The direction vector of the straight line l everyone is called nonzero vector (m, n) parallel to this line.

Let the point M 1 (x 1 , y 1) and direction vector ( m, n), then the equation of the straight line passing through the point M 1 in the direction of the vector has the form: . This equation is called the canonical equation of the line.

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

We will look for the equation of the desired straight line in the form: Ax+By+C= 0. Let's write the canonical equation of the line , transform it. Get x + y - 3 = 0

Equation of a line passing through two points

Let two points be given on the plane M 1 (x 1 , y 1) and M 2 (x 2, y 2), then the equation of a straight line passing through these points has the form: . If any of the denominators is equal to zero, the corresponding numerator should be set equal to zero.

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

Applying the above formula, we get:

Equation of a straight line from a point and a slope

If the general equation of a straight line Ah + Wu + C= 0 bring to the form: and denote , then the resulting equation is called the equation of a straight line with slope k.

Equation of a straight line in segments

If in the general equation the line Ah + Wu + C= 0 coefficient FROM¹ 0, then, dividing by C, we get: or , where

geometric sense coefficients in that the coefficient but is the coordinate of the point of intersection of the line with the axis Oh, but 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 – at+ 1 = 0. Find the equation of this straight line in segments. A = -1, B = 1, C = 1, then but = -1, b= 1. The equation of a straight line in segments will take the form .

Example. The vertices of the triangle A(0; 1), B(6; 5), C(12; -1) are given. Find the equation for the height drawn from vertex C.

We find the equation of the side AB: ;

4x = 6y– 6; 2x – 3y + 3 = 0;

The desired height equation has the form: Ax+By+C= 0 or y = kx + b.

k= . Then y= . Because the height passes through point C, then its coordinates satisfy this equation: where b= 17. Total: .

Answer: 3 x + 2y – 34 = 0.

Practice #7

Class name: Curves of the second order.

Purpose of the lesson: Learn how to make curves of the 2nd order, build them.

Preparation for the lesson: Repeat theoretical material on the topic "Curves of the 2nd order"

Literature:

1. Dadayan A.A. "Mathematics", 2004

Task for the lesson:

The order of the lesson:

1. Get permission to work
2. Complete tasks
3. Answer security questions.

1. Name, purpose of the lesson, task;
2. Completed task;
3. Answers to control questions.

test questions for offset:

1. Define curves of the second order (circle, ellipse, hyperbola, parabola), write down their canonical equations.
2. What is the eccentricity of an ellipse or hyperbola called? How to find it?
3. Write the equation of an equilateral hyperbola

APPENDIX

circumference is the set of all points of the plane equidistant from one point, called the center.

Let the center of the circle be a point ABOUT(a; b), and the distance to any point M(x;y) circle is equal to R. Then ( x-a) 2 + (y-b) 2 = R 2 – canonical equation of a circle with center ABOUT(a; b) and radius R.

Example. Find the coordinates of the center and the radius of the circle if its equation is given as: 2 x 2 + 2y 2 - 8x + 5 y – 4 = 0.

To find the coordinates of the center and radius of a circle given equation must be reduced to canonical form. To do this, select the full squares:

x 2 + y 2 – 4x + 2,5y – 2 = 0

x 2 – 4x + 4 – 4 + y 2 + 2,5y + 25/16 – 25/16 – 2 = 0

(x– 2) 2 + (y + 5/4) 2 – 25/16 – 6 = 0

(x – 2) 2 + (y + 5/4) 2 = 121/16

From here we find the coordinates of the center ABOUT(2; -5/4); radius R = 11/4.

Ellipse a set of points in a plane is called, the sum of the distances from each of which to two given points (called foci) is a constant value greater than the distance between the foci.

Focuses are indicated by letters F 1 , F from, the sum of the distances from any point of the ellipse to the foci is 2 but (2but > 2c), a- a large semi-axis; b- small semi-axis.

The canonical equation of the ellipse is: , where a, b And c related to each other by equalities: a 2 - b 2 \u003d c 2 (or b 2 - a 2 \u003d c 2).

The shape of an ellipse is determined by a characteristic that is the ratio of the focal length to the length of the major axis and is called eccentricity. or .

Because by definition 2 but> 2c, then the eccentricity is always expressed as a proper fraction, i.e. .

Example. Write an equation for an ellipse if its foci are F 1 (0; 0), F 2 (1; 1), the major axis is 2.

The ellipse equation has the form: .

Distance between focuses: 2 c= , thus, a 2 – b 2 = c 2 = . By condition 2 but= 2, so but = 1, b= The desired equation of the ellipse will take the form: .

Hyperbole called the set of points in the plane, the difference in the distances from each of which to two given points, called foci, is a constant value, less than the distance between the foci.

The canonical equation of a hyperbola has the form: or , where a, b And c linked by equality a 2 + b 2 = c 2 . The hyperbola is symmetrical with respect to the middle of the segment connecting the foci and with respect to the coordinate axes. Focuses are indicated by letters F 1 , F 2 , distance between foci - 2 from, the difference in distances from any point of the hyperbola to the foci is 2 but (2but < 2c). Axis 2 but called the real axis of the hyperbola, axis 2 b is the imaginary axis of the hyperbola. A hyperbola has two asymptotes whose equations are

The eccentricity of a hyperbola is the ratio of the distance between the foci to the length of the real axis: or. Because by definition 2 but < 2c, then the eccentricity of the hyperbola is always expressed as an improper fraction, i.e. .

If the length of the real axis is equal to the length of the imaginary axis, i.e. a = b, ε = , then the hyperbola is called equilateral.

Example. Write the canonical equation of a hyperbola if its eccentricity is 2 and the foci coincide with the foci of the ellipse with the equation

We find focal length c 2 = 25 – 9 = 16.

For hyperbole: c 2 = a 2 + b 2 = 16, ε = c/a = 2; c = 2a; c 2 = 4a 2 ; a 2 = 4; b 2 = 16 – 4 = 12.

Then - the desired equation of the hyperbola.

parabola is the set of points in a plane equidistant from given point, called the focus, and a given straight line, called the directrix.

The focus of a parabola is denoted by the letter F, director - d, the distance from the focus to the directrix is R.

The canonical equation of a parabola, the focus of which is located on the x-axis, is:

y 2 = 2px or y 2 = -2px

x = -p/2, x = p/2

The canonical equation of a parabola whose focus is on the y-axis is:

X 2 = 2py or X 2 = -2py

Directrix equations, respectively at = -p/2, at = p/2

Example. On a parabola at 2 = 8X find a point whose distance from the directrix is ​​4.

From the parabola equation we get that R = 4. r=x + p/2 = 4; Consequently:

x = 2; y 2 = 16; y= ±4. Search points: M 1 (2; 4), M 2 (2; -4).

Practice #8

Class name: Actions on complex numbers in algebraic form. Geometric interpretation of complex numbers.

Purpose of the lesson: Learn how to operate on complex numbers.

Preparation for the lesson: Repeat the theoretical material on the topic "Complex numbers".

Literature:

1. Grigoriev V.P., Dubinsky Yu.A. "Elements higher mathematics", 2008

Task for the lesson:

1. Calculate:

1) i 145 + i 147 + i 264 + i 345 + i 117 ;

2) (i 64 + i 17 + i 13 + i 82)( i 72 – i 34);

Let the straight line pass through the points M 1 (x 1; y 1) and M 2 (x 2; y 2). The equation of a straight line passing through the point M 1 has the form y- y 1 \u003d k (x - x 1), (10.6)

where k - still unknown coefficient.

Since the straight line passes through the point M 2 (x 2 y 2), then the coordinates of this point must satisfy equation (10.6): y 2 -y 1 \u003d k (x 2 -x 1).

From here we find Substituting the found value k into equation (10.6), we obtain the equation of a straight line passing through the points M 1 and M 2:

It is assumed that in this equation x 1 ≠ x 2, y 1 ≠ y 2

If x 1 \u003d x 2, then the straight line passing through the points M 1 (x 1, y I) and M 2 (x 2, y 2) is parallel to the y-axis. Its equation is x = x 1 .

If y 2 \u003d y I, then the equation of the straight line can be written as y \u003d y 1, the straight line M 1 M 2 is parallel to the x-axis.

Equation of a straight line in segments

Let the straight line intersect the Ox axis at the point M 1 (a; 0), and the Oy axis - at the point M 2 (0; b). The equation will take the form:
those.
. This equation is called the equation of a straight line in segments, because the numbers a and b indicate which segments the straight line cuts off on the coordinate axes.

Equation of a straight line passing through a given point perpendicular to a given vector

Let's find the equation of a straight line passing through a given point Mo (x O; y o) perpendicular to a given non-zero vector n = (A; B).

Take an arbitrary point M(x; y) on the straight line and consider the vector M 0 M (x - x 0; y - y o) (see Fig. 1). Since the vectors n and M o M are perpendicular, their scalar product is equal to zero: that is,

A(x - xo) + B(y - yo) = 0. (10.8)

Equation (10.8) is called equation of a straight line passing through a given point perpendicular to a given vector .

The vector n = (A; B) perpendicular to the line is called normal normal vector of this line .

Equation (10.8) can be rewritten as Ah + Wu + C = 0 , (10.9)

where A and B are the coordinates of the normal vector, C \u003d -Ax o - Vu o - free member. Equation (10.9) is the general equation of a straight line(see Fig.2).

Fig.1 Fig.2

Canonical equations of the straight line

,

Where
are the coordinates of the point through which the line passes, and
- direction vector.

Curves of the second order Circle

A circle is the set of all points of a plane equidistant from a given point, which is called the center.

Canonical equation of a circle of radius R centered on a point
:

In particular, if the center of the stake coincides with the origin, then the equation will look like:

Ellipse

An ellipse is a set of points in a plane, the sum of the distances from each of them to two given points And , which are called foci, is a constant value
, greater than the distance between the foci
.

The canonical equation of an ellipse whose foci lie on the Ox axis and whose origin is in the middle between the foci has the form
G de a the length of the major semiaxis; b is the length of the minor semiaxis (Fig. 2).

The equation of a straight line passing through t.u A(ha; wah) and having a slope k, is written in the form

y - ya \u003d k (x - xa).(5)

Equation of a line passing through two points T. A (x 1; y 1) etc. B (x 2; y 2), has the form

If the points BUT And IN define a straight line parallel to the Ox axis (y 1 \u003d y 2) or y-axis (x 1 = x 2), then the equation of such a straight line is written respectively in the form:

y = y 1 or x = x 1(7)

Normal equation of a straight line

Let a straight line C be given passing through a given point Mo(Xo; V0) and perpendicular to the vector (A; B). Any vector perpendicular to a given line is called its normal vector. Let us choose an arbitrary point M on the line (x; y). Then , which means they scalar product. This equality can be written in coordinates

A (x-x o) + B (y-y o) \u003d 0 (8)

Equation (8) is called normal equation of a straight line .

Parametric and canonical equations of a straight line

Let the line l given by starting point M 0 (x 0; y 0) and direction vector ( a 1; a 2),. Let t. M(x; y)- any point on a line l Then the vector is collinear to the vector . Therefore, = . Writing this equation in coordinates, we obtain the parametric equation of the straight line

Let us exclude the parameter t from Eq. (9). This is possible because the vector , and therefore at least one of its coordinates is nonzero.

Let and , then , and, therefore,

Equation (10) is called canonical equation of the line with guide vector

\u003d (a 1; a 2). If a 1 =0 and , then equations (9) take the form

These equations define a straight line parallel to the axis, OU and passing through the point

M 0 (x 0; y 0).

x=x 0(11)

If , , then equations (9) take the form

These equations define a straight line parallel to the O axis X and passing through the point

M 0 (x 0; y 0). The canonical equation of such a straight line has the form

y=y 0(12)

Angle between lines. The condition of parallelism and perpendicularity of two

direct

Let two straight lines given by general equations be given:

And

Then the angle φ between them is determined by the formula:

(13)

Parallel condition 2 straight lines: (14)

Perpendicular condition 2 straight lines: (15)

Parallel condition in this case has the form: (17)

Perpendicular condition straight: (18)

If two lines are given by canonical equations:

And

then the angle φ between these lines is determined by the formula:

(19)

Parallel condition straight: (20)

Perpendicular condition direct: (21)

Distance from point to line

Distance d from the point M (x 1; y 1) to straight Ax+By+C=0 calculated by the formula

(22)

Implementation example practical work

Example 1 Build a line 3 X- 2at+6=0.

Solution: To build a line, it is enough to know any two of its points, for example, the points of its intersection with the coordinate axes. Point A of the intersection of the line with the axis Ox can be obtained if we take y \u003d 0 in the equation of the line. Then we have 3 X+6=0, i.e. X=-2. In this way, BUT(–2;0).

Then IN intersection of a line with an axis OU has an abscissa X=0; hence the ordinate of the point IN is found from the equation -2 y+ 6=0, i.e. y=3. In this way, IN(0;3).

Example 2 Write the equation of a straight line that cuts off on the negative half-plane OU a segment equal to 2 units, and forms with the axis Oh angle φ =30˚.

Solution: The line crosses the axis OU at the point IN(0;–2) and has a slope k=tg φ= = . Assuming in equation (2) k= and b= –2, we obtain the desired equation

Or .

Example 3 BUT(–1; 2) and

IN(0;–3). (at testimony: the slope of the straight line is found by the formula (3))

Solution: .From here we have . Substituting the coordinates into this equation t.V, we get: , i.e. initial ordinate b= -3 . Then we get the equation.

Example 4 General equation of a straight line 2 X – 3at– 6 = 0 lead to the equation in segments.

Solution: we write this equation in the form 2 X– 3at=6 and divide both its parts by the free term: . This is the equation of this straight line in segments.

Example 5 Through the dot BUT(1;2) draw a straight line cutting off equal segments on the positive semi-axes of coordinates.

Solution: Let the equation of the desired straight line have the form By condition but=b. Therefore, the equation becomes X+ at= but. Since the point A (1; 2) belongs to this line, then its coordinates satisfy the equation X + at= but; those. 1 + 2 = but, where but= 3. So, the desired equation is written as follows: x + y = 3, or x + y - 3 = 0.

Example 6 For straight write the equation in segments. Calculate the area of ​​the triangle formed by this line and the coordinate axes.

Solution: Let's transform this equation as follows: , or .

As a result, we obtain the equation , which is the equation of the given straight line in segments. The triangle formed by the given line and the coordinate axes is right triangle with legs equal to 4 and 3, so its area is equal to S= (sq. units)

Example 7 Write an equation of a straight line passing through a point (–2; 5) and a generatrix with an axis Oh angle 45º.

Solution: Slope of the desired straight line k= tg 45º = 1. Therefore, using equation (5), we obtain y - 5 = x- (-2), or x - y + 7 = 0.

Example 8 Write the equation of a straight line passing through the points BUT(–3; 5) and IN( 7; –2).

Solution: Let's use equation (6):

, or , whence 7 X + 10at – 29 = 0.

Example 9 Check if points lie BUT(5; 2), IN(3; 1) and FROM(–1; –1) on one straight line.

Solution: Compose the equation of a straight line passing through the points BUT And FROM:

, or

Substituting into this equation the coordinates of the point IN (xB= 3 and y B = 1), we get (3–5) / (–6)= = (1–2) / (–3), i.e. we get the correct equality. Thus, point coordinates IN satisfy the straight line equation ( AC), i.e. .

Example 10: Write an equation for a straight line passing through t. A (2; -3).

Perpendicular =(-1;5)

Solution: Using formula (8), we find the equation of this line -1(x-2)+5(y+3)=0,

or finally, x - 5 y - 17 \u003d 0.

Example 11: Points given M 1(2;-1) and M 2(4; 5). Write the equation of a straight line passing through a point M 1 perpendicular to the vector Solution: The normal vector of the desired straight line has coordinates (2; 6), therefore, according to the formula (8), we obtain the equation 2(x-2)+6(y+1)=0 or x+3y +1=0.

Example 12: And .

Solution: ; .

Example 13:

Solution: a) ;

Example 14: Calculate angle between lines

Solution:

Example 15: To find out mutual arrangement direct:

Solution:

Example 16: find the angle between the lines and .

Solution: .

Example 17: find out the relative position of the lines:

Solution: a ) - lines are parallel;

b) means that the lines are perpendicular.

Example 18: Calculate the distance from the point M(6; 8) to the straight line

Solution: according to formula (22) we get: .

Tasks for practical session:

Option 1

1. Bring the general equation of the straight line 2x+3y-6=0 to the equation in segments and calculate the area of ​​the triangle cut off by this straight line from the corresponding coordinate angle;

2. In ∆ABC, the vertices have coordinates of point A (-3;4), point B (-4;-3), point C (8;1). Compose the equations of the side (AB), height (VC) and median (CM);

3. Calculate the slope of the straight line passing through the point M 0 (-2; 4) and parallel to the vector (6; -1);

4. Calculate the angle between the lines

4. Calculate the angle between the lines:

a) 2x - 3y + 7 = 0 and 3x - y + 5 = 0; b) and y = 2x – 4;

5. Determine the relative position of 2 straight lines and;

, if the coordinates of the ends of the segment t.A (18; 8) and t. B (-2; -6) are known.

Option 3

1. Bring the general equation of the straight line 4x-5y+20=0 to the equation in segments and calculate the area of ​​the triangle cut off by this straight line from the corresponding coordinate angle;

2. In ∆ABC, the vertices have coordinates of point A (3;-2), point B (7;3), points

C(0;8). Compose the equations of the side (AB), height (VC) and median (CM);

3. Calculate the slope of the straight line passing through the point M 0 (-1;-2) and

parallel to the vector (3;-5);

4. Calculate the angle between the lines

a) 3x + y - 7 = 0 and x - y + 4 = 0; b) and;

5. Determine the relative position of 2 lines and y = 5x + 3;

6. Calculate the distance from the middle of the segment AB to the straight line , if the coordinates of the ends of the segment t.A (4; -3) and t.B (-6; 5) are known.

Option 4

1. Bring the general equation of the straight line 12x-5y+60=0 to the equation in segments and calculate the length of the segment that is cut off from this straight line by the corresponding coordinate angle;

2. In ∆ABC, the vertices have the coordinates of point A (0;-2), point B (3;6), point C (1;-4). Compose the equations of the side (AB), height (VC) and median (CM);

3. Calculate the slope of the straight line passing through the point M 0 (4;4) and parallel to the vector (-2;7);

4. Calculate the angle between the lines

a) x +4 y + 8 = 0 and 7x - 3y + 5 = 0; b) and;

5. Determine the relative position of 2 straight lines and;

6. Calculate the distance from the middle of the segment AB to the straight line , if the coordinates of the ends of the segment t.A (-4; 8) and t.B (0; 4) are known.

test questions

1. Name the equations of a straight line in a plane when the point through which it passes and its directing vector are known;

2. What is the normal, general equation of a straight line on a plane;

3. Name the equation of a straight line passing through two points, the equation of a straight line in segments, the equation of a straight line with a slope;

4. List the formulas for calculating the angle between lines, given equations with an angle factor. Formulate the conditions for parallelism and perpendicularity of two lines.

5. How to find the distance from a point to a line?

Let two points be given M(X 1 ,At 1) and N(X 2,y 2). Let's find the equation of the straight line passing through these points.

Since this line passes through the point M, then according to formula (1.13) its equation has the form

At – Y 1 = K(X-x 1),

Where K is the unknown slope.

The value of this coefficient is determined from the condition that the desired straight line passes through the point N, which means that its coordinates satisfy equation (1.13)

Y 2 – Y 1 = K(X 2 – X 1),

From here you can find the slope of this line:

,

Or after conversion

(1.14)

Formula (1.14) defines Equation of a line passing through two points M(X 1, Y 1) and N(X 2, Y 2).

In the particular case when the points M(A, 0), N(0, B), BUT ¹ 0, B¹ 0, lie on the coordinate axes, equation (1.14) takes a simpler form

Equation (1.15) called Equation of a straight line in segments, here BUT And B denote segments cut off by a straight line on the axes (Figure 1.6).

Figure 1.6

Example 1.10. Write the equation of a straight line passing through the points M(1, 2) and B(3, –1).

. According to (1.14), the equation of the desired straight line has the form

2(Y – 2) = -3(X – 1).

Transferring all the terms to the left side, we finally obtain the desired equation

3X + 2Y – 7 = 0.

Example 1.11. Write an equation for a line passing through a point M(2, 1) and the point of intersection of the lines X+ Y- 1 = 0, X - y+ 2 = 0.

. We find the coordinates of the point of intersection of the lines by solving these equations together

If we add these equations term by term, we get 2 X+ 1 = 0, whence . Substituting the found value into any equation, we find the value of the ordinate At:

Now let's write the equation of a straight line passing through the points (2, 1) and :

or .

Hence or -5( Y – 1) = X – 2.

Finally, we obtain the equation of the desired straight line in the form X + 5Y – 7 = 0.

Example 1.12. Find the equation of a straight line passing through points M(2.1) and N(2,3).

Using formula (1.14), we obtain the equation

It doesn't make sense because the second denominator is zero. It can be seen from the condition of the problem that the abscissas of both points have the same value. Hence, the required line is parallel to the axis OY and its equation is: x = 2.

Comment . If, when writing the equation of a straight line according to formula (1.14), one of the denominators turns out to be zero, then the desired equation can be obtained by equating the corresponding numerator to zero.

Let's consider other ways of setting a straight line on a plane.

1. Let a non-zero vector be perpendicular to a given line L, and the point M 0(X 0, Y 0) lies on this line (Figure 1.7).

Figure 1.7

Denote M(X, Y) an arbitrary point on the line L. Vectors and Orthogonal. Using the orthogonality conditions for these vectors, we obtain or BUT(X – X 0) + B(Y – Y 0) = 0.

We have obtained the equation of a straight line passing through a point M 0 is perpendicular to the vector . This vector is called Normal vector to a straight line L. The resulting equation can be rewritten as

Oh + Wu + FROM= 0, where FROM = –(BUTX 0 + By 0), (1.16),

Where BUT And IN are the coordinates of the normal vector.

We obtain the general equation of a straight line in a parametric form.

2. A line on a plane can be defined as follows: let a non-zero vector be parallel to a given line L and dot M 0(X 0, Y 0) lies on this line. Again, take an arbitrary point M(X, y) on a straight line (Figure 1.8).

Figure 1.8

Vectors and collinear.

Let us write down the condition of collinearity of these vectors: , where T is an arbitrary number, called a parameter. Let's write this equality in coordinates:

These equations are called Parametric equations Straight. Let us exclude from these equations the parameter T:

These equations can be written in the form

. (1.18)

The resulting equation is called The canonical equation of a straight line. Vector call Direction vector straight .

Comment . It is easy to see that if is the normal vector to the line L, then its direction vector can be the vector , since , i.e. .

Example 1.13. Write the equation of a straight line passing through a point M 0(1, 1) parallel to line 3 X + 2At– 8 = 0.

Solution . The vector is the normal vector to the given and desired lines. Let's use the equation of a straight line passing through a point M 0 with a given normal vector 3( X –1) + 2(At– 1) = 0 or 3 X + 2y- 5 \u003d 0. We got the equation of the desired straight line.

Equation of a line passing through a given point in a given direction. Equation of a straight line passing through two given points. Angle between two lines. Condition of parallelism and perpendicularity of two lines. Determining the point of intersection of two lines

1. Equation of a line passing through a given point A(x 1 , y 1) in a given direction, determined by the slope k,

y - y 1 = k(x - x 1). (1)

This equation defines a pencil of lines passing through a point A(x 1 , y 1), which is called the center of the beam.

2. Equation of a straight line passing through two points: A(x 1 , y 1) and B(x 2 , y 2) is written like this:

The slope of a straight line passing through two given points is determined by the formula

3. Angle between straight lines A And B is the angle by which the first straight line must be rotated A around the point of intersection of these lines counterclockwise until it coincides with the second line B. If two lines are given by slope equations

y = k 1 x + B 1 ,