Lesson from the series "Geometric Algorithms"

Hello dear reader!

Today we will start learning algorithms related to geometry. The fact is that there are 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.

Line segment AB, which has a point BUT considered the beginning (point of application), and the point AT- the end is called a vector AB and denote either , or bold lower case, 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, ).

An arbitrary vector will have coordinates, equal differences 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 non-zero 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.

Let two points be given M 1 (x 1, y 1) and M 2 (x 2, y 2). We write the equation of a straight line in the form (5), where k as yet unknown coefficient:

Since the point M 2 belongs to a given line, then its coordinates satisfy equation (5): . Expressing from here and substituting it into equation (5), we obtain the desired equation:

If a This equation can be rewritten in a form that is easier to remember:

(6)

Example. Write the equation of a straight line passing through the points M 1 (1.2) and M 2 (-2.3)

Decision. . Using the property of proportion, and performing the necessary transformations, we obtain the general equation of a straight line:

Angle between two lines

Consider two lines l 1 and l 2:

l 1: , , and

l 2: , ,

φ is the angle between them (). Figure 4 shows: .

From here , or

Using formula (7), one of the angles between the lines can be determined. The second angle is .

Example. Two straight lines are given by the equations y=2x+3 and y=-3x+2. find the angle between these lines.

Decision. It can be seen from the equations that k 1 \u003d 2, and k 2 \u003d-3. substituting these values ​​into formula (7), we find

. So the angle between these lines is .

Conditions for parallelism and perpendicularity of two lines

If straight l 1 and l 2 are parallel, then φ=0 and tgφ=0. from formula (7) it follows that , whence k 2 \u003d k 1. Thus, the condition for the parallelism of two lines is the equality of their slopes.

If straight l 1 and l 2 perpendicular, then φ=π/2, α 2 = π/2+ α 1 . . Thus, the condition for two straight lines to be perpendicular is that their slopes are reciprocal in magnitude and opposite in sign.

Distance from point to line

Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Vy + C \u003d 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 = 2tgj= ; j = p/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 height passes through point C, then its coordinates satisfy this equation: whence b = 17. Total: .

Answer: 3x + 2y - 34 = 0.

The distance from a point to a line is determined by the length of the perpendicular dropped from the point to the line.

If the line is parallel to the projection plane (h | | P 1), then in order to determine the distance from the point BUT to straight h it is necessary to drop a perpendicular from the point BUT to the horizontal h.

Consider more complex example when the line occupies general position. Let it be necessary to determine the distance from the point M to straight a general position.

Definition task distances between parallel lines solved similarly to the previous one. A point is taken on one line, and a perpendicular is drawn from it to another line. The length of the perpendicular is equal to the distance between the parallel lines.

Curve of the second order is a line defined by an equation of the second degree with respect to the current Cartesian coordinates. In the general case, Ax 2 + 2Bxy + Su 2 + 2Dx + 2Ey + F \u003d 0,

where A, B, C, D, E, F are real numbers and at least one of the numbers A 2 + B 2 + C 2 ≠0.

Circle

Circle center- this is the locus of points in the plane equidistant from the point of the plane C (a, b).

The circle is given by the following equation:

Where x, y are the coordinates of an arbitrary point on the circle, R is the radius of the circle.

Sign of the circle equation

1. There is no term with x, y

2. Coefficients at x 2 and y 2 are equal

Ellipse

Ellipse the locus of points in a plane is called, the sum of the distances of each of which from two given points of this plane is called foci (a constant value).

Canonical equation of an ellipse:

X and y belong to an ellipse.

a is the major semiaxis of the ellipse

b is the minor semiaxis of the ellipse

The ellipse has 2 axes of symmetry OX and OY. The axes of symmetry of the ellipse are its axes, the point of their intersection is the center of the ellipse. The axis on which the foci are located is called focal axis. The point of intersection of the ellipse with the axes is the vertex of the ellipse.

Compression (stretching) ratio: ε = c/a- eccentricity (characterizes the shape of the ellipse), the smaller it is, the less the ellipse is extended along the focal axis.

If the centers of the ellipse are not in the center С(α, β)

Hyperbola

Hyperbole called the locus of points in a plane, the absolute value of the difference in distances, each of which from two given points of this plane, called foci, is a constant value different from zero.

Canonical equation of a hyperbola

A hyperbola has 2 axes of symmetry:

a - real semiaxis of symmetry

b - imaginary semiaxis of symmetry

Asymptotes of a hyperbola:

Parabola

parabola is the locus of points in a plane equidistant from a given point F, called the focus, and a given line, called the directrix.

Canonical parabola equation:

Y 2 \u003d 2px, where p is the distance from the focus to the directrix (parabola parameter)

If the vertex of the parabola is C (α, β), then the equation of the parabola (y-β) 2 \u003d 2p (x-α)

If the focal axis is taken as the y-axis, then the parabola equation will take the form: x 2 \u003d 2qy

Equation of a straight line on a plane.
The direction vector is straight. Normal vector

A straight line on a plane is one of the simplest geometric shapes, familiar to you since elementary grades, and today we will learn how to deal with it using the methods of analytical geometry. To master the material, it is necessary to be able to build a straight line; know which equation defines a straight line, in particular, a straight line passing through the origin and straight lines parallel to the coordinate axes. This information can be found in the manual. Graphs and properties of elementary functions, I created it for matan, but the section on the linear function turned out to be very successful and detailed. Therefore, dear teapots, first warm up there. In addition, you need to have basic knowledge about vectors otherwise the understanding of the material will be incomplete.

In this lesson, we will look at ways in which you can write the equation of a straight line in a plane. I recommend that you do not neglect practical examples (even if it seems very simple), as I will supply them with elementary and important facts, technical methods that will be required in the future, including in other sections of higher mathematics.

• How to write the equation of a straight line with a slope?
• How ?
• How to find the direction vector by the general equation of a straight line?
• How to write an equation of a straight line given a point and a normal vector?

and we start:

Line Equation with Slope

The well-known "school" form of the equation of a straight line is called equation of a straight line with slope factor . For example, if a straight line is given by the equation, then its slope: . Consider geometric sense given coefficient and how its value affects the location of the line:

In the course of geometry it is proved that the slope of the straight line is tangent of an angle between positive axis directionand given line: , and the corner is “unscrewed” counterclockwise.

In order not to clutter up the drawing, I drew angles for only two straight lines. Consider the "red" straight line and its slope. According to the above: (angle "alpha" is indicated by a green arc). For the "blue" line with the slope, equality is true (the angle "beta" is indicated by the brown arc). And if the tangent of the angle is known, then if necessary it is easy to find and the corner using the inverse function - arc tangent. As they say, a trigonometric table or a calculator in hand. Thus, the slope characterizes the degree of inclination of the straight line to the x-axis.

In this case, the following cases are possible:

1) If the slope is negative: , then the line, roughly speaking, goes from top to bottom. Examples are "blue" and "crimson" straight lines in the drawing.

2) If the slope is positive: , then the line goes from bottom to top. Examples are "black" and "red" straight lines in the drawing.

3) If the slope is equal to zero: , then the equation takes the form , and the corresponding line is parallel to the axis. An example is the "yellow" line.

4) For a family of straight lines parallel to the axis (there is no example in the drawing, except for the axis itself), the slope does not exist (tangent of 90 degrees not defined).

The greater the slope modulo, the steeper the line graph goes.

For example, consider two straight lines. Here , so the straight line has a steeper slope. I remind you that the module allows you to ignore the sign, we are only interested in absolute values angular coefficients.

In turn, a straight line is steeper than straight lines. .

Vice versa: the smaller the slope modulo, the straight line is flatter.

For straight lines the inequality is true, thus, the straight line is more than a canopy. Children's slide, so as not to plant bruises and bumps.

Why is this needed?

Prolong your torment Knowing the above facts allows you to immediately see your mistakes, in particular, errors when plotting graphs - if the drawing turned out “clearly something is wrong”. It is desirable that you straightaway it was clear that, for example, a straight line is very steep and goes from bottom to top, and a straight line is very flat, close to the axis and goes from top to bottom.

In geometric problems, several straight lines often appear, so it is convenient to denote them somehow.

Notation: straight lines are indicated by small with Latin letters: . A popular option is the designation of the same letter with natural subscripts. For example, the five lines that we have just considered can be denoted by .

Since any straight line is uniquely determined by two points, it can be denoted by these points: etc. The notation quite obviously implies that the points belong to the line.

Time to loosen up a bit:

How to write the equation of a straight line with a slope?

If a point is known that belongs to a certain line, and the slope of this line, then the equation of this line is expressed by the formula:

Example 1

Compose the equation of a straight line with a slope if it is known that the point belongs to this straight line.

Decision: We will compose the equation of a straight line according to the formula . In this case:

Answer:

Examination performed elementarily. First, we look at the resulting equation and make sure that our slope is in its place. Second, the coordinates of the point must satisfy the given equation. Let's plug them into the equation:

The correct equality is obtained, which means that the point satisfies the resulting equation.

Conclusion: Equation found correctly.

A more tricky example for a do-it-yourself solution:

Example 2

Write the equation of a straight line if it is known that its angle of inclination to the positive direction of the axis is , and the point belongs to this straight line.

If you're having trouble, re-read theoretical material. More precisely, more practical, I miss many proofs.

rang last call, the prom has died down, and outside the gates home school we are waiting, in fact, analytic geometry. Jokes are over... Maybe it's just getting started =)

Nostalgically we wave the handle to the familiar and get acquainted with the general equation of a straight line. Since in analytic geometry it is precisely this that is in use:

The general equation of a straight line has the form: , where are some numbers. At the same time, the coefficients simultaneously are not equal to zero, since the equation loses its meaning.

Let's dress in a suit and tie an equation with a slope. First, we move all the terms to the left side:

The term with "x" must be put in first place:

In principle, the equation already has the form , but according to the rules of mathematical etiquette, the coefficient of the first term (in this case ) must be positive. Changing signs:

Remember this technical feature! We make the first coefficient (most often ) positive!

In analytic geometry, the equation of a straight line will almost always be given in general form. Well, if necessary, it is easy to bring it to a “school” form with a slope (with the exception of straight lines parallel to the y-axis).

Let's ask ourselves what enough know to build a straight line? Two points. But about this childhood case later, now sticks with arrows rule. Each straight line has a well-defined slope, to which it is easy to "adapt" vector.

A vector that is parallel to a line is called the direction vector of that line.. Obviously, any straight line has infinitely many direction vectors, and all of them will be collinear (co-directed or not - it does not matter).

I will denote the direction vector as follows: .

But one vector is not enough to build a straight line, the vector is free and is not attached to any point of the plane. Therefore, it is additionally necessary to know some point that belongs to the line.

How to write an equation of a straight line given a point and a direction vector?

If a certain point belonging to the line and the directing vector of this line are known, then the equation of this line can be compiled by the formula:

Sometimes it is called canonical equation of the line .

What to do when one of the coordinates is zero, we will look into practical examples below. By the way, note - both at once coordinates cannot be zero, since the zero vector does not specify a specific direction.

Example 3

Write an equation of a straight line given a point and a direction vector

Decision: We will compose the equation of a straight line according to the formula. In this case:

Using the properties of proportion, we get rid of fractions:

And we bring the equation to general view:

Answer:

Drawing in such examples, as a rule, is not necessary, but for the sake of understanding:

In the drawing, we see the starting point, the original direction vector (it can be postponed from any point on the plane) and the constructed line. By the way, in many cases, the construction of a straight line is most conveniently carried out using the slope equation. Our equation is easy to convert to the form and without any problems pick up one more point to build a straight line.

As noted at the beginning of the section, a line has infinitely many direction vectors, and they are all collinear. For example, I drew three such vectors: . Whichever direction vector we choose, the result will always be the same straight line equation.

Let's compose the equation of a straight line by a point and a directing vector:

Breaking down the proportion:

Divide both sides by -2 and get the familiar equation:

Those who wish can similarly test vectors or any other collinear vector.

Now let's solve the inverse problem:

How to find the direction vector by the general equation of a straight line?

Very simple:

If a straight line is given by a general equation in a rectangular coordinate system, then the vector is the direction vector of this straight line.

Examples of finding direction vectors of straight lines:

The statement allows us to find only one direction vector from an infinite set, but we don’t need more. Although in some cases it is advisable to reduce the coordinates of the direction vectors:

So, the equation specifies a straight line that is parallel to the axis, and the coordinates of the resulting steering vector are conveniently divided by -2, getting exactly the basis vector as the steering vector. Logically.

Similarly, the equation defines a straight line parallel to the axis, and dividing the coordinates of the vector by 5, we get the ort as the direction vector.

Now let's execute check example 3. The example went up, so I remind you that in it we made up the equation of a straight line using a point and a direction vector

First of all, according to the equation of a straight line, we restore its directing vector: - everything is fine, we got the original vector (in some cases, it can turn out to be collinear to the original vector, and this is usually easy to see by the proportionality of the corresponding coordinates).

Secondly, the coordinates of the point must satisfy the equation . We substitute them into the equation:

The correct equality has been obtained, which we are very pleased with.

Conclusion: Job completed correctly.

Example 4

Write an equation of a straight line given a point and a direction vector

This is a do-it-yourself example. Solution and answer at the end of the lesson. It is highly desirable to make a check according to the algorithm just considered. Try to always (if possible) check on a draft. It is foolish to make mistakes where they can be 100% avoided.

In the event that one of the coordinates of the direction vector is zero, it is very simple to do:

Example 5

Decision: The formula is invalid because the denominator on the right side is zero. There is an exit! Using the properties of proportion, we rewrite the formula in the form , and the rest rolled along a deep rut:

Answer:

Examination:

1) Restore the direction vector of the straight line:
– the resulting vector is collinear to the original direction vector.

2) Substitute the coordinates of the point in the equation:

The correct equality is obtained

Conclusion: job completed correctly

The question arises, why bother with the formula if there is a universal version that will work anyway? There are two reasons. First, the fractional formula much better to remember. And secondly, the disadvantage of the universal formula is that markedly increased risk of confusion when substituting coordinates.

Example 6

Compose the equation of a straight line given a point and a direction vector.

This is a do-it-yourself example.

Let's return to the ubiquitous two points:

How to write the equation of a straight line given two points?

If two points are known, then the equation of a straight line passing through these points can be compiled using the formula:

In fact, this is a kind of formula, and here's why: if two points are known, then the vector will be the direction vector of this line. On the lesson Vectors for dummies we considered the simplest task– how to find the coordinates of a vector from two points. According to this problem, the coordinates of the direction vector:

Note : points can be "swapped" and use the formula . Such a decision would be equal.

Example 7

Write the equation of a straight line from two points .

Decision: Use the formula:

We comb the denominators:

And shuffle the deck:

Now is the time to get rid of fractional numbers. In this case, you need to multiply both parts by 6:

Open the brackets and bring the equation to mind:

Answer:

Examination is obvious - the coordinates of the initial points must satisfy the resulting equation:

1) Substitute the coordinates of the point:

True equality.

2) Substitute the coordinates of the point:

True equality.

Conclusion: the equation of the straight line is correct.

If a at least one of points does not satisfy the equation, look for an error.

It is worth noting that the graphical verification in this case is difficult, because to build a line and see if the points belong to it , not so easy.

I will note a couple of technical points of the solution. Perhaps in this problem it is more advantageous to use the mirror formula and, for the same points make an equation:

There are fewer fractions. If you want, you can complete the solution to the end, the result should be the same equation.

The second point is to look at the final answer and see if it can be further simplified? For example, if an equation is obtained, then it is advisable to reduce it by two: - the equation will set the same straight line. However, this is already a topic of conversation about mutual arrangement of straight lines.

Having received an answer in Example 7, just in case, I checked if ALL coefficients of the equation are divisible by 2, 3 or 7. Although, most often such reductions are made during the solution.

Example 8

Write the equation of a straight line passing through the points .

This is an example for an independent solution, which will just allow you to better understand and work out the calculation technique.

Similar to the previous paragraph: if in the formula one of the denominators (direction vector coordinate) vanishes, then we rewrite it as . And again, notice how awkward and confused she began to look. I don’t see much point in giving practical examples, since we have already actually solved such a problem (see Nos. 5, 6).

Straight line normal vector (normal vector)

What is normal? In simple words, the normal is the perpendicular. That is, the normal vector of a line is perpendicular to the given line. It is obvious that any straight line has an infinite number of them (as well as directing vectors), and all the normal vectors of the straight line will be collinear (codirectional or not - it does not matter).

Dealing with them will be even easier than with direction vectors:

If a straight line is given by a general equation in a rectangular coordinate system, then the vector is the normal vector of this straight line.

If the coordinates of the direction vector have to be carefully “pulled out” from the equation, then the coordinates of the normal vector can simply be “removed”.

The normal vector is always orthogonal to the direction vector of the line. We will verify the orthogonality of these vectors using dot product:

I will give examples with the same equations as for the direction vector:

Is it possible to write an equation of a straight line, knowing one point and a normal vector? It feels like it's possible. If the normal vector is known, then the direction of the straightest line is also uniquely determined - this is a “rigid structure” with an angle of 90 degrees.

How to write an equation of a straight line given a point and a normal vector?

If some point belonging to the line and the normal vector of this line are known, then the equation of this line is expressed by the formula:

Here everything went without fractions and other surprises. Such is our normal vector. Love it. And respect =)

Example 9

Compose the equation of a straight line given a point and a normal vector. Find the direction vector of the straight line.

Decision: Use the formula:

The general equation of the straight line is obtained, let's check:

1) "Remove" the coordinates of the normal vector from the equation: - yes, indeed, the original vector is obtained from the condition (or the vector should be collinear to the original vector).

2) Check if the point satisfies the equation:

True equality.

After we are convinced that the equation is correct, we will complete the second, easier part of the task. We pull out the direction vector of the straight line:

Answer:

In the drawing, the situation is as follows:

For the purposes of training, a similar task for an independent solution:

Example 10

Compose the equation of a straight line given a point and a normal vector. Find the direction vector of the straight line.

The final section of the lesson will be devoted to less common, but also important types of equations of a straight line in a plane

Equation of a straight line in segments.
Equation of a straight line in parametric form

The equation of a straight line in segments has the form , where are nonzero constants. Some types of equations cannot be represented in this form, for example, direct proportionality (since the free term is zero and there is no way to get one on the right side).

This is, figuratively speaking, a "technical" type of equation. The usual task is to represent the general equation of a straight line as an equation of a straight line in segments. Why is it convenient? The equation of a straight line in segments allows you to quickly find the points of intersection of a straight line with coordinate axes, which is very important in some problems of higher mathematics.

Find the point of intersection of the line with the axis. We reset the “y”, and the equation takes the form . Desired point obtained automatically: .

Same with axis is the point where the line intersects the y-axis.

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 canonical equations straight.

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 .

Decision. 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 .

Decision. 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 general equations straight line in space.

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

Decision. 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:

,

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).