Properties of a straight line in Euclidean geometry.

An infinite number of straight lines can be drawn through any point.

Through any two non-coinciding points a single straight line can be drawn.

Two divergent lines in 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 lines:

lines intersect;

lines are parallel;

straight lines intersect.

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

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

General equation straight.

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

Ax + Wu + C = 0,

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

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- a 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 represented in in various forms depending on any given

initial conditions.

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

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

perpendicular to the line given by the equation

Ax + Wu + C = 0.

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

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

Let's 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 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 line,

passing through these points:

If any of the denominators is zero, the corresponding numerator should be set equal 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 line passing through points A(1, 2) and B(3, 4).

Solution. Applying the formula written above, we get:

Equation of a straight line using a point and slope.

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

and designate , then the resulting equation is called

equation of a straight line with slope k.

Equation of a straight line from a point and a direction 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 directing 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 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 the point A(1, 2).

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

coefficients must satisfy the following 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 Ах + Ву + С = 0 С≠0, then, dividing by -С, 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- 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 line in segments.

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

Normal equation of a 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 a 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 straight line,

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

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

this straight line.

The equation of this line in segments:

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

Equation of a 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 a plane.

Definition. If two lines are given y = k 1 x + b 1 , y = 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 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 parallel when the coefficients are proportional

A 1 = λA, B 1 = λB. If also С 1 = 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 perpendicular to a given line.

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

represented by the equation:

Distance from a point to a line.

Theorem. If a point is given M(x 0, y 0), then 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 a perpendicular dropped from a point M for a given

direct. Then the distance between 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 perpendicularly

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 has been proven.

This article continues the topic of the equation of a line on a plane: we will consider this type of equation as the general equation of a line. Let us define the theorem and give its proof; Let's figure out what an incomplete general equation of a line is and how to make transitions from a general equation to other types of equations of a line. We will reinforce the entire theory with illustrations and solutions to practical problems.

Yandex.RTB R-A-339285-1

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

Theorem 1

Any equation of the first degree, having the form A x + B y + C = 0, where A, B, C are some real numbers(A and B are not equal to zero at the same time) defines a straight line in a rectangular coordinate system on a plane. In turn, any straight line in a rectangular coordinate system on a plane is determined by an equation that has the form A x + B y + C = 0 for a certain set of values A, B, C.

Proof

This theorem consists of two points; we will prove each of them.

Let us prove that the equation A x + B y + C = 0 defines a straight line on the plane.

Let there be some point M 0 (x 0 , y 0) whose coordinates correspond to the equation A x + B y + C = 0. Thus: A x 0 + B y 0 + C = 0. Subtract from the left and right sides of the equations A x + B y + C = 0 the left and right sides of the equation A x 0 + B y 0 + C = 0, we obtain a new equation that looks like A (x - x 0) + B (y - y 0) = 0 . It is equivalent to A x + B y + C = 0.

The resulting equation A (x - x 0) + B (y - y 0) = 0 is a necessary and sufficient condition for the perpendicularity of the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0 ) . Thus, the set of points M (x, y) defines a straight line in a rectangular coordinate system perpendicular to the direction of the vector n → = (A, B). We can assume that this is not so, but then the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) would not be perpendicular, and the equality A (x - x 0 ) + B (y - y 0) = 0 would not be true.

Consequently, the equation A (x - x 0) + B (y - y 0) = 0 defines a certain line in a rectangular coordinate system on the plane, and therefore the equivalent equation A x + B y + C = 0 defines the same line. This is how we proved the first part of the theorem.

Let us provide a proof that any straight line in a rectangular coordinate system on a plane can be specified by an equation of the first degree A x + B y + C = 0.

Let us define a straight line a in a rectangular coordinate system on a plane; the point M 0 (x 0 , y 0) through which this line passes, as well as the normal vector of this line n → = (A, B) .

Let there also be some point M (x, y) - a floating point on a line. In this case, the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) are perpendicular to each other, and their scalar product there is a zero:

n → , M 0 M → = A (x - x 0) + B (y - y 0) = 0

Let's rewrite the equation A x + B y - A x 0 - B y 0 = 0, define C: C = - A x 0 - B y 0 and as a final result we get the equation A x + B y + C = 0.

So, we have proved the second part of the theorem, and we have proved the entire theorem as a whole.

Definition 1

An equation of the form A x + B y + C = 0 - This general equation of a line on a plane in a rectangular coordinate systemOxy.

Based on the proven theorem, we can conclude that a straight line and its general equation defined on a plane in a fixed rectangular coordinate system are inextricably linked. In other words, the original line corresponds to its general equation; the general equation of a line corresponds to a given line.

From the proof of the theorem it also follows that the coefficients A and B for the variables x and y are the coordinates of the normal vector of the line, which is given by the general equation of the line A x + B y + C = 0.

Let's consider specific example general equation of a straight line.

Let the equation 2 x + 3 y - 2 = 0 be given, which corresponds to a straight line in a given rectangular coordinate system. The normal vector of this line is the vector n → = (2 , 3) . Let's draw the given straight line in the drawing.

We can also state the following: the straight line that we see in the drawing is determined by the general equation 2 x + 3 y - 2 = 0, since the coordinates of all points on a given straight line correspond to this equation.

We can obtain the equation λ · A x + λ · B y + λ · C = 0 by multiplying both sides of the general equation of the line by a number λ not equal to zero. The resulting equation is equivalent to the original general equation, therefore, it will describe the same straight line on the plane.

Definition 2

Complete general equation of a line– such a general equation of the straight line A x + B y + C = 0, in which the numbers A, B, C are different from zero. Otherwise the equation is incomplete.

Let us analyze all variations of the incomplete general equation of a line.

When A = 0, B ≠ 0, C ≠ 0, the general equation takes the form B y + C = 0. Such an incomplete general equation defines in a rectangular coordinate system O x y a straight line that is parallel to the O x axis, since for any real value of x the variable y will take the value - C B . In other words, the general equation of the line A x + B y + C = 0, when A = 0, B ≠ 0, specifies the locus of points (x, y), whose coordinates are equal to the same number - C B .
If A = 0, B ≠ 0, C = 0, the general equation takes the form y = 0. This incomplete equation defines the x-axis O x .
When A ≠ 0, B = 0, C ≠ 0, we obtain an incomplete general equation A x + C = 0, defining a straight line parallel to the ordinate.
Let A ≠ 0, B = 0, C = 0, then the incomplete general equation will take the form x = 0, and this is the equation of the coordinate line O y.
Finally, for A ≠ 0, B ≠ 0, C = 0, the incomplete general equation takes the form A x + B y = 0. And this equation describes a straight line that passes through the origin. In fact, the pair of numbers (0, 0) corresponds to the equality A x + B y = 0, since A · 0 + B · 0 = 0.

Let us graphically illustrate all of the above types of incomplete general equation of a straight line.

Example 1

It is known that the given straight line is parallel to the ordinate axis and passes through the point 2 7, - 11. It is necessary to write down the general equation of the given line.

Solution

A straight line parallel to the ordinate axis is given by an equation of the form A x + C = 0, in which A ≠ 0. The condition also specifies the coordinates of the point through which the line passes, and the coordinates of this point meet the conditions of the incomplete general equation A x + C = 0, i.e. the equality is true:

A 2 7 + C = 0

From it it is possible to determine C if we give A some non-zero value, for example, A = 7. In this case, we get: 7 · 2 7 + C = 0 ⇔ C = - 2. We know both coefficients A and C, substitute them into the equation A x + C = 0 and get the required straight line equation: 7 x - 2 = 0

Answer: 7 x - 2 = 0

Example 2

The drawing shows a straight line; you need to write down its equation.

Solution

The given drawing allows us to easily take the initial data to solve the problem. We see in the drawing that the given straight line is parallel to the O x axis and passes through the point (0, 3).

The straight line, which is parallel to the abscissa, is determined by the incomplete general equation B y + C = 0. Let's find the values of B and C. The coordinates of the point (0, 3), since the given line passes through it, will satisfy the equation of the line B y + C = 0, then the equality is valid: B · 3 + C = 0. Let's set B to some value other than zero. Let's say B = 1, in which case from the equality B · 3 + C = 0 we can find C: C = - 3. We use known values B and C, we obtain the required equation of the straight line: y - 3 = 0.

Answer: y - 3 = 0 .

General equation of a line passing through a given point in a plane

Let the given line pass through the point M 0 (x 0 , y 0), then its coordinates correspond to the general equation of the line, i.e. the equality is true: A x 0 + B y 0 + C = 0. Let us subtract the left and right sides of this equation from the left and right sides of the general complete equation of the line. We get: A (x - x 0) + B (y - y 0) + C = 0, this equation is equivalent to the original general one, passes through the point M 0 (x 0, y 0) and has a normal vector n → = (A, B) .

The result we obtained makes it possible to write the general equation of the straight line with known coordinates the normal vector of a line and the coordinates of a certain point on this line.

Example 3

Given a point M 0 (- 3, 4) through which a line passes, and the normal vector of this line n → = (1 , - 2) . It is necessary to write down the equation of the given line.

Solution

The initial conditions allow us to obtain the necessary data to compose the equation: A = 1, B = - 2, x 0 = - 3, y 0 = 4. Then:

A (x - x 0) + B (y - y 0) = 0 ⇔ 1 (x - (- 3)) - 2 y (y - 4) = 0 ⇔ ⇔ x - 2 y + 22 = 0

The problem could have been solved differently. The general equation of a straight line is A x + B y + C = 0. The given normal vector allows us to obtain the values of coefficients A and B, then:

A x + B y + C = 0 ⇔ 1 x - 2 y + C = 0 ⇔ x - 2 y + C = 0

Now let’s find the value of C using the point M 0 (- 3, 4) specified by the condition of the problem, through which the straight line passes. The coordinates of this point correspond to the equation x - 2 · y + C = 0, i.e. - 3 - 2 4 + C = 0. Hence C = 11. The required straight line equation takes the form: x - 2 · y + 11 = 0.

Answer: x - 2 y + 11 = 0 .

Example 4

Given a line 2 3 x - y - 1 2 = 0 and a point M 0 lying on this line. Only the abscissa of this point is known, and it is equal to - 3. It is necessary to determine the ordinate of a given point.

Solution

Let us designate the coordinates of point M 0 as x 0 and y 0 . The source data indicates that x 0 = - 3. Since the point belongs to a given line, then its coordinates correspond to the general equation of this line. Then the equality will be true:

2 3 x 0 - y 0 - 1 2 = 0

Define y 0: 2 3 · (- 3) - y 0 - 1 2 = 0 ⇔ - 5 2 - y 0 = 0 ⇔ y 0 = - 5 2

Answer: - 5 2

Transition from the general equation of a line to other types of equations of a line and vice versa

As we know, there are several types of equations for the same straight line on a plane. The choice of the type of equation depends on the conditions of the problem; it is possible to choose the one that is more convenient for solving it. The skill of converting an equation of one type into an equation of another type is very useful here.

First, let's consider the transition from the general equation of the form A x + B y + C = 0 to the canonical equation x - x 1 a x = y - y 1 a y.

If A ≠ 0, then we move the term B y to the right side of the general equation. On the left side we take A out of brackets. As a result, we get: A x + C A = - B y.

This equality can be written as a proportion: x + C A - B = y A.

If B ≠ 0, we leave only the term A x on the left side of the general equation, transfer the others to the right side, we get: A x = - B y - C. We take – B out of brackets, then: A x = - B y + C B .

Let's rewrite the equality in the form of a proportion: x - B = y + C B A.

Of course, there is no need to memorize the resulting formulas. It is enough to know the algorithm of actions when moving from a general equation to a canonical one.

Example 5

The general equation of the line 3 y - 4 = 0 is given. It is necessary to transform it into a canonical equation.

Solution

Let's write the original equation as 3 y - 4 = 0. Next, we proceed according to the algorithm: the term 0 x remains on the left side; and on the right side we put - 3 out of brackets; we get: 0 x = - 3 y - 4 3 .

Let's write the resulting equality as a proportion: x - 3 = y - 4 3 0 . Thus, we have obtained an equation of canonical form.

Answer: x - 3 = y - 4 3 0.

To transform the general equation of a line into parametric ones, first a transition is made to the canonical form, and then a transition from the canonical equation of a line to parametric equations.

Example 6

The straight line is given by the equation 2 x - 5 y - 1 = 0. Write down the parametric equations for this line.

Solution

Let us make the transition from the general equation to the canonical one:

2 x - 5 y - 1 = 0 ⇔ 2 x = 5 y + 1 ⇔ 2 x = 5 y + 1 5 ⇔ x 5 = y + 1 5 2

Now we take both sides of the resulting canonical equation equal to λ, then:

x 5 = λ y + 1 5 2 = λ ⇔ x = 5 λ y = - 1 5 + 2 λ , λ ∈ R

Answer:x = 5 λ y = - 1 5 + 2 λ , λ ∈ R

The general equation can be converted into an equation of a straight line with slope y = k · x + b, but only when B ≠ 0. For the transition, we leave the term B y on the left side, the rest are transferred to the right. We get: B y = - A x - C . Let's divide both sides of the resulting equality by B, different from zero: y = - A B x - C B.

Example 7

The general equation of the line is given: 2 x + 7 y = 0. You need to convert that equation into a slope equation.

Solution

Let's perform the necessary actions according to the algorithm:

2 x + 7 y = 0 ⇔ 7 y - 2 x ⇔ y = - 2 7 x

Answer: y = - 2 7 x .

From the general equation of a line, it is enough to simply obtain an equation in segments of the form x a + y b = 1. To make such a transition, we move the number C to the right side of the equality, divide both sides of the resulting equality by – C and, finally, transfer the coefficients for the variables 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

Example 8

It is necessary to transform the general equation of the line x - 7 y + 1 2 = 0 into the equation of the line in segments.

Solution

Let's move 1 2 to the right side: x - 7 y + 1 2 = 0 ⇔ x - 7 y = - 1 2 .

Let's divide both sides of the equality by -1/2: x - 7 y = - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y = 1 .

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

In general, the reverse transition is also easy: from other types of equations to the general one.

The equation of a line in segments and an equation with an angular coefficient can be easily converted into a general one by simply collecting all the terms on the left side of the equality:

x a + y b ⇔ 1 a x + 1 b y - 1 = 0 ⇔ A x + B y + C = 0 y = k x + b ⇔ y - k x - b = 0 ⇔ A x + B y + C = 0

The canonical equation is converted to a general one according to the following scheme:

x - x 1 a x = y - y 1 a y ⇔ a y · (x - x 1) = a x (y - y 1) ⇔ ⇔ a y x - a x y - a y x 1 + a x y 1 = 0 ⇔ A x + B y + C = 0

To move from parametric ones, first move to the canonical one, and then to the general one:

x = x 1 + a x · λ y = y 1 + a y · λ ⇔ x - x 1 a x = y - y 1 a y ⇔ A x + B y + C = 0

Example 9

The parametric equations of the line x = - 1 + 2 · λ y = 4 are given. It is necessary to write down the general equation of this line.

Solution

Let us make the transition from parametric equations to canonical ones:

x = - 1 + 2 · λ y = 4 ⇔ x = - 1 + 2 · λ y = 4 + 0 · λ ⇔ λ = x + 1 2 λ = y - 4 0 ⇔ x + 1 2 = y - 4 0

Let's move from the canonical to the general:

x + 1 2 = y - 4 0 ⇔ 0 · (x + 1) = 2 (y - 4) ⇔ y - 4 = 0

Answer: y - 4 = 0

Example 10

The equation of a straight line in the segments x 3 + y 1 2 = 1 is given. It is necessary to make a transition to general appearance equations

Solution:

We simply rewrite the equation in the required form:

x 3 + y 1 2 = 1 ⇔ 1 3 x + 2 y - 1 = 0

Answer: 1 3 x + 2 y - 1 = 0 .

Drawing up a general equation of a line

We said above that the general equation can be written with known coordinates of the normal vector and the coordinates of the point through which the line passes. Such a straight line is defined by the equation A (x - x 0) + B (y - y 0) = 0. There we also analyzed the corresponding example.

Now let's look at more complex examples, in which first you need to determine the coordinates of the normal vector.

Example 11

Given a line parallel to the line 2 x - 3 y + 3 3 = 0. The point M 0 (4, 1) through which the given line passes is also known. It is necessary to write down the equation of the given line.

Solution

The initial conditions tell us that the lines are parallel, then, as the normal vector of the line, the equation of which needs to be written, we take the direction vector of the line n → = (2, - 3): 2 x - 3 y + 3 3 = 0. Now we know all the necessary data to create the general equation of the line:

A (x - x 0) + B (y - y 0) = 0 ⇔ 2 (x - 4) - 3 (y - 1) = 0 ⇔ 2 x - 3 y - 5 = 0

Answer: 2 x - 3 y - 5 = 0 .

Example 12

The given line passes through the origin perpendicular to the line x - 2 3 = y + 4 5. It is necessary to create a general equation for a given line.

Solution

The normal vector of a given line will be the direction vector of the line x - 2 3 = y + 4 5.

Then n → = (3, 5) . The straight line passes through the origin, i.e. through point O (0, 0). Let's create a general equation for a given line:

A (x - x 0) + B (y - y 0) = 0 ⇔ 3 (x - 0) + 5 (y - 0) = 0 ⇔ 3 x + 5 y = 0

Answer: 3 x + 5 y = 0 .

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The equation of a line passing through a given point in a given direction. Equation of a line passing through two given points. The angle between two straight lines. The condition of parallelism and perpendicularity of two straight 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 beam center.

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

The angular coefficient 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 straight lines are given by equations with a slope

y = k 1 x + B 1 ,

Equation of a line on a plane.

As is known, any point on the plane is determined by two coordinates in some coordinate system. Coordinate systems can be different depending on the choice of basis and origin.

Definition. Line equation is called the relation y = f(x) between the coordinates of the points that make up this line.

Note that the equation of a line can be expressed parametrically, that is, each coordinate of each point is expressed through some independent parameter t.

A typical example is the trajectory of a moving point. In this case, the role of the parameter is played by time.

Equation of a straight line on a plane.

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

Ax + Wu + C = 0,

Moreover, the constants A and B are not equal to zero at the same time, i.e. A 2 + B 2  0. This first order equation is called general equation of a straight line.

Depending on the values constant A, B and C 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 Ox axis

B = 0, A  0, C  0 (Ax + C = 0) – straight line parallel to the Oy axis

B = C = 0, A  0 – the straight line coincides with the Oy axis

A = C = 0, B  0 – the straight line coincides with the Ox axis

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

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

Definition. In the Cartesian rectangular coordinate system, a vector with components (A, B) is perpendicular to the straight line given by the equation Ax + By + C = 0.

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

With A = 3 and B = -1, let’s compose the equation of the straight line: 3x – y + C = 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 required equation: 3x – y – 1 = 0.

Equation of a line passing through two points.

Let two points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2) be given in space, then the equation of the line passing through these points is:

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

On the 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 is called slope straight.

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

Applying the formula written above, we get:

Equation of a straight line using a point and slope.

If the general equation of the straight line Ax + By + C = 0 is reduced to the form:

and designate
, then the resulting equation is called equation of a straight line with slopek.

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

By analogy with the point considering the equation of a straight line through a normal vector, you can enter the definition of a straight line through a point and the directing vector of the straight line.

Definition. Every non-zero vector ( 1,  2), the components of which satisfy the condition A 1 + B 2 = 0 is called the directing vector of the line

Ax + Wu + C = 0.

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

We will look for the equation of the desired line in the form: Ax + By + C = 0. In accordance with the definition, the coefficients must satisfy 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:

Equation of a straight line in segments.

If in the general equation of the straight line Ах + Ву + С = 0 С 0, then, dividing by –С, 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 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 line x – y + 1 = 0 is given. Find the equation of this line in segments.

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

Normal equation of a line.

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

xcos + ysin - p = 0 –

normal equation of a line.

The sign  of the normalizing factor must be chosen so that С< 0.

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

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

equation of this line in segments:

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

normal equation of a line:

; cos = 12/13; sin = -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 of coordinates.

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

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

a = -4 is not suitable according to the conditions of the problem.

Total:
or x + y – 4 = 0.

Example. Write an equation for a straight line passing through point A(-2, -3) and the origin.

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

The angle between straight lines on a plane.

Definition. If two lines are given y = k 1 x + b 1, y = 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 = -1/k 2 .

Theorem. Direct lines Ax + Wu + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A are proportional 1 =  A, B 1 =  B. If also C 1 =  C, then the lines coincide.

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

Equation of a line passing through a given point

perpendicular to this line.

Definition. A 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:

Distance from a point to a line.

Theorem. If the point M(x) is given 0 , y 0 ), then the distance to the straight line Ах + Ву + С =0 is defined as

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

The coordinates x 1 and y 1 can be found by solving the system of equations:

The second equation of the system is the equation of a line passing through a given point M 0 perpendicular to a 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.

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

k 1 = -3; k 2 = 2 tg =
;  = /4.

Example. Show that the 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 lines are perpendicular.

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

We find the equation of side AB:
; 4x = 6y – 6;

2x – 3y + 3 = 0;

The required 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:
whence b = 17. Total:
.

Answer: 3x + 2y – 34 = 0.

Analytical geometry in space.

Equation of a line in space.

Equation of a line in space given a point and

direction vector.

Let's take an arbitrary line and a vector (m, n, p), parallel to the given line. Vector called guide vector straight.

On the straight line we take two arbitrary points M 0 (x 0 , y 0 , z 0) and M (x, y, z).

M 1

Let us denote the radius vectors of these points as And , it's obvious that - =
.

Because vectors
And are collinear, then the relation is true
= t, where t is some parameter.

In total, we can write: = + t.

Because this equation is satisfied by the coordinates of any point on the line, then the resulting equation is parametric equation of a line.

This vector equation can be represented in coordinate form:

By transforming this system and equating the values of the parameter t, we obtain the canonical equations of a straight line in space:

Definition. Direction cosines direct are the direction cosines of the vector , which can be calculated using the formulas:

;

.

From here we get: m: n: p = cos : cos : cos.

The numbers m, n, p are called angle coefficients straight. Because is a non-zero vector, then m, n and p cannot be equal to zero at the same time, but one or two of these numbers can be equal to zero. In this case, in the equation of the line, the corresponding numerators should be set equal to zero.

Equation of a straight line in space passing

through two points.

If on a straight line in space we mark two arbitrary points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), then the coordinates of these points must satisfy the straight line equation obtained above:

In addition, for point M 1 we can write:

Solving these equations together, we get:

This is the equation of a line passing through two points in space.

General equations of a straight line in space.

The equation of a straight line can be considered as the equation of the line of intersection of two planes.

As discussed above, a plane in vector form can be specified by the equation:

+ D = 0, where

- plane normal; - radius is the vector of an arbitrary point on the plane.

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

Ax + Wu + C = 0,

Moreover, the constants A and B are not equal to zero at the same time. This first order equation is called general equation of a straight line. Depending on the values of constants A, B and C, 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 Ox axis

B = 0, A ≠0, C ≠ 0 (Ax + C = 0) – straight line parallel to the Oy axis

B = C = 0, A ≠0 – the straight line coincides with the Oy axis

A = C = 0, B ≠0 – the straight line coincides with the Ox axis

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

Equation of a straight line from a point and normal vector

Definition. In the Cartesian rectangular coordinate system, a vector with components (A, B) is perpendicular to the straight line given by the equation Ax + By + C = 0.

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

Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x – y + C = 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 required equation: 3x – y – 1 = 0.

Equation of a line passing through two points

Let two points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2) be given in space, then the equation of the line passing through these points is:

If any of the denominators is equal to zero, the corresponding numerator should be equal to zero. On the plane, the equation of the line written above is simplified:

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

The fraction = k is called slope straight.

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

Solution. Applying the formula written above, we get:

Equation of a straight line from a point and slope

If the total Ax + Bu + C = 0, lead to the form:

and designate , then the resulting equation is called equation of a straight line with slopek.

Equation of a straight line from a point and a direction vector

Definition. Each non-zero vector (α 1, α 2), the components of which satisfy the condition A α 1 + B α 2 = 0 is called a directing vector of the line

Ax + Wu + C = 0.

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

Solution. We will look for the equation of the desired line in the form: Ax + By + C = 0. In accordance with the definition, the coefficients must satisfy 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. for x = 1, y = 2 we obtain C/ A = -3, i.e. required equation:

Equation of a line in segments

If in the general equation of the straight line Ах + Ву + С = 0 С≠0, then, dividing by –С, we get: or

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

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

Normal equation of a line

If both sides of the equation Ax + By + C = 0 are multiplied by the number which is called normalizing factor, then we get

xcosφ + ysinφ - p = 0 –

normal equation of a line. The sign ± of the normalizing factor must be chosen so that μ * C< 0. р – длина перпендикуляра, опущенного из начала координат на прямую, а φ - угол, образованный этим перпендикуляром с положительным направлением оси Ох.

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

equation of this line in segments:

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

; 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 of coordinates.

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

Solution. The equation of the straight line has the form: , ab /2 = 8; ab=16; a=4, a=-4. a = -4< 0 не подходит по условию задачи. Итого: или х + у – 4 = 0.

Example. Write an equation for a straight line passing through point A(-2, -3) and the origin.

Solution. The equation of the straight line is: , where x 1 = y 1 = 0; x 2 = -2; y 2 = -3.

Angle between straight lines on a plane

Definition. If two lines are given y = k 1 x + b 1, y = 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 = -1/ k 2.

Theorem. The lines Ax + Bу + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A 1 = λA, B 1 = λB are proportional. If also C 1 = λC, then the lines coincide. The coordinates of the point of intersection of two lines are found as a solution to the system of equations of these lines.

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

Definition. A 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:

Distance from point to line

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

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

(1)

The coordinates x 1 and y 1 can be found by solving the system of equations:

The second equation of the system is the equation of a line passing through a given point M 0 perpendicular to a 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.

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

k 1 = -3; k 2 = 2; tgφ = ; φ= π /4.

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

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

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

Solution. We find the equation of side AB: ; 4 x = 6 y – 6;

2 x – 3 y + 3 = 0;

The required 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: from where b = 17. Total: .

Answer: 3 x + 2 y – 34 = 0.