Let the straight line pass through the points M 1 (x 1; y 1) and M 2 (x 2; y 2). The equation of a straight line passing through the point M 1 has the form y- y 1 \u003d k (x - x 1), (10.6)

where k - still unknown coefficient.

Since the straight line passes through the point M 2 (x 2 y 2), then the coordinates of this point must satisfy equation (10.6): y 2 -y 1 \u003d k (x 2 -x 1).

From here we find Substituting the found value k into equation (10.6), we obtain the equation of a straight line passing through the points M 1 and M 2:

It is assumed that in this equation x 1 ≠ x 2, y 1 ≠ y 2

If x 1 \u003d x 2, then the straight line passing through the points M 1 (x 1, y I) and M 2 (x 2, y 2) is parallel to the y-axis. Its equation is x = x 1 .

If y 2 \u003d y I, then the equation of the straight line can be written as y \u003d y 1, the straight line M 1 M 2 is parallel to the x-axis.

Equation of a straight line in segments

Let the straight line intersect the Ox axis at the point M 1 (a; 0), and the Oy axis - at the point M 2 (0; b). The equation will take the form:
those.
. This equation is called the equation of a straight line in segments, because the numbers a and b indicate which segments the straight line cuts off on the coordinate axes.

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

Let's find the equation of a straight line passing through a given point Mo (x O; y o) perpendicular to a given non-zero vector n = (A; B).

Take an arbitrary point M(x; y) on the straight line and consider the vector M 0 M (x - x 0; y - y o) (see Fig. 1). Since the vectors n and M o M are perpendicular, their scalar product is equal to zero: that is,

A(x - xo) + B(y - yo) = 0. (10.8)

Equation (10.8) is called equation of a straight line passing through a given point perpendicular to a given vector .

The vector n = (A; B) perpendicular to the line is called normal normal vector of this line .

Equation (10.8) can be rewritten as Ah + Wu + C = 0 , (10.9)

where A and B are the coordinates of the normal vector, C \u003d -Ax o - Vu o - free member. Equation (10.9) is the general equation of a straight line(see Fig.2).

Fig.1 Fig.2

Canonical equations of the straight line

,

Where
are the coordinates of the point through which the line passes, and
- direction vector.

Curves of the second order Circle

A circle is the set of all points of a plane equidistant from a given point, which is called the center.

Canonical equation of a circle of radius R centered on a point
:

In particular, if the center of the stake coincides with the origin, then the equation will look like:

Ellipse

An ellipse is a set of points in a plane, the sum of the distances from each of them to two given points and , which are called foci, is a constant value
, greater than the distance between the foci
.

The canonical equation of an ellipse whose foci lie on the Ox axis and whose origin is in the middle between the foci has the form
G de a the length of the major semiaxis; b is the length of the minor semiaxis (Fig. 2).

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 a 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, which 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:

,

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 slope factor 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 converting an equation into general view into canonical and parametric forms.

Let an arbitrary Cartesian rectangular coordinate system be given Oxy. Consider a first degree equation or 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 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 the equation(9) passes through these points.

Answer:

Subtract (10) from (1):

We got canonical equation straight. 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.

Lesson from the series "Geometric Algorithms"

Hello dear reader!

Today we will start learning algorithms related to geometry. The fact is that there are quite a lot of Olympiad problems in computer science related to computational geometry, and the solution of such problems often causes difficulties.

In a few lessons, we will consider a number of elementary subproblems on which the solution of most problems of computational geometry is based.

In this lesson, we will write a program for finding the equation of a straight line passing through the given two dots. To solve geometric problems, we need some knowledge of computational geometry. We will devote part of the lesson to getting to know them.

Information from computational geometry

Computational geometry is a branch of computer science that studies algorithms for solving geometric problems.

The initial data for such problems can be a set of points on the plane, a set of segments, a polygon (given, for example, by a list of its vertices in clockwise order), etc.

The result can be either an answer to some question (such as does a point belong to a segment, do two segments intersect, ...), or some geometric object (for example, the smallest convex polygon connecting given points, the area of ​​a polygon, etc.) .

We will consider problems of computational geometry only on the plane and only in the Cartesian coordinate system.

Vectors and coordinates

To apply the methods of computational geometry, it is necessary to translate geometric images into the language of numbers. We will assume that a Cartesian coordinate system is given on the plane, in which the direction of rotation counterclockwise is called positive.

Now geometric objects receive an analytical expression. So, to set a point, it is enough to specify its coordinates: a pair of numbers (x; y). A segment can be specified by specifying the coordinates of its ends, a straight line can be specified by specifying the coordinates of a pair of its points.

But the main tool for solving problems will be vectors. Let me remind you, therefore, of some information about them.

Section AB, which has a point A considered the beginning (point of application), and the point V- the end is called a vector AB and denoted by either , or a bold lowercase letter, for example a .

To denote the length of a vector (that is, the length of the corresponding segment), we will use the module symbol (for example, ).

Arbitrary vector will have coordinates equal to the difference between the corresponding coordinates of its end and beginning:

,

dots here A and B have coordinates respectively.

For calculations, we will use the concept oriented angle, that is, an angle that takes into account the relative position of the vectors.

Oriented angle between vectors a and b positive if the rotation is away from the vector a to the vector b is done in the positive direction (counterclockwise) and negative in the other case. See fig.1a, fig.1b. It is also said that a pair of vectors a and b positively (negatively) oriented.

Thus, the value of the oriented angle depends on the order of enumeration of the vectors and can take values ​​in the interval .

Many computational geometry problems use the concept of vector (skew or pseudoscalar) products of vectors.

The vector product of vectors a and b is the product of the lengths of these vectors and the sine of the angle between them:

.

Vector product of vectors in coordinates:

The expression on the right is a second-order determinant:

Unlike the definition given in analytic geometry, this is a scalar.

The sign of the cross product determines the position of the vectors relative to each other:

a and b positively oriented.

If the value is , then the pair of vectors a and b negatively oriented.

The cross product of nonzero vectors is zero if and only if they are collinear ( ). This means that they lie on the same line or on parallel lines.

Let's consider some simple tasks necessary for solving more complex ones.

Let's define the equation of a straight line by the coordinates of two points.

The equation of a straight line passing through two different points given by their coordinates.

Let two non-coinciding points are given on the line: with coordinates (x1;y1) and with coordinates (x2; y2). Accordingly, the vector with the beginning at the point and the end at the point has coordinates (x2-x1, y2-y1). If P(x, y) is an arbitrary point on our line, then the coordinates of the vector are (x-x1, y - y1).

With the help of the cross product, the condition for the collinearity of the vectors and can be written as follows:

Those. (x-x1)(y2-y1)-(y-y1)(x2-x1)=0

(y2-y1)x + (x1-x2)y + x1(y1-y2) + y1(x2-x1) = 0

We rewrite the last equation as follows:

ax + by + c = 0, (1)

c = x1(y1-y2) + y1(x2-x1)

So, the straight line can be given by an equation of the form (1).

Task 1. The coordinates of two points are given. Find its representation in the form ax + by + c = 0.

In this lesson, we got acquainted with some information from computational geometry. We solved the problem of finding the equation of the line by the coordinates of two points.

In the next lesson, we will write a program to find the intersection point of two lines given by our equations.