We said that a second-order algebraic curve is defined algebraic equation second degree relative to X And at. IN general view such an equation is written as

A X 2 + B hu+ C at 2+D x+ E y+ F = 0, (6)

where A 2 + B 2 + C 2 ¹ 0 (that is, the numbers A, B, C do not vanish at the same time). Terms A X 2 , V hu, WITH at 2 are called senior terms of the equation, the number

called discriminant this equation. Equation (6) is called general equation curve of the second order.

For the previously considered curves we have:

Ellipse: Þ A = , B = 0, C = , D = E = 0, F = –1,

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

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

d = - .< 0.

Parabola: at 2 = 2pxÞ A \u003d B \u003d 0, C \u003d 1, D \u003d -2 R, E = F = 0, d = 0,

X 2 = 2RUÞ A \u003d 1B \u003d C \u003d D \u003d 0, E \u003d -2 R, F = 0, d = 0.

The curves given by equation (6) are called central curves if d¹0. If d> 0, then the curve elliptical type if d<0, то кривая hyperbolic type. Curves for which d = 0 are curves parabolic type.

It is proved that the second-order line in any Cartesian coordinate system is given by a second-order algebraic equation. Only in one system the equation has a complex form (for example, (6)), and in the other it is simpler, for example, (5). Therefore, it is convenient to consider such a coordinate system in which the curve under study is written by the simplest (for example, canonical) equation. The transition from one coordinate system, in which the curve is given by an equation of the form (6) to another, where its equation has a simpler form, is called coordinate transformation.

Consider the main types of coordinate transformations.

I. Transfer Transform coordinate axes (with preservation of direction). Let the point M in the initial XOU coordinate system have coordinates ( X, atX¢, at¢). It can be seen from the drawing that the coordinates of the point M in different systems are related by the relations

(7), or (8).

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

II. Rotate Transform coordinate axes by angle a. If in the initial XOU coordinate system the point M has coordinates ( X, at), and in the new XO¢Y coordinate system it has coordinates ( X¢, at¢). Then the relationship between these coordinates is expressed by the formulas

, (9)

or

Using the coordinate transformation, equation (6) can be reduced to one of the following canonical equations.

1) - ellipse,

2) - hyperbole,

3) at 2 = 2px, X 2 = 2RU- parabola

4) A 2 X 2 – b 2 y 2 \u003d 0 - a pair of intersecting lines (Fig. a)

5) y 2 – a 2 \u003d 0 - a pair of parallel lines (Fig. b)

6) x 2 –a 2 \u003d 0 - a pair of parallel lines (Fig. c)

7) y 2 = 0 - coinciding lines (OX axis)

8) x 2 = 0 - coinciding lines (OS axis)

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

10) imaginary ellipse

11) y 2 + a 2 = 0– pair of imaginary lines

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

Each of these equations is a second-order line equation. The lines defined by equations 4 - 12 are called degenerate curves of the second order.

Let us consider examples of transformation of the general equation of a curve to a canonical form.

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

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

.

Let's put X¢ = X – 3, at¢ = at+ 1, we get canonical equation ellipse . Equality X¢ = X – 3, at¢ = at+ 1 define the translation transformation of the coordinate system to the point (3, –1). Having built the old and new coordinate systems, it is easy to draw this ellipse.

2) 3at 2 +4X– 12at+8 = 0. Let's transform:

(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) .

Let's put X¢ = X – 1, at¢ = at– 2, we get the parabola equation at¢2 = - X¢. The chosen substitution corresponds to the transfer of the coordinate system to the point О¢(1,2).

Properties of a straight line in Euclidean geometry.

There are infinitely many lines that can be drawn through any point.

Through any two non-coinciding points, there is only one straight line.

Two non-coincident lines in the plane either intersect at a single point, or are

parallel (follows from the previous one).

In three-dimensional space, there are three options for the relative position of two lines:

• lines intersect;

• straight lines are parallel;

• straight lines intersect.

Straight line- algebraic curve of the first order: in the Cartesian coordinate system, a straight line

is given on the plane by an equation of the first degree (linear equation).

General equation of a straight line.

Definition. Any line in the plane can be given by a first order equation

Ah + Wu + C = 0,

and constant A, B not equal to zero at the same time. This first order equation is called general

straight line equation. Depending on the values ​​of the constants A, B And WITH The following special cases are possible:

. C = 0, A ≠ 0, B ≠ 0- the line passes through the origin

. A = 0, B ≠0, C ≠0 ( By + C = 0)- straight line parallel to the axis Oh

. B = 0, A ≠ 0, C ≠ 0 ( Ax + C = 0)- straight line parallel to the axis OU

. B = C = 0, A ≠ 0- the line coincides with the axis OU

. A = C = 0, B ≠ 0- the line coincides with the axis Oh

The equation of a straight line can be represented in various forms depending on any given

initial conditions.

Equation of a straight line by a point and a normal vector.

Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)

perpendicular to the line given by the equation

Ah + Wu + C = 0.

Example. Find the equation of a straight line passing through a point A(1, 2) perpendicular to the vector (3, -1).

Solution. Let's compose at A \u003d 3 and B \u003d -1 the equation of the straight line: 3x - y + C \u003d 0. To find the coefficient C

we substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore

C = -1. Total: the desired equation: 3x - y - 1 \u003d 0.

Equation of a straight line passing through two points.

Let two points be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2 , z 2), Then straight line equation,

passing through these points:

If any of the denominators is equal to zero, the corresponding numerator should be set equal to zero. On

plane, the equation of a straight line written above is simplified:

If x 1 ≠ x 2 And x = x 1, If x 1 = x 2 .

Fraction = k called slope factor straight.

Example. Find the equation of a straight line passing through the points A(1, 2) and B(3, 4).

Solution. Applying the above formula, we get:

Equation of a straight line by a point and a slope.

If the general equation of a straight line Ah + Wu + C = 0 bring to the form:

and designate , then the resulting equation is called

equation of a straight line with slope k.

The equation of a straight line on a point and a directing vector.

By analogy with the point considering the equation of a straight line through the normal vector, you can enter the task

a straight line through a point and a direction vector of a straight line.

Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition

Aα 1 + Bα 2 = 0 called direction vector of the straight line.

Ah + Wu + C = 0.

Example. Find the equation of a straight line with direction vector (1, -1) and passing through point A(1, 2).

Solution. We will look for the equation of the desired straight line in the form: Ax + By + C = 0. According to the definition,

coefficients must satisfy the conditions:

1 * A + (-1) * B = 0, i.e. A = B.

Then the equation of a straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0.

at x=1, y=2 we get C/ A = -3, i.e. desired equation:

x + y - 3 = 0

Equation of a straight line in segments.

If in the general equation of the straight line Ah + Wu + C = 0 C≠0, then, dividing by -C, we get:

or , where

The geometric meaning of the coefficients is that the coefficient a is the coordinate of the intersection point

straight with axle Oh, A b- the coordinate of the point of intersection of the line with the axis OU.

Example. The general equation of a straight line is given x - y + 1 = 0. Find the equation of this straight line in segments.

C \u003d 1, , a \u003d -1, b \u003d 1.

Normal equation of a straight line.

If both sides of the equation Ah + Wu + C = 0 divide by number , which is called

normalizing factor, then we get

xcosφ + ysinφ - p = 0 -normal equation of a straight line.

The sign ± of the normalizing factor must be chosen so that μ * C< 0.

R- the length of the perpendicular dropped from the origin to the line,

A φ - the angle formed by this perpendicular with the positive direction of the axis Oh.

Example. Given the general equation of a straight line 12x - 5y - 65 = 0. Required to write various types of equations

this straight line.

The equation of this straight line in segments:

The equation of this line with slope: (divide by 5)

Equation of a straight line:

cos φ = 12/13; sin φ= -5/13; p=5.

It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines,

parallel to the axes or passing through the origin.

Angle between lines on a plane.

Definition. If two lines are given y \u003d k 1 x + b 1, y \u003d k 2 x + b 2, then the acute angle between these lines

will be defined as

Two lines are parallel if k 1 = k 2. Two lines are perpendicular

If k 1 \u003d -1 / k 2 .

Theorem.

Direct Ah + Wu + C = 0 And A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients are proportional

A 1 \u003d λA, B 1 \u003d λB. If also С 1 \u003d λС, then the lines coincide. Coordinates of the point of intersection of two lines

are found as a solution to the system of equations of these lines.

The equation of a line passing through a given point is perpendicular to a given line.

Definition. A line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b

represented by the equation:

The distance from a point to a line.

Theorem. If a point is given M(x 0, y 0), then the distance to the line Ah + Wu + C = 0 defined as:

Proof. Let the point M 1 (x 1, y 1)- the base of the perpendicular dropped from the point M for a given

direct. Then the distance between the points M And M 1:

(1)

Coordinates x 1 And 1 can be found as a solution to the system of equations:

The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicularly

given line. If we transform the first equation of the system to the form:

A(x - x 0) + B(y - y 0) + Ax 0 + By 0 + C = 0,

then, solving, we get:

Substituting these expressions into equation (1), we find:

The theorem has been proven.

If PDCS is introduced on the plane, then any equation of the first degree with respect to the current coordinates and

, (5)

Where And simultaneously not equal to zero defines a straight line.

The converse statement is also true: in PDSC, any straight line can be given by a first-degree equation of the form (5).

An equation of the form (5) is called the general equation of a straight line .

Particular cases of equation (5) are given in the following table.

	The value of the coefficients	Equation of a straight line	Line position
			The line passes through the origin
			Straight line parallel to the axis
			Straight line parallel to the axis
			The straight line coincides with the axis
			The straight line coincides with the axis

Equation of a straight line with slope and initial ordinate.

At glom inclination straight to the axis
called the smallest angle
, on which you need to turn the abscissa axis counterclockwise until it coincides with this straight line (Fig. 6). The direction of any straight line is characterized by its slope factor , which is defined as the tangent of the slope
this straight line, i.e.

.

The only exception is a straight line perpendicular to the axis
, which has no slope.

The equation of a straight line with a slope and crossing the axis
at a point whose ordinate is (initial ordinate) , is written as

.

Equation of a straight line in segments

Equation of a straight line in segments is called an equation of the form

, (6)

Where And
respectively, the lengths of the segments cut off by a straight line on the coordinate axes, taken with certain signs.

Equation of a line passing through a given point in a given direction. bundle of straight lines

Equation of a line passing through a given point
and having a slope is written in the form

. (7)

A bunch of straight lines is the collection of lines in a plane passing through one and the same point
beam center. If the coordinates of the beam center are known, then equation (8) can be considered as the beam equation, since any straight line of the beam can be obtained from equation (8) with the corresponding value of the slope (the exception is a straight line, which is parallel to the axis
her equation
).

If the general equations of two lines belonging to the pencil are known
and (generators of the bundle), then the equation of any straight line from this bundle can be written in the form

Equation of a line passing through two points

Equation of a line passing through two given points
And
, has the form

.

If the points
And
define a line parallel to the axis

or axes

, then the equation of such a straight line is written, respectively, in the form

or
.

Mutual arrangement of two straight lines. Angle between lines. Parallel condition. Perpendicular condition

Mutual arrangement of two straight lines given by general equations

And ,

presented in the following table.

Under angle between two lines one of the adjacent angles formed at their intersection is understood. Acute angle between straight lines
m
, is determined by the formula

.

Note that if at least one of these lines is parallel to the axis
, then formula (11) does not make sense, so we will use the general equations of lines

And .

formula (11) takes the form

.

Parallel condition:

or
.

Perpendicular condition:

or
.

Normal equation of a straight line. The distance of a point from a line. Bisector equations

Normal equation of a straight line has the form

Where
the length of the perpendicular (normal) dropped from the origin to the straight line,
the angle of inclination of this perpendicular to the axis
. To give the general equation of a straight line
normal form, both parts of equality (12) must be multiplied by normalizing factor
, taken with the opposite sign of the free term .

Distance points
from straight find by formulas

. (9)

Equation of bisectors of angles between straight lines
And :

.

Task 16. Dana straight
. Write an equation for a line passing through a point
parallel to this line.

Solution. By the condition of parallel lines
. To solve the problem, we will use the equation of a straight line passing through a given point
in this direction (8):

.

Find the slope of this straight line. To do this, from the general equation of the straight line (5) we pass to the equation with the slope coefficient (6) (we express through ):

Hence,
.

Problem 17. Find a point
, symmetrical to the point
, relatively straight
.

Solution. To find a point symmetrical to a point relatively straight (Fig.7) it is necessary:

1) lower from point directly perpendicular,

2) find the base of this perpendicular
point ,

3) on the continuation of the perpendicular, set aside a segment
.

So, let's write the equation of a straight line passing through a point perpendicular to this line. To do this, we use the equation of a straight line passing through a given point in a given direction (8):

.

Substitute the coordinates of the point
:

. (11)

We find the angular coefficient from the condition of perpendicularity of the lines:

.

Slope of a given line

,

hence the slope of the perpendicular line

.

Substitute it into equation (11):

Next, let's find a point
the point of intersection of a given line and its perpendicular line. Since the point belongs to both lines, then its coordinates satisfy their equations. This means that in order to find the coordinates of the intersection point, it is required to solve a system of equations composed of the equations of these lines:

System Solution
,
, i.e.
.

Dot is the midpoint of the segment
, then from formulas (4):

,
,

find the coordinates of a point
:

Thus, the desired point
.

Problem 18.Compose the equation of a straight line that passes through a point
and cuts off a triangle with an area equal to 150 square units from the coordinate angle. (Fig.8).

Solution. To solve the problem, we will use the equation of a straight line “in segments” (7):

. (12)

Since the point
lies on the desired line, then its coordinates must satisfy the equation of this line:

.

The area of ​​a triangle cut off by a straight line from the coordinate angle is calculated by the formula:

(the module is written because And may be negative).

Thus, we got a system for finding the parameters And :

This system is equivalent to two systems:

Solution of the first system
,
And
,
.

Solution of the second system
,
And
,
.

We substitute the found values ​​into equation (12):

,
,
,
.

We write the general equations of these lines:

,
,
,
.

Problem 19. Calculate distance between parallel lines
And
.

Solution. The distance between parallel lines is equal to the distance of an arbitrary point of one line to the second line.

Let's choose on a straight line point
arbitrarily, therefore, you can set one coordinate, i.e. for example
, Then
.

Now let's find the distance of the point to straight according to formula (10):

.

Thus, the distance between the given parallel lines is equal.

Task 20. Find the equation of the line passing through the point of intersection of the lines
And
(not finding the intersection point) and

Solution. 1) Let us write down the equation of a pencil of lines with known generators (9):

Then the desired straight line has the equation

It is required to find such values
And , for which the beam line passes through the point
, i.e., its coordinates must satisfy equation (13):

Substitute found
into equation (13) and after simplification we obtain the desired straight line:

.

.

Let's use the condition of parallel lines:
. Let's find the slope coefficients of the lines And . We have that
,
.

Hence,

Substitute the found value
into equation (13) and simplify, we obtain the equation of the desired line
.

Tasks for independent solution.

Task 21. Write the equation of a straight line passing through the points
And
: 1) with a slope; 2) general; 3) "in segments".

Task 22. Write an equation for a line that passes through a point and forms with the axis
corner
, if 1)
,
; 2)
,
.

Task 23. Write the equations for the sides of a rhombus with diagonals of 10 cm and 6 cm, taking the larger diagonal as the axis
, and the smaller
per axle
.

Task 24. Equilateral triangle
with a side equal to 2 units, is located as shown in Figure 9. draw up the equations of its sides.

Problem 25. Through the dot
draw a straight line that cuts off equal segments on the positive semi-axes of coordinates.

Problem 26. Find the area of ​​a triangle that cuts off a straight line from the coordinate angle:

1)
; 2)
.

Problem 27.Write the equation of a straight line passing through a point and cutting off a triangle from the coordinate angle with an area equal to , If

1)
,
sq. units; 2)
,
sq. units

Task 28. Given the vertices of a triangle
. Find the equation of the midline parallel to the side
, If

The general equation of a second-order curve in a plane is:

Ax 2 + 2Bxy + Cy 2 + 2Dx + 2Ey + F = 0, (39)

Where A 2 + B 2 + C 2 0, (A, B, C, D, E, F) R. It defines all possible conic sections arbitrarily located on the plane.

From the coefficients of equation (39) we compose two determinants:

called discriminant of the equation(39), and - discriminant of the leading terms of the equation. At 0, equation (39) determines: > 0 - ellipse;< 0 - гиперболу; = 0 - параболу. В случае = 0 кривые вырождаются в точку или прямые линии.

From the general equation (39) one can pass to the canonical equation if the linear and cross terms are eliminated by passing to new system coordinates coinciding with the axes of symmetry of the figure. Let us replace in (39) x on x + a And y on y + b, Where a, b some constants. Let us write down the obtained coefficients for X And y and equate them to 0

(aa + bb + D)x = 0, (Cb + Ba + E)y = 0. (41)

As a result, equation (39) will take the form:

A(x) 2 + 2B(x)(y) + C(y) 2 + F = 0, (42)

where coefficients A, B, C have not changed, but F= / . The solution of the system of equations (41) will determine the coordinates of the center of symmetry of the figure:

If B= 0, then a = -D/A, b = -E/C and it is convenient to eliminate the linear terms in (39) by the method of reduction to the full square:

Ax 2 + 2Dx = A(x 2 + 2xD/A + (D/A) 2 - (D/A) 2) = A(x + D/A) 2 - D 2 /A.

In equation (42), let's rotate the coordinates through the angle a (38). We write the resulting coefficient at the cross term xy and equate it to 0

xy = 0. (44)

Condition (44) determines the required angle of rotation of the coordinate axes until they coincide with the symmetry axes of the figure and takes the form:

Equation (42) takes the form:

A+X2+ C + Y 2 + F = 0 (46)

from which it is easy to pass to the canonical equation of the curve:

Odds A + , C+ , under condition (45), can be represented as the roots of an auxiliary quadratic equation:

t 2 - (A + C)t + = 0. (48)

As a result, the position and direction of the axes of symmetry of the figure, its semiaxes are determined:

and it can be constructed geometrically.

In the case = 0 we have a parabola. If its axis of symmetry is parallel to the axis Oh, then the equation becomes:

if not, then to the form:

where the expressions in brackets, equated to 0, define the lines of the new coordinate axes: , .

Solving typical problems

Example 15 Give Equation 2 x 2 + 3y 2 - 4x + 6y- 7 = 0 to the canonical form and build a curve.

Solution. B= 0, = -72 0, = 6 > 0 ellipse.

Let's perform the reduction to the full square:

2(x - 1) 2 + 3(y + 1) 2 - 12 = 0.

Symmetry center coordinates (1; -1), linear transformation X = x - 1, Y = y+ 1 brings the equation to the canonical form .

Example 16 Give Equation 2 xy = a 2 to the canonical form and construct a curve.

Solution. B = 1, = a 2 0, = -1 < 0 гипербола .

The center of the coordinate system is located at the center of symmetry of the curve, since there are no linear terms in the equation. Let's rotate the axes through the angle a. By formula (45), we have tg2a = B/(A - C) = , i.e. a = 45°. Coefficients of the canonical equation (46) A + , C+ are determined by equation (48): t 2 = 1 or t 1,2 = 1 A + = 1, C+ = -1, i.e.
X 2 - Y 2 = a 2 or . So Equation 2 hu = A 2 describes a hyperbola with center of symmetry at (0; 0). The axes of symmetry are located along the bisectors of the coordinate angles, the asymptotes are the coordinate axes, the semiaxes of the hyperbola are equal to A.y - 9 =0;

9x 2 + y 2 - 18x + 2y + 1 = 0;

2x 2 + 4X + y - 2 = 0;

3x 2 - 6X - y + 2 = 0;

-x 2 + 4y 2 - 8x - 9y + 16 = 0;

4x 2 + 8X - y - 5 = 0;

9x 2 - y 2 + 18x + 2y - 1 = 0;

9x 2 - 4y 2 + 36x + 16y - 16 = 0.

In this article, we will consider the general equation of a straight line in a plane. Let us give examples of constructing the general equation of a straight line if two points of this straight line are known or if one point and the normal vector of this straight line are known. Let us present methods for transforming an equation in general form into canonical and parametric forms.

Let an arbitrary Cartesian rectangular coordinate system be given Oxy. Consider a first degree equation or linear equation:

Ax+By+C=0,

(1)

Where A, B, C are some constants, and at least one of the elements A And B different from zero.

We will show that a linear equation in the plane defines a straight line. Let us prove the following theorem.

Theorem 1. In an arbitrary Cartesian rectangular coordinate system on a plane, each straight line can be given by a linear equation. Conversely, each linear equation (1) in an arbitrary Cartesian rectangular coordinate system on the plane defines a straight line.

Proof. It suffices to prove that the line L is determined by a linear equation for any one Cartesian rectangular coordinate system, since then it will be determined by a linear equation and for any choice of Cartesian rectangular coordinate system.

Let a straight line be given on the plane L. We choose a coordinate system so that the axis Ox aligned with the line L, and the axis Oy was perpendicular to it. Then the equation of the line L will take the following form:

y=0.

(2)

All points on a line L will satisfy the linear equation (2), and all points outside this straight line will not satisfy the equation (2). The first part of the theorem is proved.

Let a Cartesian rectangular coordinate system be given and let linear equation (1) be given, where at least one of the elements A And B different from zero. Find the locus of points whose coordinates satisfy equation (1). Since at least one of the coefficients A And B is different from zero, then equation (1) has at least one solution M(x 0 ,y 0). (For example, when A≠0, dot M 0 (−C/A, 0) belongs to the given locus of points). Substituting these coordinates into (1) we obtain the identity

Ax 0 +By 0 +C=0.

(3)

Let us subtract identity (3) from (1):

A(x−x 0)+B(y−y 0)=0.

(4)

Obviously, equation (4) is equivalent to equation (1). Therefore, it suffices to prove that (4) defines some line.

Since we are considering a Cartesian rectangular coordinate system, it follows from equality (4) that the vector with components ( x−x 0 , y−y 0 ) is orthogonal to the vector n with coordinates ( A,B}.

Consider some line L passing through the point M 0 (x 0 , y 0) and perpendicular to the vector n(Fig.1). Let the point M(x,y) belongs to the line L. Then the vector with coordinates x−x 0 , y−y 0 perpendicular n and equation (4) is satisfied (scalar product of vectors n and equals zero). Conversely, if the point M(x,y) does not lie on a line L, then the vector with coordinates x−x 0 , y−y 0 is not orthogonal to vector n and equation (4) is not satisfied. The theorem has been proven.

Proof. Since lines (5) and (6) define the same line, the normal vectors n 1 ={A 1 ,B 1 ) and n 2 ={A 2 ,B 2) are collinear. Since the vectors n 1 ≠0, n 2 ≠ 0, then there is a number λ , What n 2 =n 1 λ . Hence we have: A 2 =A 1 λ , B 2 =B 1 λ . Let's prove that C 2 =C 1 λ . It is obvious that coinciding lines have a common point M 0 (x 0 , y 0). Multiplying equation (5) by λ and subtracting equation (6) from it we get:

Since the first two equalities from expressions (7) are satisfied, then C 1 λ −C 2=0. Those. C 2 =C 1 λ . The remark has been proven.

Note that equation (4) defines the equation of a straight line passing through the point M 0 (x 0 , y 0) and having a normal vector n={A,B). Therefore, if the normal vector of the line and the point belonging to this line are known, then the general equation of the line can be constructed using equation (4).

Example 1. A line passes through a point M=(4,−1) and has a normal vector n=(3, 5). Construct the general equation of a straight line.

Solution. We have: x 0 =4, y 0 =−1, A=3, B=5. To construct the general equation of a straight line, we substitute these values ​​into equation (4):

Answer:

Vector parallel to line L and hence is perpendicular to the normal vector of the line L. Let's construct a normal line vector L, given that the scalar product of vectors n and is equal to zero. We can write, for example, n={1,−3}.

To construct the general equation of a straight line, we use formula (4). Let us substitute into (4) the coordinates of the point M 1 (we can also take the coordinates of the point M 2) and the normal vector n:

Substituting point coordinates M 1 and M 2 in (9) we can make sure that the straight line given by the equation(9) passes through these points.

Answer:

Subtract (10) from (1):

We have obtained the canonical equation of a straight line. Vector q={−B, A) is the direction vector of the straight line (12).

See reverse transformation.

Example 3. A straight line in a plane is represented by the following general equation:

Move the second term to the right and divide both sides of the equation by 2 5.