The figure shows the angle of inclination of the straight line and indicates the value of the slope at various options the location of the line relative to the rectangular coordinate system.
Finding the slope of a straight line with 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 angular coefficient and calculate the tangent of the angle of inclination.
Example.
Find the slope of the straight line if its angle of inclination to the abscissa axis is equal to .
Solution.
According to the condition. Then, by definition of the slope of a 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 complicated. Here it is necessary to take into account the sign of the slope. When the angle of inclination of the straight line is acute and is found as . When the angle of inclination of the straight line is obtuse and can be determined by the formula 
.
Example.
Determine the angle of inclination of the straight line to the abscissa axis if its slope is equal to 3.
Solution.
Since by condition the angular coefficient is positive, the angle of inclination of the straight line to the Ox axis is acute. We calculate it using the formula.
Answer:
Example.
The slope of the straight line is . Determine the angle of inclination of the straight line to the Ox axis.
Solution.
Let's denote k is the angular coefficient of the straight line, - the angle of inclination of this straight line to the positive direction of the Ox axis. Because 
, then we use the formula to find the angle of inclination of the line of the following form . We substitute the data from the condition into it: .
Answer:
Equation of a straight line with an angular coefficient.
Equation of a straight line with slope has the form , where k is the slope of the line, b is some real number. Using the equation of a straight line with an angular coefficient, you can specify any straight line that is not parallel to the Oy axis (for a straight line parallel to the ordinate axis, the angular coefficient is not defined).
Let's understand the meaning of the phrase: “a straight line on a plane in a fixed coordinate system is given by an equation with an angular coefficient of the form “.” This means that the equation is satisfied by the coordinates of any point on the line and is not satisfied by the coordinates of any other points on the plane. Thus, if, when substituting the coordinates of a point, the correct equality is obtained, then the straight line passes through this point. Otherwise, the point does not lie on the line.
Example.
The straight line is given by an equation with a slope. Do the points also belong to this line?
Solution.
Let's substitute the coordinates of the point into the original equation of the straight line with the slope: 
. We have obtained the correct equality, therefore, point M 1 lies on the line.
When substituting the coordinates of a point, we get an incorrect equality: 
. Thus, point M 2 does not lie on a line.
Answer:
Dot M 1 belongs to the line, M 2 does not.
It should be noted that a straight line defined by the equation of a straight line with an angular coefficient passes through the point, since when we substitute its coordinates into the equation we obtain the correct equality: .
Thus, the equation of a straight line with an angular coefficient defines on the plane a straight line passing through a point and forming an angle with the positive direction of the abscissa axis, and .
As an example, let us depict a straight line defined by the equation of a straight line with an angle coefficient of the form . This line passes through a point and has a slope 
radians (60 degrees) to the positive direction of the Ox axis. Its slope is equal to .
Equation of a straight line with slope passing through a given point.
Now let's decide very important task: we obtain the equation of a straight line with a given slope k and passing through the point.
Since the line passes through the point, the equality is true 
. We don't know the number b. To get rid of it, we subtract the left and right sides of the last equality from the left and right sides of the equation of the straight line with the slope coefficient, respectively. In this case we get 
. This equality is equation of a straight line with a given slope k, which passes through a given point.
Let's look at an example.
Example.
Write the equation of a line passing through the point, the slope of this line is -2.
Solution.
From the condition we have 
. Then the equation of a straight line with an angular coefficient 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 equal to .
Solution.
First, let's calculate the slope of the line whose equation we are looking for (we solved this problem in the previous paragraph of this article). By definition 
. Now we have all the data to write the equation of a straight line with an angle coefficient:
Answer:
Example.
Write the equation of a line with an angular coefficient passing through a point parallel to the line.
Solution.
Obviously, the angles of inclination of parallel lines to the Ox axis coincide (if necessary, see the article parallelism of lines), therefore, the angular 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 equal to 2. Now we can create the required equation of a straight line with a slope:
Answer:
Transition from the equation of a line with an angle coefficient to other types of equation of a line and vice versa.
Despite all the familiarity, the equation of a straight line with an angular coefficient is not always convenient to use when solving problems. In some cases, problems are easier to solve when the equation of a line is presented in a different form. For example, the equation of a straight line with an angular coefficient 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, you should learn to move from the equation of a straight line with an angle coefficient to other types of equations of this straight line.
From the equation of a straight line with an angular coefficient it is easy to obtain the canonical equation of a straight line on a plane of the form 
. To do this, we move the term b from the right side of the equation to the left side with the opposite sign, then divide both sides of the resulting equality by the slope k: . These actions lead us from the equation of a line with an angle coefficient to the canonical equation of a line.
Example.
Give the equation of a straight line with an angle coefficient 
to the canonical form.
Solution.
Let's perform the necessary transformations: .
Answer:
Example.
A straight line is given by the equation of a straight line with an angular coefficient. Is the vector a normal vector of this line?
Solution.
To solve this problem, let's move from the equation of a straight line with an angle coefficient to the general equation of this straight line: 
. We know that the coefficients of the variables x and y in the general equation of a line are the corresponding coordinates of the normal vector of this line, that is, the normal vector of the line 
. It is obvious that the vector is collinear to the vector, since the relation is valid (if necessary, see the article). Thus, the original vector is also a normal line vector , 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 reducing the equation of a straight line on a plane to the equation of a straight line with an angle coefficient.
From general equation straight view 
, in which it is very easy to go to an equation with a slope coefficient. To do this, you need to solve the general equation of the line with respect to y. In this case we get . The resulting equality is an equation of a straight line with an angular coefficient equal to .
In the previous chapter it was shown that by choosing a certain coordinate system on the plane, we can geometric properties, which characterizes the points of the line under consideration, is expressed analytically by an equation between the current coordinates. Thus we get the equation of the line. This chapter will look at the equations of straight lines.
To create an equation for a straight line in Cartesian coordinates, you need to somehow set the conditions that determine its position relative to the coordinate axes.
First, we will introduce the concept of the angular coefficient of a line, which is one of the quantities characterizing the position of a line on a plane.
Let's call the angle of inclination of the straight line to the Ox axis the angle by which the Ox axis needs to 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 through an angle of 180° will again align it with the straight line, the angle of inclination of the straight line to the axis can not be chosen unambiguously (to within a term, a multiple of ).
The tangent of this angle is determined uniquely (since changing the angle does not change its tangent).
The tangent of the angle of inclination of the straight line to the Ox axis is called the angular coefficient of the straight line.
The angular coefficient characterizes the direction of the straight line (we do not distinguish here between two mutually opposite directions of the straight line). If the slope of a line is zero, then the line is parallel to the x-axis. With a positive angular coefficient, the angle of inclination of the straight line to the Ox axis will be acute (we are considering here the smallest positive value of the inclination angle) (Fig. 39); Moreover, the greater the angular coefficient, the greater the angle of its inclination to the Ox axis. If the angular coefficient is negative, then the angle of inclination of the straight line to the Ox axis will be obtuse (Fig. 40). Note that a straight line perpendicular to the Ox axis does not have an angular coefficient (the tangent of the angle does not exist).
Continuation of the topic, the equation of a line on a plane is based on the study of a straight line from algebra lessons. This article provides general information on the topic of equation of a straight line with a slope. Let's consider the definitions, get the equation itself, and identify the connection with other types of equations. Everything will be discussed using examples of problem solving.
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Before writing such an equation, it is necessary to define the angle of inclination of the straight line to the O x axis with their angular coefficient. Let us assume that a Cartesian coordinate system O x on the plane is given.
Definition 1
The angle of inclination of the straight line to the O x axis, 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 the line is parallel to O x or coincides in it, the angle of inclination is 0. Then the angle of inclination of the given straight line α is determined on the interval [ 0 , π) .
Definition 2
Direct slope is the tangent of the angle of inclination of a given straight line.
Standard designation is k. From the definition we find that k = t g α . When the line is parallel to Ox, they say that the slope does not exist, since it goes to infinity.
The slope is positive when the graph of the function increases and vice versa. The figure shows various layout variations right angle relative to the coordinate system with the coefficient value.
To find this angle, it is necessary to apply the definition of the angular coefficient and calculate the tangent of the angle of inclination in the plane.
Solution
From the condition we have that α = 120°. By definition, the slope must be calculated. Let's find it from the formula k = t g α = 120 = - 3.
Answer: k = - 3 .
If the angular coefficient is known, and it is necessary to find the angle of inclination to the abscissa 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 an angular coefficient of 3.
Solution
From the condition we have that the angular coefficient is positive, which means that the angle of inclination to O x is less than 90 degrees. Calculations are made using 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.
Solution
If we take the letter k as the designation of the angular coefficient, then α is the angle of inclination to a 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 π 6 .
An equation of the form y = k x + b, where k is the slope and b is some real number, is called the equation of a 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 specified by an equation with an angular coefficient that has the form y = k x + b. IN in this case means that the equation corresponds to the coordinates of any point on the line. If we substitute the coordinates of point M, M 1 (x 1, y 1) into the equation y = 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
A straight line with slope y = 1 3 x - 1 is given. Calculate whether the points M 1 (3, 0) and M 2 (2, - 2) belong to the given line.
Solution
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, which means 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 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 line is defined by the equation y = k · x + b, passing through M 1 (0, b), upon substitution we obtained an equality of the form b = k · 0 + b ⇔ b = b. From this we can conclude that the equation of a straight line with an angular coefficient 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.
Let us consider, as an example, a straight line defined using an angular coefficient specified in the form y = 3 · x - 1. We obtain 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 in the positive direction of the O x axis. This shows that the coefficient is 3.
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, you need to subtract the equation with the slope 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
Write an equation for a straight line passing through point M 1 with coordinates (4, - 1), with an angular coefficient equal to - 2.
Solution
By condition we have that x 1 = 4, y 1 = - 1, k = - 2. From here the equation of the line will be written as follows: 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 an angular coefficient that passes through the point M 1 with coordinates (3, 5), parallel to the straight line y = 2 x - 2.
Solution
According to the condition, we have that parallel lines have coinciding angles of inclination, which means that the angular coefficients are equal. To find the slope from given equation, you need to remember its basic formula y = 2 x - 2, it follows that k = 2. We compose an equation with the 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 .
Transition from a straight line equation with a slope to other types of straight line equations and back
This equation is not always applicable for solving problems, since it is not entirely convenient to write. To do this, you need to present it in a different form. For example, an equation of the form y = k · x + b does not allow us to write down the coordinates of the direction vector of a straight line or the coordinates of a normal vector. To do this, you need to learn to represent with equations of a different type.
We can get canonical equation line on a plane using the equation of a 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 line with a slope has become the canonical equation of this line.
Example 7
Bring the equation of a straight line with an angular coefficient y = - 3 x + 12 to canonical form.
Solution
Let us calculate and present it 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 for this it is necessary to make transformations: y = k · x + b ⇔ k · x - y + b = 0. A transition is made from the general equation of the line to equations of a different type.
Example 8
Given a straight line equation of the form y = 1 7 x - 2 . Find out whether the vector with coordinates a → = (- 1, 7) is a normal line vector?
Solution
To solve it is necessary to move 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 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 the 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 it is considered a normal vector for the line y = 1 7 x - 2.
Answer: Is
Let's solve the inverse problem of this one.
It is necessary to move from the general form of the equation A x + B y + C = 0, where B ≠ 0, to an equation with an angular coefficient. 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
A straight line equation of the form 2 3 x - 4 y + 1 = 0 is given. Obtain the equation of a given line with an angular coefficient.
Solution
Based on the condition, it is necessary to solve for y, then we obtain 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 .
An equation of the form x a + y b = 1 is solved in a similar way, which is called the equation of a straight line in segments, or canonical type x - x 1 a x = y - y 1 a y . We need to solve it for y, only then we get an equation with the 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 an angular coefficient. To do 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. Reduce to the form of an equation with an angular coefficient.
Solution.
Based on the condition, it is necessary to transform, then we obtain an equation of the form _formula_. Both sides of the equation must be multiplied by - 3 to obtain 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
Reduce the straight line equation of the form x - 2 2 = y + 1 5 to a form with an angular coefficient.
Solution
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 completely enable it, to do 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 problems, parametric equations of the line of the form x = x 1 + a x · λ y = y 1 + a y · λ should be reduced to the canonical equation of the line, only after this can one proceed to the equation with the slope coefficient.
Example 12
Find the slope of the line if it is given by parametric equations x = λ y = - 1 + 2 · λ.
Solution
It is necessary to transition from the parametric view to the 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 an angular coefficient. To do this, let's write it this way:
x 1 = y + 1 2 ⇔ 2 x = 1 (y + 1) ⇔ y = 2 x - 1
It follows that the slope of the line is 2. This is written as k = 2.
Answer: k = 2.
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The straight line y=f(x) will be tangent to the graph shown in the figure at point x0 if it passes through the point with coordinates (x0; f(x0)) and has an angular coefficient f"(x0). Find such a coefficient, Knowing the features of a tangent, it’s not difficult.
You will need
- - mathematical reference book;
- - a simple pencil;
- - notebook;
- - protractor;
- - compass;
- - pen.
Instructions
If the value f‘(x0) does not exist, then either there is no tangent, or it runs vertically. In view of this, the presence of a derivative of the function at the point x0 is due to the existence of a non-vertical tangent tangent to the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent will be equal to f "(x0). Thus, it becomes clear geometric meaning derivative – calculation of the slope of the tangent.
Draw additional tangents that would be in contact with the graph of the function at points x1, x2 and x3, and also mark the angles formed by these tangents with the x-axis (this 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) will be obtuse, and the third (α3) will be 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 at tg0 the result is zero.
Please note
Correctly determine the angle formed by the tangent. To do this, use a protractor.
Two inclined lines will be parallel if their angular coefficients are equal to each other; perpendicular if the product of the angular coefficients of these tangents is equal to -1.
Sources:
- Tangent to the graph of a function
Cosine, like sine, is classified as a “direct” trigonometric function. Tangent (together with cotangent) is classified as 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.
Instructions
Subtract the quotient of one by the cosine given angle, and extract the square root from the result - this will be the tangent value of the angle, expressed by its cosine: tan(α)=√(1-1/(cos(α))²). Please note that in the formula the cosine is in the denominator of the fraction. The impossibility of dividing by zero precludes the use of this expression for angles equal to 90°, as well as those differing from this value by numbers that are multiples of 180° (270°, 450°, -90°, etc.).
There is also alternative way calculating the tangent from a known cosine value. It can be used if there is no restriction on the use of others. To implement this method, first determine the angle value from a known cosine value - this can be done using the arc cosine function. Then simply calculate the tangent for the angle of the resulting value. IN general view this algorithm can be written as follows: tg(α)=tg(arccos(cos(α))).
There is also an exotic option using the definition of cosine and tangent through sharp corners right triangle. In this definition, cosine corresponds to the ratio of the length of the leg adjacent to the angle under consideration to the length of the hypotenuse. Knowing the value of the cosine, you can select the corresponding lengths of these two sides. For example, if cos(α) = 0.5, then the adjacent can be taken equal to 10 cm, and the hypotenuse - 20 cm. The specific numbers do not matter here - you will get the same and correct numbers with any values that have the same . Then, using the Pythagorean theorem, determine the length of the missing side - the opposite leg. It 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 classical definition 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 to represent the tangent is as a sine ratio angle to its cosine. This is an odd periodic and non-continuous function, each cycle of which equal to the number Pi, and the break point corresponds to half this number.
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