Equation of a straight line passing through a point, equation of a straight line passing through two points, angle between two lines, slope of a straight line. Equation of a line passing through a point in a given direction

In this article, we will learn how to write equations of a straight line passing through given point on a plane perpendicular to a given line. Let's study the theoretical information, give illustrative examples, where it is necessary to write such an equation.

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Before finding the equation of a straight line passing through a given point perpendicular to a given straight line. The theorem is considered in high school. Through a given point in a plane, one can draw a single straight line perpendicular to the given one. If there is a three-dimensional space, then the number of such lines will increase to infinity.

Definition 1

If the plane α passes through a given point M 1 perpendicular to a given line b, then the lines lying in this plane, including those passing through M 1, are perpendicular to a given line b.

From this we can conclude that the formulation of the equation of a straight line passing through a given point perpendicular to a given straight line is applicable only for the case on a plane.

Problems with three-dimensional space imply the search for the equation of a plane passing through a given point perpendicular to a given straight line.

If on a plane with a coordinate system O x y z we have a straight line b, then it corresponds to the equation of a straight line on a plane, a point with coordinates M 1 (x 1, y 1) is given, and it is necessary to compose an equation of a straight line a that passes through the point M 1, and perpendicular to the line b.

By condition, we have the coordinates of the point M 1. To write the equation of a straight line, it is necessary to have the coordinates of the directing vector of the straight line a, or the coordinates of the normal vector of the straight line a, or the slope of the straight line a.

It is necessary to obtain data from the given equation of the straight line b . By condition, the lines a and b are perpendicular, which means that the directing vector of the line b is considered to be a normal vector of the line a . From here we get that the slope coefficients are denoted as k b and k a . They are related by the relation k b · k a = - 1 .

We got that the direction vector of the straight line b has the form b → = (b x, b y) , hence the normal vector is n a → = (A 2 , B 2) , where the values A 2 = b x , B 2 = b y . Then we write general equation a straight line passing through a point with coordinates M 1 (x 1, y 1) having a normal vector n a → = (A 2 , B 2) having the form A 2 (x - x 1) + B 2 (y - y 1) = 0 .

The normal vector of the line b is defined and has the form n b → = (A 1 , B 1) , then the directing vector of the line a is the vector a → = (a x , a y) , where the values a x = A 1 , a y = B 1 . So it remains to compose a canonical or parametric equation of a straight line a passing through a point with coordinates M 1 (x 1, y 1) with a direction vector a → = (a x, a y) having the form x - x 1 a x = y - y 1 a y or x = x 1 + a x λ y = y 1 + a y λ respectively.

After finding the slope k b of the straight line b, you can calculate the slope of the straight line a . It will be equal to - 1 k b . It follows that you can write the equation of a straight line a passing through M 1 (x 1, y 1) with a slope - 1 k b in the form y - y 1 = - 1 k b · (x - x 1) .

The resulting equation of a straight line passing through a given point of the plane perpendicular to the given one. If circumstances so require, you can go to another form of this equation.

Solution of examples

Consider the formulation of the equation of a straight line passing through a given point of the plane and perpendicular to a given straight line.

Example 1

Write the equation of the line a, which passes through the point with coordinates M 1 (7, - 9) and is perpendicular to the line b, which is given by the canonical equation of the line x - 2 3 = y + 4 1.

Decision

From the condition we have that b → = (3 , 1) is the directing vector of the line x - 2 3 = y + 4 1 . The coordinates of the vector b → = 3 , 1 are the coordinates of the normal vector of the line a , since the lines a and b are mutually perpendicular. So we get n a → = (3 , 1) . Now it is necessary to write the equation of a straight line passing through the point M 1 (7, - 9) , having a normal vector with coordinates n a → = (3, 1) .

We get an equation of the form: 3 (x - 7) + 1 (y - (- 9)) = 0 ⇔ 3 x + y - 12 = 0

The resulting equation is the required one.

Answer: 3 x + y - 12 = 0 .

Example 2

Write an equation for a straight line that passes through the origin of the coordinate system O x y z, perpendicular to the straight line 2 x - y + 1 = 0.

Decision

We have that n b → = (2, - 1) is a normal vector of a given straight line. Hence a → = (2, - 1) - the coordinates of the desired directing vector of the straight line.

Let us fix the equation of a straight line passing through the origin with a directing vector a → = (2 , - 1) . We get that x - 0 2 = y + 0 - 1 ⇔ x 2 = y - 1 . The resulting expression is the equation of a straight line passing through the origin perpendicular to the straight line 2 x - y + 1 = 0 .

Answer: x 2 = y - 1.

Example 3

Write the equation of a straight line passing through a point with coordinates M 1 (5, - 3) perpendicular to the line y = - 5 2 x + 6.

Decision

From the equation y = - 5 2 x + 6 the slope is - 5 2 . The slope of a straight line that is perpendicular to it has the value - 1 - 5 2 = 2 5 . From this we conclude that the line passing through the point with coordinates M 1 (5, - 3) perpendicular to the line y \u003d - 5 2 x + 6 is equal to y - (- 3) \u003d 2 5 x - 5 ⇔ y \u003d 2 5 x - 5 .

Answer: y = 2 5 x - 5 .

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The direction vector of the straight line l any non-zero 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 With¹ 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 point of intersection of the line with the axis 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 – at+ 1 = 0. Find the equation of this straight line in segments. A = -1, B = 1, C = 1, then a = -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.

Practical lesson №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 the theoretical material on the topic "Curves of the 2nd order"

Literature:

Dadayan A.A. "Mathematics", 2004

Task for the lesson:

The order of the lesson:

Get permission to work
Complete tasks
Answer security questions.

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

Control questions for offset:

Define curves of the second order (circle, ellipse, hyperbola, parabola), write down their canonical equations.
What is the eccentricity of an ellipse or hyperbola called? How to find it?
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 O(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 O(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 the circle, this equation must be reduced to the canonical form. To do this, select 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 O(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 with, the sum of the distances from any point of the ellipse to the foci is 2 a (2a > 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 a> 2c, then the eccentricity is always expressed proper fraction, i.e. .

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

The ellipse equation has the form: .

Distance between focuses: 2 c= , thus, a 2 – b 2 = c 2 = . By condition 2 a= 2, so a = 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 with, the difference in distances from any point of the hyperbola to the foci is 2 a (2a < 2c). Axis 2 a 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 a < 2c, then the eccentricity of the hyperbola is always expressed as 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

Finding the 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 the plane equidistant from a given point, called the focus, and a given 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; hence:

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

Practice #8

Class name: Actions over 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:

Grigoriev V.P., Dubinsky Yu.A. "Elements of Higher Mathematics", 2008.

Task for the lesson:

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);

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).

There are three options in 3D space. relative position two straight lines:

lines intersect;

straight lines are parallel;

straight lines intersect.

Straight line is an algebraic curve of the first order: in Cartesian system coordinates 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).

Decision. 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 the

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).

Decision. 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).

Decision. 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 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.

y - y 1 \u003d k (x - x 1)

equation of a straight line: y \u003d kx + v

If we transform the original equation y - y 1 \u003d k (x - x 1), then we get y \u003d kx + (y 1 -kx 1) It satisfies the conditions of the straight line equation: y \u003d kx + in, because

1. its degree is the first, which means it can be straight,

2. the line passes through the point (x 1; y 1), because the coordinates of this point satisfy the equation: 0=0

3. the role of the coefficient in is played by the expression y 1 -kx 1

The straight line with the equation y - y 1 \u003d k (x - x 1) passes through 1 point. We require that the second point also lies on this line, i.e. so that the equality y 2 - y 1 \u003d k (x 2 - x 1) is fulfilled. From here we find k \u003d y 2 - y 1 ¸ x 2 - x 1 and substitute into the equation:

y - y 1 \u003d y 2 - y 1 ¸ x 2 - x 1 × (x - x 1) or

x - x 1 ¸x 2 - x 1 \u003d y - y 1 ¸y 2 - y 1

15. Angle m / y straight on the plane

Straight lines: y \u003d k 1 x + in 1, y \u003d k 2 x + in 2

In tr-ke ABC the amount of ext. angles a 1 +b is equal to the outer corner a 2 therefore b=a 2 -a 1 Obviously, tga 1 = k 1 ; tga 2 \u003d k 2. Using the formula for tg of the difference of 2 angles, we get tgb \u003d tg (a 2 -a 1) \u003d tga 2 -tga 1 ¸1+ tga 2 × tga 1

Finally, we have tgb= k 2 - k 1 ¸1+k 2 × ×k 1 By calculating the tangent, we can also find the angle b itself.

16. Conditions || and ^ lines in the plane.

The equations of lines with the angular coefficient are given. y \u003d k 1 x and y \u003d k 2 x + in 2

Conditions || direct is the equality of the angular coefficients. k 1 = k 2 (1)

Condition (1) is satisfied. and for merged lines. Angular coefficient formula. lines (tga= k 2 - k 1 ¸1+k 2 × ×k 1) can be written as: ctga= 1+k 2 × ×k 1 ¸k 2 - k 1 (this is the case if k 1 ¹k 2) . Condition ^ direct is expressed as k 2 × ×k 1 = -1. If k 1 =0 or k 2 =0, then one of the lines || axis Ox, and the second to it ^, has an equation of the form x \u003d a.

Let the lines be given by the general equation. A 1 x + B 1 y + C 1 \u003d 0, A 2 x + B 2 y + C 2 \u003d 0, If B1 \u003d B2 \u003d 0, then both lines are parallel to the Oy axis and to each other (their equations look like x \u003d a ) If В1=0, and В2¹0, then straight lines^. In the case when A2=0 (the equation is reduced to the form x=a, y=b) In the case of B1¹0 and B2¹0, you can express y in each equation. y \u003d -A1x¸B1-C1¸B1;

Y= - А2х¸В2-С2¸В2, then k1= -А1¸В1, and k2= - А2¸В2 and the condition || A1¸B1= A2¸B2 or A1¸A2= B1¸B2.

Using the equality 1+k1×k2=0, 1+ А1¸В1× А2¸В2=0. We arrive at the condition of ^ lines A1×A2+B1×B2=0.

Ellipse

An ellipse is the locus of points in a plane, the sum of the distances of which to two given points, called foci, is a constant value (greater than the distance between the foci)

The ellipse equation will take the simplest form if the foci are placed on the Ox axis to the left of the origin at an equal distance from it. F 1 F 2 - foci of the ellipse. Denote F 1 F 2 = 2c then the foci have coordinates (-c, 0) and (c, 0). Let us denote the distances about foci to the current point of the ellipse M as r 1 and r 2 . They are called focal radii. We denote the constant value r 1 + r 2 2a: r 1 + r 2 \u003d 2a. placing the point M at the points and A "it is easy to figure out that A" A \u003d 2a. The segments AA "and BB" are called the axes of the ellipse, and the segments OA and OB are called the semi-axes of the ellipse. Points A, A, B, B are called the vertices of the ellipse. Let M (x, y) be at point B, then r 1 \u003d r 2 \u003d a. From the tr-ka BOF 2 BO \u003d ÖBF 2 2 -OF 2 2 Let's denote BO \u003d in, then in \u003d Öa 2 - c 2. Through the semi-axis-ellipse a and in the equation will be written as follows:

This equation is called the canonical equation of the ellipse. A circle is a special case of an ellipse, obtained when a=b=R(R is the radix of the circle). The more semiaxes a and b differ from each other, the more oblate the ellipse will be. The degree of oblateness of an ellipse is usually measured by the eccentricity

Obviously, 0£ɛ£1. At ɛ=0 we have a circle, with an increase in ɛ the ellipse is more and more different from the circle, becoming more convex.

Hyperbola

A hyperbole is called a geome. the place of the points of the plane for which the absolute value of the difference between the distances to two given points, called foci, is a constant value, not equal to 0 and less than the distance between the foci. Focuses F 1 and F 2 will again be placed on the Ox axis at points (-c, 0), (c, 0). The segments F 1 M = r 1 and F 2 M = r 2 are called focal radii. By definition |r 1 - r 2 | is a constant value. Let's denote it 2а: |r 1 - r 2 | = 2a. Points A and A "are called the vertices of the hyperbola. It is easy to understand that AA" \u003d 2a. Indeed, for point A r 1 =AF 1 and r 2 =AF 2 . Obviously, AF 2 \u003d A "F 1, therefore r 1 - r 2 \u003d AF 1 -AF 2 \u003d AF 1 \u003d A" F 1 \u003d A "A. On the other hand, r 1 - r 2 \u003d 2a. The segment AA" is called the real axis of the hyperbola. Let v \u003d Öc 2 -a 2 Points B and B "have coordinates (0, c) and (0,-c). The segment BB" is called the imaginary axis of the hyperbola. The canonical equation of a hyperbola has the form:

the hyperbola has 2 branches, with a = b the hyperola is called isosceles. Equations y=inh¸a and y=-inh¸a. They are called asymptotes. If a point is removed along any of the branches of the hyperbola, then its distance to the corresponding asymptote tends to 0. For a hyperbola, the eccentricity takes values greater than 1.

Parabola.

A parabola is the locus of points in a plane equidistant from a given line, called the directrix, and from a given point not belonging to the directrix, called the focus. Let us denote the distance between the focus and the directrix as p. The canonical equation of a parabola has the form:

y 2 \u003d 2px and it turns out if the focus of F is placed at the point (p¸2, 0), and the straight line x = - p¸2 is taken as the directrix. The number p is called the parameter of the parabola, the point (0,0) is its vertex.

20. Plane in space: general equation, geometric meaning coefficients, the equation of a plane passing through a given point in space.

The general equation of the plane: Ax + Vy + Cz + D \u003d 0, in which at least one of coefficients A, B, C different from 0. These coefficients have a definition. Geom. meaning

We set the position of the plane using some point M 0 (x 0, y 0, z 0) and a non-zero vector N (A, B, C) perpendicular to the plane. According to these data, the plane is uniquely determined. Let M(x, y, z) be the current point of the plane. The vectors N (A, B, C) and M 0 M (x-x 0, y-y 0, z-z 0) are orthogonal, therefore their scalar product equals)

A (x-x 0) + B (y-y 0) + C (z-z 0) \u003d 0 (1)

After transformations, we get the equation:

Ah + Wu + Cz + D \u003d 0, where D \u003d -Ax 0 -B 0- Cz 0

Therefore, A, B, C are the coordinates of a vector perpendicular to the plane given by the general equation.

The set of planes described by equation (1), at a fixed point (x 0, y 0, z 0) and variable coefficients A, B, C, is called a bunch of planes. When among the conditions that define the desired plane, its point M 0 (x 0, y 0, z 0) is listed, you can start solving the problem by applying equation (1). A plane is also called a surface of the first order.

Sphere,

Sphere. The equation of a sphere whose center is at the origin is: x 2 + y 2 + z 2 = R 2 . Now let the center be located at the point M 0 (x 0, y 0, z 0)

The current point M(x, y, z) of the sphere is at a distance R from t. M.

From the equality MM 0 2 \u003d R 2 we get: (x-x 0) 2 + (y-y 0) 2 + (z-z 0) 2 \u003d R 2

Ellipsoid canonical the equation:

A, b, c - semiaxes of the ellipsoid. When a = b, an ellipsoid of revolution is obtained. This is the shape of the surface of our planet. When a=b=c, the ellipsoid turns into spheres of radius R=a

Paraboloid of revolution

In the yOz plane, consider the parabola y 2 =2pz. Surface, formed by rotation of this parabola around the Oz axis is called the paraboloid of revolution.

Let M(x, y, z) be an arbitrary point on the surface, and M 0 be a point with the same applicate z lying on the parabola y 2 =2pz. Because O "M \u003d O" M 0, then y 2 for the point M 0 can be replaced in the equation by x 2 + y 2 for the point M: x 2 + y 2 \u003d 2pz - the equation of the paraboloid of revolution

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 + With= 0, where With = –(BUTX 0 + By 0), (1.16),

Where BUT and AT 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 Canonical equation straight. 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.

Decision . 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 s given vector normals 3( X –1) + 2(At– 1) = 0 or 3 X + 2y- 5 \u003d 0. We got the equation of the desired straight line.