If the angle is acute then the coefficient. The equation of the tangent to the graph of the function. Comprehensive Guide (2019)
In the previous chapter it was shown that, by choosing a certain coordinate system on the plane, we can analytically express the geometric properties characterizing the points of the line under consideration by the equation between the current coordinates. Thus, we get the equation of the line. In this chapter, the equations of straight lines will be considered.
To formulate the equation of a straight line in Cartesian coordinates, you need to somehow set the conditions that determine its position relative to the coordinate axes.
First, we introduce the concept of the slope of a straight line, which is one of the quantities characterizing the position of a straight line on a plane.
Let's call the angle of inclination of the line to the Ox axis the angle by which the Ox axis must be rotated so that it coincides with the given line (or turns out to be parallel to it). As usual, we will consider the angle taking into account the sign (the sign is determined by the direction of rotation: counterclockwise or clockwise). Since an additional rotation of the Ox axis by an angle of 180 ° will again combine it with the straight line, the angle of inclination of the straight line to the axis can be chosen ambiguously (up to a multiple of ).
The tangent of this angle is uniquely determined (since changing the angle to does not change its tangent).
The tangent of the angle of inclination of a straight line to the x-axis is called the slope of the straight line.
Slope characterizes the direction of the straight line (here we do not distinguish between two mutually opposite directions of the straight line). If the slope is straight zero, then the line is parallel to the x-axis. With a positive slope, the angle of inclination of the straight line to the x-axis will be acute (we consider here the smallest positive value tilt angle) (Fig. 39); in this case, the larger the slope, the greater the angle of its inclination to the Ox axis. If the slope is negative, then the angle of inclination of the straight line to the x-axis will be obtuse (Fig. 40). Note that a straight line perpendicular to the x-axis does not have a slope (the tangent of an angle does not exist).
The line y \u003d f (x) will be tangent to the graph shown in the figure at the point x0 if it passes through the point with coordinates (x0; f (x0)) and has a slope f "(x0). Find such a coefficient, knowing the features of the tangent, it is not difficult.
You will need
- - mathematical reference book;
- - a simple pencil;
- - notebook;
- - protractor;
- - compass;
- - pen.
Instruction
If the value f‘(x0) does not exist, then either there is no tangent, or it passes vertically. In view of this, the presence of the derivative of the function at the point x0 is due to the existence of a non-vertical tangent that is in contact with the graph of the function at the point (x0, f(x0)). In this case, the slope of the tangent will be equal to f "(x0). Thus, it becomes clear geometric meaning derivative - calculation of the slope of the tangent.
Draw on additional tangents that would be in contact with the function graph at points x1, x2 and x3, and also mark the angles formed by these tangents with the abscissa axis (such an angle is counted in the positive direction from the axis to the tangent line). For example, the angle, that is, α1, will be acute, the second (α2) is obtuse, and the third (α3) is zero, since the tangent line is parallel to the OX axis. In this case, the tangent of an obtuse angle is negative, the tangent of an acute angle is positive, and for tg0 the result is zero.
note
Correctly determine the angle formed by the tangent. To do this, use a protractor.
Two oblique lines will be parallel if their slopes are equal to each other; perpendicular if the product of the slopes of these tangents is -1.
Sources:
- Tangent to function graph
Cosine, like sine, is referred to as "direct" trigonometric functions. The tangent (together with the cotangent) is added to another pair called "derivatives". There are several definitions of these functions that make it possible to find the tangent given by known value cosine of the same value.
Instruction
Subtract the quotient from unity by the cosine of the given angle raised to the value, and extract the square root from the result - this will be the value of the tangent from the angle, expressed by its cosine: tg (α) \u003d √ (1-1 / (cos (α)) ²) . At the same time, pay attention to the fact that in the formula the cosine is in the denominator of the fraction. The impossibility of dividing by zero excludes the use of this expression for angles equal to 90°, as well as differing from this value by multiples of 180° (270°, 450°, -90°, etc.).
There is also alternative way calculating the tangent from the known value of the cosine. It can be used if there is no restriction on the use of other . To implement this method, first determine the angle value from a known cosine value - this can be done using the arccosine function. Then simply calculate the tangent for the angle of the resulting value. In general, this algorithm can be written as follows: tg(α)=tg(arccos(cos(α))).
There is another exotic option using the definition of cosine and tangent through the acute angles of a right triangle. The cosine in this definition corresponds to the ratio of the length of the leg adjacent to the considered angle to the length of the hypotenuse. Knowing the value of the cosine, you can choose the lengths of these two sides corresponding to it. For example, if cos(α)=0.5, then the adjacent can be taken equal to 10 cm, and the hypotenuse - 20 cm. Specific numbers do not matter here - you will get the same and correct with any values \u200b\u200bthat have the same. Then, using the Pythagorean theorem, determine the length of the missing side - the opposite leg. She will be equal square root from the difference between the lengths of the squared hypotenuse and the known leg: √(20²-10²)=√300. By definition, the tangent corresponds to the ratio of the lengths of the opposite and adjacent legs (√300/10) - calculate it and get the tangent value found using the classical definition of cosine.
Sources:
- cosine through tangent formula
One of trigonometric functions, most often denoted by the letters tg, although the designations tan are also found. The easiest way is to represent the tangent as the ratio of the sine corner to its cosine. This is an odd periodic and not continuous function, each cycle of which is equal to the number Pi, and the break point corresponds to half that number.
In mathematics, one of the parameters describing the position of a straight line on the Cartesian coordinate plane is the slope of this straight line. This parameter characterizes the slope of the straight line to the x-axis. To understand how to find the slope, first recall the general form of the equation of a straight line in the XY coordinate system.
In general, any line can be represented by the expression ax+by=c, where a, b and c are arbitrary real numbers, but necessarily a 2 + b 2 ≠ 0.
With the help of simple transformations, such an equation can be brought to the form y=kx+d, in which k and d are real numbers. The number k is a slope, and the equation of a straight line of this kind is called an equation with a slope. It turns out that to find the slope, you just need to bring the original equation to the above form. For a better understanding, consider a specific example:
Task: Find the slope of the line given by the equation 36x - 18y = 108
Solution: Let's transform the original equation.
Answer: The desired slope of this line is 2.
If, during the transformation of the equation, we obtained an expression of the type x = const and as a result we cannot represent y as a function of x, then we are dealing with a straight line parallel to the X axis. The slope of such a straight line is equal to infinity.
For lines that are expressed by an equation like y = const, the slope is zero. This is typical for straight lines parallel to the x-axis. For example:
Task: Find the slope of the line given by the equation 24x + 12y - 4(3y + 7) = 4
Solution: We bring the original equation to a general form
24x + 12y - 12y + 28 = 4
It is impossible to express y from the resulting expression, therefore, the slope of this straight line is equal to infinity, and the straight line itself will be parallel to the Y axis.
geometric sense
For a better understanding, let's look at the picture:
In the figure, we see a graph of a function of the type y = kx. To simplify, we take the coefficient c = 0. In the triangle OAB, the ratio of the side BA to AO will be equal to the slope k. At the same time, the ratio VA/AO is the tangent of an acute angle α in right triangle OAV. It turns out that the slope of a straight line is equal to the tangent of the angle that this straight line makes with the x-axis of the coordinate grid.
Solving the problem of how to find the slope of a straight line, we find the tangent of the angle between it and the x-axis of the coordinate grid. The boundary cases, when the line under consideration is parallel to the coordinate axes, confirm the above. Indeed, for a straight line described by the equation y=const, the angle between it and the x-axis is equal to zero. The tangent of the zero angle is also zero and the slope is also zero.
For straight lines perpendicular to the x-axis and described by the equation x=const, the angle between them and the x-axis is 90 degrees. Tangent right angle is equal to infinity, and the slope of similar straight lines is equal to infinity, which confirms what was written above.
Tangent Slope
A common, often encountered in practice, task is also to find the slope of the tangent to the function graph at some point. The tangent is a straight line, therefore the concept of slope is also applicable to it.
To figure out how to find the slope of a tangent, we will need to recall the concept of a derivative. The derivative of any function at some point is a constant, numerically equal to tangent the angle that is formed between the tangent at the specified point to the graph of this function and the abscissa axis. It turns out that to determine the slope of the tangent at the point x 0, we need to calculate the value of the derivative of the original function at this point k \u003d f "(x 0). Let's consider an example:
Task: Find the slope of the line tangent to the function y = 12x 2 + 2xe x at x = 0.1.
Solution: Find the derivative of the original function in general form
y "(0,1) = 24 . 0.1 + 2 . 0.1 . e 0.1 + 2 . e 0.1
Answer: The desired slope at the point x \u003d 0.1 is 4.831
The continuation of the topic of the equation of a straight line on a plane is based on the study of a straight line from algebra lessons. This article gives generalized information on the topic of the equation of a straight line with a slope. Consider the definitions, get the equation itself, reveal the connection with other types of equations. Everything will be discussed on examples of problem solving.
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Before writing such an equation, it is necessary to define the angle of inclination of a straight line to the O x axis with their slope. Let us assume that a Cartesian coordinate system O x is given on the plane.
Definition 1
The angle of inclination of the straight line to the axis O x, located in the Cartesian coordinate system O x y on the plane, this is the angle that is measured from the positive direction O x to the straight line counterclockwise.
When a line is parallel to Ox or coincidence occurs in it, the angle of inclination is 0. Then the angle of inclination of the given straight line α is defined on the interval [ 0 , π) .
Definition 2
Slope of a straight line is the tangent of the slope of the given line.
The standard notation is k. From the definition we get that k = t g α . When the line is parallel to Ox, the slope is said to not exist because it goes to infinity.
The slope is positive when the graph of the function is increasing and vice versa. The figure shows various variations of the location of the right angle relative to the coordinate system with the value of the coefficient.
To find this angle, it is necessary to apply the definition of the slope coefficient and calculate the tangent of the inclination angle in the plane.
Decision
From the condition we have that α = 120 °. By definition, you need to calculate the slope. Let's find it from the formula k = t g α = 120 = - 3 .
Answer: k = - 3 .
If the angular coefficient is known, but it is necessary to find the angle of inclination to the x-axis, then the value of the angular coefficient should be taken into account. If k > 0, then the right angle is acute and is found by the formula α = a r c t g k . If k< 0 , тогда угол тупой, что дает право определить его по формуле α = π - a r c t g k .
Example 2
Determine the angle of inclination of the given straight line to O x with a slope equal to 3.
Decision
From the condition we have that the slope is positive, which means that the angle of inclination to O x is less than 90 degrees. Calculations are made according to the formula α = a r c t g k = a r c t g 3 .
Answer: α = a r c t g 3 .
Example 3
Find the angle of inclination of the straight line to the O x axis, if the slope = - 1 3 .
Decision
If we take the letter k as the designation of the slope, then α is the angle of inclination to the given straight line in the positive direction O x. Hence k = - 1 3< 0 , тогда необходимо применить формулу α = π - a r c t g k При подстановке получим выражение:
α = π - a r c t g - 1 3 = π - a r c t g 1 3 = π - π 6 = 5 π 6 .
Answer: 5 pi 6 .
An equation of the form y \u003d k x + b, where k is a slope, and b is some real number, is called the equation of a straight line with a slope. The equation is typical for any straight line that is not parallel to the O y axis.
If we consider in detail a straight line on a plane in a fixed coordinate system, which is given by an equation with a slope that looks like y = k · x + b . In this case, it means that the coordinates of any point on the line correspond to the equation. If we substitute the coordinates of the point M, M 1 (x 1, y 1), into the equation y \u003d k x + b, then in this case the line will pass through this point, otherwise the point does not belong to the line.
Example 4
Given a straight line with slope y = 1 3 x - 1 . Calculate whether the points M 1 (3 , 0) and M 2 (2 , - 2) belong to the given line.
Decision
It is necessary to substitute the coordinates of the point M 1 (3, 0) into the given equation, then we get 0 = 1 3 3 - 1 ⇔ 0 = 0 . The equality is true, so the point belongs to the line.
If we substitute the coordinates of the point M 2 (2, - 2), then we get an incorrect equality of the form - 2 = 1 3 · 2 - 1 ⇔ - 2 = - 1 3 . We can conclude that the point M 2 does not belong to the line.
Answer: M 1 belongs to the line, but M 2 does not.
It is known that the straight line is defined by the equation y = k · x + b passing through M 1 (0 , b) , substitution yielded an equality of the form b = k · 0 + b ⇔ b = b . From this we can conclude that the equation of a straight line with a slope y = k · x + b on the plane defines a straight line that passes through the point 0, b. It forms an angle α with the positive direction of the O x axis, where k = t g α .
Consider, for example, a straight line defined using a slope given by the form y = 3 · x - 1 . We get that the straight line will pass through the point with coordinate 0, - 1 with a slope of α = a r c t g 3 = π 3 radians along the positive direction of the O x axis. From this it can be seen that the coefficient is 3.
The equation of a straight line with a slope passing through a given point
It is necessary to solve a problem where it is necessary to obtain the equation of a straight line with a given slope passing through the point M 1 (x 1, y 1) .
The equality y 1 = k · x + b can be considered valid, since the line passes through the point M 1 (x 1 , y 1) . To remove the number b, it is necessary to subtract the equation with the slope coefficient from the left and right sides. It follows from this that y - y 1 = k · (x - x 1) . This equality is called the equation of a straight line with a given slope k, passing through the coordinates of the point M 1 (x 1, y 1) .
Example 5
Compose the equation of a straight line passing through the point M 1 with coordinates (4, - 1), with a slope equal to - 2.
Decision
By condition, we have that x 1 \u003d 4, y 1 \u003d - 1, k \u003d - 2. From here, the equation of the straight line will be written in this way y - y 1 = k · (x - x 1) ⇔ y - (- 1) = - 2 · (x - 4) ⇔ y = - 2 x + 7.
Answer: y = - 2 x + 7 .
Example 6
Write the equation of a straight line with a slope that passes through the point M 1 with coordinates (3, 5) parallel to the straight line y \u003d 2 x - 2.
Decision
By condition, we have that parallel lines have coinciding angles of inclination, hence the slope coefficients are equal. To find the slope from given equation, it is necessary to recall its basic formula y = 2 x - 2, hence it follows that k = 2 . We compose an equation with a slope coefficient and get:
y - y 1 = k (x - x 1) ⇔ y - 5 = 2 (x - 3) ⇔ y = 2 x - 1
Answer: y = 2 x - 1 .
The transition from the equation of a straight line with a slope to other types of equations of a straight line and vice versa
Such an equation is not always applicable for solving problems, since it has a not very convenient notation. To do this, it must be presented in a different form. For example, an equation of the form y = k · x + b does not allow you to write down the coordinates of the direction vector of the straight line or the coordinates of the normal vector. To do this, you need to learn how to represent equations of a different kind.
We can get the canonical equation of a straight line in a plane using the equation of a straight line with a slope. We get x - x 1 a x = y - y 1 a y . It is necessary to move the term b to the left side and divide by the expression of the resulting inequality. Then we get an equation of the form y = k x + b ⇔ y - b = k x ⇔ k x k = y - b k ⇔ x 1 = y - b k .
The equation of a straight line with a slope has become the canonical equation of a given straight line.
Example 7
Bring the equation of a straight line with slope y = - 3 x + 12 to canonical form.
Decision
We calculate and represent in the form of a canonical equation of a straight line. We get an equation of the form:
y = - 3 x + 12 ⇔ - 3 x = y - 12 ⇔ - 3 x - 3 = y - 12 - 3 ⇔ x 1 = y - 12 - 3
Answer: x 1 = y - 12 - 3.
The general equation of a straight line is easiest to obtain from y = k x + b, but this requires transformations: y = k x + b ⇔ k x - y + b = 0. A transition is made from the general equation of a straight line to equations of another type.
Example 8
An equation of a straight line of the form y = 1 7 x - 2 is given. Find out if the vector with coordinates a → = (- 1 , 7) is a normal straight line vector?
Decision
To solve it, it is necessary to switch to another form of this equation, for this we write:
y = 1 7 x - 2 ⇔ 1 7 x - y - 2 = 0
The coefficients in front of the variables are the coordinates of the normal vector of the straight line. Let's write it like this n → = 1 7 , - 1 , hence 1 7 x - y - 2 = 0 . It is clear that the vector a → = (- 1 , 7) is collinear to the vector n → = 1 7 , - 1 , since we have a fair relation a → = - 7 · n → . It follows that the original vector a → = - 1 , 7 is a normal vector of the line 1 7 x - y - 2 = 0 , which means that it is considered a normal vector for the line y = 1 7 x - 2 .
Answer: Is an
Let's solve the problem inverse to this one.
Need to move from general view equation A x + B y + C = 0 , where B ≠ 0 , to the slope equation. To do this, we solve the equation for y. We get A x + B y + C = 0 ⇔ - A B · x - C B .
The result is an equation with a slope equal to - A B .
Example 9
An equation of a straight line of the form 2 3 x - 4 y + 1 = 0 is given. Get the equation of a given line with a slope.
Decision
Based on the condition, it is necessary to solve for y, then we get an equation of the form:
2 3 x - 4 y + 1 = 0 ⇔ 4 y = 2 3 x + 1 ⇔ y = 1 4 2 3 x + 1 ⇔ y = 1 6 x + 1 4 .
Answer: y = 1 6 x + 1 4 .
In a similar way, an equation of the form x a + y b \u003d 1 is solved, which is called the equation of a straight line in segments, or the canonical form x - x 1 a x \u003d y - y 1 a y. It is necessary to solve it with respect to y, only then we get an equation with a slope:
x a + y b = 1 ⇔ y b = 1 - x a ⇔ y = - b a x + b .
The canonical equation can be reduced to a form with a slope. For this:
x - x 1 a x = y - y 1 a y ⇔ a y (x - x 1) = a x (y - y 1) ⇔ ⇔ a x y = a y x - a y x 1 + a x y 1 ⇔ y = a y a x x - a y a x x 1 + y 1
Example 10
There is a straight line given by the equation x 2 + y - 3 = 1 . Bring to the form of an equation with a slope.
Decision.
Based on the condition, it is necessary to transform, then we get an equation of the form _formula_. Both sides of the equation should be multiplied by -3 to get the required slope equation. Transforming, we get:
y - 3 = 1 - x 2 ⇔ - 3 y - 3 = - 3 1 - x 2 ⇔ y = 3 2 x - 3 .
Answer: y = 3 2 x - 3 .
Example 11
The straight line equation of the form x - 2 2 \u003d y + 1 5 is brought to the form with a slope.
Decision
It is necessary to calculate the expression x - 2 2 = y + 1 5 as a proportion. We get that 5 · (x - 2) = 2 · (y + 1) . Now you need to fully enable it, for this:
5 (x - 2) = 2 (y + 1) ⇔ 5 x - 10 = 2 y + 2 ⇔ 2 y = 5 x - 12 ⇔ y = 5 2 x
Answer: y = 5 2 x - 6 .
To solve such tasks, one should bring parametric equations of the straight line x = x 1 + a x λ y = y 1 + a y λ to canonical equation straight line, only after that you can proceed to the equation with the slope coefficient.
Example 12
Find the slope of the straight line if it is given by parametric equations x = λ y = - 1 + 2 · λ .
Decision
You need to transition from parametric view to slope. To do this, we find the canonical equation from the given parametric one:
x = λ y = - 1 + 2 λ ⇔ λ = x λ = y + 1 2 ⇔ x 1 = y + 1 2 .
Now it is necessary to resolve this equality with respect to y in order to obtain the equation of a straight line with a slope. To do this, we write this way:
x 1 = y + 1 2 ⇔ 2 x = 1 (y + 1) ⇔ y = 2 x - 1
It follows that the slope of the straight line is equal to 2. This is written as k = 2 .
Answer: k = 2 .
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The figure shows the angle of inclination of the straight line and the value of the slope coefficient for various options for the location of the straight line relative to the rectangular coordinate system.
Finding the slope of a straight line at a known angle of inclination to the Ox axis does not present any difficulties. To do this, it is enough to recall the definition of the slope coefficient and calculate the tangent of the slope angle.
Example.
Find the slope of the line if the angle of its inclination to the x-axis is equal to .
Decision.
By condition . Then, by definition of the slope of the straight line, we calculate 
.
Answer:
The task of finding the angle of inclination of a straight line to the x-axis with a known slope is a little more difficult. Here it is necessary to take into account the sign of the slope coefficient. When the angle of inclination of the straight line is acute and is found as . When the angle of inclination of a straight line is obtuse and can be determined by the formula 
.
Example.
Determine the angle of inclination of a straight line to the x-axis if its slope is 3.
Decision.
Since, by condition, the slope is positive, the angle of inclination of the straight line to the Ox axis is sharp. We calculate it according to the formula.
Answer:
Example.
The slope of the straight line is . Determine the angle of inclination of the straight line to the axis Ox.
Decision.
Denote k is the slope of the straight line, is the angle of inclination of this straight line to the positive direction of the Ox axis. As 
, then we use the formula for finding the angle of inclination of a straight line of the following form . We substitute the data from the condition into it: .
Answer:
Equation of a straight line with a slope.
Line Equation with Slope has the form , where k is the slope of the straight line, b is some real number. The equation of a straight line with a slope can be used to specify any straight line that is not parallel to the Oy axis (for a straight line parallel to the y-axis, the slope is not defined).
Let's look at the meaning of the phrase: "a line on a plane in a fixed coordinate system is given by an equation with a slope of the form". This means that the equation is satisfied by the coordinates of any point on the line and not by the coordinates of any other point on the plane. Thus, if the correct equality is obtained when substituting the coordinates of a point, then the line passes through this point. Otherwise, the point does not lie on a line.
Example.
The straight line is given by an equation with slope . Do the points also belong to this line?
Decision.
Substitute the coordinates of the point into the original equation of a straight line with a slope: 
. We have obtained the correct equality, therefore, the point M 1 lies on a straight line.
When substituting the coordinates of the point, we get the wrong equality: 
. Thus, the point M 2 does not lie on a straight line.
Answer:
Dot M 1 belongs to the line, M 2 does not.
It should be noted that the straight line, defined by the equation of a straight line with a slope , passes through the point, since when substituting its coordinates into the equation, we get the correct equality: .
Thus, the equation of a straight line with a slope determines a straight line on a plane passing through a point and forming an angle with the positive direction of the abscissa axis, and .
As an example, let's draw a straight line defined by the equation of a straight line with a slope of the form . This line passes through the point and has a slope 
radians (60 degrees) to the positive direction of the Ox axis. Its slope is .
The equation of a straight line with a slope passing through a given point.
Now we will solve a very important problem: we will obtain the equation of a straight line with a given slope k and passing through the point .
Since the line passes through the point , then the equality 
. The number b is unknown to us. To get rid of it, we subtract from the left and right parts of the equation of a straight line with a slope, respectively, the left and right parts of the last equality. In doing so, we get 
. This equality is equation of a straight line with a given slope k that passes through a given point.
Consider an example.
Example.
Write the equation of a straight line passing through the point, the slope of this straight line is -2.
Decision.
From the condition we have 
. Then the equation of a straight line with a slope will take the form .
Answer:
Example.
Write the equation of a straight line if it is known that it passes through a point and the angle of inclination to the positive direction of the Ox axis is .
Decision.
First, we calculate the slope of the straight line whose equation we are looking for (we solved such a problem in the previous paragraph of this article). A-priory 
. Now we have all the data to write the equation of a straight line with a slope:
Answer:
Example.
Write the equation of a line with a slope that passes through a point parallel to the line.
Decision.
It is obvious that the angles of inclination of parallel lines to the axis Ox coincide (if necessary, see the article parallel lines), therefore, the slope coefficients of parallel lines are equal. Then the slope of the straight line, the equation of which we need to obtain, is equal to 2, since the slope of the straight line is 2. Now we can compose the required equation of a straight line with a slope:
Answer:
The transition from the equation of a straight line with a slope to other types of the equation of a straight line and vice versa.
With all the familiarity, the equation of a straight line with a slope is far from always convenient to use when solving problems. In some cases, problems are easier to solve when the equation of a straight line is presented in a different form. For example, the equation of a straight line with a slope does not allow you to immediately write down the coordinates of the directing vector of the straight line or the coordinates of the normal vector of the straight line. Therefore, one should learn to move from the equation of a straight line with a slope to other types of the equation of this straight line.
From the equation of a straight line with a slope, it is easy to obtain the canonical equation of a straight line on a plane of the form 
. To do this, we transfer the term b from the right side of the equation to the left side with the opposite sign, then divide both parts of the resulting equality by the slope k:. These actions lead us from the equation of a straight line with a slope to the canonical equation of a straight line.
Example.
Give the equation of a straight line with a slope 
to the canonical form.
Decision.
Let's perform the necessary transformations: .
Answer:
Example.
The straight line is given by the equation of a straight line with slope . Is the vector a normal vector of this line?
Decision.
To solve this problem, let's move from the equation of a straight line with a slope to the general equation of this straight line: 
. We know that the coefficients in front of the variables x and y in the general equation of a straight line are the corresponding coordinates of the normal vector of this straight line, that is, the normal vector of the straight line 
. Obviously, the vector is collinear to the vector , since the relation is true (if necessary, see the article). Thus, the original vector is also a normal vector of the line , and, therefore, is a normal vector and the original line .
Answer:
Yes it is.
And now we will solve the inverse problem - the problem of bringing the equation of a straight line on a plane to the equation of a straight line with a slope.
From the general straight line equation 
, where , it is very easy to pass to the slope equation. For this you need general equation direct resolve with respect to y . At the same time, we get . The resulting equality is the equation of a straight line with a slope equal to .