Equation of a line passing through a given point in a given direction. Equation of a straight line passing through two given points. Angle between two lines. Condition of parallelism and perpendicularity of two 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 center of the beam.

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

The slope 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 lines are given by slope equations

y = k 1 x + B 1 ,

In this article, we will consider the general equation of a straight line in a plane. Let us give examples of constructing the general equation of a straight line if two points of this straight line are known or if one point and the normal vector of this straight line are known. Let us present methods for transforming an equation in general form into canonical and parametric forms.

Let an arbitrary Cartesian rectangular coordinate system be given Oxy. Consider a first degree equation or a linear equation:

Where A, B, C are some constants, and at least one of the elements A And B different from zero.

We will show that a linear equation in the plane defines a straight line. Let us prove the following theorem.

Theorem 1. In an arbitrary Cartesian rectangular coordinate system on a plane, each straight line can be given by a linear equation. Conversely, each linear equation (1) in an arbitrary Cartesian rectangular coordinate system on the plane defines a straight line.

Proof. It suffices to prove that the line L is determined by a linear equation for any one Cartesian rectangular coordinate system, since then it will be determined by a linear equation and for any choice of Cartesian rectangular coordinate system.

Let a straight line be given on the plane L. We choose a coordinate system so that the axis Ox aligned with the line L, and the axis Oy was perpendicular to it. Then the equation of the line L will take the following form:

All points on a line L will satisfy the linear equation (2), and all points outside this straight line will not satisfy the equation (2). The first part of the theorem is proved.

Let a Cartesian rectangular coordinate system be given and let a linear equation (1) be given, where at least one of the elements A And B different from zero. Find the locus of points whose coordinates satisfy equation (1). Since at least one of the coefficients A And B is different from zero, then equation (1) has at least one solution M(x 0 ,y 0). (For example, when A≠0, dot M 0 (−C/A, 0) belongs to the given locus of points). Substituting these coordinates into (1) we obtain the identity

Let us subtract identity (3) from (1):

Obviously, equation (4) is equivalent to equation (1). Therefore, it suffices to prove that (4) defines some line.

Since we are considering a Cartesian rectangular coordinate system, it follows from equality (4) that the vector with components ( x−x 0 , y−y 0 ) is orthogonal to the vector n with coordinates ( A,B}.

Consider some line L passing through the point M 0 (x 0 , y 0) and perpendicular to the vector n(Fig.1). Let the point M(x,y) belongs to the line L. Then the vector with coordinates x−x 0 , y−y 0 perpendicular n and equation (4) is satisfied (scalar product of vectors n and equals zero). Conversely, if the point M(x,y) does not lie on a line L, then the vector with coordinates x−x 0 , y−y 0 is not orthogonal to vector n and equation (4) is not satisfied. The theorem has been proven.

Proof. Since lines (5) and (6) define the same line, the normal vectors n 1 ={A 1 ,B 1 ) and n 2 ={A 2 ,B 2) are collinear. Since the vectors n 1 ≠0, n 2 ≠ 0, then there is a number λ , What n 2 =n 1 λ . Hence we have: A 2 =A 1 λ , B 2 =B 1 λ . Let's prove that C 2 =C 1 λ . It is obvious that coinciding lines have a common point M 0 (x 0 , y 0). Multiplying equation (5) by λ and subtracting equation (6) from it we get:

Since the first two equalities from expressions (7) are satisfied, then C 1 λ −C 2=0. Those. C 2 =C 1 λ . The remark has been proven.

Note that equation (4) defines the equation of a straight line passing through the point M 0 (x 0 , y 0) and having a normal vector n={A,B). Therefore, if the normal vector of the line and the point belonging to this line are known, then the general equation of the line can be constructed using equation (4).

Example 1. A line passes through a point M=(4,−1) and has a normal vector n=(3, 5). Construct the general equation of a straight line.

Solution. We have: x 0 =4, y 0 =−1, A=3, B=5. To construct the general equation of a straight line, we substitute these values ​​into equation (4):

Answer:

Vector parallel to line L and hence is perpendicular to the normal vector of the line L. Let's construct a normal line vector L, given that the scalar product of vectors n and is equal to zero. We can write, for example, n={1,−3}.

To construct the general equation of a straight line, we use formula (4). Let us substitute into (4) the coordinates of the point M 1 (we can also take the coordinates of the point M 2) and the normal vector n:

Substituting point coordinates M 1 and M 2 in (9) we can make sure that the straight line given by equation (9) passes through these points.

Answer:

Subtract (10) from (1):

We have obtained the canonical equation of a straight line. Vector q={−B, A) is the direction vector of the straight line (12).

See reverse transformation.

Example 3. A straight line in a plane is represented by the following general equation:

Move the second term to the right and divide both sides of the equation by 2 5.

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 the relation y = f(x) between the coordinates of the points that make up this line.

Note that the line equation can be expressed in a parametric way, 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, time plays the role of a parameter.

Equation of a straight line on a plane.

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

Ah + Wu + C = 0,

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

Depending on the values ​​of the constants A, B and C, the following special cases are possible:

C \u003d 0, A  0, B  0 - the line passes through the origin

A \u003d 0, B  0, C  0 (By + C \u003d 0) - the line is parallel to the Ox axis

B \u003d 0, A  0, C  0 ( Ax + C \u003d 0) - the line is parallel to the Oy axis

B \u003d C \u003d 0, A  0 - the straight line coincides with the Oy axis

A \u003d C \u003d 0, B  0 - the straight line coincides with the Ox axis

The equation of a straight line can be presented 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) is perpendicular to the line given by the equation Ax + By + C = 0.

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

Let us 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 \u003d -1.

Total: the desired 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 a straight line passing through these points:

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

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

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

Fraction
=k is called slope factor straight.

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

Applying the above formula, we get:

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

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

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

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 assignment of a straight line through a point and a directing vector of a 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

Ah + 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).

We will look for the equation of the desired straight 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 a straight line has the form: Ax + Ay + C = 0, or x + y + C/A = 0.

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

Equation of a straight line in segments.

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

, Where

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

Example. Given the general equation of the line x - y + 1 = 0. Find the equation of this line in the segments.

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

Normal equation of a straight line.

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

xcos + ysin - p = 0 –

normal equation of a straight 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. Given the general equation of the line 12x - 5y - 65 = 0. It is required to write various types of equations for this line.

the equation of this straight line in segments:

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

normal equation of a straight 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.

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

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

a = -4 does not fit the condition of the problem.

Total:
or x + y - 4 = 0.

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

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

Angle between 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. Straight lines Ax + Vy + 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. The line passing through the point M 1 (x 1, y 1) and perpendicular to the line y \u003d kx + b is represented by the equation:

The distance from a point to a line.

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

.

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

The x 1 and y 1 coordinates can be found as a solution to the system of equations:

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

If we transform the first equation of the system to the form:

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

then, solving, we get:

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

.

The theorem has been proven.

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

k 1 \u003d -3; k 2 = 2tg =
;  = /4.

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

We find: k 1 \u003d 3/5, k 2 \u003d -5/3, k 1 k 2 \u003d -1, therefore, the lines are perpendicular.

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

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

2x - 3y + 3 = 0;

The desired height equation is: 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.

Line equation in space.

The equation of a straight line in space by a point and

direction vector.

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

Let's take two arbitrary points M 0 (x 0 , y 0 , z 0) and M(x, y, z) on the straight line.

z

M1

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 straight line.

This vector equation can be represented in coordinate form:

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 by the formulas:

;

.

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

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

Equation of a straight line in space passing

through two points.

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

.

In addition, for point M 1 we can write:

.

Solving these equations together, we get:

.

This is the equation of a straight 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 a line of intersection of two planes.

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

+ D = 0, where

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

Equation of a straight line passing through two points. In the article" " I promised you to analyze the second way to solve the presented problems for finding the derivative, with a given function graph and a tangent to this graph. We will explore this method in , do not miss! Why next?

The fact is that the formula of the equation of a straight line will be used there. Of course, one could simply show this formula and advise you to learn it. But it is better to explain where it comes from (how it is derived). It's necessary! If you forget it, then quickly restore itwill not be difficult. Everything is detailed below. So, we have two points A on the coordinate plane(x 1; y 1) and B (x 2; y 2), a straight line is drawn through the indicated points:

Here is the direct formula:

*That is, when substituting the specific coordinates of the points, we get an equation of the form y=kx+b.

** If this formula is simply “memorized”, then there is a high probability of getting confused with indices when X. In addition, indexes can be denoted in different ways, for example:

That is why it is important to understand the meaning.

Now the derivation of this formula. Everything is very simple!

Triangles ABE and ACF are similar in terms of an acute angle (the first sign of the similarity of right triangles). It follows from this that the ratios of the corresponding elements are equal, that is:

Now we simply express these segments in terms of the difference in the coordinates of the points:

Of course, there will be no error if you write the relationships of the elements in a different order (the main thing is to keep the correspondence):

The result is the same equation of a straight line. This is all!

That is, no matter how the points themselves (and their coordinates) are designated, understanding this formula, you will always find the equation of a straight line.

The formula can be deduced using the properties of vectors, but the principle of derivation will be the same, since we will talk about the proportionality of their coordinates. In this case, the same similarity of right triangles works. In my opinion, the conclusion described above is more understandable)).

View output via vector coordinates >>>

Let a straight line be constructed on the coordinate plane passing through two given points A (x 1; y 1) and B (x 2; y 2). Let us mark an arbitrary point C on the line with coordinates ( x; y). We also denote two vectors:

It is known that for vectors lying on parallel lines (or on one line), their corresponding coordinates are proportional, that is:

- we write the equality of the ratios of the corresponding coordinates:

Consider an example:

Find the equation of a straight line passing through two points with coordinates (2;5) and (7:3).

You can not even build the line itself. We apply the formula:

It is important that you catch the correspondence when drawing up the ratio. You can't go wrong if you write:

Answer: y=-2/5x+29/5 go y=-0.4x+5.8

In order to make sure that the resulting equation is found correctly, be sure to check it - substitute the data coordinates into it in the condition of the points. You should get correct equalities.

That's all. I hope the material was useful to you.

Sincerely, Alexander.

P.S: I would be grateful if you tell about the site in social networks.

The canonical equations of a straight line in space are equations that define a straight line passing through a given point collinearly to a direction vector.

Let a point and a direction vector be given. An arbitrary point lies on a line l only if the vectors and are collinear, i.e., they satisfy the condition:

.

The above equations are the canonical equations of the line.

Numbers m , n And p are projections of the direction vector onto the coordinate axes. Since the vector is non-zero, then all numbers m , n And p cannot be zero at the same time. But one or two of them may be zero. In analytical geometry, for example, the following notation is allowed:

,

which means that the projections of the vector on the axes Oy And Oz are equal to zero. Therefore, both the vector and the straight line given by the canonical equations are perpendicular to the axes Oy And Oz, i.e. planes yOz .

Example 1 Compose equations of a straight line in space perpendicular to a plane and passing through the point of intersection of this plane with the axis Oz .

Solution. Find the point of intersection of the given plane with the axis Oz. Since any point on the axis Oz, has coordinates , then, assuming in the given equation of the plane x=y= 0 , we get 4 z- 8 = 0 or z= 2 . Therefore, the point of intersection of the given plane with the axis Oz has coordinates (0; 0; 2) . Since the desired line is perpendicular to the plane, it is parallel to its normal vector. Therefore, the normal vector can serve as the directing vector of the straight line given plane.

Now we write the desired equations of the straight line passing through the point A= (0; 0; 2) in the direction of the vector :

Equations of a straight line passing through two given points

A straight line can be defined by two points lying on it And In this case, the directing vector of the straight line can be the vector . Then the canonical equations of the line take the form

.

The above equations define a straight line passing through two given points.

Example 2 Write the equation of a straight line in space passing through the points and .

Solution. We write the desired equations of the straight line in the form given above in the theoretical reference:

.

Since , then the desired line is perpendicular to the axis Oy .

Straight as a line of intersection of planes

A straight line in space can be defined as a line of intersection of two non-parallel planes and, i.e., as a set of points that satisfy a system of two linear equations

The equations of the system are also called the general equations of the straight line in space.

Example 3 Compose canonical equations of a straight line in the space given by general equations

Solution. To write the canonical equations of a straight line or, what is the same, the equation of a straight line passing through two given points, you need to find the coordinates of any two points on the straight line. They can be the points of intersection of a straight line with any two coordinate planes, for example yOz And xOz .

Point of intersection of a line with a plane yOz has an abscissa x= 0 . Therefore, assuming in this system of equations x= 0 , we get a system with two variables:

Her decision y = 2 , z= 6 together with x= 0 defines a point A(0; 2; 6) of the desired line. Assuming then in the given system of equations y= 0 , we get the system

Her decision x = -2 , z= 0 together with y= 0 defines a point B(-2; 0; 0) intersection of a line with a plane xOz .

Now we write the equations of a straight line passing through the points A(0; 2; 6) and B (-2; 0; 0) :

,

or after dividing the denominators by -2:

,