Oh-oh-oh-oh ... Well, tin, as if you read it myself \u003d) however, then relaxation will help, especially since today I bought suitable accessories. Therefore, I will proceed to the first section, I hope, by the end of the article I preserve the vigorous arrangement of the Spirit.

Mutual location of two straight lines

The case when the hall sits the choir. Two straight lines can:

1) coincide;

2) be parallel :;

3) or intersect in a single point :.

Help for teapots : Please remember the mathematical sign of the intersection, it will meet very often. The entry denotes that the direct intersects with a straight point at the point.

How to determine the mutual location of two straight lines?

Let's start from the first time:

Two straight line coincide, then and only if their respective coefficients are proportional, that is, there is an such number "Lambda", which is performed equality

Consider direct and make three equations from the respective coefficients :. It follows from each equation that, therefore, the direct data coincide.

Indeed, if all the coefficients of the equation Multiply to -1 (change marks), and all equation coefficients Reduce 2, then the same equation will be obtained :.

The second case is when straight parallel to:

Two straight parallels then and only if their coefficients are proportional to the variables: , but.

As an example, consider two straight. Check the proportionality of the corresponding coefficients with variables:

However, it is quite obvious that.

And the third case, when the straight line intersect:

Two straight lines intersect, then and only if their coefficients are not proportional to variables, that is, there is no such meaning of "lambda" to be carried out equal

So, for directly make a system:

From the first equation it follows that, and from the second equation:, it means the system is incomplete (No solutions). Thus, the coefficients with variables are not proportional.

Conclusion: Straight intersect

In practical tasks, you can only use the solution scheme. She, by the way, quite reminds the algorithm for checking vectors for the collinearity, which we considered in the lesson The concept of linear (no) vectors dependences. Basis vectors. But there is more civilized packaging:

Example 1.

Find out the mutual location of direct:

Decision Based on the study of direct vectors of direct:

a) from the equations will find direct vectors: .

So, vectors are not collinear and straight intersect.

Just in case, put a stone with pointers to the crossroads:

The rest jump the stone and follow the next, straight to the idleness of the immortal \u003d)

b) We will find direct vectors direct:

Straight have the same guide vector, it means that they are either parallel or coincide. Here and the determinant is not necessary.

Obviously, the coefficients at unknown are proportional to, with this.

We find out whether equality is true:

In this way,

c) We find direct vectors direct:

Calculate the determinant compiled from the data coordinates of the vectors:
Therefore, the guide vectors collinear. Direct either parallel or coincide.

The ratio of the proportionality of "Lambda" is not difficult to see directly from the ratio of collinear vectors. However, it can be found through the coefficients of the equations themselves: .

Now find out whether equality is true. Both free member zero, so:

The obtained value satisfies this equation (it satisfies any number in general).

Thus, direct coincide.

Answer:

Very soon you will learn (or have already learned) to solve the considered task orally literally in seconds. In this regard, I see no sense to offer anything for self-decide, It is better to launch another important brick in a geometric foundation:

How to build a straight parallel to this?

For ignorance of this simplest problem, the nightingale-robber is severely punishable.

Example 2.

Direct is given by the equation. Make the equation of a parallel direct, which passes through the point.

Decision: Denote by an unknown direct letter. What is said about her in the condition? Direct passes through the point. And if straight parallels, it is obvious that the direct "CE" guide vector is suitable for building a straight line "DE".

Pull out the guide vector from the equation:

Answer:

The example geometry looks uncomfortable:

Analytical check consists in the following steps:

1) We check that the same guide vector (if the direct equation is not simplified properly, then the vectors will be collinear).

2) We check whether the point obtained equation satisfies.

Analytical check in most cases is easy to perform orally. Look at the two equations, and many of you will quickly determine the parallelism of direct without any drawing.

Examples for an independent solution today will be creative. Because you still have to take a Baba Yaga, and she, you know, a lover of all kinds of mysteries.

Example 3.

Make the equation of direct passing through a point parallel to the line if

There is a rational and not very rational solution. The shortest path is at the end of the lesson.

With parallel straight, they worked a little and come back to them. The case of coinciding straight lines is more interesting, so consider the task that is familiar to you from the school program:

How to find the intersection point of two straight lines?

If straight intersect at the point, its coordinates are a decision Systems of linear equations

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

Here I am the geometric meaning of the system of two linear equations with two unknown - These are two intersecting (most often) straight on the plane.

Example 4.

Find a point of intersection of direct

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

The graphic method is to simply draw the data direct and learn the intersection point directly from the drawing:

Here is our point :. To check, it is necessary to substitute its coordinates in each equation direct, they must come out there and there. In other words, the coordinates of the point are the solution of the system. In fact, we reviewed a graphical solution systems of linear equations With two equations, two unknowns.

The graphic method, of course, is not bad, but there are noticeable cons. No, it's not that the seventh graders decide that, the fact is that the right and accurate drawing will take time. In addition, some direct build is not so simple, and the intersection point itself may be somewhere in the thirtieth kingdom outside the airtal sheet.

Therefore, the point of intersection is more expedient to look for an analytical method. Resolving the system:

To solve the system, the method of reassembly of equations is used. To work out the appropriate skills, visit the lesson How to solve the system of equations?

Answer:

Check trivial - the coordinates of the intersection point must satisfy each equation of the system.

Example 5.

Find the point of intersection direct if they intersect.

This is an example for an independent solution. The task is convenient to smash into several stages. The analysis of the condition suggests that it is necessary:
1) Make the equation direct.
2) Make a direct equation.
3) Find out the mutual location of the straight lines.
4) If direct intersects, find the intersection point.

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

Complete solution and answer at the end of the lesson:

Stoptan and pair of shoes, as we got to the second lesson section:

Perpendicular straight lines. Distance from point to straight.
The angle between straight

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, and now the hut on curious legs will unfold 90 degrees:

How to build a straight, perpendicular to this?

Example 6.

Direct is given by the equation. Make the equation perpendicular to the direct pass passing through the point.

Decision: Under the condition it is known that. It would be nice to find the guide vector straight. Since straight perpendicular, focus is simple:

From the equation "Remove" the vector of normal: which will be a direct line.

The equation is direct to be on the point and the guide vector:

Answer:

We will launch a geometric etude:

M-yes ... Orange Sky, Orange Sea, Orange Camel.

Analytical Solution Check:

1) from the equations pull out the guide vectors and with help scalar product vectors We conclude that straight lines are really perpendicular :.

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

2) Checking whether the point of the obtained equation satisfies .

Check, again, easily perform orally.

Example 7.

Find the intersection point perpendicular direct, if the equation is known and point.

This is an example for an independent solution. In the task several actions, so the solution is convenient to place on points.

Our fascinating journey continues:

Distance from point to direct

We have a direct strip of river and our task is to reach it with the shortest way. There are no obstacles, and the most optimal route will move on perpendicular. That is, the distance from the point to the line is the length of the perpendicular segment.

The distance in geometry is traditionally denoted greek letter "RO", for example: - distance from the point "EM" to a straight line "DE".

Distance from point to direct Formula is expressed

Example 8.

Find the distance from point to direct

Decision: All you need, it is gently substituting the numbers in the formula and carry out computation:

Answer:

Perform a drawing:

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

Consider another task on the same drawing:

The task is to find the coordinates of the point that is symmetric about the direct point . I propose to perform actions yourself, but I denote the solution algorithm with intermediate results:

1) Find straight, which is perpendicular to the straight line.

2) Find the intersection point of direct: .

Both actions are disassembled in detail within the framework of this lesson.

3) The point is a middle of the segment. We know the coordinates of the middle and one of the ends. By mid-segment coordinate formulas Find.

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

Difficulties here may arise in the calculations, but in the tower greatly cuts out a microcalculator that allows you to count ordinary fractions. Repeatedly advised, advise and again.

How to find the distance between two parallel straight?

Example 9.

Find the distance between two parallel straight

This is another example for an independent decision. I will tell you a little: there are infinitely many ways to solve. Halfing the flights at the end of the lesson, but better try to guess yourself, I think your smelter managed to disperse well.

The angle between two straight

Nothing a corner, then jamb:


In geometry, a smaller angle is accepted for the angle between two direct, from which it automatically follows that it cannot be blunt. In the picture, the angle marked with a red arc is not considered an angle between intersecting straight. And it is considered such a "green" neighbor or oppositely oriented "Raspberry" corner.

If direct is perpendicular, then by the angle between them you can take any of the 4 corners.

What is the difference between the angles? Orientation. First, it is fundamentally important to the direction of "scrolling" angle. Secondly, a negatively oriented angle is recorded with a minus sign, for example, if.

Why did I tell it? It seems possible to do and the usual concept of angle. The fact is that in the formulas for which we will find corners, it may easily be a negative result, and this should not find you surprise. The angle with the "minus" sign is no worse, and has a completely concrete geometric meaning. In the drawing for a negative angle, it is necessary to specify the arrow of its orientation (clockwise).

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

Example 10.

Find the corner between straight

Decision and Fashion first

Consider two direct specified by the equations in general:

If straight not perpendicularT. orienteed The angle between them can be calculated using the formula:

The closest attention is paid to the denominator - it is exactly scalar product Direct vectors direct:

If, the denominator of the formula is drawn to zero, and the vectors will be orthogonal and direct perpendicular. That is why a reservation is made about the imperpendacularity of direct in the wording.

Based on the foregoing, the solution is convenient to arrange two steps:

1) Calculate the scalar product of direct vectors of direct:
So straight is not perpendicular.

2) The angle between direct will find by the formula:

Using the reverse function, it is easy to find an angle itself. At the same time, we use the oddness of Arctangent (see Charts and properties of elementary functions):

Answer:

In response, specify the exact value, as well as the approximate value (preferably in degrees, and in radians) calculated using the calculator.

Well, minus, so minus, nothing terrible. Here is a geometric illustration:

It is not surprising that the angle turned out to be a negative orientation, because in terms of the task, the first number goes straight and "rejuvenation" of the angle began with it.

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

Formulation of the problem. Find the point coordinates, symmetrical point relative to the plane.

Plan solution.

1. Find the equation direct, which is perpendicular to this plane and passes through the point . So direct perpendicular to the specified plane, the vector of the plane normal can be taken as its guide vector, i.e.

.

Therefore, the direct equation will

.

2. Find a point intersections are direct and planes (see task 13).

3. Point is the middle of the segment where the point is a point symmetrical point , so

Task 14.. Find a point, symmetric point plane.

The equation is direct, which passes through the point perpendicular to the specified plane will be:

.

Find a direct and plane intersection point.

From - point of intersection of a straight and plane. The middle of the segment is therefore

Those. .

Uniform coordinates of the plane. Affine conversion on the plane.

Let be M. h. and w.

M.(h., w.ME (h., w., 1) in space (Fig. 8).

ME (h., w.

ME (h., w. hu.

(HX, HY, H), H  0,

Comment

h. (eg, h.

In fact, counting h.

Comment

Example 1.

b.) At corner (Fig. 9).

1st step.

2nd step. Rotate angle 

matrix of the appropriate conversion.

3rd step. Transfer to Vector A (A, b)

matrix of the appropriate conversion.

Example 3.

 Along the abscissa axis and

1st step.

matrix of the appropriate conversion.

2nd step.

3rd step.

we get finally

Comment

[R], [d], [m], [t],

Let be M. - arbitrary point plane with coordinates h. and w.calculated relative to a given straightforward coordinate system. The homogeneous coordinates of this point are any triple of at the same time unequal zero of the numbers x 1, x 2, x 3 associated with the specified numbers x and in the following ratios:

When solving computer graphics tasks, homogeneous coordinates are usually entered like this: an arbitrary point M.(h., w.) Plane is put in line with the point ME (h., w., 1) in space (Fig. 8).

Note that an arbitrary point on a straight line connecting the origin of the coordinates, point 0 (0, 0, 0), with a point ME (h., w., 1), can be set by the three numbers of the form (HX, HY, H).

The vector with HX, HY coordinates, is a straight line guide vector connecting 0 (0, 0, 0) and ME (h., w., one). This straight line crosses the plane z \u003d 1 at the point (x, y, 1), which uniquely determines the point (x, y) of the coordinate plane hu.

Thus, between an arbitrary point with coordinates (x, y) and a plurality of triple numbers of the form

(HX, HY, H), H  0,

set (mutually unambiguous) correspondence that allows you to count the numbers HX, HY, H new coordinates of this point.

Comment

Uniform coordinates widely used in projective geometry make it possible to effectively describe the so-called immunity elements (essentially those that the projective plane differs from the usual Euclidean plane). In more detail about the new features provided by the introduced homogeneous coordinates, the fourth section of this chapter is said.

In projective geometry for homogeneous coordinates, the following designation is taken:

x: y: 1, or, more general, x 1: x 2: x 3

(Recall that it certainly requires that numbers x 1, x 2, x 3 at the same time, did not appeal to zero).

The use of homogeneous coordinates is convenient already when solving the simplest tasks.

Consider, for example, issues related to changes in scale. If the display device works only with integers (or if it is necessary to work only with integers), then for arbitrary value h. (eg, h. \u003d 1) point with homogeneous coordinates

it is impossible to submit. However, with a reasonable choice of H, it can be achieved that the coordinates of this point be integers. In particular, with H \u003d 10 for the example under consideration we have

Consider another case. So that the results of the transformation are not led to arithmetic overflow, for a point with coordinates (8000 40000 1000), you can take, for example, H \u003d 0.001. As a result, we obtain (80 40 1).

The above examples show the usefulness of the use of homogeneous coordinates during calculations. However, the main purpose of introducing homogeneous coordinates in computer graphics is their undoubted convenience in applying geometric transformations.

With the help of triples of homogeneous coordinates and third-order matrices, any affine plane transformation can be described.

In fact, counting h. \u003d 1, compare two records: a marked symbol * and the following, matrix:

It is easy to see that after moving the expressions on the right side of the last relationship, we obtain both formulas (*) and the correct numeric equality 1 \u003d 1.

Comment

Sometimes another entry is used in the literature - recording on columns:

Such an entry is equivalent to the above records on lines (and is obtained from it with transposition).

The elements of an arbitrary matrix of affine transform are not in itself a clearly pronounced geometrical meaning. Therefore, to implement this or that mapping, that is, finding the elements of the corresponding matrix according to a given geometric description, special techniques are needed. Usually, the construction of this matrix in accordance with the complexity of the problem under consideration and with the special cases described above are divided into several stages.

At each stage, a matrix is \u200b\u200bsearched for a matrix corresponding to one or another of the above cases A, B, B or G, which have well-pronounced geometric properties.

Drink the corresponding third-order matrices.

A. Rotation Matrix, (Rotation)

B. Stretching Matrix (Dilatation)

V. Matrix Reflection (Reflection)

Translation Matrix (Translation)

Consider examples of affine plane transformations.

Example 1.

Build a turn matrix around the point A (A,b.) At corner (Fig. 9).

1st step. Transfer to vector - A (-A, -B) to combine the center of rotation with the beginning of the coordinates;

matrix of the appropriate conversion.

2nd step. Rotate angle 

matrix of the appropriate conversion.

3rd step. Transfer to Vector A (A, b) To return the center of rotation to the previous position;

matrix of the appropriate conversion.

Matching the matrix in the same order as they are written:

As a result, we obtain that the desired transformation (in the matrix record) will look like this:

Elements of the resulting matrix (especially in the last row) are not so easy to remember. At the same time, each of the three variable matrices on the geometric description of the corresponding display is easily built.

Example 3.

Build a stretching matrix with stretching coefficients Along the abscissa axis and Along the axis of the ordinate and with the center at point A (A, B).

1st step. Transfer to vector -A (-A -B) to combine the center of stretching with the start of coordinates;

matrix of the appropriate conversion.

2nd step. Stretching along the coordinate axes with coefficients  and , respectively; The conversion matrix is

3rd step. Transfer to vector A (A, B) to return the center of stretching to the previous position; Matrix of the appropriate conversion -

Aligning .matizals in the same order

we get finally

Comment

Arguing in a similar way, that is, breaking the proposed transformation to steps supported by matrices[R], [d], [m], [t], you can construct a matrix of any affine transformation according to its geometric description.

The shift is implemented by adding, and scaling and rotation - multiplication.

Scaling transformation (dilatation) relative to the start of the coordinate looks:

or in matrix form:

where D.xD.y.- scaling coefficients on the axes, and

- Scaling matrix.

With d\u003e 1-extension, at 0<=D<1- сжатие

Turn conversion Regarding the start of the coordinate looks:

or in matrix form:

where φ is the angle of rotation, and

- Turn matrix.

Comment:Columns and rows of the rotation matrix are mutually orthogonal single vectors. In fact, the squares of the string lengths are equal to one:

cosφ · cosφ + sinφ · sinφ \u003d 1 and (-sinφ) · (-sinφ) + cosφ · cosφ \u003d 1,

and the scalar product of the row vectors is

cosφ · (-sinφ) + sinφ · cosφ \u003d 0.

Since the scalar product of vectors A. · B. = |A.| ·| B.| · Cosψ, where | A.| - Length vector A., |B.| - Length vector B., and ψ - the smallest positive corner between them, then from equality 0 of the scalar product of two vector strings 1 it follows that the angle between them is 90 °.

Direct in space can always be determined as a line of intersection of two non-parallel planes. If the equation of one plane, the equation of the second plane, then the direct equation is given by the form

here nekollynearin
. These equations are called common equations Direct in space.

Canonical equations are direct

Any nonzero vector lying on this direct or parallel to it is called the guide vector of this straight.

If a point is known
straight and its guide vector
, then the canonical equations are the form:

. (9)

Parametric equations are direct

Let the canonical equations are given

.

From here, we obtain parametric equations direct:

(10)

These equations are convenient when the direct and plane intersection points are located.

The equation is direct passing through two points
and
it has the form:

.

The angle between straight

The angle between straight

and

equal to the corner between their guide vectors. Consequently, it can be calculated by formula (4):

Condition of parallelism direct:

.

The condition perpendicularity of the planes:

Distance point from straight

P ust Dana Point
and straight

.

From canonical equations directly known dot
belonging direct and its guide vector
. Then distance is the point
from straight equal to height of a parallelogram built in vectors and
. Hence,

.

The intersection condition of direct

Two non-parallel straight lines

,

intersect then and only when

.

Mutual arrangement of the straight and plane.

Let be given straight
and plane. Angle between them can be found by the formula

.

Task 73. Write canonical equations direct

(11)

Decision. In order to record the canonical equations of straight (9), you need to know any point belonging to the straight line, and the direct vector direct.

Find vector , parallel given direct. Since it should be perpendicular to normal vector data vectors, i.e.

,
T.

.

From the general equations direct we have that
,
. Then

.

Since point
any point is straight, then its coordinates must satisfy the equations direct and one of them can be set, for example,
, Two other coordinates will find from the system (11):

Hence
.

Thus, the canonical equations of the desired direct have the form:

or
.

Task 74.

and
.

Decision. From the canonical equations of the first straight, the coordinates of the point are known.
belonging to the line and coordinates of the guide vector
. From the canonical equations of the second straight, the coordinates of the point are also known.
and coordinates of the guide vector
.

The distance between parallel straight is equal to the distance of the point
from the second direct. This distance is calculated by the formula

.

Find the coordinates of the vector
.

Calculate vector art
:

.

Task 75. Find a point symmetric point
related

.

Decision. Write the equation of the plane perpendicular to this direct and passing through the point . As its normal vector you can take a straight guide vector. Then
. Hence,

Find a point
point of intersection of this direct and plane P. To do this, write the parametric equations direct using equations (10), we get

Hence,
.

Let be
point symmetrical point
regarding this direct. Then point
mid-cut
. To find the coordinates of the point using the formulas of the middle segment coordinates:

,
,
.

So,
.

Task 76. Write the equation of the plane passing through the straight
and

a) via the point
;

b) perpendicular to the plane.

Decision. We write the general equations of this direct. To do this, consider two equalities:

This means that the desired plane belongs to the beam of planes with forming and its equation can be recorded in the form (8):

a) Find
and from the condition that the plane passes through the point
Therefore, its coordinates must satisfy the equation of the plane. We substitute the coordinates of the point
the plane beam equation:

Found value
substitute to equation (12). We obtain the equation of the desired plane:

b) Find
and from the condition that the desired plane is perpendicular to the plane. Vector normal of this plane
, vector normal is the desired plane (see the plane beam equation (12).

Two vectors are perpendicular if and only if their scalar product is zero. Hence,

Substitute the value found
in the plane beam equation (12). We obtain the equation of the desired plane:

Tasks for self solutions

Task 77. To lead to the canonical type of equation direct:

1)
2)

Task 78. Write parametric equations direct
, if a:

1)
,
; 2)
,
.

Task 79.. Write the equation of the plane passing through the point
perpendicular to direct

Task 80. Write equations direct passing point
perpendicular to the plane.

Task 81. Find the angle between straight:

1)
and
;

2)
and

Task 82. Prove the direct parallelism:

and
.

Task 83. Prove the perpendicularity of direct:

and

Task 84. Calculate the distance point
from straight:

1)
; 2)
.

Task 85. Calculate the distance between parallel straight:

and
.

Task 86.. In equations are direct
determine the parameter so that this direct intersects with direct and find the point of their intersection.

Task 87.. Show that straight
parallel plane
, and straight
lies in this plane.

Task 88.. Find a point symmetric point relative to the plane
, if a:

1)
, ;

2)
, ;.

Task 89. Write the perpendicular equation, lowered from the point
on straight
.

Task 90.. Find a point symmetric point
related
.

Let some direct, given by a linear equation, and the point specified by its coordinates (X0, Y0) and not lying on this straight line. It is necessary to detect a point that would be symmetrical to this point regarding this line, that is, it would coincide with it if the plane mentally bent in the pressure in this straight line.

Instruction

1. It is clear that both points are specified and desirable - obliged to lie on one straight line, and this direct must be perpendicular to this. Thus, the first part of the problem is, in order to detect the equation direct, which would be perpendicular to some direct line and at the same time passed through this point.

2. Direct can be set by two methods. The canonical equation direct looks like this: AX + BY + C \u003d 0, where A, B, and C - constants. Also, the direct is allowed to determine using a linear function: y \u003d kx + b, where K is an angular figure, b - displacement. These two methods are interchangeable, and from any reason to go to another. If AX + BY + C \u003d 0, then y \u003d - (ax + c) / b. In other words, in the linear function y \u003d kx + b angle k \u003d -a / b, and the b \u003d -c / b offset. For the task, it is more comfortable to argue, based on the canonical equation, direct.

3. If two direct is perpendicular to each other, and the equation of the first straight line AX + BY + C \u003d 0, then the equation 2-th straight should look like BX - AY + D \u003d 0, where D is a constant. In order to detect a certain value of D, it is necessary to know addingly, through which point is the perpendicular straight line. In this case, it is a point (x0, y0). The D must satisfy the equality: BX0 - AY0 + D \u003d 0, that is, D \u003d AY0 - BX0.

4. Later, the perpendicular direct is detected, it is necessary to calculate the coordinates of the point of its intersection with this. To do this, it is required to solve the system of linear equations: AX + BY + C \u003d 0, BX - AY + AY0 - BX0 \u003d 0. The solution will give numbers (x1, y1) that serve as the coordinates of the intersection point of direct.

5. The desired point should lie on the detected straight, and its distance to the intersection point should be equal to the distance from the intersection point to the point (x0, y0). The coordinates of the point, symmetric point (x0, y0), are allowed, thus detect, solving the system of equations: BX - AY + AY0 - BX0 \u003d 0 ,? ((x1 - x0) ^ 2 + (y1 - y0) ^ 2 \u003d ? ((x - x1) ^ 2 + (y - y1) ^ 2).

6. But it is easier to do it easier. If the points (x0, y0) and (x, y) are at equal distances from the point (x1, y1), and all three points lie on one straight line, then: x - x1 \u003d x1 - x0, y - y1 \u003d y1 - Y0. Landy, x \u003d 2 × 1 - x0, y \u003d 2y1 - y0. Substituting these values \u200b\u200binto the second equation of the first system and simplifying the expression, it is easy to make sure that its right part becomes the same left. Extremely consider the first equation, it makes no sense that it is believed that the points (x0, y0) and (x1, y1) are satisfied with it, and the point (X, Y) is obviously on the same direct.

The task is to find the coordinates of the point that is symmetric about the direct point . I propose to perform actions yourself, but I denote the solution algorithm with intermediate results:

1) Find straight, which is perpendicular to the straight line.

2) Find the intersection point of direct: .

Both actions are disassembled in detail within the framework of this lesson.

3) The point is a middle of the segment. We know the coordinates of the middle and one of the ends. By mid-segment coordinate formulas Find.

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

Difficulties here may arise in the calculations, but the microcalculator helps in the tower, which allows us to consider ordinary fractions. Repeatedly advised, advise and again.

How to find the distance between two parallel straight?

Example 9.

Find the distance between two parallel straight

This is another example for an independent decision. I will tell you a little: there are infinitely many ways to solve. Halfing the flights at the end of the lesson, but better try to guess yourself, I think your smelter managed to disperse well.

The angle between two straight

Nothing a corner, then jamb:


In geometry, a smaller angle is accepted for the angle between two direct, from which it automatically follows that it cannot be blunt. In the picture, the angle marked with a red arc is not considered an angle between intersecting straight. And it is considered such a "green" neighbor or oppositely oriented "Raspberry" corner.

If direct is perpendicular, then by the angle between them you can take any of the 4 corners.

What is the difference between the angles? Orientation. First, it is fundamentally important to the direction of "scrolling" angle. Secondly, a negatively oriented angle is recorded with a minus sign, for example, if.

Why did I tell it? It seems possible to do and the usual concept of angle. The fact is that in the formulas for which we will find corners, it may easily be a negative result, and this should not find you surprise. The angle with the "minus" sign is no worse, and has a completely concrete geometric meaning. In the drawing for a negative angle, it is necessary to specify the arrow of its orientation (clockwise).

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

Example 10.

Find the corner between straight

Decision and Fashion first

Consider two straight lines given by equations in general form:

If straight not perpendicularT. orienteed The angle between them can be calculated using the formula:

The closest attention is paid to the denominator - it is exactly scalar product Direct vectors direct:

If, the denominator of the formula is drawn to zero, and the vectors will be orthogonal and direct perpendicular. That is why a reservation is made about the imperpendacularity of direct in the wording.

Based on the foregoing, the solution is convenient to arrange two steps:

1) Calculate the scalar product of direct vectors of direct:

2) The angle between direct will find by the formula:

Using the reverse function, it is easy to find an angle itself. At the same time, we use the oddness of Arctangent (see Charts and properties of elementary functions):

Answer:

In response, specify the exact value, as well as the approximate value (preferably in degrees, and in radians) calculated using the calculator.

Well, minus, so minus, nothing terrible. Here is a geometric illustration:

It is not surprising that the angle turned out to be a negative orientation, because in terms of the task, the first number goes straight and "rejuvenation" of the angle began with it.

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

I will not clean it, I myself select direct in the order to turn out to be positive. So more beautiful, but no more.

To verify the solution, you can take the vehicle and measure the angle.

Method of the second

If direct is given by equations with an angular coefficient and not perpendicularT. orienteed The angle between them can be found using the formula:

The condition of perpendicularity of direct is expressed by equality, from where, by the way, it is necessary for a very useful relationship of the angular coefficients perpendicular direct: which is used in some tasks.

The solution algorithm is similar to the previous item. But first rewrite our straight in the right form:

Thus, angular coefficients:

1) Check whether direct is perpendicular:
So straight is not perpendicular.

2) We use the formula:

Answer:

The second way is appropriate to be used when the equations are directly specified with the angular coefficient. It should be noted that if at least one straight parallel to the axis of the ordinate, the formula is not applicable in general, since for such direct angular coefficients is not defined (see article Direct equation on the plane).

There is a third solution to solutions. The idea is to calculate the angle between the guide vectors of the direct vectors using the formula considered in the lesson. Scalar product vectors:

Here, we are not talking about oriented angle, but "just about the coal", that is, the result will be deliberately positive. The snag is that it may turn out a stupid angle (not the one that is needed). In this case, it will have to make a reservation that the angle between direct is a smaller angle, and from the "PI" radian (180 degrees) to deduct the resulting Arkkosinus.

Those who wish can break the task to the third way. But I still recommend to stick to the first approach with an oriented angle, for the reason that it is widespread.

Example 11.

Find the angle between straight.

This is an example for an independent solution. Try to solve it in two ways.

Somehow stalled the fairy tale along the way. Because there is no idiot of the immortal. I have, and not very smelled. To be honest, I thought, the article would be much longer. But still take a recently acquired hat with glasses and go swimming in the September laservo water. Perfectly relieves fatigue and negative energy.

See you soon!

And remember, nobody canceled Babu Yagu \u003d)

Solutions and answers:

Example 3:Decision : Find the line guide vector :

Equation of the desired direct form on the point and a guide vector . Since one of the coordinates of the guide vector zero, equation rewrite in the form:

Answer :

Example 5:Decision :
1) equation direct make up two points :

2) equation direct make up two points :

3) Relevant coefficients for variables Not proportional to: So direct intersect.
4) Find a point :


Note : Here, the first equation of the system is multiplied by 5, then from the 1st equation the 2nd equation is renewed.
Answer :