Equation direct online calculator. General equation of the straight line: description, examples, problem solving

Consider the equation of a straight line passing through a point and a normal vector. Let a point and a nonzero vector be given in the coordinate system (Fig. 1).

Definition

As you can see, there is a single straight line that passes through a point perpendicular to the direction of the vector (in this case, it is called normal vector straight ).

Figure: 1

Let us prove that the linear equation

this is the equation of a straight line, that is, the coordinates of each point of the straight line satisfy equation (1), but the coordinates of a point that does not lie on, equation (1) does not.

To prove it, let us pay attention that the scalar product of vectors and \u003d in coordinate form coincides with the left side of equation (1).

Then we use the obvious property of the straight line: vectors and are perpendicular if and only if the point lies on. And provided that both vectors are perpendicular, their scalar product (2) turns into for all points that lie on, and only for them. Hence, (1) is the equation of the straight line.

Definition

Equation (1) is called the equation of the straight line that passes through a given point with normal vector \u003d.

Let us transform equation (1)

Denoting \u003d, we get

Thus, a straight line corresponds to a linear equation of the form (3). On the contrary, for a given equation of the form (3), where at least one of the coefficients is not equal to zero, a straight line can be constructed.

Indeed, let a pair of numbers satisfy equation (3), that is,

Subtracting the latter from (3), we obtain a relation that defines the line behind the vector and point.

Study of the general equation of a line

It is useful to know the peculiarities of the placement of a straight line in individual cases when one or two of the numbers are equal to zero.

1. The general equation looks like this:. The point satisfies it, which means that the straight line passes through the origin. It can be written: \u003d - x (see Fig. 2).

Figure: 2

We believe that:

If we put, then, we get one more point (see Fig. 2).

2., then the equation looks like this, where \u003d -. The normal vector lies on the axis, the straight line. Thus, a straight line is perpendicular at a point, or parallel to the axis (see Fig. 3). In particular, if and, then the equation is the equation of the ordinate axis.

Figure: 3

3. Similarly, for the equation is written, where. The vector belongs to the axis. A straight line at a point (Fig. 4).

If, then the equation of the axis.

The study can be formulated in the following form: the straight line is parallel to the coordinate axis, the change of which is absent in the general equation of the straight line.

For example:

Let's construct a straight line according to the general equation, provided that - are not equal to zero. To do this, it is enough to find two points that lie on this straight line. It is sometimes more convenient to find such points on the coordinate axes.

Then we put \u003d -.

For, then \u003d -.

We denote - \u003d, - \u003d. The points and are found. Set aside on the axes and draw a straight line through them (see Fig. 5).

Figure: five

From the general, you can go to the equation, which will include numbers and:

And then it turns out:

Or, according to the notation, we get the equation

Which is called the equation of the line in the segments... The numbers and, with accuracy to the sign, are equal to the segments that are cut off by a straight line on the coordinate axes.

Equation of a straight line with a slope

To find out what the equation of a straight line with a slope is, consider equation (1):

Denoting - \u003d, we get

equation of a straight line that passes through a point in a given direction. The geometric content of the coefficient is clear from Fig. 6.

B \u003d \u003d, where is the smallest angle by which you need to rotate the positive direction of the axis around a common point until it is aligned with a straight line. Obviously, if the angle is sharp, then title \u003d "(! LANG: Rendered by QuickLaTeX.com" height="17" width="97" style="vertical-align: -4px;">; если же – тупой угол, тогда .!}

Let's expand the brackets in (5) and simplify it:

where. Relation (6) - equation slope straight... When, is a segment that cuts off a straight line on the axis (see Fig. 6).

Note!

To go from the general equation of a straight line to an equation with a slope, you must first solve for.

Figure: 6

\u003d - x + - \u003d

where \u003d -, \u003d - is denoted. If, then it is already known from the study of the general equation that such a straight line is perpendicular to the axis.

Consider the canonical equation of a line using an example.

Let a point and a nonzero vector be given in the coordinate system (Fig. 7).

Figure: 7

It is necessary to form an equation of a line that passes through a point parallel to a vector, which is called a direction vector. An arbitrary point belongs to this line if and only if. Since the vector is given, and the vector, then according to the parallelism condition, the coordinates of these vectors are proportional, that is:

Definition

Relation (7) is called the equation of a straight line that passes through a given point in a given direction or the canonical equation of a straight line.

Note that one can go to an equation of the form (7), for example, from the equation of a pencil of straight lines (4)

or from the equation of a straight line through a point and a normal vector (1):

Above, it was assumed that the direction vector is non-zero, but it may happen that one of its coordinates, for example,. Then expression (7) will be formally written:

which doesn't make sense at all. However, the equation of a straight line perpendicular to the axis is accepted and obtained. Indeed, it can be seen from the equation that the straight line is defined by a point and a direction vector perpendicular to the axis. If we get rid of the denominator in this equation, then we get:

Or - the equation of a straight line perpendicular to the axis. The same would be obtained for the vector.

Parametric equation of a straight line

To understand what a parametric equation of a straight line is, it is necessary to return to equation (7) and equate each fraction (7) to a parameter. Since at least one of the denominators in (7) is not equal to zero, and the corresponding numerator can acquire arbitrary values, then the range of parameter change is the entire number axis.

Definition

Equation (8) is called the parametric equation of the line.

Examples of tasks on a straight line

Of course, it is difficult to solve anything solely by definitions, because you need to solve at least a few examples or tasks on your own that will help consolidate the material covered. Therefore, let's analyze the main tasks in a straight line, since similar tasks often come across on exams and tests.

Canonical and parametric equation

Example 1

On the straight line given by the equation, find the point that are 10 units from the point of this straight line.

Decision:

Let be sought point of a straight line, then we write down for distance. Given that . Since the point belongs to a straight line, which has a normal vector, then the equation of the straight line can be written: \u003d \u003d and then it turns out:

Then the distance. Provided, or. From a parametric equation:

Example 2

A task

The point moves uniformly with speed in the direction of the vector from the starting point. Find the coordinates of a point through from the beginning of the movement.

Decision

First you need to find the unit vector. Its coordinates are direction cosines:

Then the velocity vector:

X \u003d x \u003d.

The canonical equation of the straight line will now be written:

\u003d \u003d, \u003d Is a parametric equation. After that, you need to use the parametric equation of the line at.

Decision:

The equation of a straight line that passes through a point is found by the formula for a pencil of straight lines, where slope for a straight line and \u003d for a straight line.

Considering the figure, where it is seen that between the straight lines and are two angles: one is acute and the other is obtuse. According to formula (9), this is the angle between the straight lines and by which you need to turn the straight line counterclockwise relative to their intersection point until it is aligned with the straight line.

So, we remembered the formula, sorted out the angles, and now we can return to our example. Hence, taking into account formula (9), we first find the equations of the leg.

Since rotation of a straight line by an angle counterclockwise relative to a point leads to alignment with a straight line, then in formula (9), a. From the equation:

By the formula of the pencil of the equation, the straight line is written:

Similarly, we find, and,

Straight line equation:

Equation of a straight line - types of equation of a straight line: passing through a point, general, canonical, parametric, etc. updated: November 22, 2019 by the author: Scientific Articles.Ru

Definition.In a Cartesian rectangular coordinate system, a vector with components (A, B) is perpendicular to the straight line given by the equation Ax + Vy + C \u003d 0.

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

Decision... With 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 \u003d 0, hence C \u003d -1. Total: the required equation: 3x - y - 1 \u003d 0.

Equation of a straight 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 straight line passing through these points:

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

if x 1 ≠ x 2 and x \u003d x 1, if x 1 \u003d x 2.

Fraction \u003d k is called slopestraight.

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

If the general equation of the straight line Ax + Vy + C \u003d 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 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 specification of a straight line through a point and a direction vector of a straight line.

Definition.Each nonzero vector (α 1, α 2) whose components satisfy the condition А α 1 + В α 2 \u003d 0 is called the directing vector of the line

Ax + Wu + C \u003d 0.

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

Decision. The equation of the desired straight line will be sought in the form: Ax + By + C \u003d 0. According to the definition, the coefficients must satisfy the conditions:

1 * A + (-1) * B \u003d 0, i.e. A \u003d B.

Then the equation of the straight line has the form: Ax + Ay + C \u003d 0, or x + y + C / A \u003d 0. for x \u003d 1, y \u003d 2 we obtain C / A \u003d -3, i.e. required equation:

Equation of a straight line in segments

If in the general equation of the straight line Ax + Vu + C \u003d 0 C ≠ 0, then, dividing by –C, we get: or

The geometric meaning of the coefficients is that the coefficient and 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. A general equation of the straight line x - y + 1 \u003d 0 is given. Find the equation of this straight line in segments.

C \u003d 1, a \u003d -1, b \u003d 1.

Normal Equation of a Line

If both sides of the equation Ax + Vy + C \u003d 0 divided by the number which is called normalizing factor, then we get

xcosφ + ysinφ - p \u003d 0 -

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

Example... A general equation of the straight line 12x - 5y - 65 \u003d 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 line with slope: (divide by 5)

normal equation of the line:

; cos φ \u003d 12/13; sin φ \u003d -5/13; p \u003d 5.

It should be noted that not every line can be represented by an equation in segments, for example, lines parallel to 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 \u200b\u200bthe triangle formed by these segments is 8 cm 2.

Decision. The straight line equation has the form:, ab / 2 \u003d 8; a \u003d 4; -4. a \u003d -4 does not match the problem statement. Total: or x + y - 4 \u003d 0.

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

Decision. The straight line equation has the form: , where x 1 \u003d y 1 \u003d 0; x 2 \u003d -2; y 2 \u003d -3.

Let's consider how to draw up the equation of a straight line passing through two points using examples.

Example 1.

Make an equation of the straight line passing through the points A (-3; 9) and B (2; -1).

Method 1 - compose the equation of a straight line with a slope.

The equation of a straight line with a slope has the form. Substituting the coordinates of points A and B into the equation of the straight line (x \u003d -3 and y \u003d 9 - in the first case, x \u003d 2 and y \u003d -1 - in the second), we obtain a system of equations from which we find the values \u200b\u200bof k and b:

Adding the 1st and 2nd equations term by term, we get: -10 \u003d 5k, whence k \u003d -2. Substituting k \u003d -2 into the second equation, we find b: -1 \u003d 2 · (-2) + b, b \u003d 3.

Thus, y \u003d -2x + 3 is the desired equation.

Method 2 - compose the general equation of the straight line.

The general equation of the straight line is. Substituting the coordinates of points A and B into the equation, we get the system:

Since the number of unknowns is greater than the number of equations, the system is not solvable. But you can express all the variables through one. For example, through b.

Multiplying the first equation of the system by -1 and adding term by term with the second:

we get: 5a-10b \u003d 0. Hence a \u003d 2b.

Substitute the resulting expression into the second equation: 2 · 2b -b + c \u003d 0; 3b + c \u003d 0; c \u003d -3b.
Substitute a \u003d 2b, c \u003d -3b into the equation ax + by + c \u003d 0:

2bx + by-3b \u003d 0. It remains to split both parts by b:

The general equation of a straight line is easily reduced to the equation of a straight line with a slope:

Method 3 - compose the equation of a straight line passing through 2 points.

The equation of a straight line passing through two points has:

Substitute in this equation the coordinates of points A (-3; 9) and B (2; -1)

(that is, x 1 \u003d -3, y 1 \u003d 9, x 2 \u003d 2, y 2 \u003d -1):

and simplify:

whence 2x + y-3 \u003d 0.

In the school course, the equation of a straight line with a slope is most often used. But the easiest way is to derive and use the formula for the equation of a straight line passing through two points.

Comment.

If, when substituting the coordinates of the given points, one of the denominators of the equation

turns out to be equal to zero, then the desired equation is obtained by equating to zero of the corresponding numerator.

Example 2.

Make an equation of a straight line passing through two points C (5; -2) and D (7; -2).

Substitute into the equation of a straight line passing through 2 points, the coordinates of points C and D.

This article received equation of a straight line passing through two given points in a rectangular Cartesian coordinate system on a plane, and also derived the equations of a straight line that passes through two specified points in a rectangular coordinate system in three-dimensional space. After the presentation of the theory, solutions of typical examples and problems are shown in which it is required to compose equations of a straight line of various types, when the coordinates of two points of this straight line are known.

Page navigation.

Equation of a straight line passing through two given points on a plane.

Before we get the equation of a straight line passing through two given points in a rectangular coordinate system on a plane, let's recall some facts.

One of the axioms of geometry says that a single straight line can be drawn through two non-coinciding points on the plane. In other words, by specifying two points on a plane, we uniquely define a straight line that passes through these two points (if necessary, refer to the section on ways to define a straight line on a plane).

Let Oxy be fixed on the plane. In this coordinate system, any straight line corresponds to some equation of a straight line on a plane. The directing vector of the straight line is inextricably linked with this line. This knowledge is quite enough to compose the equation of a straight line passing through two given points.

Let us formulate the condition of the problem: write the equation of the straight line a, which in the rectangular Cartesian coordinate system Oxy passes through two non-coincident points and.

Let's show the simplest and most universal solution to this problem.

We know that the canonical equation of a line in the plane of the form specifies in the rectangular Oxy coordinate system a straight line passing through a point and having a direction vector.

Let us write the canonical equation of the line a passing through two given points and.

Obviously, the direction vector of the straight line a, which passes through points M 1 and M 2, is a vector, it has coordinates (see article if necessary). Thus, we have all the necessary data to write the canonical equation of the straight line a - the coordinates of its direction vector and the coordinates of the point (s) lying on it. It has the form (or ).

We can also write parametric equations of a straight line on a plane passing through two points and. They look like or .

Let's look at the example solution.

Example.

Write the equation of a straight line that passes through two given points .

Decision.

We found out that the canonical equation of a straight line passing through two points with coordinates and has the form .

From the condition of the problem we have ... We substitute this data into the equation ... We get .

Answer:

.

If we need not a canonical equation of a straight line and not parametric equations of a straight line passing through two given points, but an equation of a straight line of a different kind, then we can always come to it from the canonical equation of a straight line.

Example.

Write the general equation of the straight line, which in the rectangular coordinate system Oxy on the plane passes through two points and.

Decision.

First, we write the canonical equation of a straight line passing through two given points. It looks like. Now let us bring the resulting equation to the required form:.

Answer:

.

This is where we can finish with the equation of a straight line passing through two given points in a rectangular coordinate system on a plane. But I would like to recall how we solved such a problem in high school in algebra lessons.

At school, we only knew the equation of a straight line with a slope of the form. Let us find the value of the slope k and the number b at which the equation defines in the rectangular coordinate system Oxy on the plane a straight line passing through the points and at. (If x 1 \u003d x 2, then the slope of the straight line is infinite, and the straight line М 1 М 2 is determined by the general incomplete equation of the straight line of the form x-x 1 \u003d 0).

Since points М 1 and М 2 lie on a straight line, the coordinates of these points satisfy the equation of a straight line, that is, equalities and are true. Solving a system of equations of the form with respect to unknown variables k and b, we find or ... For these values \u200b\u200bof k and b, the equation of the straight line passing through two points and takes the form or .

It makes no sense to memorize these formulas; when solving examples, it is easier to repeat the indicated actions.

Example.

Write the equation of a line with a slope if this line passes through the points and.

Decision.

In the general case, the equation of a straight line with a slope has the form. Let us find k and b for which the equation corresponds to a straight line passing through two points and.

Since points М 1 and М 2 lie on a straight line, their coordinates satisfy the equation of a straight line, that is, the equalities are true and. The values \u200b\u200bof k and b are found as a solution to the system of equations (refer to the article if necessary):

It remains to substitute the found values \u200b\u200binto the equation. Thus, the required equation of the straight line passing through two points and has the form.

Colossal work, isn't it?

It is much easier to write down the canonical equation of a straight line passing through two points and, it has the form , and from it go to the equation of the straight line with the slope:.

Answer:

Equation of a straight line that passes through two given points in three-dimensional space.

Let a rectangular coordinate system Oxyz be fixed in three-dimensional space, and two non-coincident points and through which the line M 1 M 2 passes. Let's get the equations of this straight line.

We know that the canonical equations of a straight line in a space of the form and parametric equations of a straight line in space of the form set in the rectangular coordinate system Oxyz a straight line that passes through the point with coordinates and has a direction vector .

The direction vector of the line M 1 M 2 is a vector, and this line passes through the point (and ), then the canonical equations of this line have the form (or ), and parametric equations - (or ).

.

If you need to set a straight line M 1 M 2 using the equations of two intersecting planes, then you should first compose the canonical equations of a straight line passing through two points and , and from these equations obtain the necessary equations of the planes.

List of references.

• Atanasyan L.S., Butuzov V.F., Kadomtsev S.B., Poznyak E.G., Yudina I.I. Geometry. 7 - 9 grades: textbook for educational institutions.
• Atanasyan L.S., Butuzov V.F., Kadomtsev S.B., Kiseleva L.S., Poznyak E.G. Geometry. Textbook for grades 10-11 of secondary school.
• Pogorelov A.V., Geometry. Textbook for grades 7-11 of educational institutions.
• Bugrov Y.S., Nikolsky S.M. Higher mathematics. Volume One: Elements of Linear Algebra and Analytic Geometry.
• Ilyin V.A., Poznyak E.G. Analytic geometry.

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

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

Theorem 1

Any equation of the first degree, having the form A x + B y + C \u003d 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 \u003d 0 for a certain set of values \u200b\u200bA, B, C.

Evidence

this theorem consists of two points, we will prove each of them.

1. Let us prove that the equation A x + B y + C \u003d 0 defines a straight line on the plane.

Let there exist some point М 0 (x 0, y 0), the coordinates of which correspond to the equation A x + B y + C \u003d 0. Thus: A x 0 + B y 0 + C \u003d 0. Subtract from the left and right sides of the equations A x + B y + C \u003d 0 the left and right sides of the equation A x 0 + B y 0 + C \u003d 0, we obtain a new equation of the form A (x - x 0) + B (y - y 0) \u003d 0. It is equivalent to A x + B y + C \u003d 0.

The resulting equation A (x - x 0) + B (y - y 0) \u003d 0 is a necessary and sufficient condition for the vectors n → \u003d (A, B) and M 0 M → \u003d (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 → \u003d (A, B). We can assume that this is not so, but then the vectors n → \u003d (A, B) and M 0 M → \u003d (x - x 0, y - y 0) would not be perpendicular, and the equality A (x - x 0 ) + B (y - y 0) \u003d 0 would not be true.

Therefore, the equation A (x - x 0) + B (y - y 0) \u003d 0 defines some straight line in a rectangular coordinate system on the plane, and hence the equivalent equation A x + B y + C \u003d 0 defines the same straight line. This is how we proved the first part of the theorem.

1. Let us give a proof that any straight line in a rectangular coordinate system on a plane can be defined by an equation of the first degree A x + B y + C \u003d 0.

Let us set the straight line a in a rectangular coordinate system on the plane; point M 0 (x 0, y 0), through which this line passes, as well as the normal vector of this line n → \u003d (A, B).

Let there also exist some point M (x, y) - a floating point of a straight line. In this case, the vectors n → \u003d (A, B) and M 0 M → \u003d (x - x 0, y - y 0) are perpendicular to each other, and their scalar product is zero:

n →, M 0 M → \u003d A (x - x 0) + B (y - y 0) \u003d 0

Rewrite the equation A x + B y - A x 0 - B y 0 \u003d 0, define C: C \u003d - A x 0 - B y 0 and in the end we get the equation A x + B y + C \u003d 0.

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

Definition 1

An equation of the form A x + B y + C \u003d 0 - this is general equation of the line on a plane in a rectangular coordinate system O x y.

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

It also follows from the proof of the theorem that the coefficients A and B for the variables x and y are the coordinates of the normal vector of the straight line, which is given by the general equation of the straight line A x + B y + C \u003d 0.

Consider a specific example of a general equation of a straight line.

Let the equation 2 x + 3 y - 2 \u003d 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 → \u003d (2, 3). Draw a given straight line in the drawing.

You can also say the following: the straight line that we see in the drawing is determined by the general equation 2 x + 3 y - 2 \u003d 0, since the coordinates of all points of a given straight line correspond to this equation.

We can obtain the equation λ · A x + λ · B y + λ · C \u003d 0 by multiplying both sides of the general equation of the line by a nonzero number λ. The resulting equation is equivalent to the original general equation, therefore, will describe the same straight line on the plane.

Definition 2

Complete general equation of the line - such a general equation of the straight line A x + B y + C \u003d 0, in which the numbers A, B, C are nonzero. Otherwise the equation is incomplete.

Let us consider all the variations of the incomplete general equation of the line.

1. When A \u003d 0, B ≠ 0, C ≠ 0, the general equation becomes B y + C \u003d 0. Such an incomplete general equation defines a straight line in a rectangular coordinate system O x y 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 straight line A x + B y + C \u003d 0, when A \u003d 0, B ≠ 0, specifies the locus of points (x, y), whose coordinates are equal to the same number - C B.
2. If A \u003d 0, B ≠ 0, C \u003d 0, the general equation takes the form y \u003d 0. This incomplete equation defines the abscissa axis O x.
3. When A ≠ 0, B \u003d 0, C ≠ 0, we obtain an incomplete general equation A x + C \u003d 0, defining a straight line parallel to the ordinate axis.
4. Let A ≠ 0, B \u003d 0, C \u003d 0, then the incomplete general equation will take the form x \u003d 0, and this is the equation of the coordinate line O y.
5. Finally, for A ≠ 0, B ≠ 0, C \u003d 0, the incomplete general equation takes the form A x + B y \u003d 0. And this equation describes a straight line that passes through the origin. Indeed, the pair of numbers (0, 0) corresponds to the equality A x + B y \u003d 0, since A · 0 + B · 0 \u003d 0.

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

Example 1

It is known that a 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 a given straight line.

Decision

A straight line parallel to the ordinate axis is given by an equation of the form A x + C \u003d 0, in which A ≠ 0. Also, the condition sets 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 \u003d 0, i.e. the equality is true:

A · 2 7 + C \u003d 0

It is possible to determine C from it by giving A some non-zero value, for example, A \u003d 7. In this case, we get: 7 · 2 7 + C \u003d 0 ⇔ C \u003d - 2. We know both coefficients A and C, substitute them into the equation A x + C \u003d 0 and obtain the required equation of the straight line: 7 x - 2 \u003d 0

Answer: 7 x - 2 \u003d 0

Example 2

The drawing shows a straight line, it is necessary to write down its equation.

Decision

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

The straight line, which is parallel to the eyes of the abscissas, determines the incomplete general equation B y + C \u003d 0. Let's find the values \u200b\u200bof B and C. The coordinates of the point (0, 3), since a given straight line passes through it, will satisfy the equation of the straight line B y + C \u003d 0, then the equality is valid: B · 3 + C \u003d 0. Let's set for B some value other than zero. Suppose B \u003d 1, in this case from the equality B 3 + C \u003d 0 we can find C: C \u003d - 3. We use the known values \u200b\u200bof B and C, we obtain the required equation of the straight line: y - 3 \u003d 0.

Answer: y - 3 \u003d 0.

General equation of a straight line passing through a given point of the plane

Let the given straight line pass through the point М 0 (x 0, y 0), then its coordinates correspond to the general equation of the straight line, i.e. the equality is true: A x 0 + B y 0 + C \u003d 0. We 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 \u003d 0, this equation is equivalent to the original general, passes through the point М 0 (x 0, y 0) and has a normal vector n → \u003d (A, B).

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

Example 3

Given a point М 0 (- 3, 4), through which a straight line passes, and a normal vector of this straight line n → \u003d (1, - 2). It is necessary to write down the equation of a given straight line.

Decision

The initial conditions allow us to obtain the necessary data for drawing up the equation: A \u003d 1, B \u003d - 2, x 0 \u003d - 3, y 0 \u003d 4. Then:

A (x - x 0) + B (y - y 0) \u003d 0 ⇔ 1 (x - (- 3)) - 2 y (y - 4) \u003d 0 ⇔ ⇔ x - 2 y + 22 \u003d 0

The problem could have been solved differently. The general equation of the line is A x + B y + C \u003d 0. A given normal vector allows you to get the values \u200b\u200bof the coefficients A and B, then:

A x + B y + C \u003d 0 ⇔ 1 x - 2 y + C \u003d 0 ⇔ x - 2 y + C \u003d 0

Now we 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 \u003d 0, i.e. - 3 - 2 4 + C \u003d 0. Hence C \u003d 11. The required equation of the straight line takes the form: x - 2 y + 11 \u003d 0.

Answer: x - 2 y + 11 \u003d 0.

Example 4

A straight line 2 3 x - y - 1 2 \u003d 0 and a point М 0 lying on this straight line are given. Only the abscissa of this point is known, and it is equal to - 3. It is necessary to determine the ordinate of the given point.

Decision

Let's set the designation of the coordinates of the point М 0 as x 0 and y 0. The initial data indicates that x 0 \u003d - 3. Since a point belongs to a given straight line, then its coordinates correspond to the general equation of this straight line. Then the equality will be true:

2 3 x 0 - y 0 - 1 2 \u003d 0

Determine y 0: 2 3 (- 3) - y 0 - 1 2 \u003d 0 ⇔ - 5 2 - y 0 \u003d 0 ⇔ y 0 \u003d - 5 2

Answer: - 5 2

The transition from the general equation of a straight line to other types of equations of a straight line and vice versa

As we know, there are several types of equations for the same straight line on the 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. This is where the skill of converting an equation of one kind into an equation of another kind comes in handy.

To begin with, consider the transition from the general equation of the form A x + B y + C \u003d 0 to the canonical equation x - x 1 a x \u003d y - y 1 a y.

If А ≠ 0, then we transfer the term B y to the right-hand side of the general equation. On the left side, place A outside the brackets. As a result, we get: A x + C A \u003d - B y.

This equality can be written as a proportion: x + C A - B \u003d y A.

If В ≠ 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 \u003d - B y - C. We take out - B outside the brackets, then: A x \u003d - B y + C B.

Let's rewrite equality as a proportion: x - B \u003d y + C B A.

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

Example 5

The general equation of the straight line is given: 3 y - 4 \u003d 0. It is necessary to transform it into a canonical equation.

Decision

Rewrite the original equation as 3 y - 4 \u003d 0. Next, we act according to the algorithm: on the left side there remains the term 0 x; and on the right side we place - 3 outside the brackets; we get: 0 x \u003d - 3 y - 4 3.

Let's write the resulting equality as a proportion: x - 3 \u003d y - 4 3 0. So, we got an equation of the canonical form.

Answer: x - 3 \u003d y - 4 3 0.

To transform the general equation of the straight line into parametric ones, one first makes the transition to the canonical form, and then the transition from the canonical equation of the straight line to the parametric equations.

Example 6

The straight line is given by the equation 2 x - 5 y - 1 \u003d 0. Write down the parametric equations of this straight line.

Decision

Let's make the transition from the general equation to the canonical one:

2 x - 5 y - 1 \u003d 0 ⇔ 2 x \u003d 5 y + 1 ⇔ 2 x \u003d 5 y + 1 5 ⇔ x 5 \u003d y + 1 5 2

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

x 5 \u003d λ y + 1 5 2 \u003d λ ⇔ x \u003d 5 λ y \u003d - 1 5 + 2 λ, λ ∈ R

Answer: x \u003d 5 λ y \u003d - 1 5 + 2 λ, λ ∈ R

The general equation can be transformed into an equation of a straight line with a slope y \u003d k x + b, but only if B ≠ 0. For the transition on the left, we leave the term B y, the rest are transferred to the right. We get: B y \u003d - A x - C. Divide both sides of the resulting equality by B, different from zero: y \u003d - A B x - C B.

Example 7

The general equation of the straight line is given: 2 x + 7 y \u003d 0. You must convert that equation to a slope equation.

Decision

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

2 x + 7 y \u003d 0 ⇔ 7 y - 2 x ⇔ y \u003d - 2 7 x

Answer: y \u003d - 2 7 x.

From the general equation of a straight line, it is enough to simply obtain an equation in segments of the form x a + y b \u003d 1. To make such a transition, we transfer the number C to the right-hand side of the equality, divide both sides of the resulting equality by - С and, finally, transfer the coefficients for the variables x and y to the denominators:

A x + B y + C \u003d 0 ⇔ A x + B y \u003d - C ⇔ ⇔ A - C x + B - C y \u003d 1 ⇔ x - C A + y - C B \u003d 1

Example 8

It is necessary to transform the general equation of the line x - 7 y + 1 2 \u003d 0 into the equation of the line in segments.

Decision

Move 1 2 to the right side: x - 7 y + 1 2 \u003d 0 ⇔ x - 7 y \u003d - 1 2.

Divide both sides of the equality by -1/2: x - 7 y \u003d - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y \u003d 1.

Answer: x - 1 2 + y 1 14 \u003d 1.

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

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

x a + y b ⇔ 1 a x + 1 b y - 1 \u003d 0 ⇔ A x + B y + C \u003d 0 y \u003d k x + b ⇔ y - k x - b \u003d 0 ⇔ A x + B y + C \u003d 0

The canonical equation is transformed to the general one as follows:

x - x 1 ax \u003d y - y 1 ay ⇔ ay (x - x 1) \u003d ax (y - y 1) ⇔ ⇔ ayx - axy - ayx 1 + axy 1 \u003d 0 ⇔ A x + B y + C \u003d 0

To move from parametric, first, the transition to the canonical is carried out, and then to the general:

x \u003d x 1 + a x λ y \u003d y 1 + a y λ ⇔ x - x 1 a x \u003d y - y 1 a y ⇔ A x + B y + C \u003d 0

Example 9

Parametric equations of the straight line x \u003d - 1 + 2 · λ y \u003d 4 are given. It is necessary to write down the general equation of this straight line.

Decision

Let's make the transition from parametric equations to canonical ones:

x \u003d - 1 + 2 λ y \u003d 4 ⇔ x \u003d - 1 + 2 λ y \u003d 4 + 0 λ ⇔ λ \u003d x + 1 2 λ \u003d y - 4 0 ⇔ x + 1 2 \u003d y - 4 0

Let's move from the canonical to the general:

x + 1 2 \u003d y - 4 0 ⇔ 0 (x + 1) \u003d 2 (y - 4) ⇔ y - 4 \u003d 0

Answer: y - 4 \u003d 0

Example 10

The equation of a straight line in segments x 3 + y 1 2 \u003d 1 is given. It is necessary to make a transition to the general form of the equation.

Decision:

Let's just rewrite the equation as needed:

x 3 + y 1 2 \u003d 1 ⇔ 1 3 x + 2 y - 1 \u003d 0

Answer: 1 3 x + 2 y - 1 \u003d 0.

Drawing up the general equation of a straight line

Above, we said that the general equation can be written with the known coordinates of the normal vector and the coordinates of the point through which the straight line passes. Such a straight line is determined by the equation A (x - x 0) + B (y - y 0) \u003d 0. We also analyzed the corresponding example there.

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

Example 11

A straight line parallel to the straight line 2 x - 3 y + 3 3 \u003d 0 is given. Also known is the point M 0 (4, 1), through which the given line passes. It is necessary to write down the equation of a given straight line.

Decision

The initial conditions tell us that the straight lines are parallel, then, as the normal vector of the straight line, the equation of which is to be written, we take the direction vector of the straight line n → \u003d (2, - 3): 2 x - 3 y + 3 3 \u003d 0. Now we know all the necessary data to compose the general equation of the straight line:

A (x - x 0) + B (y - y 0) \u003d 0 ⇔ 2 (x - 4) - 3 (y - 1) \u003d 0 ⇔ 2 x - 3 y - 5 \u003d 0

Answer: 2 x - 3 y - 5 \u003d 0.

Example 12

The specified line passes through the origin perpendicular to the line x - 2 3 \u003d y + 4 5. It is necessary to draw up a general equation for a given line.

Decision

The normal vector of the given line will be the direction vector of the line x - 2 3 \u003d y + 4 5.

Then n → \u003d (3, 5). The straight line passes through the origin, i.e. through the point O (0, 0). Let's compose the general equation of a given straight line:

A (x - x 0) + B (y - y 0) \u003d 0 ⇔ 3 (x - 0) + 5 (y - 0) \u003d 0 ⇔ 3 x + 5 y \u003d 0

Answer: 3 x + 5 y \u003d 0.

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