The underground level is straight on the square. The parallel alignment of curves of a different order. The alignment of straight lines in sections

Curve of a different order- geometric location of a point on a plane, rectangular coordinates

who are satisfied with the zeal for the sight:

in which case one of the coefficients is hired a 11, a 12, a 22 not equal to zero.

Invariant curves of a different order.

The type of curve lies under 4 invariants, pointing below:

Invariant to rotation and rotation of the coordinate system:

Invariant to rotation of the coordinate system ( napivinvariant):

To transform curves of a different order, we look at the solid A*S.

Zagalne equal to the curve of a different order looks like this:

Ax 2 +2Bxy+Cy 2 +2Dx+2Ey+F=0

Yakshcho A*C > 0 eliptical type. Be more eliptical

rivnyannya - the rank of either a primary ellipse, or a degenerate ellipse (point), or a manifest

ellipse (in this case, the alignment does not signify a geometric image on the surface);

Yakshcho A*C< 0 , then jealousy emerges in the form of jealousy hyperbolic type. Be more hyperbolic

The line expresses either a simple hyperbole or a degenerate hyperbole (two straight lines that interchange);

Yakshcho A * C = 0, then the line of a different order will not be central. This type of jealousy is called

equals parabolic type and express on a plane either a simple parabola or 2 parallel

(either avoid) straight lines, or bend on the surface of a given geometric image;

Yakshcho A*C ≠ 0, the curve will be of a different order

This article continues the theme of straight leveling on a flat surface: let’s look at this type of straight leveling, such as straight leveling. Let us state the theorem and prove it; Let's figure out what is different about direct alignment and how to make transitions from backhand alignment to other types of direct alignment. The entire theory is reinforced by illustrations and the most practical instructions.

Let the plane be given a rectilinear coordinate system O x y .

Theorem 1

Whether it is equal to the first step, it looks like A x + B y + C = 0 de A, B, C – the active numbers (A and B are not equal to zero) mean a straight line in a rectangular coordinate system on a plane. In its own way, be it a straight line in a rectangular coordinate system on a plane, it is indicated by equals, which looks like A x + B y + C = 0 with a given set of values ​​A, B, C.

Finished

The theorem is stated in two points, and we will prove it from them.

1. Let us prove that the level A x + B y + C = 0 means straight on the plane.

Let's start with the point M 0 (x 0 , y 0) whose coordinates indicate the level A x + B y + C = 0 . Otzhe: A x 0 + B y 0 + C = 0. Visible from the left and right sides of the level A x + B y + C = 0 from the left and right side of the level A x 0 + B y 0 + C = 0 a new level is removed, which looks like A (x - x 0) + B (y - y 0) = 0 . This is equivalent to A x + B y + C = 0.

Removing the alignment A (x - x 0) + B (y - y 0) = 0 є necessary and sufficient mental perpendicularity of vectors n → = (A , B) and M 0 M → = (x - x 0 , y - y 0 ). Thus, the pointless point M (x, y) defines a straight line in the rectilinear coordinate system, perpendicular to the direction of the vector n → = (A, B). We can assume that this is not the case, but that the vectors n → = (A , B) і M 0 M → = (x - x 0 , y - y 0) were not perpendicular, and equal to A (x - x 0) + B(y - y 0) = 0 would not be true.

Therefore, the line A (x - x 0) + B (y - y 0) = 0 means the action of the straight line in the rectilinear coordinate system on the plane, and then, equivalently, the line A x + B y + C = 0 means the same straight line . This is how we completed the first part of the theorem.

1. Let us prove that, directly in a rectangular coordinate system on a plane, it is possible to set the levels of the first stage A x + B y + C = 0.

Defined in a rectangular coordinate system on a rectangular plane a ; point M 0 (x 0 , y 0), through which this straight line passes, and direct the normal vector of this straight line n → = (A, B) .

Let us also assume that the point M (x, y) is a straight floating point. In this case, the vectors n → = (A, B) і M 0 M → = (x - x 0, y - y 0) are perpendicular to each other, and their scalar solid is zero:

n → , M 0 M → = A (x - x 0) + B (y - y 0) = 0

We rewrite the equation A x + B y - A x 0 - B y 0 = 0, significant C: C = - A x 0 - B y 0 and in the final result the equation A x + B y + C = 0 is removed.

This is how we explained both part of the theorem and the whole theorem as a whole.

Viznachennya 1

Rivnyanna, what it looks like A x + B y + C = 0 – tse back-room straight on a plane in a rectilinear coordinate systemOxy.

Based on the above theorem, we can develop a conclusion that the tasks on the plane of a fixed rectilinear coordinate system, a straight line and a parallel line are inextricably linked. Otherwise, the output direct corresponds to the outer line; The direct line is given to the direct line.

The proof of the theorem also shows that the coefficients A and B, when x and y change, are the coordinates of the normal vector of the line, which is given to the parallel levels of the line A x + B y + C = 0.

Let's take a look at the specific butt of the straight line.

Let the level 2 x + 3 y - 2 = 0 be set, which is indicated by a straight line in a given rectangular coordinate system. The normal vector of a straight line is a vector n → = (2, 3) . A straight line on the chair is clearly defined.

It can also be confirmed in the following way: the straight line, as we sit on the chair, is indicated by the parallel lines 2 x + 3 y - 2 = 0. The coordinates of all points of the given straight line indicate this line.

We can remove the equation λ · A x + λ · B y + λ · C = 0 by multiplying the offending parts of the direct equation by the number λ, which is not equal to zero. The derivation of the level is equivalent to the exiting level, therefore, we describe the same straight line on the surface.

Vicennia 2

Outside the hallway straight line– this is the direct line A x + B y + C = 0, in which the numbers A, B are subtracted from zero. In other cases, equalization is incomprehensible.

Let's take a look at all the variations of the non-vocal literal straight line.

1. If A = 0, ≠ 0, C ≠ 0, the hidden equation looks like B y + C = 0. Such an unusual alignment sets the straight coordinate system O x y in a straight line, which is parallel to the O x axis, leaving behind any effective value x change y value in future -C B. Otherwise, seemingly, the opposite level of the straight line A x + B y + C = 0, if A = 0, B ≠ 0 specifies the geometric location of the point (x, y), the coordinates of which are equal to the same number -C B.
2. If A = 0, B ≠ 0, C = 0, the unequal equation looks like y = 0. Such unequal equalization means the entire abscissa O x.
3. If A ≠ 0, B = 0, C ≠ 0, it is impossible to determine the exact alignment of A x + C = 0, which defines a straight line parallel to the ordinate axis.
4. Let A ≠ 0, B = 0, C = 0, then the uneven alignment will look like x = 0, and this is the alignment of the coordinate line O y.
5. Let's say, when A ≠ 0, B ≠ 0, C = 0, the out-of-the-world equation looks like A x + B y = 0. And ceremonial means straight, like going through the coordinates. In fact, a pair of numbers (0, 0) demonstrates the equality A x + B y = 0, and the fragments A · 0 + B · 0 = 0.

We graphically illustrate all the types of uneven direct alignment.

Butt 1

It is clear that a straight line is given that is parallel to the ordinate axis and passes through point 2 7 - 11. It is necessary to write down the exact value of the given line.

Decision

A straight line parallel to the ordinate axis is defined as A x + C = 0, in which A ≠ 0. Also, the mental task of the coordinates of the point through which the straight line passes, and the coordinates of this point indicate to the minds of the uneven equation A x + C = 0, then. true zeal:

A 2 7 + C = 0

You can value C if you give A a non-zero value, for example, A = 7. In this case we can deduce: 7 · 2 7 + C = 0 ⇔ C = - 2 . We are aware of the difference between the coefficients A and C, we substitute them with the equation A x + C = 0 and remove the necessary equation of the straight line: 7 x - 2 = 0

Subject: 7 x - 2 = 0

Butt 2

The chair is depicted directly, it is necessary to write down its position.

Decision

The positioned chair allows us to easily take the output data for the most important task. Mi bachimo on a chair, which is given a straight line parallel to the O x axis to pass through the point (0, 3) .

Directly, since the abscis is parallel to the eye, it means that the opposite direction is equal to B y + C = 0. The values ​​of B and C are known. The coordinates of the point (0, 3), as long as a given straight line passes through it, will be satisfied with the alignment of the straight line B y + C = 0, then the fair alignment is: · 3 + C = 0. Set for any value, other than zero. Let's say, Y = 1, if we equal 3 + Z = 0, we can know Z: Z = - 3. Based on the value of i 3, it is necessary to equalize the straight line: y - 3 = 0.

Subject: y - 3 = 0.

A straight line that passes through a given point on the plane

If a straight line is given to pass through the point M 0 (x 0 , y 0) then its coordinates suggest a straight line, then. Correctness of equality: A x 0 + B y 0 + C = 0. It is clear that the left and right parts of the total alignment are straight. Rejected: A (x - x 0) + B (y - y 0) + C = 0 the center is equivalent to the output halal, passing through the point M 0 (x 0, y 0) and the normal vector n → = (A, B ).

The result that we have taken makes it possible to write down the direct alignment of the straight line with the given coordinates of the normal vector of the straight line and the coordinates of the point of the straight line.

Butt 3

Given a point M 0 (- 3, 4), through which a straight line passes, and the normal vector of the straight line n → = (1, - 2) . It is necessary to write down the alignment of the given line.

Decision

The outputs allow you to extract the necessary data for the calculation: A = 1, B = - 2, x 0 = - 3, y 0 = 4. Todi:

A (x - x 0) + B (y - y 0) = 0 ⇔ 1 (x - (- 3)) - 2 y (y - 4) = 0 ⇔ ⇔ x - 2 y + 22 = 0

The story could have been written differently. The outer line of the straight line looks like A x + B y + C = 0. The normal vector allows you to subtract the values ​​of coefficients A and B as follows:

A x + B y + C = 0 ⇔ 1 x - 2 y + C = 0 ⇔ x - 2 y + C = 0

Now we know the value of Z, and the point M 0 (- 3, 4) is given by the mind, which is to pass through the straight line. The coordinates of this point indicate the alignment x - 2 · y + C = 0, then. - 3 - 2 · 4 + C = 0. Star Z = 11. The necessary straight line looks like: x - 2 · y + 11 = 0.

Subject: x - 2 y + 11 = 0.

Butt 4

Given a straight line 2 3 x - y - 1 2 = 0 and a point M 0 that lies on this straight line. There are more abscise points, and there is a corresponding one - 3. It is necessary to calculate the ordinate of the given point.

Decision

Specify the coordinates of point M0 as x0 and y0. The output data indicates that x0 = -3. If the point lies on a given straight line, then its coordinates indicate the correct alignment of the straight line. Then there will be zeal:

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

Significantly y 0: 2 3 · (- 3) - y 0 - 1 2 = 0 ⇔ - 5 2 - y 0 = 0 ⇔ y 0 = - 5 2

Subject: - 5 2

Transition from the legal level of direct to other types of level of direct and back

As we know, most of the types of leveling are the same and the same straight on the plane. The choice of jealousy to lie in the minds of the task; You can choose the one that is most convenient for you. Here you really need the skill of converting the equalness of one species into the equalness of another species.

Let's first look at the transition from the literal level to the form A x + B y + C = 0 to the canonical level x - x 1 a x = y - y 1 a y .

If A ≠ 0 then it is transferable additionally B y to the right part of the halal level. On the left side, carry A by the arms. The result can be deduced: A x + C A = - B y.

This jealousy can be written as a proportion: x + C A - B = y A.

When B ≠ 0, only the addition of A x is removed from the left side of the literal equation, and the rest is transferred to the right side, subtracted: A x = - B y - C . We carry - by the arms, then: A x = - B y + C B.

Let's rewrite zeal as a proportion: x - B = y + C B A .

Of course, there is no need to memorize the formula. It is enough to know the algorithm that operates during the transition from the official level to the canonical one.

Butt 5

The line level of the straight line 3 y - 4 = 0 is installed. It is necessary to remake them from the canonical level.

Decision

We write the output as 3 y - 4 = 0. The next step is to follow the algorithm: the left side is deprived of the additional input 0 x; and on the right side there is a charge - 3 for the arms; derivable: 0 x = - 3 y - 43.

Let’s write down the equation as a proportion: x - 3 = y - 430. This is how we took the canonical form of the equation.

Version: x - 3 = y - 4 3 0.

To turn a straight line into a parametric line, first create a transition to the canonical look, and then move from a canonical straight line to a parametric line.

Butt 6

The direct line is given to the equals 2 x – 5 y – 1 = 0 . Write down parametric equalization of direct values.

Decision

It is possible to transition from the galal level to the canonical one:

2 x - 5 y - 1 = 0 ⇔ 2 x = 5 y + 1 ⇔ 2 x = 5 y + 1 5 ⇔ x 5 = y + 1 5 2

Now we accept the offended parts of the removed canonical equal relationship with equals:

x 5 = λ y + 1 5 2 = λ ⇔ x = 5 λ y = - 1 5 + 2 λ , λ ∈ R

Subject:x = 5 λ y = - 1 5 + 2 λ , λ ∈ R

The linear alignment can be converted into a direct alignment with the cut coefficient y = k x + b, or only if B ≠ 0. To move to the left side, additional data must not be transferred to the right. Removable: B y = - A x - C. We can separate the offending parts of the offset on B subtracted from zero: y = - A B x - C B .

Butt 7

The line level of the straight line is set: 2 x + 7 y = 0. It is necessary to turn this relationship into a relationship with the global coefficient.

Decision

Here are the specific steps required for the algorithm:

2 x + 7 y = 0 ⇔ 7 y - 2 x ⇔ y = - 2 7 x

Subject: y = - 2 7 x.

From the direct equation, you can simply remove the equation from the sections in the form x a + y b = 1. To make such a transition, we transfer the number C from the right part of the jealousy, separating the offending parts of the achieved jealousy to – C and, accordingly, transferred from the sign of the coefficient when x and y change:

A x + B y + C = 0 ⇔ A x + B y = - C ⇔ ⇔ A - C x + B - C y = 1 ⇔ x - C A + y - C B = 1

Butt 8

It is necessary to change the horizontal alignment of the straight line x - 7 y + 1 2 = 0 alignment of the straight line in the sections.

Decision

Moved 1 2 to the right side: x - 7 y + 1 2 = 0 ⇔ x - 7 y = - 1 2 .

Divided into -1/2 equal parts: x - 7 y = - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y = 1.

Subject: x-1 2 + y 1 14 = 1 .

It is difficult to carry out the gateway transition: from other levels to the gateway.

Straight raspberry in cuts and raisin with a cut factor can be easily converted at the counter by simply collecting all the warehouses on the left side of the ravine:

x a + y b ⇔ 1 a x + 1 b y - 1 = 0 ⇔ A x + B y + C = 0 y = k x + b ⇔ y - k x - b = 0 ⇔ A x + B y + C = 0

The canonical battle is recreated in the kitchen using the following scheme:

x - x 1 a x = y - y 1 a y ⇔ a y · (x - x 1) = a x (y - y 1) ⇔ ⇔ a y x - a x y - a y x 1 + a x y 1 = 0 ⇔ A x + B y + C =

To transition from parametric ones, you need to go to the canonical one, and then to the initial one:

x = x 1 + a x · λ y = y 1 + a y · λ ⇔ x - x 1 a x = y - y 1 a y ⇔ A x + B y + C = 0

Butt 9

Specify the parametric alignment of the straight line x = - 1 + 2 · y = 4. It is necessary to write down the hidden value of the direct line.

Decision

It is possible to transition from parametric levels to canonical:

x = - 1 + 2 · λ y = 4 ⇔ x = - 1 + 2 · λ y = 4 + 0 · λ ⇔ λ = x + 1 2 λ = y - 4 0 ⇔ x + 1 2 = y - 4 0

Let's move from canonical to colloquial:

x + 1 2 = y - 4 0 ⇔ 0 · (x + 1) = 2 (y - 4) ⇔ y - 4 = 0

Subject: y - 4 = 0

Butt 10

The alignment of the straight line is specified in the sections x 3 + y 1 2 = 1. It is necessary to make the transition to a formal form of alignment.

Decision:

Let’s just rewrite the tribute to the necessary view:

x 3 + y 1 2 = 1 ⇔ 1 3 x + 2 y - 1 = 0

Subject: 1 3 x + 2 y - 1 = 0.

Folding straight line

Most of all, we talked about those that can be written down using the given coordinates of the normal vector and the coordinates of the point through which a straight line passes. This direct line is equal to A(x – x 0) + B (y – y 0) = 0. There we picked up an excellent buttstock.

Now let's take a look at folding stocks, for which it is first necessary to determine the coordinates of the normal vector.

Butt 11

Given a straight line parallel to the straight line 2 x - 3 y + 3 3 = 0 . Also, from the point M 0 (4, 1), a straight line is given to pass through. It is necessary to write down the alignment of the given line.

Decision

The main thing is to tell us about those that are directly parallel, just like a normal vector is straight, which needs to be written down, let’s take a straight vector n → = (2 , - 3) : 2 x - 3 y + 3 3 = 0 . Now we know all the necessary data in order to make the comparison straight:

A (x - x 0) + B (y - y 0) = 0 ⇔ 2 (x - 4) - 3 (y - 1) = 0 ⇔ 2 x - 3 y - 5 = 0

Subject: 2 x - 3 y - 5 = 0.

Butt 12

A straight line is given to pass through the coordinate root perpendicularly to the straight line x - 2 3 = y + 4 5. It is necessary to bend the line of the given straight line.

Decision

The normal vector of a given line will be the normal vector of lines x - 2 3 = y + 4 5.

Todi n → = (3, 5) . Go straight through the cob of coordinates, then. through point O (0, 0). The folded line level is given by the straight line:

A (x - x 0) + B (y - y 0) = 0 ⇔ 3 (x - 0) + 5 (y - 0) = 0 ⇔ 3 x + 5 y = 0

Vіdpovid: 3 x + 5 y = 0

If you have marked a favor in the text, please see it and press Ctrl+Enter

As shown above, direct alignment can be written in three forms: direct alignment, parametric direct alignment and canonical straight alignment. Let's take a look at the transition from straight line in one type to straight line in another look.

First of all, it is important to note that when you specify the alignment of a straight line in parametric form, then you specify the point through which the straight line and the direct vector of the straight line pass. It is not important for him to write down the line in canonical form.

butt.

The straight line is given in parametric form

Decision.

Pass straight through a point
and is the direct vector
. Well, the canonical righteousness is clearly visible

.

The problem of transition from canonical straight lines to parametric straight lines is solved in a similar way.

The transition from canonical levels to direct levels can be seen below on the butt.

butt.

Given the canonical straight line

.

Write down the direct line.

Decision.

Let us write the canonical ranks directly in the form of a system of two ranks

.

By eliminating the banners by multiplying both parts of the first level by 6, and the other level by 4, we cancel the system

.

.

The straight line system has been removed.

Let's take a look at the transition from parametric straight lines to parametric and canonical straight lines. To write canonical and parametric directions of lines, you need to know the point through which the line passes and the line vector. How to determine the coordinates of two points
і
, which lies on a straight line, then as a direct vector m we can take the vector
. The coordinates of two points that lie on a straight line can be deduced as a solution to the alignment system, which means the opposite alignment of the straight line. Yaku is a point, there is a straight line to go through Yaku, you can take it from the point
і
. The above is illustrated from the butt.

butt.

Given a direct line

.

Decision.

We know the coordinates of two points that lie on a straight line, as a result of this alignment system. Respectfully
, we cancel the level system

.

Virus of this system, we know
. Ozhe, period
lie on a straight line. Respectfully
, we cancel the level system

,

Apparently we know
. Ozhe, go straight through the point
. Todi yak direct vector you can take the vector

.

Ozhe, go straight through the point
and is the direct vector
. Well, the parametric straight line looks like

.

Then the canonical equalities of the direct will be written as

.

Another way to find a direct vector is directly based on the direct alignment of the planes, which means the normals to these planes.

Let the dark side of the straight line loom

і - normals to the first and other planes, obviously. Todi vector
You can take it as a straight vector. In fact, a straight line, being a line of the crossbar of these planes, is immediately perpendicular to the vectors. і . Well, that's colinear to the vector.
And this vector can be taken as a straight vector. Let's take a look at the butt.

butt.

Given a direct line

.

Write down the parametric and canonical straight lines.

Decision.

Straight line between planes and normals
і
. Beremo yak direct vector direct vector

We know the point that lies on a straight line. We know the point that lies on a straight line. Let's go
. Then we remove the system

.

Virus the system, we know
.Ozhe, period
lie on a straight line. Then the parametric equalization of the direct line can be written as

.

The canonical straight lines are visible

.

Ok, you can go to the canonical levels by turning on one of the levels in one level, and then another change. Let's look at this method from the butt.

butt.

Given a direct line

.

Record canonical equalities directly.

Decision.

Included from another equal measure, adding to the next first, multiplied by more. Cannot be rejected

.

.

Now we can turn it off from another level. , Having added to the new level, multiplied by two. Cannot be rejected

.

.

The direct line is clearly canonical

.

.

.

We were told that the curve of an algebra of a different order is indicated by the algebraic equations of a different level of order Xі at. The zagalom is written like this:

A X 2+V xy+ Z at 2+D x+ E y+ F = 0, (6)

Moreover, A 2 + B 2 + C 2 ¹ 0 (so the numbers A, B, C do not become zero overnight). Dodanki A X 2 , V xy, W at 2 are called senior members of the rank, number

called discriminant Whose jealousy. Rivnyanna (6) is called Zagalnym Rivnyany a curve of a different order.

For the previously reviewed curves:

Elips: Þ A = , B = 0, C = , D = E = 0, F = -1,

colo X 2 + at 2 = A 2 Þ A = C = 1, B = D = E = 0, F = – A 2 d = 1>0;

Hyperbole: Þ A = , B = 0, C = - , D = E = 0, F = -1,

d = -.< 0.

Parabola: at 2 = 2pxÞ A = B = 0, C = 1, D = -2 R, E = F = 0, d = 0,

X 2 = 2RU A = 1B = C = D = 0, E = -2 R, F = 0, d = 0

Curves assigned to equals (6) are called central krivimi, yakscho d10. If d>0, then the curve eliptic type, like d<0, то кривая hyperbolic type. Curves for which d = 0 are curves parabolic type.

It has been reported that the line is of a different order in come what may The Cartesian coordinate system is given to algebraic elements of a different order. Some systems have a folded look (for example, (6)), while others have a simpler look, for example, (5). Therefore, it is easy to look at such a coordinate system, in which the curve that flows is written in the most simple (for example, canonical) terms. The transition from one coordinate system, in which the curve is given to the equal view (6) to another, where the equal view has a simpler view, is called reorganization of coordinates.

Let's take a look at the main types of coordinates.

I. Re-creation of the transfer coordinate axes (with savings directly). Let the output coordinate system XOU point M have coordinates ( X, atX¢, at¢). From the chair you can see that the coordinates of point M in different systems are connected by connections

(7), or (8).

Formulas (7) and (8) are called coordinate transformation formulas.

ІІ. Turning around coordinate axes on cut a. Since the output coordinate system XOU point M has coordinates ( X, at), and in the new coordinate system XO¢U there are coordinates ( X¢, at¢). Then the connections between these coordinates are expressed by the formulas

, (9)

or else

With the help of reversing the coordinates, the level (6) can be brought to one of the advancing canonical Rivnyan.

1) - ellipse,

2) - hyperbole,

3) at 2 = 2px, X 2 = 2RU– parabola

4) A 2 X 2 – b 2 y 2 = 0 – a pair of straight lines that overlap (Fig. a)

5) y 2 – a 2 = 0 - a pair of parallel lines (Fig. b)

6) x 2 –a 2 = 0 - a pair of parallel lines (Fig. c)

7) y 2 = 0 - avoid straight lines (all OX)

8) x 2 = 0 - straight lines are avoided (all op-amps)

9) a 2 X 2 + b 2 y 2 = 0 - point (0, 0)

10) obvious elipse

11)y 2 + a 2 = 0 - a pair of obvious straight lines

12) x 2 + a 2 = 0 pair of straight lines.

Each of these lines is related to the lines of a different order. Lines that are designated by ranks 4 – 12 are called virogenimi curves of a different order.

Let's take a look at the butt of transforming the zagalny curve of the curve to its canonical form.

1) 9X 2 + 4at 2 – 54X + 8at+ 49 = 0 Þ (9 X 2 – 54X) + (4at 2 + 8at) + 49 = 0 Þ

9(X 2 – 6X+ 9) + 4(at 2 + 2at+ 1) - 81 - 4 + 49 = 0 Þ 9( X –3) 2 + 4(at+ 1) = 36, Þ

.

We agree X¢ = X – 3, at¢ = at+ 1, the canonical alignment of the ellipse is rejected . Zealousness X¢ = X – 3, at¢ = at+ 1 indicates the transformation of the transfer of the coordinate system at the point (3, -1). Having used the old and new coordinate systems, it is not important to depict the ellipse.

2) 3at 2 +4X– 12at+8 = 0. Convertible:

(3at 2 – 12at)+ 4 X+8 = 0

3(at 2 – 4at+4) - 12 + 4 X +8 = 0

3(y – 2) 2 + 4(X –1) = 0

(at – 2) 2 = – (X – 1) .

We agree X¢ = X – 1, at¢ = at– 2, the level of the parabola is eliminated at¢ 2 = – X¢. The selected replacement indicates a transfer of the coordinate system to the point O¢(1,2).

If the PDSC has been introduced on the area, then the level of the first stage should be based on exact coordinates and

, (5)

de і immediately does not reach zero, which means straight.

The correct and the turning point: in the PDSC, it can be directly assigned to the first-level peers of the form (5).

Reverence to the species (5) is called direct .

Around the drop of the level (5) is indicated at the next table.

	coefficient values	Rivnyannya direct	Straight position
			Go straight through the coordinate cob
			Directly parallel to the axis
			Directly parallel to the axis
			Goes straight to the chase
			Goes straight to the chase

The straight line with the cut coefficient and the cob ordinate.

U glom nahilu straight to the axis
called the smallest kut
, which requires turning the entire abscissa against the year arrow until it runs from the straight line (Fig. 6). Directly, whatever direct characterizes it Kutovym coefcient , Which is indicated by the tangent of the kuta nahilu
This is straight forward, then.

.

The screws become no longer straight, perpendicular to the axis
there is no cutoff coefficient.

Direct line, which has a cut coefficient and everything is shaking
at the point, the ordinate of which is ancient (Pochatkov ordinate) , make an appointment with the viewer

.

Rivnyannya straight from the cut-outs

Rivnyanyam direct from the cuttings is called jealousy of appearance

, (6)

de і
Obviously, there are at least two sections that intersect with a straight line on the coordinate axes taken from the symbols.

Rivnyannya is straight, to pass through any point in any direction. Straight bundle

Rivnyannya straight line to pass through this point
and there is a cut coefficient Sign up with Viglyada

. (7)

Straight bundle called the collection of straight planes that pass through one point
beam center. If we know the coordinates of the center of the beam, then the straight line (8) can be seen as a straight beam, fragments of any straight beam can be removed from the straight line (8) for the corresponding value of the cut coefficient (Set the axis straight, parallel to the axis
її jealousy
).

As we see the hidden alignment of two straight lines that lie in a bundle
and (they create a beam), then the relationship of any direct line from which beam can be written down at a glance

Rivnyannya straight, to pass through two points

A straight line that passes through two given points
і
, may look

.

Yakscho points
і
mean straight line parallel to the axis

or axes

, then the similarity of such a direct line is clearly written in the form

or else
.

Mutual separation of two straight lines. Cut between straight lines. Mind the parallels. Mind perpendicularity

Mutual retouching of two direct lines, given by the hidden equalities

і ,

presented in the following table.

Pid where there are two straight lines This is, of course, one of the contiguous cuts created during their transition. Gostriy kut between straight lines
m
is indicated by the formula

.

Dear, that you would like one of these straight lines to be parallel to the axis
, then formula (11) does not make sense, so we will vikoristuvat zagalny equal straight lines

ta .

Formula (11) I can see

.

Mind the parallels:

or else
.

Mind perpendicularity:

or else
.

Normal alignment is straight. Place the points in line with the straight line. Rivnyannya bisectors

Normal alignment is straight I can see

de
the dove of the perpendicular (normal) lowered from the beginning of the coordinates on the straight line,
the nearest perpendicular to the axis
. To bring the underground level straight
to look normal, you need to multiply the offensive part of jealousy (12) by normal multiplier
taking the sign protilegous to the sign of a free member .

Vidstan specks
straight view know the formulas

. (9)

Rivnyannya of bisectors of cutives between straight lines
і :

.

Zavdannya 16. Dana is straight
. Linear slopes are straight, like passing through a point
parallel to the direct one.

Decision. Behind the mental parallelism of straight lines
. For the main task, we will use the straight line to pass through this point
at tsomu directly (8):

.

We know the exact coefficient of the price. For which type of legal alignment straight (5) let’s move on to alignment with the cut coefficient (6) (variably through ):

Otje,
.

Zavdannya 17. Know the point
, symmetrical point
It's straight forward
.

Decision. In order to know the point symmetrical to the point It's straight forward (Fig.7) it is necessary:

1) drop the dots directly perpendicular,

2) know the basis of this perpendicular
point ,

3) on the extended perpendicular to the section of the cut
.

So, let’s write down a straight line to pass through a point perpendicular to the given line. For this purpose, the speed of a straight line is to pass through a given point in a given direction (8):

.

Substitute the coordinates of the point
:

. (11)

The cut coefficient is known from the perpendicularity of straight lines:

.

Cut coefficient of this direct data

,

Well, the cut coefficient of the perpendicular straight line

.

Let's put him in the equation (11):

Let's find out the point
the point of the crossbar of a given straight line and a perpendicular straight line. So that's the point To be both straight, their coordinates satisfy their equals. So, to find the coordinates of the crossbar point, you need to create a system of lines, composed of lines of these lines:

System solutions
,
, then.
.

Krapka є in the middle of the cut
, Todi from formulas (4):

,
,

we know the coordinates of the point
:

In this manner, shukana period
.

Zavdannya 18.Slopes are straight, like passing through a point
and it rises from the coordinate point of the trikutnik with an area that is more than 150 sq.d. (Fig.8).

Decision. To solve the problem, you can use the straight line “in the sections” (7):

. (12)

Bo point
lies on a straight line, then its coordinates will satisfy the alignment of the straight line:

.

The area of ​​the tricubitus, which appears as a straight line from the coordinate point, is calculated using the formula:

(record module, fragments і may be negative).

In this manner, the system was set up for adjusting the parameters і :

This system is equivalent to two systems:

First system solutions
,
і
,
.

Solutions for another system
,
і
,
.

Explicitly found values ​​of equalization (12):

,
,
,
.

Let's write down the following lines:

,
,
,
.

Zavdannya 19. Calculate the distance between parallel lines
і
.

Decision. Stand between the parallel lines and the distance between the points of one straight line and the other straight line.

Vibermo direct point
Of course, you can also specify one coordinate, so for example
then
.

Now we know where the points are to straight line following formula (10):

.

In this manner, stand between these parallel straight lines.

Zavdannya 20. Find the straight line to pass through the point of the straight line
і
(I don’t know the point of the cross) that

Decision. 1) Let’s write down the alignment of the bundle of straight lines with the known adjusters (9):

Then there’s a real jealousy

It is necessary to know such values
і , in which straight beam will pass through the point
, i.e., the coordinates of the culprit will satisfy the level (13):

Possibly known
to the level (13) and after simplifying it is removed directly:

.

.

The mental parallelism of straight lines speeds up:
. Let's find out what the coefficients are for direct і . Maemo, what?
,
.

Otje,

The meaning is imaginable
in line (13) and simply, we remove the line of the straight line
.

Preservation for independent virtuousness.

Zavdannya 21. Straight slopes to pass through points
і
: 1) with a coefficient; 2) underground; 3) "at the cutting edges".

Zavdannya 22. Linear slopes are straight, like passing through a point and gets everything done
kut
, as 1)
,
; 2)
,
.

Zavdannya 23. Write the equal sides of a rhombus with diagonals of 10 cm and 6 cm, taking the great diagonal for the entire
, and menshu
for the whole
.

Zavdannya 24. Even-sided tricutnik
On the other side, which is equal to 2 units, spread out as shown in small 9. The folds of the same side.

Zavdannya 25. Through the point
draw a straight line that appears on the positive side of the coordinates of the cutting plane.

Zavdannya 26. Find the area of ​​the tricubitus, which is located in front of the coordinate line:

1)
; 2)
.

Zavdannya 27.Write a straight line that passes through a point And it appears from the coordinate point of the trikutnik, which is flatter than the previous one. , yakscho

1)
,
sq. od.; 2)
,
sq. od.

Zavdannya 28. Given the apex of the tricutaneous
. Find the level of the middle line, parallel to the side
, yakscho