Determine the distance from a point to a given line. Distance from a point to a line - definition

Oh-oh-oh-oh-oh... well, it’s tough, as if he was reading out a sentence to himself =) However, relaxation will help later, especially since today I bought the appropriate accessories. Therefore, let's proceed to the first section, I hope that by the end of the article I will maintain a cheerful mood.

The relative position of two straight lines

This is the case when the audience sings along in chorus. Two straight lines can:

1) match;

2) be parallel: ;

3) or intersect at a single point: .

Help for dummies : Please remember the mathematical intersection sign, it will appear very often. The notation means that the line intersects with the line at 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 corresponding coefficients are proportional, that is, there is a number “lambda” such that the equalities are satisfied

Let's consider the straight lines and create 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 reduce all coefficients of the equation 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 of 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 quite obvious 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 NO such value of “lambda” that the equalities are satisfied

So, for straight lines we will create a system:

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

Conclusion: lines intersect

In practical problems, you can use the solution scheme just discussed. By the way, it is very reminiscent of the algorithm for checking vectors for collinearity, which we looked at in class The concept of linear (in)dependence of vectors. Basis of vectors. But there is a more civilized packaging:

Example 1

Find out the relative position of the lines:

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

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

, which means that the vectors are not collinear and the lines intersect.

Just in case, I’ll put a stone with signs at the crossroads:

The rest jump over the stone and follow further, straight to Kashchei the Immortal =)

b) Find the direction vectors of the lines:

The lines have the same direction vector, which means they are either parallel or coincident. There is no need to count the determinant here.

It is obvious that the coefficients of the unknowns are proportional, and .

Let's find out whether the equality is true:

Thus,

c) Find the direction vectors of the lines:

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

The proportionality coefficient “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 whether 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 problem discussed verbally literally in a matter of seconds. In this regard, I see no point in offering anything for independent decision, it’s better to lay another important brick in the geometric foundation:

How to construct 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.

Solution: Let's denote the unknown line by the letter . What does the condition say about her? The straight line passes through the point. And if the lines are parallel, then it is obvious that the direction vector of the straight line “tse” is also suitable for constructing the straight line “de”.

We take the direction vector out of the equation:

Answer:

The example geometry looks simple:

Analytical testing consists of the following steps:

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

2) Check whether the point satisfies the resulting equation.

In most cases, analytical testing can be easily performed orally. Look at the two equations, and many of you will quickly determine the parallelism of the lines without any drawing.

Examples for independent solutions today will be creative. Because you will still have to compete with Baba Yaga, and she, you know, is a lover of all sorts 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 so rational way to solve it. Most shortcut- at the end of the lesson.

We worked a little 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 familiar to you from school curriculum:

How to find the point of intersection of two lines?

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

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

Here you go geometric meaning systems of two linear equations with two unknowns- these are two intersecting (most often) lines on a plane.

Example 4

Find the point of intersection of lines

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

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

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

The graphical method is, of course, 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 create a correct and ACCURATE drawing. In addition, some straight lines are not so easy to construct, and the point of intersection itself may be located somewhere in the thirtieth kingdom outside the notebook sheet.

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

To solve the system, the method of term-by-term addition of equations was used. To develop relevant skills, take a lesson How to solve a system of equations?

Answer:

The check 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 an example for you to solve on your own. It is convenient to split the task into several stages. Analysis of the condition suggests that it is necessary:
1) Write down the equation of the straight line.
2) Write down the equation of the 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.

Complete solution and the answer at the end of the lesson:

Not even a pair of shoes were worn out before we got to the second section of the lesson:

Perpendicular lines. Distance from a point to a line.
Angle between straight 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 this one, and now the hut on chicken legs will turn 90 degrees:

How to construct a line perpendicular to a given one?

Example 6

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

Solution: By condition it is known that . It would be nice to find the directing vector of the 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.

Let's compose the equation of a straight line using a point and a direction vector:

Answer:

Let's expand the geometric sketch:

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

Analytical verification of the solution:

1) We take out the direction vectors from the equations and with the help scalar product of vectors we come to the conclusion that the lines are indeed perpendicular: .

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

2) Check whether the point satisfies the resulting equation .

The test, again, is easy to perform orally.

Example 7

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

This is an example for you to solve on your own. The problem has several actions, so it is convenient to formulate the solution point by point.

Is our an amusing trip continues:

Distance from point to line

In front of us is a straight strip of the river and our task is to get to it by the shortest route. There are no obstacles, and the most optimal route will be to move along the perpendicular. That is, the distance from a point to a line is the length of the perpendicular segment.

Distance in geometry is traditionally denoted by the Greek letter “rho”, for example: – the distance from the point “em” to the straight line “de”.

Distance from point to line expressed by the formula

Example 8

Find the distance from a point to a line

Solution: all you need to do is carefully substitute the numbers into the formula and carry out the calculations:

Answer:

Let's make the drawing:

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

Let's consider another task based on the same drawing:

The task is to find the coordinates of a point that is symmetrical to the point relative to the straight line . I suggest performing the steps yourself, but I will outline the solution algorithm with intermediate results:

1) Find a line that is perpendicular to the 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 midpoint of a segment we find .

It would be a good idea to check that the distance is also 2.2 units.

Difficulties may arise in calculations here, but a microcalculator is a great help in the tower, allowing you to count common fractions. I have advised you many times and will recommend you again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for you to decide on your own. I’ll give you a little hint: there are infinitely many ways to solve this. Debriefing at the end of the lesson, but it’s better to try to guess for yourself, I think your ingenuity was well developed.

Angle between two straight lines

Every corner is a jamb:


In geometry, the angle between two straight lines is taken to be 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 the angle between intersecting lines. And his “green” neighbor or oppositely oriented"raspberry" 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. Firstly, the direction in which the angle is “scrolled” is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example if .

Why did I tell you this? It seems that we can get by with the usual concept of an angle. The fact is that the formulas by which we will find angles can easily result in a negative result, 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, be sure to indicate its orientation with an arrow (clockwise).

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

Example 10

Find the angle between lines

Solution And Method one

Consider two straight lines given by the equations in general view:

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

Let us pay close attention to the denominator - this is exactly scalar product directing vectors of straight lines:

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

Based on the above, it is convenient to formalize the solution in two steps:

1) Let's calculate the scalar product of the direction vectors of the lines:
, which means the lines are not perpendicular.

2) Find the angle between straight lines using the formula:

By using inverse function It's easy to find the corner itself. In this case, we use the oddness of the arctangent (see. Graphs and properties of elementary functions):

Answer:

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

Well, minus, minus, no big deal. Here is a geometric illustration:

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

If you really want to get a positive angle, you need to swap the 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 .

The distance from a point to a line is the length of the perpendicular drawn from the point to the line. In descriptive geometry, it is determined graphically using the algorithm given below.

Algorithm

1. The straight line is moved to a position in which it will be parallel to any projection plane. For this purpose, methods of transforming orthogonal projections are used.
2. From a point a perpendicular is drawn to a line. This construction is based on the projection theorem right angle.
3. The length of a perpendicular is determined by transforming its projections or using the right triangle method.

The following figure shows a complex drawing of point M and line b, defined by segment CD. You need to find the distance between them.

According to our algorithm, the first thing to do is to move the line to a position parallel to the projection plane. It is important to understand that after the transformations have been carried out, the actual distance between the point and the line should not change. That is why it is convenient here to use the plane replacement method, which does not involve moving figures in space.

The results of the first stage of construction are shown below. The figure shows how an additional frontal plane P 4 is introduced parallel to b. In the new system (P 1, P 4), points C"" 1, D"" 1, M"" 1 are at the same distance from the X 1 axis as C"", D"", M"" from the axis X.

Carrying out the second part of the algorithm, from M"" 1 we lower the perpendicular M"" 1 N"" 1 to the straight line b"" 1, since the right angle MND between b and MN is projected onto the plane P 4 in full size. Using the communication line, we determine the position of point N" and carry out the projection M"N" of the segment MN.

On final stage you need to determine the size of the segment MN from its projections M"N" and M"" 1 N"" 1. For this we are building right triangle M"" 1 N"" 1 N 0, whose leg N"" 1 N 0 is equal to the difference (Y M 1 – Y N 1) of the distance of points M" and N" from the X 1 axis. The length of the hypotenuse M"" 1 N 0 of the triangle M"" 1 N"" 1 N 0 corresponds to the desired distance from M to b.

Second solution

• Parallel to CD, we introduce a new frontal plane P 4. It intersects P 1 along the X 1 axis, and X 1 ∥C"D". In accordance with the method of replacing planes, we determine the projections of points C"" 1, D"" 1 and M"" 1, as shown in the figure.
• Perpendicular to C"" 1 D"" 1 we build an additional horizontal plane P 5, onto which straight line b is projected to point C" 2 = b" 2.
• The distance between point M and line b is determined by the length of the segment M" 2 C" 2, indicated in red.

Similar tasks:

Let a rectangular coordinate system be fixed in three-dimensional space Oxyz, given point , straight line a and you need to find the distance from the point A to a straight line a.

We will show two methods that allow you to calculate the distance from a point to a line in space. In the first case, finding the distance from a point M 1 to a straight line a comes down to finding the distance from the point M 1 to the point H 1 , Where H 1 - the base of a perpendicular dropped from a point M 1 directly a. In the second case, we will find the distance from the point to the plane as the height of the parallelogram.

So let's get started.

The first way to find the distance from a point to a line a in space.

Since by definition the distance from a point M 1 to a straight line a is the length of the perpendicular M 1 H 1 , then, having determined the coordinates of the point H 1 , we can calculate the required distance as the distance between points And according to the formula .

Thus, the problem comes down to finding the coordinates of the base of the perpendicular constructed from the point M 1 to a straight line a. This is quite simple to do: period H 1 is the point of intersection of the line a with a plane passing through a point M 1 perpendicular to the line a.

Hence, algorithm that allows you to determine the distance from a point to a straight linea in space, is:

The second method allows you to find the distance from a point to a line a in space.

Since in the problem statement we are given a straight line a, then we can determine its direction vector and the coordinates of some point M 3 , lying on the straight line a. Then, according to the coordinates of the points and we can calculate the coordinates of a vector: (if necessary, refer to the article coordinates of a vector through the coordinates of its start and end points).

Let's put aside the vectors and from the point M 3 and construct a parallelogram on them. In this parallelogram we draw the height M 1 H 1 .

Obviously the height M 1 H 1 of the constructed parallelogram is equal to the required distance from the point M 1 to a straight line a. Let's find it.

On one side, the area of ​​the parallelogram (let us denote it S) can be found through the cross product of vectors and according to the formula . On the other hand, the area of ​​a parallelogram is equal to the product of the length of its side and its height, that is, , Where - vector length , equal to the length of the side of the parallelogram in question. Therefore, the distance from a given point M 1 to a given straight line a can be found from the equality How .

So, to find the distance from a point to a straight linea in space needed

Solving problems of finding the distance from a given point to a given line in space.

Let's look at the example solution.

Example.

Find the distance from the point to a straight line .

Solution.

First way.

Let's write the equation of the plane passing through the point M 1 perpendicular to a given line:

Find the coordinates of the point H 1 - points of intersection of the plane and a given straight line. To do this, let us make the transition from the canonical equations of a straight line to the equations of two intersecting planes

after which we solve the system of linear equations Cramer's method:

Thus, .

It remains to calculate the required distance from a point to a line as the distance between points And : .

Second way.

The numbers in the denominators of fractions in the canonical equations of a line represent the corresponding coordinates of the direction vector of this line, that is, - direct vector . Let's calculate its length: .

Obviously straight passes through a point , then a vector with origin at point and end at a point There is . Let's find the vector product of vectors And :
then the length of this vector product is .

Now we have all the data to use the formula to calculate the distance from a given point to a given plane: .

Answer:

The relative position of lines in space

Formula for calculating the distance from a point to a line on a plane

If the equation of the line Ax + By + C = 0 is given, then the distance from the point M(M x , M y) to the line can be found using the following formula

Examples of problems for calculating the distance from a point to a line on a plane

Example 1.

Find the distance between the line 3x + 4y - 6 = 0 and the point M(-1, 3).

Solution. Let's substitute the coefficients of the line and the coordinates of the point into the formula

Answer: the distance from the point to the line is 0.6.

equation of a plane passing through points perpendicular to a vectorGeneral equation of a plane

A nonzero vector perpendicular to a given plane is called normal vector (or, in short, normal ) for this plane.

Let the following be given in coordinate space (in a rectangular coordinate system):

a) point ;

b) non-zero vector (Fig. 4.8, a).

You need to create an equation for a plane passing through a point perpendicular to the vector End of proof.

Let's now consider Various types equations of a straight line on a plane.

1) General equation of the planeP .

From the derivation of the equation it follows that at the same time A, B And C are not equal to 0 (explain why).

The point belongs to the plane P only if its coordinates satisfy the equation of the plane. Depending on the odds A, B, C And D plane P occupies one position or another:

- the plane passes through the origin of the coordinate system, - the plane does not pass through the origin of the coordinate system,

- plane parallel to the axis X,

X,

- plane parallel to the axis Y,

- the plane is not parallel to the axis Y,

- plane parallel to the axis Z,

- the plane is not parallel to the axis Z.

Prove these statements yourself.

Equation (6) is easily derived from equation (5). Indeed, let the point lie on the plane P. Then its coordinates satisfy the equation. Subtracting equation (7) from equation (5) and grouping the terms, we obtain equation (6). Let us now consider two vectors with coordinates respectively. From formula (6) it follows that their scalar product is equal to zero. Therefore, the vector is perpendicular to the vector. The beginning and end of the last vector are located, respectively, at points that belong to the plane P. Therefore, the vector is perpendicular to the plane P. Distance from point to plane P, whose general equation determined by the formula The proof of this formula is completely similar to the proof of the formula for the distance between a point and a line (see Fig. 2).
Rice. 2. To derive the formula for the distance between a plane and a straight line.

Indeed, the distance d between a straight line and a plane is equal

where is a point lying on the plane. From here, as in lecture No. 11, the above formula is obtained. Two planes are parallel if their normal vectors are parallel. From here we obtain the condition for parallelism of two planes - odds general equations planes. Two planes are perpendicular if their normal vectors are perpendicular, hence we obtain the condition for the perpendicularity of two planes if their general equations are known

Corner f between two planes equal to angle between their normal vectors (see Fig. 3) and can therefore be calculated using the formula
Determining the angle between planes.

(11)

Distance from a point to a plane and methods for finding it

Distance from point to plane– the length of the perpendicular dropped from a point onto this plane. There are at least two ways to find the distance from a point to a plane: geometric And algebraic.

With the geometric method You must first understand how the perpendicular from a point to a plane is located: maybe it lies in some convenient plane, is a height in some convenient (or not so convenient) triangle, or maybe this perpendicular is generally a height in some pyramid.

After this first and most complex stage, the problem breaks down into several specific planimetric problems (perhaps in different planes).

With the algebraic method in order to find the distance from a point to a plane, you need to enter a coordinate system, find the coordinates of the point and the equation of the plane, and then apply the formula for the distance from a point to a plane.

St. Petersburg State Marine Technical University

Department computer graphics and information support

LESSON 3

PRACTICAL TASK No. 3

Determining the distance from a point to a straight line.

You can determine the distance between a point and a straight line by performing the following constructions (see Fig. 1):

· from point WITH lower the perpendicular to a straight line A;

· mark a point TO intersection of a perpendicular with a straight line;

measure the length of the segment KS, the beginning of which is a given point, and the end is the marked intersection point.

Fig.1. Distance from a point to a line.

The basis for solving problems of this type is the right angle projection rule: a right angle is projected without distortion if at least one of its sides is parallel to the projection plane(i.e. occupies a private position). Let's start with just such a case and consider constructions for determining the distance from a point WITH to a straight line segment AB.

There are no test cases in this task, but options for execution individual tasks shown in table1 and table2. The solution to the problem is described below, and the corresponding constructions are shown in Fig. 2.

1. Determining the distance from a point to a particular line.

First, projections of a point and a segment are constructed. Projection A1B1 parallel to the axis X. This means that the segment AB parallel to the plane P2. If from point WITH draw perpendicular to AB, then the right angle is projected without distortion onto the plane P2. This allows you to draw a perpendicular from a point C2 to projection A2B2.

Dropdown menu Drawing-Segment (Draw- Line) . Place cursor at point C2 and fix it as the first point of the segment. Move the cursor in the direction of the normal to the segment A2B2 and fix the second point on it at the moment the hint appears Normal (Perpendicular) . Mark the constructed point K2. Enable mode ORTHO(ORTHO) , and from the point K2 draw a vertical connection line until it intersects with the projection A1 B1. Designate the intersection point by K1. Dot TO, lying on the segment AB, is the intersection point of the perpendicular drawn from the point WITH, with segment AB. Thus, the segment KS is the required distance from the point to the line.

From the constructions it is clear that the segment KS occupies a general position and, therefore, its projections are distorted. When talking about distance, we always mean true value of the segment, expressing the distance. Therefore, we need to find the true value of the segment KS, by rotating it to a particular position, for example, KS|| P1. The result of the constructions is shown in Fig. 2.

From the constructions shown in Fig. 2, we can conclude: the particular position of the line (the segment is parallel P1 or P2) allows you to quickly build projections of the distance from a point to a line, but they are distorted.

Fig.2. Determining the distance from a point to a particular line.

2. Determining the distance from a point to a line general position.

Not always in initial condition the segment occupies a particular position. With a general initial position, the following constructions are performed to determine the distance from a point to a line:

a) using the drawing transformation method, convert a segment from a general position to a particular one - this will allow constructing distance projections (distorted);

b) using the method again, convert the segment corresponding to the required distance to a particular position - we obtain a projection of the distance in magnitude equal to the actual one.

Consider the sequence of constructions to determine the distance from a point A to a segment in general position Sun(Fig. 3).

On first spin it is necessary to obtain the particular position of the segment INC. To do this in the layer TMR need to connect the dots AT 2, C2 And A2. Using the command Change-Rotate (Modify – Rotate) triangle В2С2А2 rotate around a point C2 to the position where the new projection B2*C2 will be located strictly horizontally (point WITH is motionless and, therefore, its new projection coincides with the original one and the designation C2* And C1* may not be shown on the drawing). As a result, new projections of the segment will be obtained B2*C2 and points: A2*. Next from points A2* And AT 2* vertical ones are carried out, and from the points IN 1 And A1 horizontal communication lines. The intersection of the corresponding lines will determine the position of the points of the new horizontal projection: the segment B1*C1 and dots A1*.

In the resulting particular position, we can construct distance projections for this: from the point A1* the normal to B1*C1. The point of their mutual intersection is K1*. A vertical connection line is drawn from this point until it intersects with the projection B2*C2. A point is marked K2*. As a result, the projections of the segment were obtained AK, which is the required distance from the point A to a straight line segment Sun.

Next, it is necessary to construct distance projections in the initial condition. To do this from the point K1* it is convenient to draw a horizontal line until it intersects with the projection В1С1 and mark the intersection point K1. Then a point is constructed K2 on the frontal projection of the segment and projections are carried out A1K1 And A2K2. As a result of the constructions, projections of the distance were obtained, but both in the initial and in the new partial position of the segment sun, line segment AK occupies a general position, and this leads to the fact that all its projections are distorted.

On the second rotation it is necessary to rotate the segment AK to a particular position, which will allow us to determine the true value of the distance - projection A2*K2**. The result of all constructions is shown in Fig. 3.

TASK No. 3-1. WITH to a straight line of particular position specified by a segment AB. Give the answer in mm (Table 1).Remove projection lenses

Table 1

TASK No. 3-2. Find the true distance from a point M to a straight line in general position given by the segment ED. Give the answer in mm (table 2).

table 2

Checking and passing completed TASK No. 3.