Let some affine coordinate system OXY be given.

Theorem 2.1. Any straight l coordinate system ОX is given by a linear equation of the form

A x+ B y+ C = O, (1)

where А, В, С R and А 2 + В 2 0. Conversely, any equation of the form (1) defines a straight line.

Equation of the form (1) - general equation of the line .

Let in equation (1) all coefficients A, B and C are nonzero. Then

Ax-By = -C, and.

Let's denote -C / A = a, -C / B = b. We get

-line segment equation .

Indeed, the numbers | a | and | b | indicate the values ​​of the segments cut off by the straight line l on the OX and OY axes, respectively.

Let it be straight l is given by the general equation (1) in a rectangular coordinate system and let the points M 1 (x 1, y 1) and M 2 (x 2, y 2) belong to l... Then

A x 1 + B at 1 + C = A NS 2 + B at 2 + C, that is, A ( x 1 -x 2) + B ( at 1 -at 2) = 0.

The last equality means that the vector = (A, B) is orthogonal to the vector = (x 1 -x 2, y 1 -y 2). those. The vector (A, B) is called normal vector of the line l.

Consider a vector = (- B, A). Then

A (-B) + BA = 0. those. ^.

Therefore, the vector = (- B, A) is the directing vector of the spicy l.

Parametric and canonical equations of a straight line

Equation of a straight line passing through two given points

Let a straight line be given in the affine coordinate system (0, X, Y) l, its direction vector = (m, n) and point M 0 ( x 0 ,y 0) owned l... Then for an arbitrary point M ( x,at) of this line we have

and so how .

If we denote and

The radius vectors of the points M and M 0, respectively, then

- equation of a straight line in vector form.

Since = ( NS,at), =(NS 0 ,at 0), then

x= x 0 + mt,

y= y 0 + nt

- parametric equation of a straight line .

Hence it follows that

- canonical equation of the line .

Finally, if on a straight line l two points M 1 ( NS 1 ,at 1) and

M 2 ( x 2 ,at 2), then vector = ( NS 2 -NS 1 ,y 2 -at 1) is guiding vector straight line l... Then

- equation of a straight line passing through two given points.

The relative position of two straight lines.

Let the straight lines l 1 and l 2 are given by their general equations

l 1: A 1 NS+ B 1 at+ С 1 = 0, (1)

l 2: A 2 NS+ B 2 at+ C 2 = 0.

Theorem... Let the straight lines l 1 and l 2 are given by equations (1). Then and only then:

1) straight lines intersect when there is no number λ such that

A 1 = λA 2, B 1 = λB 2;

2) the lines coincide when there is a number λ such that

А 1 = λA 2, B 1 = λB 2, С 1 = λС 2;

3) the lines are distinct and parallel when there is a number λ such that

А 1 = λA 2, В 1 = λВ 2, С 1 λС 2.

Bundle of straight lines

A bunch of straight lines is called the set of all lines in the plane passing through a point, called center beam.

To set the beam equation, it is enough to know any two straight lines l 1 and l 2 passing through the center of the beam.

Let in the affine coordinate system the lines l 1 and l 2 are given by the equations

l 1: A 1 x+ B 1 y+ C 1 = 0,

l 2: A 2 x+ B 2 y+ C 2 = 0.

The equation:

A 1 x+ B 1 y+ C + λ (A 2 NS+ B 2 y+ C) = 0

- the equation of the pencil of straight lines, defined by the equations l 1 and l 2.

In what follows, by a coordinate system we mean a rectangular coordinate system .

Conditions for parallelism and perpendicularity of two straight lines

Let the straight lines l 1 and l 2. their general equations; = (A 1, B 1), = (A 2, B 2) - normal vectors of these lines; k 1 = tgα 1, k 2 = tgα 2 - slopes; = ( m 1 ,n 1), (m 2 ,n 2) - direction vectors. Then, straight l 1 and l 2 are parallel if and only if one of the following conditions is met:

or either k 1 =k 2 or.

Let the straight lines now l 1 and l 2 are perpendicular. Then, obviously, that is, A 1 A 2 + B 1 B 2 = 0.

If straight l 1 and l 2 are given by the equations

l 1: at=k 1 x+ b 1 ,

l 2: at=k 2 x+ b 2 ,

then tgα 2 = tg (90º + α) = .

Hence it follows that

Finally, if and are the direction vectors of straight lines, then ^, that is

m 1 m 2 + n 1 n 2 = 0

The latter relations express the necessary and sufficient condition for the perpendicularity of two planes.

Angle between two straight lines

At an angle φ between two straight lines l 1 and l 2 we will understand the smallest angle through which one straight line must be rotated so that it becomes parallel to another straight line or coincides with it, that is, 0 £ φ £

Let the straight lines be given by general equations. It's obvious that

cosφ =

Let the straight lines now l 1 and l 2 is given by equations with slope coefficients k 1 in k 2 respectively. Then

It is obvious that, that is ( NS-NS 0) + B ( at-at 0) + C ( z-z 0) = 0

Let's expand the brackets and denote D = -A x 0 - B at 0 - C z 0. We get

A x+ B y+ C z+ D = 0 (*)

- general equation of the plane or general equation of the plane.

Theorem 3.1 The linear equation (*) (A 2 + B 2 + C 2 ≠ 0) is the equation of the plane and vice versa, any equation of the plane is linear.

1) D = 0, then the plane passes through the origin.

2) A = 0, then the plane is parallel to the OX axis

3) A = 0, B = 0, then the plane is parallel to the OXY plane.

Let all coefficients in the equation be nonzero.

- plane equation in line segments... The numbers | a |, | b |, | c | indicate the values ​​of the line segments cut off by the plane on the coordinate axes.

If in the general equation of the straight line Ax + Vy + C = 0 C ¹ 0, then, dividing by –C, we get: or

The geometric meaning of the coefficients is that the coefficient a is the coordinate of the point of intersection of the straight line with the Ox axis, and b- the coordinate of the point of intersection of the straight line with the Oy axis.

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

C = 1, a = -1, b = 1.

Normal equation of a straight line.

If both sides of the equation Ax + Vy + C = 0 are divided by a number called normalizing factor, then we get

Xcosj + ysinj - p = 0 -

normal equation of a straight line.

The ± sign of the normalizing factor should be chosen so that m × С< 0.

p is the length of the perpendicular dropped from the origin to the straight line, and j is the angle formed by this perpendicular with the positive direction of the Ox axis.

Example. A general equation of the straight line 12x - 5y - 65 = 0 is given. It is required to write various types of equations of this straight line.

the equation of this straight line in segments:

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

normal equation of the line:

; cosj = 12/13; sinj = -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.

Example. The straight line cuts off equal positive segments on the coordinate axes. Make a straight line equation if the area of ​​the triangle formed by these segments is 8 cm 2.

The straight line equation has the form:, a = b = 1; ab / 2 = 8; a = 4; -4.

a = -4 does not match the problem statement.

Total: or x + y - 4 = 0.

Example. Draw up the equation of the straight line passing through point A (-2, -3) and the origin.

The straight line equation has the form:, where x 1 = y 1 = 0; x 2 = -2; y 2 = -3.

Equation of a straight line passing through a given point

Perpendicular to the given line.

Definition. The straight line passing through the point M 1 (x 1, y 1) and perpendicular to the straight line y = kx + b is represented by the equation:

The angle between straight lines on the plane.

Definition. If two straight lines y = k 1 x + b 1, y = k 2 x + b 2 are given, then the acute angle between these straight lines will be defined as

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

Two straight lines are perpendicular if k 1 = -1 / k 2.

Theorem. Straight lines Ax + Vy + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the proportional coefficients A 1 = lA, B 1 = lB. If also С 1 = lС, then the lines coincide.

The coordinates of the point of intersection of two straight lines are found as a solution to the system of equations of these straight lines.

Distance from point to line.

Theorem. If a point M (x 0, y 0) is given, then the distance to the straight line Ax + Vy + C = 0 is determined as

Proof. Let point M 1 (x 1, y 1) be the base of the perpendicular dropped from point M onto a given straight line. Then the distance between points M and M 1:

The coordinates x 1 and y 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 perpendicular to a given straight 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 is proved.

Example . Determine the angle between the straight lines: y = -3x + 7; y = 2x + 1.

k 1 = -3; k 2 = 2 tgj =; j = p / 4.

Example. Show that the straight lines 3x - 5y + 7 = 0 and 10x + 6y - 3 = 0 are perpendicular.

We find: k 1 = 3/5, k 2 = -5/3, k 1 k 2 = -1, therefore, the straight lines are perpendicular.

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 required height equation is: Ax + By + C = 0 or y = kx + b.

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

Answer: 3x + 2y - 34 = 0.

Curves of the second order.

A second-order curve can be given by the equation

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

There is a coordinate system (not necessarily rectangular Cartesian), in which this equation can be represented in one of the forms below.

1) - the equation of the ellipse.

2) - the equation of the "imaginary" ellipse.

3) - the equation of the hyperbola.

4) a 2 x 2 - c 2 y 2 = 0 - the equation of two intersecting lines.

5) y 2 = 2px - parabola equation.

6) y 2 - a 2 = 0 is the equation of two parallel lines.

7) y 2 + a 2 = 0 is the equation of two “imaginary” parallel lines.

8) y 2 = 0 is a pair of coinciding straight lines.

9) (x - a) 2 + (y - b) 2 = R 2 is the equation of the circle.

Circle.

In the circle (x - a) 2 + (y - b) 2 = R 2, the center has coordinates (a; b).

Example. Find the coordinates of the center and the radius of the circle, if its equation is given in the form:

2x 2 + 2y 2 - 8x + 5y - 4 = 0.

To find the coordinates of the center and the radius of the circle, this equation must be reduced to the form indicated above in paragraph 9. To do this, select the complete 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 O (2; -5/4); R = 11/4.

Ellipse.

Definition. Ellipse called the curve given by the equation.

Definition. Focuses such two points are called, the sum of the distances from which to any point of the ellipse is a constant.

F 1, F 2 - focuses. F 1 = (c; 0); F 2 (-c; 0)

c - half the distance between the foci;

a - semi-major axis;

b - semi-minor axis.

Theorem. The focal length and semiaxes of the ellipse are related by the ratio:

a 2 = b 2 + c 2.

Proof: If point M is at the intersection of the ellipse with the vertical axis, r 1 + r 2= 2 (by the Pythagorean theorem). If point M is at the intersection of the ellipse with the horizontal axis, r 1 + r 2 = a - c + a + c. Because by definition the amount r 1 + r 2 Is a constant value, then, equating, we get:

a 2 = b 2 + c 2

r 1 + r 2 = 2a.

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

Because with< a, то е < 1.

Definition. The quantity k = b / a is called compression ratio ellipse, and the quantity 1 - k = (a - b) / a is called squeezing ellipse.

The compression ratio and eccentricity are related by the ratio: k 2 = 1 - e 2.

If a = b (c = 0, e = 0, foci merge), then the ellipse turns into a circle.

If for a point M (x 1, y 1) the condition is satisfied:, then it is inside the ellipse, and if, then the point is outside the ellipse.

Theorem. For an arbitrary point M (x, y) belonging to an ellipse, the following relations hold::

R 1 = a - ex, r 2 = a + ex.

Proof. It was shown above that r 1 + r 2 = 2a. In addition, for geometric reasons, you can write:

After squaring and reducing similar terms:

It can be proved similarly that r 2 = a + ex. The theorem is proved.

Two straight lines are connected to the ellipse, called directors... Their equations are:

X = a / e; x = -a / e.

Theorem. For a point to lie on an ellipse, it is necessary and sufficient that the ratio of the distance to the focus to the distance to the corresponding directrix be equal to the eccentricity e.

Example. Equate the straight line passing through the left focus and the bottom vertex of the ellipse given by the equation:

1) Coordinates of the lower vertex: x = 0; y 2 = 16; y = -4.

2) Coordinates of the left focus: c 2 = a 2 - b 2 = 25 - 16 = 9; c = 3; F 2 (-3; 0).

3) Equation of a straight line passing through two points:

Example. Draw up the equation of 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:

2c =, so a 2 - b 2 = c 2 = ½

by condition 2a = 2, therefore a = 1, b =

Hyperbola.

Definition. Hyperbole is called the set of points of the plane for which the modulus of the difference between the distances from two given points, called tricks there is a constant value less than the distance between the foci.

By definition ïr 1 - r 2 ï = 2a. F 1, F 2 - foci of hyperbola. F 1 F 2 = 2c.

Let us choose an arbitrary point M (x, y) on the hyperbola. Then:

denote c 2 - a 2 = b 2 (geometrically, this value is the minor semiaxis)

Received the canonical hyperbole equation.

The hyperbola is symmetrical about the midpoint of the segment connecting the foci and about the coordinate axes.

Axis 2a is called the real axis of the hyperbola.

Axis 2b is called the imaginary axis of the hyperbola.

The hyperbola has two asymptotes, the equations of which are

Definition. The relationship is called eccentricity hyperbolas, where c is half the distance between the foci, and is the real semiaxis.

Considering that c 2 - a 2 = b 2:

If a = b, e =, then the hyperbola is called isosceles (equilateral).

Definition. Two straight lines perpendicular to the real axis of the hyperbola and located symmetrically about the center at a distance a / e from it are called directors hyperbole. Their equations are:.

Theorem. If r is the distance from an arbitrary point M of the hyperbola to any focus, d is the distance from the same point to the directrix corresponding to this focus, then the ratio r / d is a constant value equal to the eccentricity.

Proof. Let's sketch a hyperbola.

From the obvious geometric relationships, you can write:

a / e + d = x, therefore d = x - a / e.

(x - c) 2 + y 2 = r 2

From the canonical equation:, taking into account b 2 = c 2 - a 2:

Then since c / a = e, then r = ex - a.

For the left branch of the hyperbola, the proof is similar. The theorem is proved.

Example. Find the equation of a hyperbola, the vertices and foci of which are at the corresponding vertices and foci of the ellipse.

For an ellipse: c 2 = a 2 - b 2.

For hyperbole: c 2 = a 2 + b 2.

Hyperbola equation:.

Example. Write the equation of the hyperbola if its eccentricity is 2, and the foci coincide with the foci of the ellipse with the equation of the parabola parameter. Let us derive the canonical equation of the parabola.

From geometric relationships: AM = MF; AM = x + p / 2;

MF 2 = y 2 + (x - p / 2) 2

(x + p / 2) 2 = y 2 + (x - p / 2) 2

x 2 + xp + p 2/4 = y 2 + x 2 - xp + p 2/4

Directrix equation: x = -p / 2.

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

From the parabola equation we find that p = 4.

r = x + p / 2 = 4; hence:

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

Example. The equation of the curve in the polar coordinate system is:

Find the equation of a curve in a Cartesian rectangular coordinate system, determine the type of the curve, find foci and eccentricity. Build a curve schematically.

Let's use the connection between the Cartesian rectangular and polar coordinate systems:;

Received the canonical hyperbole equation. From the equation it is seen that the hyperbola is shifted along the Ox axis by 5 to the left, the major semiaxis a is equal to 4, the minor semiaxis b is equal to 3, from which we obtain c 2 = a 2 + b 2; c = 5; e = c / a = 5/4.

Focuses F 1 (-10; 0), F 2 (0; 0).

Let's plot this hyperbola.

Properties of a straight line in Euclidean geometry.

You can draw infinitely many straight lines through any point.

A single straight line can be drawn through any two non-coinciding points.

Two mismatched straight lines on a 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 straight lines:

• straight lines intersect;

• straight lines are parallel;

• straight lines intersect.

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

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

General equation of the straight line.

Definition... Any straight line on a plane can be given by a first-order equation

Ax + Wu + C = 0,

with constant A, B are not equal to zero at the same time. This first-order equation is called common

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

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

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

The equation of a straight line can be presented in different forms, depending on any given

initial conditions.

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

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

perpendicular to the straight line given by the equation

Ax + Wu + C = 0.

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

Solution... At A = 3 and B = -1, we compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C

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

C = -1. Total: the required equation: 3x - y - 1 = 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 equation of a straight line,

passing through these points:

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

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

if x 1 ≠ x 2 and x = x 1, if x 1 = x 2 .

Fraction = k called slope straight.

Example... Find the equation of the 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 point and slope.

If the general equation of the straight line Ax + Wu + C = 0 lead to the form:

and designate , then the resulting equation is called

equation of a straight line with slope k.

Equation of a straight line along a point and a direction vector.

By analogy with the paragraph 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 nonzero vector (α 1, α 2) whose components satisfy the condition

Аα 1 + Вα 2 = 0 called directing vector of a straight line.

Ax + Wu + C = 0.

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

Solution... The equation of the required straight line will be sought in the form: Ax + By + C = 0. According to the definition,

the coefficients must meet the conditions:

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

Then the equation of the 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. required equation:

x + y - 3 = 0

Equation of a straight line in segments.

If in the general equation of the straight line Ax + Vy + 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 axis Oh, a b- the coordinate of the point of intersection of the straight line with the axis OU.

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

C = 1, a = -1, b = 1.

Normal equation of a straight line.

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

normalizing factor, then we get

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

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

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

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

Example... A general equation of the straight line is given 12x - 5y - 65 = 0... Required to write different types of equations

this straight line.

The equation of this line in segments:

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.

The angle between straight lines on the plane.

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

will be defined as

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

if k 1 = -1 / k 2 .

Theorem.

Direct Ax + Wu + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients are proportional

А 1 = λА, В 1 = λВ... If also С 1 = λС, then the straight lines coincide. Coordinates of the point of intersection of two lines

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

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

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

is represented by the equation:

Distance from point to line.

Theorem... If a point is given M (x 0, y 0), the distance to the straight line Ax + 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

straight line. Then the distance between the points M and M 1:

(1)

Coordinates x 1 and at 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 perpendicular to

a given straight 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 is proved.

Equation of a straight line, where a and b- some real numbers other than zero, called the equation of the straight line in the segments... This name is not accidental, since the absolute values ​​of the numbers a and b are equal to the lengths of the segments that the straight line cuts off on the coordinate axes Ox and Oy respectively (segments are measured from the origin). Thus, the equation of a straight line in segments makes it easy to build this straight line in the drawing. To do this, mark in a rectangular coordinate system on the plane points with coordinates and, and use a ruler to connect them with a straight line.

As an example, let's build a straight line given by an equation in segments of the form. We mark the points and connect them.

You can get detailed information about this kind of equation of a straight line on a plane in the article the equation of a straight line in segments.

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All topics in this section:

Definition of differentiability
The operation of finding the derivative is called function differentiation. A function is called differentiable at some point if it has a finite derivative at this point, and

Differentiation rule
Corollary 1. The constant factor can be moved outside the sign of the derivative:

The geometric meaning of the derivative. Tangent equation
The angle of inclination of the straight line y = kx + b is the angle measured from the position

The geometric meaning of the derivative of a function at a point
Consider the cutting line AB of the graph of the function y = f (x) such that points A and B, respectively, have coordinates

Solution
The function is defined for all real numbers. Since (-1; -3) is the point of contact, then

Necessary conditions for an extremum and sufficient conditions for an extremum
Determination of an increasing function. The function y = f (x) increases on the interval X if for any

Sufficient criteria for the extremum of a function
To find the maxima and minima of the function, you can use any of the three sufficient signs of an extremum. Although the most common and convenient is the first one.

Newton-Leibniz formula (with proof)
Newton-Leibniz formula. Let the function y = f (x) be continuous on an interval and F (x) be one of the antiderivatives of the function on this interval, then it is true that

We continue to study the section "Equation of a straight line on a plane" and in this article we will analyze the topic "Equation of a straight line in segments". Let us sequentially consider the form of the equation of a straight line in segments, the construction of a straight line, which is given by this equation, the transition from the general equation of a straight line to the equation of a straight line in segments. All this will be accompanied by examples and analysis of problem solving.

Let a rectangular coordinate system O x y be located on the plane.

A straight line on a plane in the Cartesian coordinate system O x y is given by an equation of the form x a + y b = 1, where a and b are some real numbers other than zero, the values ​​of which are equal to the lengths of the segments cut off by a straight line on the axes O x and O y. The lengths of the line segments are counted from the origin.

As we know, the coordinates of any of the points belonging to a straight line given by the equation of a straight line satisfy the equation of this straight line. Points a, 0 and 0, b belong to this straight line, since a a + 0 b = 1 ⇔ 1 ≡ 1 and 0 a + b b = 1 ⇔ 1 ≡ 1. Points a, 0 and b, 0 are located on the axes of coordinates O x and O y and are removed from the origin by a and b units. The direction in which you want to postpone the length of the segment is determined by the sign that stands in front of the numbers a and b. The "-" sign indicates that the length of the line segment must be plotted in the negative direction of the coordinate axis.

Let us explain all of the above by placing the straight lines relative to the fixed Cartesian coordinate system O x y in the schematic drawing. The equation of a straight line in segments x a + y b = 1 is used to construct a straight line in the Cartesian coordinate system O x y. To do this, we need to mark the points a, 0 and b, 0 on the axes, and then connect these points with a line using a ruler.

The drawing shows the cases when the numbers a and b have different signs, and, therefore, the lengths of the segments are plotted in different directions of the coordinate axes.

Let's look at an example.

Example 1

A straight line is given by the equation of a straight line in segments of the form x 3 + y - 5 2 = 1. It is necessary to build this straight line on a plane in the Cartesian coordinate system O x y.

Solution

Using the equation of a straight line in segments, we determine the points through which the straight line passes. This is 3, 0, 0, - 5 2. Let's mark them and draw a line.

Reduction of the general equation of a straight line to an equation of a straight line in segments

The transition from a given equation of a straight line to an equation of a straight line in segments makes it easier for us to solve various problems. Having a complete general equation of the line, we can get the equation of the line in segments.

The complete general equation of a straight line in a plane is A x + B y + C = 0, where A, B and C are not zero. We transfer the number C to the right side of the equality, divide both sides of the resulting equality by - С. In this case, we send the coefficients of x and y to the denominators:

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

To make the last transition, we used the equality p q = 1 q p, p ≠ 0, q ≠ 0.

As a result, we made the transition from the general equation of the straight line A x + B y + C = 0 to the equation of the straight line in the segments x a + y b = 1, where a = - C A, b = - C B.

Let's take a look at the following example.

Example 2

Let us carry out the transition to the equation of a straight line in segments, having the general equation of a straight line x - 7 y + 1 2 = 0.

Solution

Move the other half to the right side of the equality x - 7 y + 1 2 = 0 ⇔ x - 7 y = - 1 2.

Divide both sides of the equality by - 1 2: x - 7 y = - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y = 1.

Convert the resulting equality to the required form: 1 - 1 2 x - 7 - 1 2 y = 1 ⇔ x - 1 2 + y 1 14 = 1.

We got the equation of a straight line in segments.

Answer: x - 1 2 + y 1 14 = 1

In cases where a straight line is given by the canonical or parametric equation of a straight line on a plane, then first we go to the general equation of a straight line, and then to the equation of a straight line in segments.

To pass from the equation of a straight line in segments and the general equation of a straight line is simple: we transfer the unit from the right side of the equation of a straight line in segments of the form x a + y b = 1 to the left side with the opposite sign, select the coefficients in front of the unknowns x and y.

x a + y b = 1 ⇔ x a + y b - 1 = 0 ⇔ 1 a x + 1 b y - 1 = 0

We get the general equation of a straight line, from which you can go to any other kind of equation of a straight line on a plane. We have analyzed the transition process in detail in the topic "Reducing the general equation of a straight line to other types of equations of a straight line."

Example 3

The equation of a straight line in segments has the form x 2 3 + y - 12 = 1. It is necessary to write the general equation of a straight line on a plane.

Solution

Acts according to the previously described algorithm:

x 2 3 + y - 12 = 1 ⇔ 1 2 3 x + 1 - 12 y - 1 = 0 ⇔ ⇔ 3 2 x - 1 12 y - 1 = 0

Answer: 3 2 x - 1 12 y - 1 = 0

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