Concept of line equation. Defining a line using the equation condition for parallelism of lines

defines a curve on the plane. A group of terms is called a quadratic form, – linear form. If a quadratic form contains only squares of variables, then this form is called canonical, and the vectors of an orthonormal basis in which the quadratic form has a canonical form are called the principal axes of the quadratic form.
Matrix is called a matrix of quadratic form. Here a 1 2 =a 2 1. To reduce matrix B to diagonal form, it is necessary to take the eigenvectors of this matrix as a basis, then , where λ 1 and λ 2 are the eigenvalues ​​of matrix B.
In the basis of the eigenvectors of the matrix B, the quadratic form will have the canonical form: λ 1 x 2 1 +λ 2 y 2 1 .
This operation corresponds to the rotation of the coordinate axes. Then the origin of coordinates is shifted, thereby getting rid of the linear shape.
The canonical form of the second-order curve: λ 1 x 2 2 +λ 2 y 2 2 =a, and:
a) if λ 1 >0; λ 2 >0 is an ellipse, in particular, when λ 1 =λ 2 it is a circle;
b) if λ 1 >0, λ 2<0 (λ 1 <0, λ 2 >0) we have a hyperbole;
c) if λ 1 =0 or λ 2 =0, then the curve is a parabola and after rotating the coordinate axes it has the form λ 1 x 2 1 =ax 1 +by 1 +c (here λ 2 =0). Complementing to a complete square, we have: λ 1 x 2 2 =b 1 y 2.

Example. The equation of the curve 3x 2 +10xy+3y 2 -2x-14y-13=0 is given in the coordinate system (0,i,j), where i =(1,0) and j =(0,1).
1. Determine the type of curve.
2. Bring the equation to canonical form and construct a curve in the original coordinate system.
3. Find the corresponding coordinate transformations.

Solution. We bring the quadratic form B=3x 2 +10xy+3y 2 to the main axes, that is, to the canonical form. The matrix of this quadratic form is . We find the eigenvalues ​​and eigenvectors of this matrix:

Characteristic equation:
; λ 1 =-2, λ 2 =8. Type of quadratic form: .
The original equation defines a hyperbola.
Note that the form of the quadratic form is ambiguous. You can write 8x 1 2 -2y 1 2 , but the type of curve remains the same - a hyperbola.
We find the principal axes of the quadratic form, that is, the eigenvectors of matrix B. .
Eigenvector corresponding to the number λ=-2 at x 1 =1: x 1 =(1,-1).
As a unit eigenvector we take the vector , where is the length of the vector x 1 .
The coordinates of the second eigenvector corresponding to the second eigenvalue λ=8 are found from the system
.
1 ,j 1).
According to formulas (5) of paragraph 4.3.3. Let's move on to a new basis:
or

; . (*)

We enter the expressions x and y into the original equation and, after transformations, we get: .
Selecting complete squares: .
We carry out a parallel translation of the coordinate axes to a new origin: , .
If we introduce these relations into (*) and resolve these equalities for x 2 and y 2, we obtain: , . In the coordinate system (0*, i 1, j 1) this equation has the form: .
To construct a curve, we construct a new one in the old coordinate system: the x 2 =0 axis is specified in the old coordinate system by the equation x-y-3=0, and the y 2 =0 axis by the equation x+y-1=0. The origin of the new coordinate system 0 * (2,-1) is the intersection point of these lines.
To simplify perception, we will divide the process of constructing a graph into 2 stages:
1. Transition to a coordinate system with axes x 2 =0, y 2 =0, specified in the old coordinate system by the equations x-y-3=0 and x+y-1=0, respectively.

2. Construction of a graph of the function in the resulting coordinate system.

The final version of the graph looks like this (see. Solution:Download solution

Exercise. Establish that each of the following equations defines an ellipse, and find the coordinates of its center C, semi-axis, eccentricity, directrix equations. Draw an ellipse on the drawing, indicating the axes of symmetry, foci and directrixes.
Solution.

Let us consider a relation of the form F(x, y)=0, connecting variables x And at. We will call equality (1) equation with two variables x, y, if this equality is not true for all pairs of numbers X And at. Examples of equations: 2x + 3y = 0, x 2 + y 2 – 25 = 0,

sin x + sin y – 1 = 0.

If (1) is true for all pairs of numbers x and y, then it is called identity. Examples of identities: (x + y) 2 - x 2 - 2xy - y 2 = 0, (x + y)(x - y) - x 2 + y 2 = 0.

We will call equation (1) equation of a set of points (x; y), if this equation is satisfied by the coordinates X And at any point of the set and are not satisfied by the coordinates of any point that does not belong to this set.

An important concept in analytical geometry is the concept of the equation of a line. Let a rectangular coordinate system and a certain line be given on the plane α.

Definition. Equation (1) is called the line equation α (in the created coordinate system), if this equation is satisfied by the coordinates X And at any point lying on the line α , and do not satisfy the coordinates of any point not lying on this line.

If (1) is the equation of the line α, then we will say that equation (1) defines (sets) line α.

Line α can be determined not only by an equation of the form (1), but also by an equation of the form

F (P, φ) = 0 containing polar coordinates.

• equation of a straight line with an angular coefficient;

Let some straight line, not perpendicular, to the axis be given OH. Let's call inclination angle given straight line to the axis OH corner α , to which the axis needs to be rotated OH so that the positive direction coincides with one of the directions of the straight line. Tangent of the angle of inclination of the straight line to the axis OH called slope this line and is denoted by the letter TO.

	K=tg α
		(1)

Let us derive the equation of this line if we know its TO and the value in the segment OB, which it cuts off on the axis OU.

(2)

y=kx+b

Let us denote by M"plane point (x; y). If we draw straight BN And N.M., parallel to the axes, then r BNM – rectangular. T. MC C BM <=>, when the values N.M. And BN satisfy the condition: . But NM=CM-CN=CM-OB=y-b, BN=x=> taking into account (1), we obtain that the point M(x;y)C on this line<=>, when its coordinates satisfy the equation: =>

Equation (2) is called equation of a straight line with an angular coefficient. If K=0, then the straight line is parallel to the axis OH and its equation is y = b.

• equation of a line passing through two points;

(4)

Let two points be given M 1 (x 1; y 1) And M 2 (x 2; y 2). Taking at (3) point M(x;y) behind M 2 (x 2; y 2), we get y 2 -y 1 =k(x 2 - x 1). Defining k from the last equality and substituting it into equation (3), we obtain the desired equation of the line: . This is the equation if y 1 ≠ y 2, can be written as:

If y 1 = y 2, then the equation of the desired line has the form y = y 1. In this case, the straight line is parallel to the axis OH. If x 1 = x 2, then the straight line passing through the points M 1 And M 2, parallel to the axis OU, its equation has the form x = x 1.

• equation of a straight line passing through a given point with a given slope;

(3)

Аx + Вy + С = 0

Theorem. In a rectangular coordinate system Ohoo any straight line is given by an equation of the first degree:

and, conversely, equation (5) for arbitrary coefficients A, B, C (A And B ≠ 0 simultaneously) defines a certain straight line in a rectangular coordinate system Ooh.

Proof.

First, let's prove the first statement. If the line is not perpendicular Oh, then it is determined by the equation of the first degree: y = kx + b, i.e. equation of the form (5), where

A = k, B = -1 And C = b. If the line is perpendicular Oh, then all its points have the same abscissa, equal to the value α segment cut off by a straight line on the axis Oh.

The equation of this line has the form x = α, those. is also a first degree equation of the form (5), where A = 1, B = 0, C = - α. This proves the first statement.

Let us prove the converse statement. Let equation (5) be given, and at least one of the coefficients A And B ≠ 0.

If B ≠ 0, then (5) can be written in the form . Flat , we get the equation y = kx + b, i.e. an equation of the form (2) that defines a straight line.

If B = 0, That A ≠ 0 and (5) takes the form . Denoting by α, we get

x = α, i.e. equation of a line perpendicular Oh.

Lines defined in a rectangular coordinate system by an equation of the first degree are called first order lines.

Equation of the form Ax + Wu + C = 0 is incomplete, i.e. Some of the coefficients are equal to zero.

1) C = 0; Ah + Wu = 0 and defines a straight line passing through the origin.

2) B = 0 (A ≠ 0); the equation Ax + C = 0 OU.

3) A = 0 (B ≠ 0); Wu + C = 0 and defines a straight line parallel Oh.

Equation (6) is called the equation of a straight line “in segments”. Numbers A And b are the values ​​of the segments that the straight line cuts off on the coordinate axes. This form of the equation is convenient for the geometric construction of a straight line.

• normal equation of a line;

Аx + Вy + С = 0 is the general equation of a certain line, and (5) x cos α + y sin α – p = 0(7)

its normal equation.

Since equations (5) and (7) define the same straight line, then ( A 1x + B 1y + C 1 = 0 And

A 2x + B 2y + C 2 = 0 => ) the coefficients of these equations are proportional. This means that by multiplying all terms of equation (5) by a certain factor M, we obtain the equation MA x + MV y + MS = 0, coinciding with equation (7) i.e.

MA = cos α, MB = sin α, MC = - P(8)

To find the factor M, we square the first two of these equalities and add:

M 2 (A 2 + B 2) = cos 2 α + sin 2 α = 1

(9)

§ 9. The concept of the equation of a line.

Defining a line using an equation

Equality of the form F (x, y) = 0 called an equation in two variables x, y, if it is not true for all pairs of numbers x, y. They say two numbers x = x 0 , y=y 0, satisfy some equation of the form F(x, y)=0, if when substituting these numbers instead of variables X And at in the equation, its left side vanishes.

The equation of a given line (in a designated coordinate system) is an equation with two variables that is satisfied by the coordinates of each point lying on this line, and not satisfied by the coordinates of each point not lying on it.

In what follows, instead of the expression “the equation of the line is given F(x, y) = 0" we will often say in short: given a line F (x, y) = 0.

If the equations of two lines are given F(x, y) = 0 And Ф(x, y) = Q, then the joint solution of the system

Gives all their intersection points. More precisely, each pair of numbers that is a joint solution of this system determines one of the intersection points.

1)X 2 +y 2 = 8, x-y = 0;

2) X 2 +y 2 -16x+4at+18 = 0, x + y= 0;

3) X 2 +y 2 -2x+4at -3 = 0, X 2 + y 2 = 25;

4) X 2 +y 2 -8x+10у+40 = 0, X 2 + y 2 = 4.

163. Points are given in the polar coordinate system

Determine which of these points lie on the line defined by the equation in polar coordinates  = 2 cos , and which do not lie on it. Which line is determined by this equation? (Draw it on the drawing:)

164. On the line defined by the equation  =
, find points whose polar angles are equal to the following numbers: a) ,b) - ,c) 0, d) . Which line is defined by this equation?

(Build it on the drawing.)

165. On the line defined by the equation  =
, find points whose polar radii are equal to the following numbers: a) 1, b) 2, c)
. Which line is defined by this equation? (Build it on the drawing.)

166. Establish which lines are determined in polar coordinates by the following equations (construct them on the drawing):

1)  = 5; 2)  = ; 3)  = ; 4)  cos  = 2; 5)  sin  = 1;

6)  = 6 cos ; 7)  = 10 sin ; 8) sin  =

Consider the function given by the formula (equation)

This function, and therefore equation (11), corresponds to a well-defined line on the plane, which is the graph of this function (see Fig. 20). From the definition of the graph of a function it follows that this line consists of those and only those points of the plane whose coordinates satisfy equation (11).

Let it now

The line, which is the graph of this function, consists of those and only those points of the plane whose coordinates satisfy equation (12). This means that if a point lies on the specified line, then its coordinates satisfy equation (12). If the point does not lie on this line, then its coordinates do not satisfy equation (12).

Equation (12) is resolved with respect to y. Consider an equation containing x and y and not solved for y, such as the equation

Let us show that this equation in the plane also corresponds to a line, namely a circle with a center at the origin and a radius equal to 2. Let us rewrite the equation in the form

Its left side is the square of the distance of the point from the origin (see § 2, paragraph 2, formula 3). From equality (14) it follows that the square of this distance is equal to 4.

This means that any point whose coordinates satisfy equation (14), and therefore equation (13), is located at a distance of 2 from the origin.

The geometric location of such points is a circle with a center at the origin and radius 2. This circle will be the line corresponding to equation (13). The coordinates of any of its points obviously satisfy equation (13). If the point does not lie on the circle we found, then the square of its distance from the origin will be either greater or less than 4, which means that the coordinates of such a point do not satisfy equation (13).

Let now, in the general case, be given the equation

on the left side of which there is an expression containing x and y.

Definition. The line defined by equation (15) is the geometric locus of points in the plane whose coordinates satisfy this equation.

This means that if the line L is determined by an equation, then the coordinates of any point L satisfy this equation, but the coordinates of any point in the plane lying outside L do not satisfy equation (15).

Equation (15) is called the line equation

Comment. One should not think that any equation determines any line. For example, the equation does not define any line. In fact, for any real values ​​of and y, the left side of this equation is positive and the right side is equal to zero, and therefore, this equation cannot be satisfied by the coordinates of any point in the plane

A line can be defined on a plane not only by an equation containing Cartesian coordinates, but also by an equation in polar coordinates. A line defined by an equation in polar coordinates is the geometric locus of points in the plane whose polar coordinates satisfy this equation.

Example 1. Construct an Archimedes spiral at .

Solution. Let's make a table for some values ​​of the polar angle and the corresponding values ​​of the polar radius.

We construct a point in the polar coordinate system, which obviously coincides with the pole; then, drawing the axis at an angle to the polar axis, we construct a point with a positive coordinate on this axis, after which we similarly construct points with positive values ​​of the polar angle and polar radius (the axes for these points are not indicated in Fig. 30).

By connecting the points, we get one branch of the curve, indicated in Fig. 30 with a bold line. When changing from 0 to this branch of the curve consists of an infinite number of turns.

An equality of the form F(x, y) = 0 is called an equation with two variables x, y if it is not true for all pairs of numbers x, y. They say that two numbers x = x 0, y = y 0 satisfy some equation of the form F(x, y) = 0 if, when substituting these numbers instead of the variables x and y into the equation, its left side becomes zero.

The equation of a given line (in a designated coordinate system) is an equation with two variables that is satisfied by the coordinates of every point lying on this line and not satisfied by the coordinates of every point not lying on it.

In what follows, instead of the expression “given the equation of the line F(x, y) = 0,” we will often say more briefly: given the line F(x, y) = 0.

If the equations of two lines are given: F(x, y) = 0 and Ф(x, y) = 0, then the joint solution of the system

F(x,y) = 0, Ф(x, y) = 0

gives all their intersection points. More precisely, each pair of numbers that is a joint solution of this system determines one of the intersection points,

157. Given points *) M 1 (2; -2), M 2 (2; 2), M 3 (2; - 1), M 4 (3; -3), M 5 (5; -5), M 6 (3; -2). Determine which of the given points lie on the line defined by the equation x + y = 0 and which do not lie on it. Which line is defined by this equation? (Draw it on the drawing.)

158. On the line defined by the equation x 2 + y 2 = 25, find points whose abscissas are equal to the following numbers: 1) 0, 2) -3, 3) 5, 4) 7; on the same line find points whose ordinates are equal to the following numbers: 5) 3, 6) -5, 7) -8. Which line is defined by this equation? (Draw it on the drawing.)

159. Determine which lines are determined by the following equations (construct them on the drawing): 1)x - y = 0; 2) x + y = 0; 3) x - 2 = 0; 4)x + 3 = 0; 5) y - 5 = 0; 6) y + 2 = 0; 7) x = 0; 8) y = 0; 9) x 2 - xy = 0; 10) xy + y 2 = 0; 11) x 2 - y 2 = 0; 12) xy = 0; 13) y 2 - 9 = 0; 14) x 2 - 8x + 15 = 0; 15) y 2 + by + 4 = 0; 16) x 2 y - 7xy + 10y = 0; 17) y - |x|; 18) x - |y|; 19) y + |x| = 0; 20) x + |y| = 0; 21) y = |x - 1|; 22) y = |x + 2|; 23) x 2 + y 2 = 16; 24) (x - 2) 2 + (y - 1) 2 = 16; 25 (x + 5) 2 + (y-1) 2 = 9; 26) (x - 1) 2 + y 2 = 4; 27) x 2 + (y + 3) 2 = 1; 28) (x - 3) 2 + y 2 = 0; 29) x 2 + 2y 2 = 0; 30) 2x 2 + 3y 2 + 5 = 0; 31) (x - 2) 2 + (y + 3) 2 + 1 = 0.

160. Given lines: l)x + y = 0; 2)x - y = 0; 3)x 2 + y 2 - 36 = 0; 4) x 2 + y 2 - 2x + y = 0; 5) x 2 + y 2 + 4x - 6y - 1 = 0. Determine which of them pass through the origin.

161. Given lines: 1) x 2 + y 2 = 49; 2) (x - 3) 2 + (y + 4) 2 = 25; 3) (x + 6) 2 + (y - Z) 2 = 25; 4) (x + 5) 2 + (y - 4) 2 = 9; 5) x 2 + y 2 - 12x + 16y - 0; 6) x 2 + y 2 - 2x + 8y + 7 = 0; 7) x 2 + y 2 - 6x + 4y + 12 = 0. Find their points of intersection: a) with the Ox axis; b) with the Oy axis.

162. Find the intersection points of two lines:

1) x 2 + y 2 - 8; x - y =0;

2) x 2 + y 2 - 16x + 4y + 18 = 0; x + y = 0;

3) x 2 + y 2 - 2x + 4y - 3 = 0; x 2 + y 2 = 25;

4) x 2 + y 2 - 8y + 10y + 40 = 0; x 2 + y 2 = 4.

163. In the polar coordinate system, the points M 1 (l; π/3), M 2 (2; 0), M 3 (2; π/4), M 4 (√3; π/6) and M 5 ( 1; 2/3π). Determine which of these points lie on the line defined in polar coordinates by the equation p = 2cosΘ, and which do not lie on it. Which line is determined by this equation? (Draw it on the drawing.)

164. On the line defined by the equation p = 3/cosΘ, find points whose polar angles are equal to the following numbers: a) π/3, b) - π/3, c) 0, d) π/6. Which line is defined by this equation? (Build it on the drawing.)

165. On the line defined by the equation p = 1/sinΘ, find points whose polar radii are equal to the following numbers: a) 1 6) 2, c) √2. Which line is defined by this equation? (Build it on the drawing.)

166. Establish which lines are determined in polar coordinates by the following equations (construct them on the drawing): 1) p = 5; 2) Θ = π/2; 3) Θ = - π/4; 4) p cosΘ = 2; 5) p sinΘ = 1; 6.) p = 6cosΘ; 7) p = 10 sinΘ; 8) sinΘ = 1/2; 9) sinp = 1/2.

167. Construct the following Archimedes spirals on the drawing: 1) p = 20; 2) p = 50; 3) p = Θ/π; 4) p = -Θ/π.

168. Construct the following hyperbolic spirals on the drawing: 1) p = 1/Θ; 2) p = 5/Θ; 3) p = π/Θ; 4) р= - π/Θ

169. Construct the following logarithmic spirals on the drawing: 1) p = 2 Θ; 2) p = (1/2) Θ.

170. Determine the lengths of the segments into which the Archimedes spiral p = 3Θ is cut by a beam emerging from the pole and inclined to the polar axis at an angle Θ = π/6. Make a drawing.

171. On the Archimedes spiral p = 5/πΘ, point C is taken, the polar radius of which is 47. Determine how many parts this spiral cuts the polar radius of point C. Make a drawing.

172. On a hyperbolic spiral P = 6/Θ, find a point P whose polar radius is 12. Make a drawing.

173. On a logarithmic spiral p = 3 Θ, find a point P whose polar radius is 81. Make a drawing.