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Drawing up equations reflecting the chemical interaction of an oxidizing agent and a reducing agent comes down to determining the coefficients in the formulas of the starting substances and reaction products, the composition of which is determined from experience.

It is recommended to compile equations to determine the number of criteria so that each of the equations includes three variable quantities ab a2, a3, and the remaining quantities a4 and i are included in the equations one by one.

Drawing up equations is possible only for the simplest objects. More complex objects, which include most oil industry objects, are still being studied experimentally. The object properties used in the study of automatic control systems are self-leveling, capacitance and delay.

We will compile equations in difference form for a conducting medium and for a dielectric, as well as for one-dimensional and two-dimensional problems in which the change in field values ​​over distance occurs in one or two coordinate directions, respectively.

The composition of equations for virtual variations is demonstrated using the example of taking into account nonholonomic connections. It is shown that the holonomic coupling equation with a parameter is an ideal coupling when it describes the envelope. The rules for virtual variation of connections for two independent variables are discussed.

Drawing up equations has a lot in common with this kind of translation. In mild cases, the verbal formulation breaks down almost mechanically into a number of successive parts, each of which can be directly expressed in mathematical symbols. In more difficult cases, the condition consists of parts that cannot be directly translated into mathematical symbols. In this case, we should pay less attention to the verbal formulation and focus our attention on the meaning of this formulation. Before we begin to write mathematically, we may have to formulate the conditions differently, always keeping in mind the mathematical means for writing this new formulation.

Drawing up equations for such chemical processes does not present any difficulties.

The composition of variational equations in general form is discussed below.

Drawing up an equation for twist angles Q and determining its derivatives.

Drawing up equations is possible only for the simplest objects. More complex objects, which include most oil industry objects, are still being studied experimentally. The object properties used in the study of automatic control systems are self-leveling, capacitance and delay.

Drawing up equations analytically is possible only for relatively simple objects, the processes or physical phenomena in which have been sufficiently well studied. In the general case, the dynamic properties of controlled objects are described by differential equations expressing the dependence between output and input quantities in time. These equations are compiled on the basis of physical laws that determine transient processes in objects.

Drawing up equations (6 - 58) and solving them for A and B. A general method for solving this problem can be indicated provided that A and B enter the equation linearly.

Solutions to word problems involving composing equations will be useful primarily for schoolchildren. The curriculum for grades 9 and 10 covers a wide class of problems that require identifying unknowns, creating an equation and solving them. Below is only a small part of the possible problems and the methodology for their calculations.

Example 1. The first cyclist travels 50 meters less every minute than the second, so he spends 2 hours more on the 120 km journey than the second. Find the speed of the second cyclist (in km per hour).
Solution: The task is difficult for many, but in reality everything is simple.
Hidden under the phrase “It travels 50 meters less every minute” is a speed of 50 m/min. Since the rest of the data is in km and hours, we convert 50 m/min to km/h.
50/1000*60=3000/1000=3 (km/h).
Let us denote the speed of the second cyclist by V, and the time of movement by t.
Multiplying the speed by the time of movement we get the path
V*t=120.
The first cyclist rides slower and therefore takes longer. We compose the corresponding equation of motion
(V-3)(t+2)=120.
We have a system of two equations with two unknowns.
From the first equation we express the time of movement and substitute it into the second
t=120/V; (V-3)(120/V+2)=120.
After multiplying by V/2 and grouping similar terms, we can obtain the following quadratic equation
V^2-3V-180=0.
We calculate the discriminant of the equation
D=9+4*180=729=27*27
and roots
V=(3+27)/2=15;
V=(3-27)/2=-12.
We reject the second one; it has no physical meaning. The found value V = 15 km/h is the speed of the second cyclist.
Answer: 15 km/h.

Example 2. Sea water contains 5% salt by weight. How much fresh water must be added to 30 kg of sea water to reduce the salt concentration by 70%?
Solution: Find how much salt is in 30 kg of sea water
30*5/100=1.5 (kg).
In the new solution this will be
(100%-70%) = 30% of 5%, make up the proportions
5% – 100%
X – 30%.
Carrying out calculations
X=5*30/100=150/100=1.5%.
Thus, 1.5 kg of salt corresponds to 1.5% in the new solution. Adding up the proportions again
1.5 – 1.5% Y – 100% .
Finding the mass of sea water solution
Y=1.5*100/1.5=100 (kg).
Subtract the mass of salt water to find the amount of fresh water
100-30=70 (kg).
Answer: 70 kg of fresh water.

Example 3. The motorcyclist was delayed at the barrier for 24 minutes. After that, increasing his speed by 10 kilometers per hour, he made up for being late on the 80 km stretch. Determine the speed of the motorcyclist before slowing down (in km per hour).
Solution: Problem of composing an equation for speed. Let's denote the initial speed of the motorcyclist by V, and the time for which he had to travel by t. There are two unknowns, therefore there must also be 2 equations. According to the condition, during this time he had to travel 80 km.
V*t=80 (km) .
Delayed means the time has decreased by 24 minutes. It is also worth noting that in such problems, time must be converted into hours or minutes (depending on the condition) and then solved. We compose the equation of motion taking into account less time and higher speed
(V+10)(t-24/60)=80.
There are two equations for determining time and speed. Since the problem asks for speed, we will express the time from the first equation and substitute it into the second
t=80/V;
(V+10)(80/V-24/60)=80.
Our goal is to teach you how to create equations for problems from which you can determine the required quantities.
Therefore, without going into details, the resulting equation by multiplying by 60 * V and dividing by 24 can be reduced to the following quadratic equation
V^2+10*V-2000=0.
Find the discriminant and roots of the equation yourself. You should get the value
V=-50;
V=40.
We discard the first value; it has no physical meaning. The second V = 40 km/h is the desired speed of the motorcyclist.
Answer: 40 km/h.

Example 4. The freight train was delayed on the way for 12 minutes, and then, at a distance of 112 kilometers, it made up for lost time, increasing its speed by 10 km/h. Find the initial speed of the train (in km/h).
Solution: We have a problem in which the unknowns are the train speed V and travel time t.
Since the problem according to the equation scheme corresponds to the previous one, we write two equations for the unknowns
V*t=112;
(V+10)*(t-12/60)=112.
Equations should be written in exactly this notation. This allows us to express time in a simple way from the first equation
t =112/V
and, substituting into the second one, get the equation only for speed
(V+10)*(112/ V -12/60)=112.
If you choose the wrong notation, you can get an equation for the unknowns of this type
V*(t+12)=112;
(V+10)*t=112.
Here t corresponds to the time after increasing the speed by 10 km/h, but that is not the point. The given equations are also correct, but not convenient from a calculation point of view.
Try solving the first two equations and the last ones and you will understand that the second scheme should be avoided when composing equations. Therefore, think carefully about what notation you enter to minimize the number of calculations.
The resulting equation
(V+10)*(112/ V -12/60)=112.
reduce to a quadratic equation (multiply by 60*V/12)
V^2+10*V-5600=0.
Without going into intermediate calculations, the roots will be
V=-80;
V=70.
In problems of this type we always get a negative root (V=-80) that needs to be discarded. The speed of the train is 70 km/h.

Example 5. Having departed from the bus station 10 minutes later, the bus drove to the first stop at a speed of 16 km/h more than scheduled and arrived on time. What speed (in km/hour) should the bus have according to the schedule if the distance from the bus station to the first stop is 16 kilometers?
Solution: The unknowns are the bus speed V and time t.
We create an equation, taking into account that the delay time is specified in minutes, not hours
V * t = 16 - this is how the bus should have traveled as usual;
(V + 16) (t-10/60) = 16 is the equation of motion due to the late departure of the bus.
There are two equations and two unknowns.
Let's express time from the first equation and substitute it into the second
t=16/V;
(V+16)(16/V-1/6)=16.
The resulting equation for speed is reduced to quadratic (*6*V)
V^2+16*V-1536=0.
The roots of a quadratic equation are
V=32; V=-48.
The required speed of the bus is 32 km/h.
Answer: 32 km/h.

Example 6. The driver of the car stopped to change a tire for 12 minutes. After that, by increasing the speed by 15 km/h, he made up for the time spent on 60 kilometers. At what speed (in km/h) was he moving after stopping?
Solution: The algorithm for solving the problem was given several times in previous examples. We standardly denote speed and time by V, t.
When writing the equation, remember to convert minutes to hours. The system of equations will look like
V*t=60;
(V+15)(t-12/60)=60.
You should also know or memorize further manipulations.
t=60/V;
(V+15)(60/V -12/60)=60.
This equation can be reduced to a quadratic equation
V^2+15*V-4500=0.
Having solved the quadratic equation, we obtain the following speed values
V=60; V=-75.
Speed ​​cannot be negative, so the only correct answer is V=60 km/h.

Example 7. A certain two-digit number is 4 times the sum and 3 times the product of its digits. Find this number.
Solution: Problems with numbers occupy an important place among problems for composing equations and can be no less interesting in constructing solutions than tasks with speed. All you need is to understand the problem well. Let's denote the number by ab, that is, the number is equal to 10 * a + b. Based on the condition, we create a system of equations
10*a+b=4*(a+b);
10*a+b=3*a*b.
Since the unknowns enter into the first equation linearly, we write it out and express one of the unknowns through the other
10*a+b-4*a-4*b=0;
6*a-3*b=0; b=2*a.
Substitute b = 2 * a into the second equation
10*a+2*a=3*a*2*a;
6*a2-12*a=0; a(a-2)=0.
Hence a=0; a=2 . There is no point in considering the first value; if a=2, the second digit is equal to b=2*a=2*2=4, and the required number is 24.
Answer: the number is 24.

Let's talk about how to create a chemical equation, because they are the main elements of this discipline. Thanks to a deep understanding of all the patterns of interactions and substances, you can control them and apply them in various fields of activity.

Theoretical features

Compiling chemical equations is an important and responsible stage, considered in the eighth grade of secondary schools. What should precede this stage? Before the teacher tells his students how to create a chemical equation, it is important to introduce schoolchildren to the term “valence” and teach them to determine this value for metals and non-metals using the periodic table of elements.

Compilation of binary formulas by valency

In order to understand how to create a chemical equation by valence, you first need to learn how to create formulas for compounds consisting of two elements using valency. We propose an algorithm that will help cope with the task. For example, you need to create a formula for sodium oxide.

First, it is important to take into account that the chemical element that is mentioned last in the name should be in first place in the formula. In our case, sodium will be written first in the formula, oxygen second. Let us recall that oxides are binary compounds in which the last (second) element must be oxygen with an oxidation state of -2 (valency 2). Next, using the periodic table, it is necessary to determine the valence of each of the two elements. To do this we use certain rules.

Since sodium is a metal that is located in the main subgroup of group 1, its valence is a constant value, it is equal to I.

Oxygen is a non-metal, since it is the last one in the oxide; to determine its valence, we subtract 6 from eight (the number of groups) (the group in which oxygen is located), we obtain that the valency of oxygen is II.

Between certain valences we find the least common multiple, then divide it by the valency of each of the elements to obtain their indices. We write down the finished formula Na 2 O.

Instructions for composing an equation

Now let's talk in more detail about how to write a chemical equation. First, let's look at the theoretical aspects, then move on to specific examples. So, composing chemical equations presupposes a certain procedure.

• 1st stage. After reading the proposed task, you need to determine which chemicals should be present on the left side of the equation. A “+” sign is placed between the original components.
• 2nd stage. After the equal sign, you need to create a formula for the reaction product. When performing such actions, you will need the algorithm for composing formulas for binary compounds, which we discussed above.
• 3rd stage. We check the number of atoms of each element before and after chemical interaction, if necessary, we put additional coefficients in front of the formulas.

Example of a combustion reaction

Let's try to figure out how to create a chemical equation for the combustion of magnesium using an algorithm. On the left side of the equation we write the sum of magnesium and oxygen. Do not forget that oxygen is a diatomic molecule, so it must be given an index of 2. After the equal sign, we compose the formula for the product obtained after the reaction. It will be in which magnesium is written first, and oxygen is written second in the formula. Next, using the table of chemical elements, we determine the valencies. Magnesium, which is in group 2 (the main subgroup), has a constant valency II; for oxygen, by subtracting 8 - 6 we also get valency II.

The process record will look like: Mg+O 2 =MgO.

In order for the equation to comply with the law of conservation of mass of substances, it is necessary to arrange the coefficients. First, we check the amount of oxygen before the reaction, after the process is completed. Since there were 2 oxygen atoms, but only one was formed, a coefficient of 2 must be added on the right side before the magnesium oxide formula. Next, we count the number of magnesium atoms before and after the process. As a result of the interaction, 2 magnesium was obtained, therefore, on the left side in front of the simple substance magnesium, a coefficient of 2 is also required.

The final type of reaction: 2Mg+O 2 =2MgO.

Example of a substitution reaction

Any chemistry summary contains a description of different types of interactions.

Unlike a compound, in a substitution there will be two substances on both the left and right sides of the equation. Let's say we need to write the reaction of interaction between zinc and We use the standard writing algorithm. First, on the left side we write zinc and hydrochloric acid through the sum, and on the right side we write the formulas for the resulting reaction products. Since zinc is located before hydrogen in the electrochemical voltage series of metals, in this process it displaces molecular hydrogen from the acid and forms zinc chloride. As a result, we get the following entry: Zn+HCL=ZnCl 2 +H 2.

Now we move on to equalizing the number of atoms of each element. Since there was one atom on the left side of chlorine, and after the interaction there were two, it is necessary to put a factor of 2 in front of the formula of hydrochloric acid.

As a result, we obtain a ready-made reaction equation corresponding to the law of conservation of mass of substances: Zn+2HCL=ZnCl 2 +H 2 .

Conclusion

A typical chemistry note necessarily contains several chemical transformations. Not a single section of this science is limited to a simple verbal description of transformations, processes of dissolution, evaporation; everything is necessarily confirmed by equations. The specificity of chemistry lies in the fact that all processes that occur between different inorganic or organic substances can be described using coefficients and indices.

How else does chemistry differ from other sciences? Chemical equations help not only to describe the transformations that occur, but also to carry out quantitative calculations based on them, thanks to which it is possible to carry out laboratory and industrial production of various substances.

54. Problems involving composing equations with one unknown:

We can apply equation solving skills to problem solving. The following examples will show you how to do this.

Task 1. The house was for sale. One buyer had an amount of money equal to ¾ of its value, and the other had an amount equal to 5/6 of its value. If they were added together, they would have a surplus of 7,000 rubles. What is the cost of the house?

Let's assume that the house costs x rubles. Then (in accordance with the beginning of the problem) the first buyer had (x · ¾) rubles. or, which is the same thing, 3x/4 rubles, and the second had 5x/6 rubles. The next phrase is the condition of the problem, namely, “if they were added together, they would have a surplus of 7,000 rubles.” - is an equation expressed in words: it is now necessary to express it not in words, but in mathematical symbols. First, let’s take a similar phrase in a simplified form: “if you add the numbers a and b, then the resulting sum will give an excess of m against the number c” - this phrase can be rewritten in mathematical symbols like this: a + b = c + m.

The equation in our problem can be written in exactly the same way: if we add the numbers 3x/4 and 5x/6, the resulting sum will give a surplus of 7000 over the number x, or
3x/4 + 5x/6 = x + 7000.

The resulting equation should be simplified: 1) multiply both sides of the equation by the common denominator 12 - we get

9x + 10x = 12x + 84000

2) Move the unknown terms to the left side:

9x + 10x – 12x = 84000

Now we can answer the problem:

The cost of the house was 12,000 rubles.

Task 2. There were 13 students absent from class on Monday and 5 students absent from class on Tuesday. The ratio of the number of students present on Monday to the number of students present on Tuesday was 7/9. How many students were there in this class?

Let's assume that there are x students in total in the class. Then on Monday there were (x – 13) students present, and on Tuesday (x – 5) students. The phrase "the ratio of the number of students present on Monday to the number present on Tuesday was 7/9" is an equation expressed in words and can be rewritten in mathematical symbols:

(x – 13) / (x – 5) = 7/9.

Let's solve this equation:

9(x – 13) = 7(x – 5) or 9x – 117 = 7x – 35.

From here we get: 2x = 82 and x = 41.
So, there were 41 students in this class.

Task 3. Find a fraction whose denominator is 3 more than the numerator and which becomes 4/5 if you subtract 1 from its numerator and denominator.

This task is somewhat different from the previous ones. It requires “finding a fraction,” but it would be impossible to start solving the problem the way they did in the 1st and 2nd problems: let’s assume that the required fraction is equal to x. It would be impossible to start like this because the problem deals separately with the numerator and separately with the denominator: you have to subtract 1 separately from the numerator and separately from the denominator. Therefore, it is necessary to designate the fraction in such a way that both its numerator and its denominator are visible. Since it is said that the denominator is 3 more than the numerator, we can denote by the letter x either the numerator or the denominator - then it is easy to find an expression for the other member of the fraction and for the fraction itself.

Here is the solution to the problem.

Let us assume that the numerator of the desired fraction is equal to x. Then its denominator is x + 3, and the desired fraction is x/(x+3). The phrase "which (i.e., a fraction) becomes 4/5 when 1 is subtracted from its numerator and denominator" is an equation and can be written mathematically:
(x – 1) / (x + 3 – 1) = 4/5 or (x – 1) / (x + 2) = 4/5.

5(x – 1) = 4(x + 2); 5x – 5 = 4x + 8; 5x – 4x = 5 + 8; x = 13.

Then the denominator of the fraction is 16 and the desired fraction is 13/16.

Task 4. One brother is 14 years older than the other, and in 6 years he will be 2 times older. How old is each brother?

Here you need to give two answers: how old is the younger brother and how old is the older one, but the problem can be solved using an equation with 1 unknown, since it is said that the older brother is 14 years older than the younger one. Let's solve the problem like this:

Let's assume that the younger brother is x years old; then the oldest is (x + 14) years old.

In 6 years, the younger brother will be (x + 6) years old, and the older brother will be (x + 14 + 6) years old or (x + 20) years old.

It is said that the elder will then (in 6 years) be 2 times older than the younger, i.e. the number x + 20 must be 2 times greater than x + 6, and this can be written as

(x + 20) / (x + 6) = 2 or x + 20 = 2 (x + 6) or (x + 20) / 2 = x + 6.

The most natural notation is the first: to find out how many times one number is greater than another, you need to divide; we need to find out how many times the number (x + 20) is greater than the number (x + 6) - for this we need to divide (x + 20) by (x + 6), and tell us the answer “twice”. Therefore, we write that from this division we get the number 2, i.e. (x + 20) / (x + 6) = 2.

The second entry can be explained as follows: we are told that the number (x + 20) must be 2 times the number (x + 6). To equalize these numbers, it is therefore necessary to multiply the smaller of them, i.e. x + 6, by 2. Then x + 20 = 2(x + 6).

Then the notation is explained as follows: in order to equalize the numbers x + 20 and x + 6, you need to reduce the larger of them by 2 times, and then (x + 20) / 2 = x + 6.

If we take the 1st entry

(x + 20) / (x + 6) = 2

and multiply both sides of the equation by x + 6, we get

x + 20 = 2(x + 6)

i.e. the second entry. It is also easy to get the 2nd or 1st entry from the 3rd entry, etc.

In any case, after freeing the equation from fractions, we get

x + 20 = 2(x + 6)

and easily solve the equation:

x + 20 = 2x + 12; 20 – 12 = 2x – x; 8 = x or x = 8.

So, the younger brother is 8 years old, and the older brother is 8 + 14 = 22 years old.

Task 5. We bought sugar and coffee, totaling £28; for a pound of sugar they paid 15 kopecks, and for a pound of coffee 80 kopecks, but for the entire purchase they paid 12 rubles. How much sugar did you buy and how much coffee did you buy?

The difficulty here may be that in the conditions of the problem the numbers are given either in kopecks or in rubles. It must be established in advance in what units, in rubles or kopecks, the decision will be made. Let's solve the problem in rubles. Then the solution is:

Let's say you bought x pounds of sugar. Then we bought (28 – x) pounds of coffee.

For sugar they paid (15x) kopecks or (3/20)x rubles (since 15 kopecks are equal to 3/20 rubles), and for coffee they paid 80(28 – x) kopecks. or 4/5 (28 – x) rub. (since 80 kopecks = 4/5 rubles).
The phrase “they paid 12 rubles for the entire purchase.” can be written:

3x/20 + 4(28x – x)/5 = 12

[If solved in pennies, the equation would be 15x + 80(28 – x) = 1200].

Let's free the equation from fractions, for which we multiply both parts by 20, and we get:

3x + 16(28 – x) = 240

3x + 448 – 16x = 240

3x – 16x = 240 – 448

–13x = –208,

So, we bought 16 pounds of sugar and 12 pounds of coffee (28 – 16 = 12).

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 figures, familiar to you from elementary school, and today we will learn how to deal with it using the methods of analytical geometry. To master the material, you must be able to build a straight line; know what equation defines a straight line, in particular, a straight line passing through the origin of coordinates 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 Mathan, but the section about the linear function turned out to be very successful and detailed. Therefore, dear teapots, warm up there first. 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 create an equation of a straight line on a plane. I recommend not to neglect practical examples (even if it seems very simple), since I will provide them with elementary and important facts, technical techniques that will be required in the future, including in other sections of higher mathematics.

• How to write an equation of a straight line with an angle coefficient?
• How ?
• How to find a direction vector using 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 begin:

Equation of a straight line with slope

The well-known “school” form of a straight line equation is called equation of a straight line with slope. For example, if a straight line is given by the equation, then its slope is: . Let's consider the geometric meaning of this coefficient and how its value affects the location of the line:

In a geometry course it is proven that the slope of the straight line is equal to tangent of the angle between positive axis directionand this line: , and the angle “unscrews” counterclockwise.

In order not to clutter the drawing, I drew angles only for two straight lines. Let's consider the “red” line and its slope. According to the above: (the “alpha” angle is indicated by a green arc). For the “blue” straight line with the angle coefficient, the equality is true (the “beta” angle is indicated by a brown arc). And if the tangent of the angle is known, then if necessary it is easy to find and the corner itself using the inverse function - arctangent. As they say, a trigonometric table or a microcalculator in your hands. Thus, the angular coefficient characterizes the degree of inclination of the straight line to the abscissa axis.

The following cases are possible:

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

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

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

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

The greater the slope coefficient in absolute value, the steeper the straight line graph goes..

For example, consider two straight lines. Here, therefore, the straight line has a steeper slope. Let me 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 .

Conversely: the smaller the slope coefficient in absolute value, the flatter the straight line.

For straight lines the inequality is true, thus the straight line is flatter. Children's slide, so as not to give yourself bruises and bumps.

Why is this necessary?

Prolong your torment Knowledge of the above facts allows you to immediately see your mistakes, in particular, errors when constructing graphs - if the drawing turns out to be “obviously something wrong.” It is advisable that you straightaway it was clear that, for example, the straight line is very steep and goes from bottom to top, and the straight line is very flat, pressed close to the axis and goes from top to bottom.

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

Designations: straight lines are designated in small Latin letters: . A popular option is to designate them using the same letter with natural subscripts. For example, the five lines we just looked at can be denoted by .

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

It's time to warm up a little:

How to write an equation of a straight line with an angle coefficient?

If a point belonging to a certain line and the angular coefficient of this line are known, then the equation of this line is expressed by the formula:

Example 1

Write an equation for a line with slope if it is known that the point belongs to the given line.

Solution: Let's compose the equation of the straight line using the formula . In this case:

Answer:

Examination is done simply. First, we look at the resulting equation and make sure that our slope is in place. Secondly, the coordinates of the point must satisfy this equation. Let's plug them into the equation:

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

Conclusion: The equation was found correctly.

A more tricky example to solve on your own:

Example 2

Write an equation for 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 have any difficulties, re-read the theoretical material. More precisely, more practical, I skip a lot of evidence.

The last bell has rung, the graduation ceremony has ended, and outside the gates of our native school, analytical geometry itself awaits us. The jokes are over... Or maybe they are just beginning =)

We nostalgically wave our pen to the familiar and get acquainted with the general equation of a straight line. Because in analytical geometry this is exactly what is used:

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 the equation with the slope coefficient. First, let's 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 analytical geometry, the equation of a straight line will almost always be given in general form. Well, if necessary, it can be easily reduced to the “school” form with an angular coefficient (with the exception of straight lines parallel to the ordinate axis).

Let's ask ourselves what enough know to construct a straight line? Two points. But more about this childhood incident, now sticks with arrows rule. Each straight line has a very specific slope, which is easy to “adapt” to. vector.

A vector that is parallel to a line is called the direction vector of that line. It is obvious that any straight line has an infinite number of direction vectors, and all of them will be collinear (co-directional or not - it doesn’t matter).

I will denote the direction vector as follows: .

But one vector is not enough to construct a straight line; the vector is free and not tied to any point on 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 using a point and a direction vector?

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

Sometimes it is called canonical equation of the line .

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

Example 3

Write an equation for a straight line using a point and a direction vector

Solution: Let's compose the equation of a straight line using the formula. In this case:

Using the properties of proportion we get rid of fractions:

And we bring the equation to its general form:

Answer:

As a rule, there is no need to make a drawing in such examples, but for the sake of understanding:

In the drawing we see the starting point, the original direction vector (it can be plotted from any point on the plane) and the constructed straight line. By the way, in many cases it is most convenient to construct a straight line using an equation with an angular coefficient. It’s easy to transform our equation into form and easily select another point to construct a straight line.

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

Let's create an equation of a straight line using a point and a direction vector:

Resolving the proportion:

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

Those interested can test vectors in the same way or any other collinear vector.

Now let's solve the inverse problem:

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

Very simple:

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

Examples of finding direction vectors of straight lines:

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

Thus, the equation specifies a straight line that is parallel to the axis and the coordinates of the resulting direction vector are conveniently divided by –2, obtaining exactly the basis vector as the direction vector. Logical.

Similarly, the equation specifies a straight line parallel to the axis, and by dividing the coordinates of the vector by 5, we obtain the unit vector as the direction vector.

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

Firstly, using the equation of the straight line we reconstruct its direction vector: – everything is fine, we have received the original vector (in some cases the result may be a collinear vector to the original one, and this is usually easy to notice 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 was obtained, which we are very happy about.

Conclusion: The task was completed correctly.

Example 4

Write an equation for a straight line using a point and a direction vector

This is an example for you to solve on your own. The solution and answer are at the end of the lesson. It is highly advisable to check using the algorithm just discussed. Try to always (if possible) check on a draft. It’s stupid to make mistakes where they can be 100% avoided.

In the event that one of the coordinates of the direction vector is zero, proceed very simply:

Example 5

Solution: The formula is not suitable since 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 directing vector of the straight line:
– the resulting vector is collinear to the original direction vector.

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

The correct equality is obtained

Conclusion: task completed correctly

The question arises, why bother with the formula if there is a universal version that will work in any case? There are two reasons. First, the formula is in the form of a fraction much better remembered. And secondly, the disadvantage of the universal formula is that the risk of getting confused increases significantly when substituting coordinates.

Example 6

Write an equation for a straight line using a point and a direction vector.

This is an example for you to solve on your own.

Let's return to the ubiquitous two points:

How to write an equation of a straight line using 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 type of formula and here's why: if two points are known, then the vector will be the direction vector of the given line. At the lesson Vectors for dummies we considered the simplest problem - how to find the coordinates of a vector from two points. According to this problem, the coordinates of the direction vector are:

Note : the points can be “swapped” and the formula can be used . Such a solution will be equivalent.

Example 7

Write an equation of a straight line using two points .

Solution: We use the formula:

Combing the denominators:

And shuffle the deck:

Now is the time to get rid of fractional numbers. In this case, you need to multiply both sides 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 line is written correctly.

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

It is worth noting that graphical verification in this case is difficult, since construct a straight line and see whether the points belong to it , not so simple.

I’ll note a couple more technical aspects of the solution. Perhaps in this problem it is more profitable to use the mirror formula and, at the same points make an equation:

Fewer fractions. If you want, you can carry out the solution to the end, the result should be the same equation.

The second point is to look at the final answer and figure out whether it can be simplified further? For example, if you get the equation , then it is advisable to reduce it by two: – the equation will define the same straight line. However, this is already a topic of conversation about relative position of lines.

Having received the answer in Example 7, just in case, I checked whether 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 an equation for a line passing through the points .

This is an example for an independent solution, which will allow you to better understand and practice calculation techniques.

Similar to the previous paragraph: if in the formula one of the denominators (the coordinate of the direction vector) becomes zero, then we rewrite it in the form . Again, notice how awkward and confused she looks. I don’t see much point in giving practical examples, since we have already actually solved this problem (see No. 5, 6).

Direct normal vector (normal vector)

What is normal? In simple words, a normal is a perpendicular. That is, the normal vector of a line is perpendicular to a given line. Obviously, any straight line has an infinite number of them (as well as direction vectors), and all the normal vectors of the straight line will be collinear (codirectional or not, it makes no difference).

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

If a line is given by a general equation in a rectangular coordinate system, then the vector is the normal vector of this 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 be simply “removed”.

The normal vector is always orthogonal to the direction vector of the line. Let us 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 construct an equation of a straight line given one point and a normal vector? I feel it in my gut, it’s possible. If the normal vector is known, then the direction of the straight line itself is clearly defined - 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 a certain point belonging to a line and the normal vector of this line are known, then the equation of this line is expressed by the formula:

Here everything worked out without fractions and other surprises. This is our normal vector. Love him. And respect =)

Example 9

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

Solution: We use the formula:

The general equation of the straight line has been obtained, let’s check:

1) “Remove” the coordinates of the normal vector from the equation: – yes, indeed, the original vector was obtained from the condition (or a collinear vector should be obtained).

2) Let's check whether the point satisfies the equation:

True equality.

After we are convinced that the equation is composed correctly, we will complete the second, easier part of the task. We take out the directing vector of the straight line:

Answer:

In the drawing the situation looks like this:

For training purposes, a similar task for solving independently:

Example 10

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

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

Equation of a straight line in segments.
Equation of a 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 equal to zero and there is no way to get one on the right side).

This is, figuratively speaking, a “technical” type of equation. A common task is to represent the general equation of a line as an equation of a line in segments. How is it convenient? The equation of a line in segments allows you to quickly find the points of intersection of a line with coordinate axes, which can be very important in some problems of higher mathematics.

Let's find the point of intersection of the line with the axis. We reset the “y” to zero, and the equation takes the form . The desired point is obtained automatically: .

Same with the axis – the point at which the straight line intersects the ordinate axis.