This article continues the topic of the equation of a line on a plane: we will consider this type of equation as the general equation of a line. Let us define the theorem and give its proof; Let's figure out what an incomplete general equation of a line is and how to make transitions from a general equation to other types of equations of a line. We will reinforce the entire theory with illustrations and solutions to practical problems.
Let a rectangular coordinate system O x y be specified on the plane.
Theorem 1
Any equation of the first degree, having the form A x + B y + C = 0, where A, B, C are some real numbers(A and B are not equal to zero at the same time) defines a straight line in rectangular system coordinates on the plane. In turn, any straight line in a rectangular coordinate system on a plane is determined by an equation that has the form A x + B y + C = 0 for a certain set of values A, B, C.
Proof
This theorem consists of two points; we will prove each of them.
- Let us prove that the equation A x + B y + C = 0 defines a straight line on the plane.
Let there be some point M 0 (x 0 , y 0) whose coordinates correspond to the equation A x + B y + C = 0. Thus: A x 0 + B y 0 + C = 0. Subtract from the left and right sides of the equations A x + B y + C = 0 the left and right sides of the equation A x 0 + B y 0 + C = 0, we obtain a new equation that looks like A (x - x 0) + B (y - y 0) = 0 . It is equivalent to A x + B y + C = 0.
The resulting equation A (x - x 0) + B (y - y 0) = 0 is a necessary and sufficient condition for the perpendicularity of the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0 ) . Thus, the set of points M (x, y) defines a straight line in a rectangular coordinate system perpendicular to the direction of the vector n → = (A, B). We can assume that this is not so, but then the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) would not be perpendicular, and the equality A (x - x 0 ) + B (y - y 0) = 0 would not be true.
Consequently, the equation A (x - x 0) + B (y - y 0) = 0 defines a certain line in a rectangular coordinate system on the plane, and therefore the equivalent equation A x + B y + C = 0 defines the same line. This is how we proved the first part of the theorem.
- Let us provide a proof that any straight line in a rectangular coordinate system on a plane can be specified by an equation of the first degree A x + B y + C = 0.
Let us define a straight line a in a rectangular coordinate system on a plane; the point M 0 (x 0 , y 0) through which this line passes, as well as the normal vector of this line n → = (A, B) .
Let there also be some point M (x, y) - a floating point on a line. In this case, the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) are perpendicular to each other, and their scalar product is zero:
n → , M 0 M → = A (x - x 0) + B (y - y 0) = 0
Let's rewrite the equation A x + B y - A x 0 - B y 0 = 0, define C: C = - A x 0 - B y 0 and as a final result we get the equation A x + B y + C = 0.
So, we have proved the second part of the theorem, and we have proved the entire theorem as a whole.
Definition 1
An equation of the form A x + B y + C = 0 - This general equation of a line on a plane in a rectangular coordinate systemOxy.
Based on the proven theorem, we can conclude that a straight line and its general equation defined on a plane in a fixed rectangular coordinate system are inextricably linked. In other words, the original line corresponds to its general equation; the general equation of a line corresponds to a given line.
From the proof of the theorem it also follows that the coefficients A and B for the variables x and y are the coordinates of the normal vector of the line, which is given by the general equation of the line A x + B y + C = 0.
Let's consider specific example general equation of a straight line.
Let the equation 2 x + 3 y - 2 = 0 be given, which corresponds to a straight line in a given rectangular coordinate system. The normal vector of this line is the vector n → = (2 , 3) . Let's draw the given straight line in the drawing.
We can also state the following: the straight line that we see in the drawing is determined by the general equation 2 x + 3 y - 2 = 0, since the coordinates of all points on a given straight line correspond to this equation.
We can obtain the equation λ · A x + λ · B y + λ · C = 0 by multiplying both sides of the general equation of the line by a number λ not equal to zero. The resulting equation is equivalent to the original general equation, therefore, it will describe the same straight line on the plane.
Definition 2Complete general equation of a line– such a general equation of the straight line A x + B y + C = 0, in which the numbers A, B, C are different from zero. Otherwise the equation is incomplete.
Let us analyze all variations of the incomplete general equation of a line.
- When A = 0, B ≠ 0, C ≠ 0, the general equation takes the form B y + C = 0. Such an incomplete general equation defines in a rectangular coordinate system O x y a straight line that is parallel to the O x axis, since for any real value of x the variable y will take the value - C B . In other words, the general equation of the line A x + B y + C = 0, when A = 0, B ≠ 0, specifies the locus of points (x, y), whose coordinates are equal to the same number - C B .
- If A = 0, B ≠ 0, C = 0, the general equation takes the form y = 0. This is not complete equation defines the abscissa axis O x .
- When A ≠ 0, B = 0, C ≠ 0, we obtain an incomplete general equation A x + C = 0, defining a straight line parallel to the ordinate.
- Let A ≠ 0, B = 0, C = 0, then the incomplete general equation will take the form x = 0, and this is the equation of the coordinate line O y.
- Finally, for A ≠ 0, B ≠ 0, C = 0, the incomplete general equation takes the form A x + B y = 0. And this equation describes a straight line that passes through the origin. In fact, the pair of numbers (0, 0) corresponds to the equality A x + B y = 0, since A · 0 + B · 0 = 0.
Let us graphically illustrate all of the above types of incomplete general equation of a straight line.
Example 1
It is known that the given straight line is parallel to the ordinate axis and passes through the point 2 7, - 11. It is necessary to write down the general equation of the given line.
Solution
A straight line parallel to the ordinate axis is given by an equation of the form A x + C = 0, in which A ≠ 0. The condition also specifies the coordinates of the point through which the line passes, and the coordinates of this point meet the conditions of the incomplete general equation A x + C = 0, i.e. the equality is true:
A 2 7 + C = 0
From it it is possible to determine C if we give A some non-zero value, for example, A = 7. In this case, we get: 7 · 2 7 + C = 0 ⇔ C = - 2. We know both coefficients A and C, substitute them into the equation A x + C = 0 and get the required straight line equation: 7 x - 2 = 0
Answer: 7 x - 2 = 0
Example 2
The drawing shows a straight line; you need to write down its equation.
Solution
The given drawing allows us to easily take the initial data to solve the problem. We see in the drawing that the given straight line is parallel to the O x axis and passes through the point (0, 3).
The straight line, which is parallel to the abscissa, is determined by the incomplete general equation B y + C = 0. Let's find the values of B and C. The coordinates of the point (0, 3), since the given line passes through it, will satisfy the equation of the line B y + C = 0, then the equality is valid: B · 3 + C = 0. Let's set B to some value other than zero. Let's say B = 1, in which case from the equality B · 3 + C = 0 we can find C: C = - 3. Using the known values of B and C, we obtain the required equation of the straight line: y - 3 = 0.
Answer: y - 3 = 0 .
General equation of a line passing through a given point in a plane
Let the given line pass through the point M 0 (x 0 , y 0), then its coordinates correspond to the general equation of the line, i.e. the equality is true: A x 0 + B y 0 + C = 0. Let us subtract the left and right sides of this equation from the left and right sides of the general complete equation of the line. We get: A (x - x 0) + B (y - y 0) + C = 0, this equation is equivalent to the original general one, passes through the point M 0 (x 0, y 0) and has a normal vector n → = (A, B) .
The result that we obtained makes it possible to write down the general equation of a line with known coordinates of the normal vector of the line and the coordinates of a certain point of this line.
Example 3
Given a point M 0 (- 3, 4) through which a line passes, and the normal vector of this line n → = (1 , - 2) . It is necessary to write down the equation of the given line.
Solution
The initial conditions allow us to obtain the necessary data to compile the equation: A = 1, B = - 2, x 0 = - 3, y 0 = 4. Then:
A (x - x 0) + B (y - y 0) = 0 ⇔ 1 (x - (- 3)) - 2 y (y - 4) = 0 ⇔ ⇔ x - 2 y + 22 = 0
The problem could have been solved differently. The general equation of a straight line is A x + B y + C = 0. The given normal vector allows us to obtain the values of coefficients A and B, then:
A x + B y + C = 0 ⇔ 1 x - 2 y + C = 0 ⇔ x - 2 y + C = 0
Now let’s find the value of C using the point M 0 (- 3, 4) specified by the problem condition, through which the straight line passes. The coordinates of this point correspond to the equation x - 2 · y + C = 0, i.e. - 3 - 2 4 + C \u003d 0. Hence C = 11. The required straight line equation takes the form: x - 2 · y + 11 = 0.
Answer: x - 2 y + 11 = 0 .
Example 4
Given a line 2 3 x - y - 1 2 = 0 and a point M 0 lying on this line. Only the abscissa of this point is known, and it is equal to - 3. It is necessary to determine the ordinate of a given point.
Solution
Let us designate the coordinates of point M 0 as x 0 and y 0 . The source data indicates that x 0 = - 3. Since the point belongs to a given line, then its coordinates correspond to the general equation of this line. Then the equality will be true:
2 3 x 0 - y 0 - 1 2 = 0
Define y 0: 2 3 · (- 3) - y 0 - 1 2 = 0 ⇔ - 5 2 - y 0 = 0 ⇔ y 0 = - 5 2
Answer: - 5 2
Transition from the general equation of a line to other types of equations of a line and vice versa
As we know, there are several types of equations for the same straight line on a plane. The choice of the type of equation depends on the conditions of the problem; it is possible to choose the one that is more convenient for solving it. The skill of converting an equation of one type into an equation of another type is very useful here.
First, let's consider the transition from the general equation of the form A x + B y + C = 0 to the canonical equation x - x 1 a x = y - y 1 a y.
If A ≠ 0, then we move the term B y to the right side of the general equation. On the left side we take A out of brackets. As a result, we get: A x + C A = - B y.
This equality can be written as a proportion: x + C A - B = y A.
If B ≠ 0, we leave only the term A x on the left side of the general equation, transfer the others to the right side, we get: A x = - B y - C. We take – B out of brackets, then: A x = - B y + C B .
Let's rewrite the equality in the form of a proportion: x - B = y + C B A.
Of course, there is no need to memorize the resulting formulas. It is enough to know the algorithm of actions when moving from a general equation to a canonical one.
Example 5
The general equation of the line 3 y - 4 = 0 is given. It is necessary to transform it into a canonical equation.
Solution
Let's write the original equation as 3 y - 4 = 0. Next, we proceed according to the algorithm: the term 0 x remains on the left side; and on the right side we put - 3 out of brackets; we get: 0 x = - 3 y - 4 3 .
Let's write the resulting equality as a proportion: x - 3 = y - 4 3 0 . Thus, we have obtained an equation of canonical form.
Answer: x - 3 = y - 4 3 0.
To transform the general equation of a line into parametric ones, first a transition is made to the canonical form, and then a transition from the canonical equation of a line to parametric equations.
Example 6
The straight line is given by the equation 2 x - 5 y - 1 = 0. Write down the parametric equations for this line.
Solution
Let us make the transition from the general equation to the canonical one:
2 x - 5 y - 1 = 0 ⇔ 2 x = 5 y + 1 ⇔ 2 x = 5 y + 1 5 ⇔ x 5 = y + 1 5 2
Now we take both sides of the resulting canonical equation equal to λ, then:
x 5 = λ y + 1 5 2 = λ ⇔ x = 5 λ y = - 1 5 + 2 λ , λ ∈ R
Answer:x = 5 λ y = - 1 5 + 2 λ , λ ∈ R
The general equation can be converted into an equation of a straight line with slope y = k · x + b, but only when B ≠ 0. For the transition, we leave the term B y on the left side, the rest are transferred to the right. We get: B y = - A x - C . Let's divide both sides of the resulting equality by B, different from zero: y = - A B x - C B.
Example 7
The general equation of the line is given: 2 x + 7 y = 0. You need to convert that equation into a slope equation.
Solution
Let's perform the necessary actions according to the algorithm:
2 x + 7 y = 0 ⇔ 7 y - 2 x ⇔ y = - 2 7 x
Answer: y = - 2 7 x .
From the general equation of a line, it is enough to simply obtain an equation in segments of the form x a + y b = 1. To make such a transition, we move the number C to the right side of the equality, divide both sides of the resulting equality by – C and, finally, transfer the coefficients for the variables x and y to the denominators:
A x + B y + C = 0 ⇔ A x + B y = - C ⇔ ⇔ A - C x + B - C y = 1 ⇔ x - C A + y - C B = 1
Example 8
It is necessary to transform the general equation of the line x - 7 y + 1 2 = 0 into the equation of the line in segments.
Solution
Let's move 1 2 to the right side: x - 7 y + 1 2 = 0 ⇔ x - 7 y = - 1 2 .
Let's divide both sides of the equality by -1/2: x - 7 y = - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y = 1 .
Answer: x - 1 2 + y 1 14 = 1 .
In general, the reverse transition is also easy: from other types of equations to the general one.
The equation of a line in segments and an equation with an angular coefficient can be easily converted into a general one by simply collecting all the terms on the left side of the equality:
x a + y b ⇔ 1 a x + 1 b y - 1 = 0 ⇔ A x + B y + C = 0 y = k x + b ⇔ y - k x - b = 0 ⇔ A x + B y + C = 0
The canonical equation is converted to a general one according to the following scheme:
x - x 1 a x = y - y 1 a y ⇔ a y · (x - x 1) = a x (y - y 1) ⇔ ⇔ a y x - a x y - a y x 1 + a x y 1 = 0 ⇔ A x + B y + C = 0
To move from parametric ones, first move to the canonical one, and then to the general one:
x = x 1 + a x · λ y = y 1 + a y · λ ⇔ x - x 1 a x = y - y 1 a y ⇔ A x + B y + C = 0
Example 9
The parametric equations of the line x = - 1 + 2 · λ y = 4 are given. It is necessary to write down the general equation of this line.
Solution
Let us make the transition from parametric equations to canonical ones:
x = - 1 + 2 · λ y = 4 ⇔ x = - 1 + 2 · λ y = 4 + 0 · λ ⇔ λ = x + 1 2 λ = y - 4 0 ⇔ x + 1 2 = y - 4 0
Let's move from the canonical to the general:
x + 1 2 = y - 4 0 ⇔ 0 · (x + 1) = 2 (y - 4) ⇔ y - 4 = 0
Answer: y - 4 = 0
Example 10
The equation of a straight line in the segments x 3 + y 1 2 = 1 is given. It is necessary to transition to the general form of the equation.
Solution:
We simply rewrite the equation in the required form:
x 3 + y 1 2 = 1 ⇔ 1 3 x + 2 y - 1 = 0
Answer: 1 3 x + 2 y - 1 = 0 .
Drawing up a general equation of a line
We said above that the general equation can be written with known coordinates of the normal vector and the coordinates of the point through which the line passes. Such a straight line is defined by the equation A (x - x 0) + B (y - y 0) = 0. There we also analyzed the corresponding example.
Now let's look at more complex examples, in which first you need to determine the coordinates of the normal vector.
Example 11
Given a line parallel to the line 2 x - 3 y + 3 3 = 0. The point M 0 (4, 1) through which the given line passes is also known. It is necessary to write down the equation of the given line.
Solution
The initial conditions tell us that the lines are parallel, then, as the normal vector of the line, the equation of which needs to be written, we take the direction vector of the line n → = (2, - 3): 2 x - 3 y + 3 3 = 0. Now we know all the necessary data to create the general equation of the line:
A (x - x 0) + B (y - y 0) = 0 ⇔ 2 (x - 4) - 3 (y - 1) = 0 ⇔ 2 x - 3 y - 5 = 0
Answer: 2 x - 3 y - 5 = 0 .
Example 12
The given line passes through the origin perpendicular to the line x - 2 3 = y + 4 5. It is necessary to create a general equation for a given line.
Solution
The normal vector of a given line will be the direction vector of the line x - 2 3 = y + 4 5.
Then n → = (3, 5) . The straight line passes through the origin, i.e. through point O (0, 0). Let's create a general equation for a given line:
A (x - x 0) + B (y - y 0) = 0 ⇔ 3 (x - 0) + 5 (y - 0) = 0 ⇔ 3 x + 5 y = 0
Answer: 3 x + 5 y = 0 .
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Chemistry is the science of substances, their properties and transformations
.
That is, if nothing happens to the substances around us, then this does not apply to chemistry. But what does “nothing happens” mean? If a thunderstorm suddenly caught us in the field, and we were all wet, as they say, “to the skin,” then isn’t this a transformation: after all, the clothes were dry, but they became wet.
If, for example, you take an iron nail, file it, and then assemble iron filings (Fe) , then isn’t this also a transformation: there was a nail - it became powder. But if you then assemble the device and carry out obtaining oxygen (O 2): heat up potassium permanganate(KMpo 4) and collect oxygen in a test tube, and then place these red-hot iron filings into it, then they will flare up with a bright flame and after combustion will turn into a brown powder. And this is also a transformation. So where is the chemistry? Despite the fact that in these examples the shape (iron nail) and the condition of the clothing (dry, wet) change, these are not transformations. The fact is that the nail itself was a substance (iron), and remained so, despite its different shape, and our clothes absorbed the water from the rain and then evaporated it into the atmosphere. The water itself has not changed. So what are transformations from a chemical point of view?
From a chemical point of view, transformations are those phenomena that are accompanied by a change in the composition of a substance. Let's take the same nail as an example. It doesn’t matter what shape it took after being filed, but after the pieces collected from it iron filings placed in an oxygen atmosphere - it turned into iron oxide(Fe 2 O 3 ) . So, something has changed after all? Yes, it has changed. There was a substance called a nail, but under the influence of oxygen a new substance was formed - element oxide gland. Molecular equation This transformation can be represented by the following chemical symbols:
4Fe + 3O 2 = 2Fe 2 O 3 (1)
For someone uninitiated in chemistry, questions immediately arise. What is "molecular equation", what is Fe? Why are the numbers “4”, “3”, “2”? What are the little numbers “2” and “3” in the formula Fe 2 O 3? This means it’s time to sort everything out in order.
Signs chemical elements.
Despite the fact that chemistry begins to be studied in the 8th grade, and some even earlier, many people know the great Russian chemist D.I. Mendeleev. And of course, his famous “Periodic Table of Chemical Elements”. Otherwise, more simply, it is called the “Periodical Table”.
In this table, the elements are arranged in the appropriate order. To date, about 120 of them are known. The names of many elements have been known to us for a long time. These are: iron, aluminum, oxygen, carbon, gold, silicon. Previously, we used these words without thinking, identifying them with objects: an iron bolt, an aluminum wire, oxygen in the atmosphere, a gold ring, etc. etc. But in fact, all these substances (bolt, wire, ring) consist of their corresponding elements. The whole paradox is that the element cannot be touched or picked up. How so? They are in the periodic table, but you can’t take them! Yes exactly. A chemical element is an abstract (that is, abstract) concept, and is used in chemistry, as well as in other sciences, for calculations, drawing up equations, and solving problems. Each element differs from the other in that it has its own characteristic electronic configuration of an atom. The number of protons in the nucleus of an atom is equal to the number of electrons in its orbitals. For example, hydrogen is element No. 1. Its atom consists of 1 proton and 1 electron. Helium is element #2. Its atom consists of 2 protons and 2 electrons. Lithium is element #3. Its atom consists of 3 protons and 3 electrons. Darmstadtium – element No. 110. Its atom consists of 110 protons and 110 electrons.
Each element is designated by a certain symbol, Latin letters, and has a certain reading translated from Latin. For example, hydrogen has the symbol "N", read as "hydrogenium" or "ash". Silicon has the symbol "Si" read as "silicium". Mercury has a symbol "Hg" and is read as "hydrargyrum". And so on. All these notations can be found in any 8th grade chemistry textbook. The main thing for us now is to understand that when composing chemical equations, it is necessary to operate with the indicated symbols of the elements.
Simple and complex substances.
Denoting various substances with single symbols of chemical elements (Hg mercury, Fe iron, Cu copper, Zn zinc, Al aluminum) we essentially denote simple substances, that is, substances consisting of atoms of the same type (containing the same number of protons and neutrons in an atom). For example, if the substances iron and sulfur interact, then the equation will take the following writing form:
Fe + S = FeS (2)
Simple substances include metals (Ba, K, Na, Mg, Ag), as well as non-metals (S, P, Si, Cl 2, N 2, O 2, H 2). Moreover, one should pay attention
Special attention to the fact that all metals are designated by single symbols: K, Ba, Ca, Al, V, Mg, etc., and non-metals are either simple symbols: C, S, P or may have different indices that indicate their molecular structure: H 2, Cl 2, O 2, J 2, P 4, S 8. In the future this will have a very great importance when writing equations. It is not at all difficult to guess that complex substances are substances formed from atoms different types, For example,
1). Oxides:
aluminium oxide Al 2 O 3, 
sodium oxide Na2O,
copper oxide CuO,
zinc oxide ZnO,
titanium oxide Ti2O3,
carbon monoxide or carbon monoxide (+2) CO
sulfur oxide (+6) SO 3
2). Reasons:
iron hydroxide(+3) Fe(OH) 3,
copper hydroxide Cu(OH)2,
potassium hydroxide or alkali potassium KOH,
sodium hydroxide NaOH.
3). Acids:
hydrochloric acid HCl,
sulfurous acid H2SO3,
Nitric acid HNO3
4). Salts:
sodium thiosulfate Na 2 S 2 O 3 ,
sodium sulfate or Glauber's salt Na2SO4,
calcium carbonate or limestone CaCO 3,
copper chloride CuCl2
5). organic matter:
sodium acetate CH 3 COONa,
methane CH 4,
acetylene C 2 H 2,
glucose C 6 H 12 O 6
Finally, after we figured out the structure various substances, you can begin to compile chemical equations.
Chemical equation.
The word “equation” itself is derived from the word “equalize”, i.e. divide something into equal parts. In mathematics, equations constitute almost the very essence of this science. For example, you can give a simple equation in which the left and right sides will be equal to “2”:
40: (9 + 11) = (50 x 2) : (80 – 30);
And in chemical equations the same principle: the left and right sides of the equation must correspond to the same numbers of atoms and elements participating in them. Or, if given ionic equation, then in it number of particles must also meet this requirement. A chemical equation is a conditional representation of a chemical reaction using chemical formulas and mathematical symbols. A chemical equation inherently reflects one or another chemical reaction, that is, the process of interaction of substances, during which new substances arise. For example, it is necessary write a molecular equation reactions in which they take part barium chloride BaCl 2 and sulfuric acid H 2 SO 4. As a result of this reaction, an insoluble precipitate is formed - barium sulfate BaSO 4 and hydrochloric acid Hcl:
BaCl 2 + H 2 SO 4 = BaSO 4 + 2HCl (3)
First of all, it is necessary to understand that the large number “2” standing in front of the substance HCl is called a coefficient, and the small numbers “2”, “4” under the formulas BaCl 2, H 2 SO 4, BaSO 4 are called indices. Both coefficients and indices in chemical equations act as multipliers, not summands. To write a chemical equation correctly, you need assign coefficients in the reaction equation. Now let's start counting the atoms of the elements on the left and right sides of the equation. On the left side of the equation: the substance BaCl 2 contains 1 barium atom (Ba), 2 chlorine atoms (Cl). In the substance H 2 SO 4: 2 hydrogen atoms (H), 1 sulfur atom (S) and 4 oxygen atoms (O). On the right side of the equation: in the BaSO 4 substance there is 1 barium atom (Ba), 1 sulfur atom (S) and 4 oxygen atoms (O), in the HCl substance: 1 hydrogen atom (H) and 1 chlorine atom (Cl). It follows that on the right side of the equation the number of hydrogen and chlorine atoms is half as much as on the left side. Therefore, before the HCl formula on the right side of the equation, it is necessary to put the coefficient “2”. If we now add up the numbers of atoms of the elements participating in this reaction, both on the left and on the right, we obtain the following balance:
In both sides of the equation, the numbers of atoms of the elements participating in the reaction are equal, therefore it is composed correctly.
Chemical equation and chemical reactions
As we have already found out, chemical equations are a reflection of chemical reactions. Chemical reactions are those phenomena during which the transformation of one substance into another occurs. Among their diversity, two main types can be distinguished:
1). Compound reactions
2). Decomposition reactions.
The overwhelming majority of chemical reactions belong to addition reactions, since changes in its composition can rarely occur with an individual substance if it is not exposed to external influences (dissolution, heating, exposure to light). Nothing characterizes a chemical phenomenon or reaction better than the changes that occur during the interaction of two or more substances. Such phenomena can occur spontaneously and be accompanied by an increase or decrease in temperature, light effects, color changes, sediment formation, release of gaseous products, and noise.
For clarity, we present several equations reflecting the processes of compound reactions, during which we obtain sodium chloride(NaCl), zinc chloride(ZnCl2), silver chloride precipitate(AgCl), aluminum chloride(AlCl 3)
Cl 2 + 2Nа = 2NaCl (4)
CuCl 2 + Zn = ZnCl 2 + Cu (5)
AgNO 3 + KCl = AgCl + 2KNO 3 (6)
3HCl + Al(OH) 3 = AlCl 3 + 3H 2 O (7)
Among the reactions of the compound, special mention should be made of the following: : substitution (5), exchange (6), and as a special case of an exchange reaction - the reaction neutralization (7).
Substitution reactions include those in which atoms of a simple substance replace atoms of one of the elements in a complex substance. In example (5), zinc atoms replace copper atoms from the CuCl 2 solution, while zinc passes into the soluble salt ZnCl 2, and copper is released from the solution in the metallic state.
Exchange reactions include those reactions in which two complex substances exchange their constituent parts. In the case of reaction (6), the soluble salts AgNO 3 and KCl, when both solutions are merged, form an insoluble precipitate of the AgCl salt. At the same time, they exchange their constituent parts - cations and anions. Potassium cations K + are added to the NO 3 anions, and silver cations Ag + are added to the Cl - anions.
A special, special case of exchange reactions is the neutralization reaction. Neutralization reactions include those reactions in which acids react with bases, resulting in the formation of salt and water. In example (7), hydrochloric acid HCl reacts with the base Al(OH) 3 to form the salt AlCl 3 and water. In this case, aluminum cations Al 3+ from the base are exchanged with Cl - anions from the acid. What happens in the end neutralization of hydrochloric acid.
Decomposition reactions include those in which two or more new simple or complex substances, but of a simpler composition, are formed from one complex substance. Examples of reactions include those in the process of which 1) decomposes. potassium nitrate(KNO 3) with the formation of potassium nitrite (KNO 2) and oxygen (O 2); 2). Potassium permanganate(KMnO 4): potassium manganate (K 2 MnO 4) is formed, manganese oxide(MnO 2) and oxygen (O 2); 3). Calcium carbonate or marble; in the process are formed carbonicgas(CO2) and calcium oxide(CaO)
2KNO 3 = 2KNO 2 + O 2 (8)
2KMnO 4 = K 2 MnO 4 + MnO 2 + O 2 (9)
CaCO 3 = CaO + CO 2 (10)
In reaction (8), one complex and one simple substance are formed from a complex substance. In reaction (9) there are two complex and one simple. In reaction (10) there are two complex substances, but simpler in composition
All classes of complex substances are subject to decomposition:
1). Oxides: silver oxide 2Ag 2 O = 4Ag + O 2 (11)
2). Hydroxides: iron hydroxide 2Fe(OH) 3 = Fe 2 O 3 + 3H 2 O (12)
3). Acids: sulfuric acid H 2 SO 4 = SO 3 + H 2 O (13)
4). Salts: calcium carbonate CaCO 3 = CaO + CO 2 (14)
5). Organic matter: alcoholic fermentation of glucose
C 6 H 12 O 6 = 2C 2 H 5 OH + 2CO 2 (15)
According to another classification, all chemical reactions can be divided into two types: reactions that release heat are called exothermic, and reactions that occur with the absorption of heat - endothermic. The criterion for such processes is thermal effect of the reaction. As a rule, exothermic reactions include oxidation reactions, i.e. interaction with oxygen, for example methane combustion:
CH 4 + 2O 2 = CO 2 + 2H 2 O + Q (16)
and to endothermic reactions - decomposition reactions already given above (11) - (15). The Q sign at the end of the equation indicates whether heat is released (+Q) or absorbed (-Q) during the reaction:
CaCO 3 = CaO+CO 2 - Q (17)
You can also consider all chemical reactions according to the type of change in the degree of oxidation of the elements involved in their transformations. For example, in reaction (17), the elements participating in it do not change their oxidation states:
Ca +2 C +4 O 3 -2 = Ca +2 O -2 +C +4 O 2 -2 (18)
And in reaction (16), the elements change their oxidation states:
2Mg 0 + O 2 0 = 2Mg +2 O -2
Reactions of this type are redox . They will be considered separately. To compose equations for reactions of this type, you must use half-reaction method and apply electronic balance equation.
After presenting the various types of chemical reactions, you can proceed to the principle of composing chemical equations, or, in other words, selecting coefficients on the left and right sides.
Mechanisms for composing chemical equations.
Whatever type this or that chemical reaction, its recording (chemical equation) must correspond to the condition of equality of the number of atoms before the reaction and after the reaction.
There are equations (17) that do not require equalization, i.e. placement of coefficients. But in most cases, as in examples (3), (7), (15), it is necessary to take actions aimed at equalizing the left and right sides of the equation. What principles should be followed in such cases? Is there any system for selecting odds? There is, and not only one. Such systems include:
1). Selection of coefficients according to given formulas.
2). Compilation by valences of reacting substances.
3). Arrangement of reacting substances according to oxidation states.
In the first case, it is assumed that we know the formulas of the reacting substances both before and after the reaction. For example, given the following equation:
N 2 + O 2 →N 2 O 3 (19)
It is generally accepted that until equality is established between the atoms of the elements before and after the reaction, the equal sign (=) is not placed in the equation, but is replaced by an arrow (→). Now let's get down to the actual adjustment. On the left side of the equation there are 2 nitrogen atoms (N 2) and two oxygen atoms (O 2), and on the right side there are two nitrogen atoms (N 2) and three oxygen atoms (O 3). There is no need to equalize it in terms of the number of nitrogen atoms, but in terms of oxygen it is necessary to achieve equality, since before the reaction there were two atoms involved, and after the reaction there were three atoms. Let's make the following diagram:
before reaction after reaction
O 2 O 3
Let's determine the smallest multiple between the given numbers of atoms, it will be “6”.
O 2 O 3
\ 6 /
Let's divide this number on the left side of the oxygen equation by “2”. We get the number “3” and put it into the equation to be solved:
N 2 + 3O 2 →N 2 O 3
We also divide the number “6” for the right side of the equation by “3”. We get the number “2”, and also put it in the equation to be solved:
N 2 + 3O 2 → 2N 2 O 3
The numbers of oxygen atoms on both the left and right sides of the equation became equal, respectively, 6 atoms each:
But the number of nitrogen atoms on both sides of the equation will not correspond to each other:
The left one has two atoms, the right one has four atoms. Therefore, in order to achieve equality, it is necessary to double the amount of nitrogen on the left side of the equation, setting the coefficient to “2”:
Thus, equality in nitrogen is observed and, in general, the equation takes the form:
2N 2 + 3О 2 → 2N 2 О 3
Now in the equation you can put an equal sign instead of an arrow:
2N 2 + 3О 2 = 2N 2 О 3 (20)
Let's give another example. The following reaction equation is given:
P + Cl 2 → PCl 5
On the left side of the equation there is 1 phosphorus atom (P) and two chlorine atoms (Cl 2), and on the right side there is one phosphorus atom (P) and five oxygen atoms (Cl 5). There is no need to equalize it in terms of the number of phosphorus atoms, but in terms of chlorine it is necessary to achieve equality, since before the reaction there were two atoms involved, and after the reaction there were five atoms. Let's make the following diagram:
before reaction after reaction
Cl 2 Cl 5
Let's determine the smallest multiple between the given numbers of atoms, it will be “10”.
Cl 2 Cl 5
\ 10 /
Divide this number on the left side of the chlorine equation by “2”. Let’s get the number “5” and put it into the equation to be solved:
P + 5Cl 2 → PCl 5
We also divide the number “10” for the right side of the equation by “5”. We get the number “2”, and also put it in the equation to be solved:
P + 5Cl 2 → 2РCl 5
The numbers of chlorine atoms on both the left and right sides of the equation became equal, respectively, 10 atoms each:
But the number of phosphorus atoms on both sides of the equation will not correspond to each other:
Therefore, in order to achieve equality, it is necessary to double the amount of phosphorus on the left side of the equation by setting the coefficient “2”:
Thus, equality for phosphorus is observed and, in general, the equation takes the form:
2Р + 5Cl 2 = 2РCl 5 (21)
When composing equations by valencies must be given valency determination and set values for the most famous elements. Valence is one of the previously used concepts, currently in a number of school programs not used. But with its help it is easier to explain the principles of drawing up equations of chemical reactions. Valence is understood as number chemical bonds, which one or another atom can form with another, or other atoms . Valency does not have a sign (+ or -) and is indicated by Roman numerals, usually above the symbols of chemical elements, for example:
Where do these values come from? How to use them when writing chemical equations? The numerical values of the valences of the elements coincide with their group number of the Periodic Table of Chemical Elements by D.I. Mendeleev (Table 1).
For other elements valence values may have other values, but never greater than the number of the group in which they are located. Moreover, for even group numbers (IV and VI), the valences of elements take only even values, and for odd ones they can have both even and odd values (Table 2).
Of course, there are exceptions to the valence values for some elements, but in each specific case these points are usually specified. Now let's consider general principle compiling chemical equations based on given valences for certain elements. Most often, this method is acceptable in the case of drawing up equations of chemical reactions of compounds of simple substances, for example, when interacting with oxygen ( oxidation reactions). Let's say you need to display an oxidation reaction aluminum. But let us recall that metals are designated by single atoms (Al), and non-metals in the gaseous state are designated by the indices “2” - (O 2). First we'll write general scheme reactions:
Al + О 2 →AlО
At this stage it is not yet known which correct writing should be aluminum oxide. And it is precisely at this stage that knowledge of the valences of elements will come to our aid. For aluminum and oxygen, let’s put them above the expected formula of this oxide:
III II
Al O
After that, “cross”-on-“cross” for these element symbols we will put the corresponding indices at the bottom:
III II
Al 2 O 3
Composition of a chemical compound Al 2 O 3 determined. The further diagram of the reaction equation will take the form:
Al+ O 2 →Al 2 O 3
All that remains is to equalize its left and right parts. Let us proceed in the same way as in the case of composing equation (19). Let's equalize the numbers of oxygen atoms by finding the smallest multiple:
before reaction after reaction
O 2 O 3
\ 6 /
Let's divide this number on the left side of the oxygen equation by “2”. Let’s get the number “3” and put it into the equation being solved. We also divide the number “6” for the right side of the equation by “3”. We get the number “2”, and also put it in the equation to be solved:
Al + 3O 2 → 2Al 2 O 3
To achieve equality in aluminum, it is necessary to adjust its quantity on the left side of the equation by setting the coefficient to “4”:
4Al + 3O 2 → 2Al 2 O 3
Thus, equality for aluminum and oxygen is observed and, in general, the equation will take its final form:
4Al + 3O 2 = 2Al 2 O 3 (22)
Using the valence method, you can predict what substance is formed during a chemical reaction and what its formula will look like. Let’s assume that the compound reacted with nitrogen and hydrogen with the corresponding valences III and I. Let’s write the general reaction scheme:
N 2 + N 2 → NH
For nitrogen and hydrogen, let’s put the valencies above the expected formula of this compound:
As before, “cross”-on-“cross” for these element symbols, let’s put the corresponding indices below:
III I
NH 3
The further diagram of the reaction equation will take the form:
N 2 + N 2 → NH 3
Equating in a well-known way, through the smallest multiple for hydrogen equal to “6”, we obtain the required coefficients and the equation as a whole:
N 2 + 3H 2 = 2NH 3 (23)
When composing equations according to oxidation states reactants, it is necessary to recall that the oxidation state of a particular element is the number of electrons accepted or given up during a chemical reaction. Oxidation state in compounds Basically, it numerically coincides with the valence values of the element. But they differ in sign. For example, for hydrogen, the valence is I, and the oxidation state is (+1) or (-1). For oxygen, the valence is II, and the oxidation state is -2. For nitrogen, the valences are I, II, III, IV, V, and the oxidation states are (-3), (+1), (+2), (+3), (+4), (+5), etc. . The oxidation states of the elements most often used in equations are given in Table 3.
In the case of compound reactions, the principle of compiling equations by oxidation states is the same as when compiling by valences. For example, let us give the equation for the oxidation of chlorine with oxygen, in which chlorine forms a compound with an oxidation state of +7. Let's write down the proposed equation:
Cl 2 + O 2 → ClO
Let us place the oxidation states of the corresponding atoms over the proposed compound ClO:
As in previous cases, we establish that the required compound formula will take the form:
7 -2
Cl 2 O 7
The reaction equation will take the following form:
Cl 2 + O 2 → Cl 2 O 7
Equating for oxygen, finding the smallest multiple between two and seven, equal to “14,” we ultimately establish the equality:
2Cl 2 + 7O 2 = 2Cl 2 O 7 (24)
A slightly different method must be used with oxidation states when composing exchange, neutralization, and substitution reactions. In some cases, it is difficult to find out: what compounds are formed during the interaction of complex substances?
How to find out: what will happen in the reaction process?
Indeed, how do you know what reaction products may arise during a particular reaction? For example, what is formed when barium nitrate and potassium sulfate react?
Ba(NO 3) 2 + K 2 SO 4 → ?
Maybe BaK 2 (NO 3) 2 + SO 4? Or Ba + NO 3 SO 4 + K 2? Or something else? Of course, during this reaction the following compounds are formed: BaSO 4 and KNO 3. How is this known? And how to write the formulas of substances correctly? Let's start with what is most often overlooked: the very concept of “exchange reaction.” This means that in these reactions substances change their constituent parts with each other. Since exchange reactions are mostly carried out between bases, acids or salts, the parts with which they will be exchanged are metal cations (Na +, Mg 2+, Al 3+, Ca 2+, Cr 3+), H + ions or OH -, anions - acid residues, (Cl -, NO 3 2-, SO 3 2-, SO 4 2-, CO 3 2-, PO 4 3-). IN general view The exchange reaction can be given in the following notation:
Kt1An1 + Kt2An1 = Kt1An2 + Kt2An1 (25)
Where Kt1 and Kt2 are metal cations (1) and (2), and An1 and An2 are their corresponding anions (1) and (2). In this case, it is necessary to take into account that in compounds before and after the reaction, cations are always installed in first place, and anions are in second place. Therefore, if the reaction occurs potassium chloride And silver nitrate, both in dissolved state
KCl + AgNO 3 →
then in its process the substances KNO 3 and AgCl are formed and the corresponding equation will take the form:
KCl + AgNO 3 =KNO 3 + AgCl (26)
During neutralization reactions, protons from acids (H +) will combine with hydroxyl anions (OH -) to form water (H 2 O):
HCl + KOH = KCl + H 2 O (27)
The oxidation states of metal cations and the charges of anions of acidic residues are indicated in the table of solubility of substances (acids, salts and bases in water). The horizontal line shows metal cations, and the vertical line shows the anions of acid residues.
Based on this, when drawing up an equation for an exchange reaction, it is first necessary to establish on the left side the oxidation states of the particles receiving in this chemical process. For example, you need to write an equation for the interaction between calcium chloride and sodium carbonate. Let’s create the initial diagram of this reaction:
CaCl + NaCO 3 →
Ca 2+ Cl - + Na + CO 3 2- →
Having performed the already known “cross”-on-“cross” action, we determine the real formulas of the starting substances:
CaCl 2 + Na 2 CO 3 →
Based on the principle of exchange of cations and anions (25), we will establish preliminary formulas for the substances formed during the reaction:
CaCl 2 + Na 2 CO 3 → CaCO 3 + NaCl
Let us place the corresponding charges above their cations and anions:
Ca 2+ CO 3 2- + Na + Cl -
Substance formulas written correctly, in accordance with the charges of cations and anions. Let's create a complete equation, equalizing its left and right sides for sodium and chlorine:
CaCl 2 + Na 2 CO 3 = CaCO 3 + 2NaCl (28)
As another example, here is the equation for the neutralization reaction between barium hydroxide and phosphoric acid:
VaON + NPO 4 →
Let us place the corresponding charges over the cations and anions:
Ba 2+ OH - + H + PO 4 3- →
Let's determine the real formulas of the starting substances:
Ba(OH) 2 + H 3 PO 4 →
Based on the principle of exchange of cations and anions (25), we will establish preliminary formulas for the substances formed during the reaction, taking into account that during an exchange reaction one of the substances must necessarily be water:
Ba(OH) 2 + H 3 PO 4 → Ba 2+ PO 4 3- + H 2 O
Let us determine the correct notation for the formula of the salt formed during the reaction:
Ba(OH) 2 + H 3 PO 4 → Ba 3 (PO 4) 2 + H 2 O
Let's equalize the left side of the equation for barium:
3Ba (OH) 2 + H 3 PO 4 → Ba 3 (PO 4) 2 + H 2 O
Since on the right side of the equation the orthophosphoric acid residue is taken twice, (PO 4) 2, then on the left it is also necessary to double its amount:
3Ba (OH) 2 + 2H 3 PO 4 → Ba 3 (PO 4) 2 + H 2 O
It remains to match the number of hydrogen and oxygen atoms on the right side of water. Since on the left the total number of hydrogen atoms is 12, on the right it must also correspond to twelve, therefore before the formula of water it is necessary set the coefficient“6” (since the water molecule already has 2 hydrogen atoms). For oxygen, equality is also observed: on the left is 14 and on the right is 14. So, the equation has the correct written form:
3Ba (OH) 2 + 2H 3 PO 4 → Ba 3 (PO 4) 2 + 6H 2 O (29)
Possibility of chemical reactions
The world consists of a great variety of substances. The number of variants of chemical reactions between them is also incalculable. But can we, having written this or that equation on paper, say that a chemical reaction will correspond to it? There is a misconception that if it is correct set the odds in the equation, then it will be feasible in practice. For example, if we take sulfuric acid solution and put it in it zinc, then you can observe the process of hydrogen evolution:
Zn+ H 2 SO 4 = ZnSO 4 + H 2 (30)
But if copper is dropped into the same solution, then the process of gas evolution will not be observed. The reaction is not feasible.
Cu+ H 2 SO 4 ≠
If concentrated sulfuric acid is taken, it will react with copper:
Cu + 2H 2 SO 4 = CuSO 4 + SO 2 + 2H 2 O (31)
In reaction (23) between the gases nitrogen and hydrogen, we observe thermodynamic equilibrium, those. how many molecules ammonia NH 3 is formed per unit time, the same amount of them will decompose back into nitrogen and hydrogen. Shift in chemical equilibrium can be achieved by increasing pressure and decreasing temperature
N 2 + 3H 2 = 2NH 3
If you take potassium hydroxide solution and pour it on him sodium sulfate solution, then no changes will be observed, the reaction will not be feasible:
KOH + Na 2 SO 4 ≠
Sodium chloride solution when interacting with bromine, it will not form bromine, despite the fact that this reaction can be classified as a substitution reaction:
NaCl + Br 2 ≠
What are the reasons for such discrepancies? The point is that it is not enough just to correctly determine compound formulas, it is necessary to know the specifics of the interaction of metals with acids, skillfully use the table of solubility of substances, and know the rules of substitution in the activity series of metals and halogens. This article outlines only the most basic principles of how assign coefficients in reaction equations, How write molecular equations, How determine the composition of a chemical compound.
Chemistry, as a science, is extremely diverse and multifaceted. The above article reflects only a small part of the processes occurring in real world. Types, thermochemical equations, electrolysis, processes of organic synthesis and much, much more. But more on that in future articles.
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To compose an equation means to express in mathematical form the relationship between the data (known) of a problem and its desired (unknown) quantities. Sometimes this connection is so clearly contained in the formulation of the problem that drawing up an equation is simply a literal retelling of the problem, in the language of mathematical symbols.
Example 1. Petrov received 160 rubles for work. more than half the amount that Ivanov received. Together they received 1120 rubles. How much did Petrov and Ivanov receive for their work? Let's denote Ivanov's earnings by x. Half of his earnings are 0.5x; Petrov’s monthly earnings are 0.5x + 160; together they earn 1120 rubles; the mathematical notation of the last phrase would be
(0.5x + 160) + x = 1120.
The equation is complete. Solving it according to the once established rules, we find that Ivanov’s earnings x = 640 rubles; Petrov’s earnings are 0.5x + 160 = 480 (rub.).
More often than not, it happens that the connection between the data and the required quantities is not directly indicated in the problem; it must be established based on the conditions of the problem. IN practical problems This is what happens almost always. The example just given is far-fetched; in life one almost never encounters such problems.
Therefore, completely comprehensive instructions cannot be given for composing the equation. However, at first it is useful to be guided by the following. Let us take some randomly chosen number (or several numbers) as the value of the desired quantity (or several quantities) and set ourselves the task of checking whether we have guessed the correct solution to the problem or not. If we were able to carry out this check and discover either that our guess is correct, or that it is incorrect (the latter is most likely to happen, of course), then we can immediately create the necessary equation (or several equations). Namely, we will write down the very actions that we performed for verification, only instead of a randomly chosen number we will enter an alphabetic sign of an unknown value. We get the required equation.
Example 2. A piece of copper and zinc alloy with a volume of 1 dm3 weighs 8.14 kg. How much copper is in the alloy? (specific weight of copper 8.9 kg/dm3; zinc - 7.0 kg/dm3).
Let's take a number at random that expresses the required volume of copper, for example 0.3 dm3. Let's check if we got this number successfully. Since 1 kg/dm3 of copper weighs 8.9 kg, then 0.3 dm3 weighs 8.9 * 0.3 = 2.67 (kg). The volume of zinc in the alloy is 1 - 0.3 = 0.7 (dm3). Its weight is 7.0 0.7 = 4.9 (kg). The total weight of zinc and copper is 2.67 + + 4.9 = 7.57 (kg). Meanwhile, the weight of our piece, according to the conditions of the problem, is 8.14 kg. Our guess is unfounded. But then we will immediately get an equation whose solution will give the correct answer. Instead of the randomly chosen number 0.3 dm3, let’s denote the volume of copper (in dm3) by x. Instead of the product 8.9 0.3 = 2.67, we take the products 8.9 x. This is the weight of copper in the alloy. Instead of 1 - 0.3 = 0.7 we take 1 - x; this is the volume of zinc. Instead of 7.0 0.7 = 4.9 we take 7.0 (1 - x); this is the weight of zinc. Instead of 2.67 + 4.9 we take 8.9 x + 7.0 (1 - x); this is the total weight of zinc and copper. According to the condition, it is equal to 8.14 kg; that means 8.9 x + 7.0 (1 - x) = 8.14.
Solving this equation gives x = 0.6. Checking a random decision can be done in various ways; Accordingly, it is possible to obtain different types of equations for the same problem; all of them, however, will give the same solution for the desired value; such equations are called equivalent to each other.
Of course, after gaining skills in composing equations, there is no need to check a number taken at random: for the value of the desired quantity, you can take not a number, but some letter (x, y, etc.) and act as if this letter ( unknown) was the number we are going to check.
Properties of a straight line in Euclidean geometry.
An infinite number of straight lines can be drawn through any point.
Through any two non-coinciding points a single straight line can be drawn.
Two divergent lines in a plane either intersect at a single point or are
parallel (follows from the previous one).
IN three-dimensional space there are three options relative position two straight lines:
- lines intersect;
- lines are parallel;
- straight lines intersect.
Straight line— algebraic curve of the first order: a straight line in the Cartesian coordinate system
is given on the plane by an equation of the first degree (linear equation).
General equation of a straight line.
Definition. Any straight line on the plane can be specified by a first-order equation
Ax + Wu + C = 0,
and constant A, B are not equal to zero at the same time. This first order equation is called general
equation of a straight line. Depending on the values of the constants A, B And WITH The following special cases are possible:
. C = 0, A ≠0, B ≠ 0- a straight line passes through the origin
. A = 0, B ≠0, C ≠0 (By + C = 0)- straight line parallel to the axis Oh
. B = 0, A ≠0, C ≠ 0 (Ax + C = 0)- straight line parallel to the axis OU
. B = C = 0, A ≠0- the straight line coincides with the axis OU
. A = C = 0, B ≠0- the straight line coincides with the axis Oh
The equation of a straight line can be presented in different forms depending on any given
initial conditions.
Equation of a straight line from a point and a normal vector.
Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)
perpendicular to the line given by the equation
Ax + Wu + C = 0.
Example. Find the equation of a line passing through a point A(1, 2) perpendicular to the vector (3, -1).
Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C
Let's substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore
C = -1. Total: the required equation: 3x - y - 1 = 0.
Equation of a line passing through two points.
Let two points be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2, z 2), Then straight line equation,
passing through these points:
If any of the denominators is zero, the corresponding numerator should be set equal to zero. On
plane, the equation of the straight line written above is simplified:
If x 1 ≠ x 2 And x = x 1, If x 1 = x 2 .
Fraction = k called slope straight.
Example. Find the equation of the line passing through points A(1, 2) and B(3, 4).
Solution. Applying the formula written above, we get:
Equation of a straight line using a point and slope.
If the general equation of the line Ax + Wu + C = 0 lead to:
and designate 
, then the resulting equation is called
equation of a straight line with slope k.
Equation of a straight line from a point and a direction vector.
By analogy with the point considering the equation of a straight line through the normal vector, you can enter the task
a straight line through a point and a directing vector of a straight line.
Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition
Aα 1 + Bα 2 = 0 called directing vector of a straight line.
Ax + Wu + C = 0.
Example. Find the equation of a straight line with a direction vector (1, -1) and passing through the point A(1, 2).
Solution. We will look for the equation of the desired line in the form: Ax + By + C = 0. According to the definition,
coefficients must satisfy the following conditions:
1 * A + (-1) * B = 0, i.e. A = B.
Then the equation of the straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0.
at x = 1, y = 2 we get C/A = -3, i.e. required equation:
x + y - 3 = 0
Equation of a straight line in segments.
If in the general equation of the straight line Ах + Ву + С = 0 С≠0, then, dividing by -С, we get:
or where
Geometric meaning coefficients is that coefficient a is the coordinate of the intersection point
straight with axis Oh, A b- coordinate of the point of intersection of the line with the axis OU.
Example. The general equation of a straight line is given x - y + 1 = 0. Find the equation of this line in segments.
C = 1, , a = -1, b = 1.
Normal equation of a line.
If both sides of the equation Ax + Wu + C = 0 divide by number 
, which is called
normalizing factor, then we get
xcosφ + ysinφ - p = 0 -normal equation of a line.
The sign ± of the normalizing factor must be chosen so that μ*C< 0.
R- the length of the perpendicular dropped from the origin to the straight line,
A φ - the angle formed by this perpendicular with the positive direction of the axis Oh.
Example. The general equation of the line is given 12x - 5y - 65 = 0. Required to write Various types equations
this straight line.
The equation of this line in segments:
The equation of this line with the slope: (divide by 5)
Equation of a line:
cos φ = 12/13; sin φ= -5/13; p = 5.
It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines,
parallel to the axes or passing through the origin.
The angle between straight lines on a plane.
Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2, That sharp corner between these lines
will be defined as
Two lines are parallel if k 1 = k 2. Two straight lines are perpendicular,
If k 1 = -1/ k 2 .
Theorem.
Direct Ax + Wu + C = 0 And A 1 x + B 1 y + C 1 = 0 parallel when the coefficients are proportional
A 1 = λA, B 1 = λB. If also С 1 = λС, then the lines coincide. Coordinates of the point of intersection of two lines
are found as a solution to the system of equations of these lines.
The equation of a line passing through a given point perpendicular to a given line.
Definition. Line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b
represented by the equation:
Distance from a point to a line.
Theorem. If a point is given M(x 0, y 0), then the distance to the straight line Ax + Wu + C = 0 defined as:
Proof. Let the point M 1 (x 1, y 1)- the base of a perpendicular dropped from a point M for a given
direct. Then the distance between points M And M 1:
(1)
Coordinates x 1 And at 1 can be found as a solution to the system of equations:
The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicularly
given 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.