General equation of a straight line on points. Equation of a straight line passing through two points

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

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

At – Y 1 = K(X-x 1),

Where K– unknown slope.

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

Y 2 – Y 1 = K(X 2 – X 1),

From here you can find the slope of this line:

,

Or after conversion

(1.14)

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

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

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

Figure 1.6

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

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

2(Y – 2) = -3(X – 1).

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

3X + 2Y – 7 = 0.

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

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

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

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

or .

Hence or -5( Y – 1) = X – 2.

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

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

Using formula (1.14), we obtain the equation

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

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

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

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

Figure 1.7

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

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

Oh + Wu + WITH= 0, where WITH = –(AX 0 + By 0), (1.16),

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

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

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

Figure 1.8

Vectors and collinear.

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

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

These equations can be written in the form

. (1.18)

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

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

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

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

Properties of a straight line in Euclidean geometry.

There are infinitely many lines that can be drawn through any point.

Through any two non-coinciding points, there is only one straight line.

Two non-coincident lines in the plane either intersect at a single point, or are

parallel (follows from the previous one).

In three-dimensional space, there are three options for the relative position of two lines:

• lines intersect;

• straight lines are parallel;

• straight lines intersect.

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

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

General equation of a straight line.

Definition. Any line in the plane can be given by a first order equation

Ah + Wu + C = 0,

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

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

. C = 0, A ≠ 0, B ≠ 0- the line passes through the origin

. A = 0, B ≠0, C ≠0 ( By + C = 0)- straight line parallel to the axis Oh

. B = 0, A ≠ 0, C ≠ 0 ( Ax + C = 0)- straight line parallel to the axis OU

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

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

The equation of a straight line can be represented in various forms depending on any given

initial conditions.

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

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

perpendicular to the line given by the equation

Ah + Wu + C = 0.

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

Solution. Let's compose at A \u003d 3 and B \u003d -1 the equation of the straight line: 3x - y + C \u003d 0. To find the coefficient C

we substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C \u003d 0, therefore

C = -1. Total: the desired equation: 3x - y - 1 \u003d 0.

Equation of a straight line passing through two points.

Let two points be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2 , z 2), Then straight line equation,

passing through these points:

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

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

If x 1 ≠ x 2 And x = x 1, If x 1 = x 2 .

Fraction = k called slope factor straight.

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

Solution. Applying the above formula, we get:

Equation of a straight line by a point and a slope.

If the general equation of a straight line Ah + Wu + C = 0 bring to the form:

and designate , then the resulting equation is called

equation of a straight line with slope k.

The equation of a straight line on a point and a directing vector.

By analogy with the point considering the equation of a straight line through the normal vector, you can enter the task

a straight line through a point and a direction vector of a straight line.

Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition

Aα 1 + Bα 2 = 0 called direction vector of the straight line.

Ah + Wu + C = 0.

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

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

coefficients must satisfy the conditions:

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

Then the equation of a straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0.

at x=1, y=2 we get C/ A = -3, i.e. desired equation:

x + y - 3 = 0

Equation of a straight line in segments.

If in the general equation of the straight line Ah + Wu + C = 0 C≠0, then, dividing by -C, we get:

or , where

geometric sense coefficients in that the coefficient a is the coordinate of the intersection point

straight with axle Oh, A b- the coordinate of the point of intersection of the line with the axis OU.

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

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

Normal equation of a straight line.

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

normalizing factor, then we get

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

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

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

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

Example. Given the general equation of a straight line 12x - 5y - 65 = 0. Required to write Various types equations

this straight line.

The equation of this straight line in segments:

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

Equation of a straight line:

cos φ = 12/13; sin φ= -5/13; p=5.

It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines,

parallel to the axes or passing through the origin.

Angle between lines on a plane.

Definition. If two lines are given y \u003d k 1 x + b 1, y \u003d k 2 x + b 2, That sharp corner between these lines

will be defined as

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

If k 1 \u003d -1 / k 2 .

Theorem.

Direct Ah + Wu + C = 0 And A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients are proportional

A 1 \u003d λA, B 1 \u003d λB. If also С 1 \u003d λС, then the lines coincide. Coordinates of the point of intersection of two lines

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

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

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

represented by the equation:

The distance from a point to a line.

Theorem. If a point is given M(x 0, y 0), then the distance to the line Ah + Wu + C = 0 defined as:

Proof. Let the point M 1 (x 1, y 1)- the base of the perpendicular dropped from the point M for a given

direct. Then the distance between the points M And M 1:

(1)

Coordinates x 1 And 1 can be found as a solution to the system of equations:

The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicularly

given line. If we transform the first equation of the system to the form:

A(x - x 0) + B(y - y 0) + Ax 0 + By 0 + C = 0,

then, solving, we get:

Substituting these expressions into equation (1), we find:

The theorem has been proven.

The line passing through the point K(x 0; y 0) and parallel to the line y = kx + a is found by the formula:

y - y 0 \u003d k (x - x 0) (1)

Where k is the slope of the straight line.

Alternative formula:
The line passing through the point M 1 (x 1 ; y 1) and parallel to the line Ax+By+C=0 is represented by the equation

A(x-x 1)+B(y-y 1)=0 . (2)

Example #2. Write the equation of a straight line parallel to the straight line 2x + 5y = 0 and forming, together with the coordinate axes, a triangle whose area is 5.
Solution . Since the lines are parallel, the equation of the required line is 2x + 5y + C = 0. Area right triangle, where a and b are its legs. Find the points of intersection of the desired line with the coordinate axes:
;
.
So, A(-C/2,0), B(0,-C/5). Substitute in the formula for the area: . We get two solutions: 2x + 5y + 10 = 0 and 2x + 5y - 10 = 0 .

Example #3. Write the equation of the line passing through the point (-2; 5) and the parallel line 5x-7y-4=0 .
Solution. This straight line can be represented by the equation y = 5/7 x – 4/7 (here a = 5/7). The equation of the desired line is y - 5 = 5 / 7 (x - (-2)), i.e. 7(y-5)=5(x+2) or 5x-7y+45=0 .

Example #4. Solving example 3 (A=5, B=-7) using formula (2), we find 5(x+2)-7(y-5)=0.

Example number 5. Write the equation of a straight line passing through the point (-2;5) and a parallel straight line 7x+10=0.
Solution. Here A=7, B=0. Formula (2) gives 7(x+2)=0, i.e. x+2=0. Formula (1) is not applicable, since this equation cannot be solved with respect to y (this straight line is parallel to the y-axis).

This article reveals the derivation of the equation of a straight line passing through two given points in a rectangular coordinate system located on a plane. We derive the equation of a straight line passing through two given points in a rectangular coordinate system. We will visually show and solve several examples related to the material covered.

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Before obtaining the equation of a straight line passing through two given points, it is necessary to pay attention to some facts. There is an axiom that says that through two non-coincident points on a plane it is possible to draw a straight line and only one. In other words, two given points of the plane are determined by a straight line passing through these points.

If the plane is given by the rectangular coordinate system Oxy, then any straight line depicted in it will correspond to the equation of the straight line on the plane. There is also a connection with the directing vector of the straight line. These data are sufficient to draw up the equation of a straight line passing through two given points.

Consider an example of solving a similar problem. It is necessary to formulate the equation of a straight line a passing through two mismatched points M 1 (x 1, y 1) and M 2 (x 2, y 2) located in the Cartesian coordinate system.

In the canonical equation of a straight line on a plane, having the form x - x 1 a x \u003d y - y 1 a y , a rectangular coordinate system O x y is specified with a straight line that intersects with it at a point with coordinates M 1 (x 1, y 1) with a guide vector a → = (a x , a y) .

It is necessary to compose the canonical equation of the straight line a, which will pass through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2) .

The straight line a has a directing vector M 1 M 2 → with coordinates (x 2 - x 1, y 2 - y 1), since it intersects the points M 1 and M 2. We have obtained the necessary data in order to transform the canonical equation with the coordinates of the direction vector M 1 M 2 → = (x 2 - x 1, y 2 - y 1) and the coordinates of the points M 1 lying on them (x 1, y 1) and M 2 (x 2 , y 2) . We get an equation of the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 or x - x 2 x 2 - x 1 = y - y 2 y 2 - y 1 .

Consider the figure below.

Following the calculations, we write the parametric equations of a straight line in a plane that passes through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2) . We get an equation of the form x \u003d x 1 + (x 2 - x 1) λ y \u003d y 1 + (y 2 - y 1) λ or x \u003d x 2 + (x 2 - x 1) λ y \u003d y 2 + (y 2 - y 1) λ.

Let's take a closer look at a few examples.

Example 1

Write the equation of a straight line passing through 2 given points with coordinates M 1 - 5 , 2 3 , M 2 1 , - 1 6 .

Solution

The canonical equation for a straight line intersecting at two points with coordinates x 1 , y 1 and x 2 , y 2 takes the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 . According to the condition of the problem, we have that x 1 \u003d - 5, y 1 \u003d 2 3, x 2 \u003d 1, y 2 \u003d - 1 6. Need to substitute numerical values into the equation x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 . From here we get that the canonical equation will take the form x - (- 5) 1 - (- 5) = y - 2 3 - 1 6 - 2 3 ⇔ x + 5 6 = y - 2 3 - 5 6 .

Answer: x + 5 6 = y - 2 3 - 5 6 .

If it is necessary to solve a problem with a different type of equation, then for a start you can go to the canonical one, since it is easier to come to any other from it.

Example 2

Compose the general equation of a straight line passing through points with coordinates M 1 (1, 1) and M 2 (4, 2) in the O x y coordinate system.

Solution

First you need to write down the canonical equation of a given line that passes through the given two points. We get an equation of the form x - 1 4 - 1 = y - 1 2 - 1 ⇔ x - 1 3 = y - 1 1 .

We bring the canonical equation to the desired form, then we get:

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

Answer: x - 3 y + 2 = 0 .

Examples of such tasks were considered in school textbooks at algebra lessons. School tasks differed in that the equation of a straight line with a slope coefficient was known, having the form y \u003d k x + b. If you need to find the value of the slope k and the number b, at which the equation y \u003d k x + b defines a line in the O x y system that passes through the points M 1 (x 1, y 1) and M 2 (x 2, y 2) , where x 1 ≠ x 2 . When x 1 = x 2 , then the slope takes on the value of infinity, and the straight line M 1 M 2 is defined by a general incomplete equation of the form x - x 1 = 0 .

Because the dots M 1 And M 2 are on a straight line, then their coordinates satisfy the equation y 1 = k x 1 + b and y 2 = k x 2 + b. It is necessary to solve the system of equations y 1 = k x 1 + b y 2 = k x 2 + b with respect to k and b.

To do this, we find k \u003d y 2 - y 1 x 2 - x 1 b \u003d y 1 - y 2 - y 1 x 2 - x 1 x 1 or k \u003d y 2 - y 1 x 2 - x 1 b \u003d y 2 - y 2 - y 1 x 2 - x 1 x 2 .

With such values ​​of k and b, the equation of a straight line passing through given two points takes the following form y = y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 1 or y \u003d y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 2.

Memorizing such a huge number of formulas at once will not work. To do this, it is necessary to increase the number of repetitions in solving problems.

Example 3

Write the equation of a straight line with a slope passing through points with coordinates M 2 (2, 1) and y = k x + b.

Solution

To solve the problem, we use a formula with a slope that has the form y \u003d k x + b. The coefficients k and b must take such a value that this equation corresponds to a straight line passing through two points with coordinates M 1 (- 7 , - 5) and M 2 (2 , 1) .

points M 1 And M 2 located on a straight line, then their coordinates should invert the equation y = k x + b true equality. From here we get that - 5 = k · (- 7) + b and 1 = k · 2 + b. Let's combine the equation into the system - 5 = k · - 7 + b 1 = k · 2 + b and solve.

Upon substitution, we get that

5 = k - 7 + b 1 = k 2 + b ⇔ b = - 5 + 7 k 2 k + b = 1 ⇔ b = - 5 + 7 k 2 k - 5 + 7 k = 1 ⇔ ⇔ b = - 5 + 7 k k = 2 3 ⇔ b = - 5 + 7 2 3 k = 2 3 ⇔ b = - 1 3 k = 2 3

Now the values ​​k = 2 3 and b = - 1 3 are substituted into the equation y = k x + b . We get that the desired equation passing through the given points will be an equation that has the form y = 2 3 x - 1 3 .

This way of solving predetermines spending a large number time. There is a way in which the task is solved literally in two steps.

We write the canonical equation of a straight line passing through M 2 (2, 1) and M 1 (- 7, - 5) , having the form x - (- 7) 2 - (- 7) = y - (- 5) 1 - (- 5) ⇔ x + 7 9 = y + 5 6 .

Now let's move on to the slope equation. We get that: x + 7 9 = y + 5 6 ⇔ 6 (x + 7) = 9 (y + 5) ⇔ y = 2 3 x - 1 3 .

Answer: y = 2 3 x - 1 3 .

If in three-dimensional space there is a rectangular coordinate system O x y z with two given non-coinciding points with coordinates M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), the straight line M passing through them 1 M 2 , it is necessary to obtain the equation of this line.

We have that canonical equations of the form x - x 1 a x = y - y 1 a y = z - z 1 a z and parametric equations x = x 1 + a x λ y = y 1 + a y λ z = z 1 + a z λ are able to set a line in the O x y z coordinate system passing through points having coordinates (x 1, y 1, z 1) with a directing vector a → = (a x, a y, a z) .

Straight M 1 M 2 has a direction vector of the form M 1 M 2 → = (x 2 - x 1 , y 2 - y 1 , z 2 - z 1) , where the line passes through the point M 1 (x 1 , y 1 , z 1) and M 2 (x 2, y 2, z 2), hence the canonical equation can be of the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 or x - x 2 x 2 - x 1 \u003d y - y 2 y 2 - y 1 \u003d z - z 2 z 2 - z 1, in turn, parametric x \u003d x 1 + (x 2 - x 1) λ y \u003d y 1 + (y 2 - y 1) λ z = z 1 + (z 2 - z 1) λ or x = x 2 + (x 2 - x 1) λ y = y 2 + (y 2 - y 1) λ z \u003d z 2 + (z 2 - z 1) λ.

Consider a figure that shows 2 given points in space and the equation of a straight line.

Example 4

Write the equation of a straight line defined in a rectangular coordinate system O x y z of three-dimensional space, passing through the given two points with coordinates M 1 (2, - 3, 0) and M 2 (1, - 3, - 5) .

Solution

We need to find the canonical equation. Because we are talking about three-dimensional space, which means that when a straight line passes through given points, the desired canonical equation will take the form x - x 1 x 2 - x 1 \u003d y - y 1 y 2 - y 1 \u003d z - z 1 z 2 - z 1.

By condition, we have that x 1 = 2, y 1 = - 3, z 1 = 0, x 2 = 1, y 2 = - 3, z 2 = - 5. It follows that the necessary equations can be written as follows:

x - 2 1 - 2 = y - (- 3) - 3 - (- 3) = z - 0 - 5 - 0 ⇔ x - 2 - 1 = y + 3 0 = z - 5

Answer: x - 2 - 1 = y + 3 0 = z - 5.

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This article continues the topic of the equation of a straight line on a plane: consider such a type of equation as the general equation of a straight line. Let's 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.

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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 line in a rectangular coordinate system on the 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.

1. Let us prove that the equation A x + B y + C = 0 defines a 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 \u003d 0 the left and right sides of the equation A x 0 + B y 0 + C \u003d 0, we get 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 in a rectangular coordinate system a straight line 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.

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

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

Let's set a 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 → = (A , B) .

Let there also exist some point M (x , y) - a floating point of the 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 is 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 finally get the equation A x + B y + C = 0 .

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

Definition 1

An equation that looks like A x + B y + C = 0 - This general equation of a straight line on a plane in a rectangular coordinate systemO x y .

Based on the proved theorem, we can conclude that a straight line given on a plane in a fixed rectangular coordinate system and its general equation are inextricably linked. In other words, the original 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 = 0 .

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) ​​. Draw a given straight line in the drawing.

The following can also be argued: 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 of a given straight line correspond to this equation.

We can get the equation λ · A x + λ · B y + λ · C = 0 by multiplying both sides of the general straight line equation by a non-zero number λ. The resulting equation is equivalent to the original general equation, therefore, it will describe the same line in the plane.

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

Let us analyze all variations of the incomplete general equation of the straight 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 the rectangular coordinate system O x y that is parallel to the axis O x , because for any actual value x variable y will take the value - C B . In other words, the general equation of the line A x + B y + C \u003d 0, when A \u003d 0, B ≠ 0, defines 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 becomes y \u003d 0. Such an incomplete equation defines the x-axis O x .
3. When A ≠ 0, B \u003d 0, C ≠ 0, we get an incomplete general equation A x + C \u003d 0, defining a straight line parallel to the y-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, when 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 = 0 , since A · 0 + B · 0 = 0 .

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

Example 1

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

Solution

A straight line parallel to the y-axis is given by an equation of the form A x + C \u003d 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 correspond to the conditions of the incomplete general equation A x + C = 0 , i.e. equality is correct:

A 2 7 + C = 0

It is possible to determine C from it by giving A some non-zero value, for example, A = 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 = 0 and get the required equation of the line: 7 x - 2 = 0

Answer: 7 x - 2 = 0

Example 2

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

Solution

The given 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 abscissa, is determined by the incomplete general equation B y + С = 0. Find the values ​​of B and C . The coordinates of the point (0, 3), since the given straight line passes through it, will satisfy the equation of the straight line B y + С = 0, then the equality is valid: В · 3 + С = 0. Let's set B to some value other than zero. Let's say B \u003d 1, in this case, from the equality B · 3 + C \u003d 0 we can find C: C \u003d - 3. Using the known values ​​of B and C, we obtain the required equation of the straight line: y - 3 = 0.

Answer: y - 3 = 0 .

General equation of a straight line passing through a given point of the 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 . Subtract the left and right sides of this equation from the left and right sides of the general complete equation of the straight line. We get: A (x - x 0) + B (y - y 0) + C \u003d 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 → \u003d (A, B) .

The result that we have obtained makes it possible to write the general equation of a straight line for 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 M 0 (- 3, 4) through which the line passes, and the normal vector of this line n → = (1 , - 2) . It is necessary to write down the equation of a given straight line.

Solution

The initial conditions allow us to obtain the necessary data for compiling 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) = 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 has the form A x + B y + C = 0 . The given normal vector allows you to get the values ​​of the coefficients A and B , then:

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

Now find the value of C using given by the condition problem point M 0 (- 3 , 4) through which the line passes. The coordinates of this point correspond to the equation x - 2 · y + C = 0 , i.e. - 3 - 2 4 + C \u003d 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 the given point.

Solution

Let's set the designation of the coordinates of the point M 0 as x 0 and y 0 . The initial data indicates that x 0 \u003d - 3. Since the point belongs to a given line, then its coordinates correspond to the general equation of this line. Then the following 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 straight line to other types of equations of a straight line and vice versa

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

First, 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 transfer 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, we transfer the others to the right side, we get: A x \u003d - B y - C. We take out - B out of brackets, then: A x \u003d - B y + C B.

Let's rewrite the equality as 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 during the transition from the general equation to the canonical one.

Example 5

The general equation of the line 3 y - 4 = 0 is given. It needs to be converted to a canonical equation.

Solution

We write the original equation as 3 y - 4 = 0 . Next, we act according to the algorithm: the term 0 x remains on the left side; and on the right side we take out - 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 the canonical form.

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

To transform the general equation of a straight line into parametric ones, first, the transition to the canonical form is carried out, and then the transition from the canonical equation of the straight 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 of this line.

Solution

Let's 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 let's take both parts 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 to a straight line equation with a slope y \u003d k x + b, but only when B ≠ 0. For the transition on the left side, we leave the term B y , the rest are transferred to the right. We get: B y = - A x - C . Let's divide both parts of the resulting equality by B , which is different from zero: y = - A B x - C B .

Example 7

The general equation of a straight line is given: 2 x + 7 y = 0 . You need to convert that equation to 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 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 side of the equality, divide both parts 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 = 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 convert the general equation of the straight line x - 7 y + 1 2 = 0 into the equation of a straight line in segments.

Solution

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

Divide by -1/2 both sides of the equation: 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 straight line in segments and the equation with a slope can be easily converted into a general one by simply collecting all the terms on the left side of the equation:

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 the 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 pass from the parametric, first the transition to the canonical is carried out, 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 straight line x = - 1 + 2 · λ y = 4 are given. It is necessary to write down the general equation of this line.

Solution

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

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 canonical to 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 segments x 3 + y 1 2 = 1 is given. It is necessary to make the transition to general view equations.

Solution:

Let's just 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 straight line

Above, we said 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 . In the same place we analyzed the corresponding example.

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

Example 11

Given a line parallel to the line 2 x - 3 y + 3 3 = 0 . 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.

Solution

The initial conditions tell us that the lines are parallel, whereas, as a normal vector of the line whose equation needs to be written, we take the directing vector of the 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 a straight 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 write the general equation of a given straight line.

Solution

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

Then n → = (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) = 0 ⇔ 3 (x - 0) + 5 (y - 0) = 0 ⇔ 3 x + 5 y = 0

Answer: 3 x + 5 y = 0 .

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