients at the variables are not proportional.

Conclusion: lines intersect

In practical problems, the solution scheme just considered can be used. By the way, it is very similar to the algorithm for checking vectors for collinearity, which we considered in the lesson. The concept of linear (non) dependence of vectors. Vector basis. But there is a more civilized package:

Example 1

Find out the relative position of the lines:

Decision based on the study of directing vectors of straight lines:

a) From the equations we find the direction vectors of the lines: .

, so the vectors are not collinear and the lines intersect.

Just in case, I will put a stone with pointers at the crossroads:

The rest jump over the stone and follow on, straight to Kashchei the Deathless =)

b) Find the direction vectors of the lines:

The lines have the same direction vector, which means they are either parallel or the same. Here the determinant is not necessary.

Obviously, the coefficients of the unknowns are proportional, while .

Let's find out if the equality is true:

Thus,

c) Find the direction vectors of the lines:

Let's calculate the determinant, composed of the coordinates of these vectors:
, therefore, the direction vectors are collinear. The lines are either parallel or coincide.

The proportionality factor "lambda" is easy to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: .

Now let's find out if the equality is true. Both free terms are zero, so:

The resulting value satisfies this equation (any number generally satisfies it).

Thus, the lines coincide.

Answer:

Very soon you will learn (or even have already learned) to solve the considered problem verbally literally in a matter of seconds. In this regard, I see no reason to offer something for an independent solution, it is better to lay one more important brick in the geometric foundation:

How to draw a line parallel to a given one?

For ignorance of this simplest task, the Nightingale the Robber severely punishes.

Example 2

The straight line is given by the equation . Write an equation for a parallel line that passes through the point.

Decision: Denote the unknown line by the letter . What does the condition say about it? The line passes through the point. And if the lines are parallel, then it is obvious that the directing vector of the line "ce" is also suitable for constructing the line "de".

We take out the direction vector from the equation:

Answer:

The geometry of the example looks simple:

Analytical verification consists of the following steps:

1) We check that the lines have the same direction vector (if the equation of the line is not properly simplified, then the vectors will be collinear).

2) Check if the point satisfies the resulting equation.

Analytical verification in most cases is easy to perform orally. Look at the two equations and many of you will quickly figure out how the lines are parallel without any drawing.

Examples for self-solving today will be creative. Because you still have to compete with Baba Yaga, and she, you know, is a lover of all kinds of riddles.

Example 3

Write an equation for a line passing through a point parallel to the line if

There is a rational and not very rational way to solve. The shortest way is at the end of the lesson.

We did a little work with parallel lines and will return to them later. The case of coinciding lines is of little interest, so let's consider a problem that is well known to you from the school curriculum:

How to find the point of intersection of two lines?

If straight intersect at the point , then its coordinates are the solution systems of linear equations

How to find the point of intersection of lines? Solve the system.

Here's to you geometric meaning of a system of two linear equations with two unknowns are two intersecting (most often) straight lines on a plane.

Example 4

Find the point of intersection of lines

Decision: There are two ways to solve - graphical and analytical.

The graphical way is to simply draw the given lines and find out the point of intersection directly from the drawing:

Here is our point: . To check, you should substitute its coordinates into each equation of a straight line, they should fit both there and there. In other words, the coordinates of a point are the solution of the system . In fact, we considered a graphical way to solve systems of linear equations with two equations, two unknowns.

The graphical method, of course, is not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide this way, the point is that it will take time to make a correct and EXACT drawing. In addition, some lines are not so easy to construct, and the intersection point itself may be somewhere in the thirtieth kingdom outside the notebook sheet.

Therefore, it is more expedient to search for the intersection point by the analytical method. Let's solve the system:

To solve the system, the method of termwise addition of equations was used. To develop the relevant skills, visit the lesson How to solve a system of equations?

Answer:

The verification is trivial - the coordinates of the intersection point must satisfy each equation of the system.

Example 5

Find the point of intersection of the lines if they intersect.

This is a do-it-yourself example. It is convenient to divide the problem into several stages. Analysis of the condition suggests that it is necessary:
1) Write the equation of a straight line.
2) Write the equation of a straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.

The development of an action algorithm is typical for many geometric problems, and I will repeatedly focus on this.

Full solution and answer at the end of the tutorial:

A pair of shoes has not yet been worn out, as we got to the second section of the lesson:

Perpendicular lines. The distance from a point to a line.
Angle between lines

Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to the given one, and now the hut on chicken legs will turn 90 degrees:

How to draw a line perpendicular to a given one?

Example 6

The straight line is given by the equation . Write an equation for a perpendicular line passing through a point.

Decision: It is known by assumption that . It would be nice to find the direction vector of the straight line. Since the lines are perpendicular, the trick is simple:

From the equation we “remove” the normal vector: , which will be the directing vector of the straight line.

We compose the equation of a straight line by a point and a directing vector:

Answer:

Let's unfold the geometric sketch:

Hmmm... Orange sky, orange sea, orange camel.

Analytical verification of the solution:

1) Extract the direction vectors from the equations and with the help dot product of vectors we conclude that the lines are indeed perpendicular: .

By the way, you can use normal vectors, it's even easier.

2) Check if the point satisfies the resulting equation .

Verification, again, is easy to perform verbally.

Example 7

Find the point of intersection of perpendicular lines, if the equation is known and dot.

This is a do-it-yourself example. There are several actions in the task, so it is convenient to arrange the solution point by point.

Our exciting journey continues:

Distance from point to line

Before us is a straight strip of the river and our task is to reach it in the shortest way. There are no obstacles, and the most optimal route will be movement along the perpendicular. That is, the distance from a point to a line is the length of the perpendicular segment.

The distance in geometry is traditionally denoted by the Greek letter "ro", for example: - the distance from the point "em" to the straight line "de".

Distance from point to line is expressed by the formula

Example 8

Find the distance from a point to a line

Decision: all you need is to carefully substitute the numbers into the formula and do the calculations:

Answer:

Let's execute the drawing:

The distance found from the point to the line is exactly the length of the red segment. If you make a drawing on checkered paper on a scale of 1 unit. \u003d 1 cm (2 cells), then the distance can be measured with an ordinary ruler.

Consider another task according to the same drawing:

The task is to find the coordinates of the point , which is symmetrical to the point with respect to the line . I propose to perform the actions on your own, however, I will outline the solution algorithm with intermediate results:

1) Find a line that is perpendicular to a line.

2) Find the point of intersection of the lines: .

Both actions are discussed in detail in this lesson.

3) The point is the midpoint of the segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the middle of the segment find .

It will not be superfluous to check that the distance is also equal to 2.2 units.

Difficulties here may arise in calculations, but in the tower a microcalculator helps out a lot, allowing you to count ordinary fractions. Have advised many times and will recommend again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for an independent solution. A little hint: there are infinitely many ways to solve. Debriefing at the end of the lesson, but better try to guess for yourself, I think your ingenuity was well dispersed.

Angle between two lines

Whatever the corner, then the jamb:


In geometry, the angle between two straight lines is taken as the SMALLER angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered to be the angle between intersecting lines. And its “green” neighbor or oppositely oriented crimson corner.

If the lines are perpendicular, then any of the 4 angles can be taken as the angle between them.

How are the angles different? Orientation. First, the direction of "scrolling" the corner is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example, if .

Why did I say this? It seems that you can get by with the usual concept of an angle. The fact is that in the formulas by which we will find the angles, a negative result can easily be obtained, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing for a negative angle, it is imperative to indicate its orientation (clockwise) with an arrow.

How to find the angle between two lines? There are two working formulas:

Example 10

Find the angle between lines

Decision and Method one

Consider two straight lines given by equations in general view:

If straight not perpendicular, then oriented the angle between them can be calculated using the formula:

Let's pay close attention to the denominator - this is exactly scalar product direction vectors of straight lines:

If , then the denominator of the formula vanishes, and the vectors will be orthogonal and the lines will be perpendicular. That is why a reservation was made about the non-perpendicularity of the lines in the formulation.

Based on the foregoing, the solution is conveniently formalized in two steps:

1) Calculate the scalar product of directing vectors of straight lines:
so the lines are not perpendicular.

2) We find the angle between the lines by the formula:

Using the inverse function, it is easy to find the angle itself. In this case, we use the oddness of the arc tangent (see Fig. Graphs and properties of elementary functions):

Answer:

In the answer, we indicate the exact value, as well as the approximate value (preferably both in degrees and in radians), calculated using a calculator.

Well, minus, so minus, it's okay. Here is a geometric illustration:

It is not surprising that the angle turned out to be of a negative orientation, because in the condition of the problem the first number is a straight line and the “twisting” of the angle began precisely from it.

If you really want to get a positive angle, you need to swap the straight lines, that is, take the coefficients from the second equation , and take the coefficients from the first equation . In short, you need to start with a direct .