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In mathematics, one of the parameters describing the position of a line on the Cartesian coordinate plane is slope this straight line. This parameter characterizes the slope of the straight line to the abscissa 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 straight line can be represented by the expression ax+by=c, where a, b and c are arbitrary real numbers, but a 2 + b 2 ≠ 0.
Using 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 the slope, and the equation of a line of this type is called an equation with a slope. It turns out that to find the slope, you simply need to reduce the original equation to the form indicated above. For a more complete understanding, consider a specific example:
Problem: Find the slope of the line given by the equation 36x - 18y = 108
Solution: Let's transform the original equation.
Answer: The required slope of this line is 2.
If, during the transformation of the equation, we received an expression like 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 angular coefficient of such a straight line is equal to infinity.
For lines expressed by an equation like y = const, the slope is zero. This is typical for straight lines parallel to the abscissa axis. For example:
Problem: Find the slope of the line given by the equation 24x + 12y - 4(3y + 7) = 4
Solution: Let's bring the original equation to its general form
24x + 12y - 12y + 28 = 4
It is impossible to express y from the resulting expression, therefore the angular coefficient of this line is equal to infinity, and the line itself will be parallel to the Y axis.
Geometric meaning
For better understanding Let's look at the picture:
In the figure we see a graph of a function like y = kx. To simplify, let’s take the coefficient c = 0. In the triangle OAB, the ratio of side BA to AO will be equal to the angular coefficient k. At the same time, the ratio VA/AO is the tangent acute angleα in right triangle OAV. It turns out that the angular coefficient of the straight line is equal to the tangent of the angle that this straight line makes with the abscissa axis of the coordinate grid.
Solving the problem of how to find the angular coefficient of a straight line, we find the tangent of the angle between it and the X axis of the coordinate grid. Boundary cases, when the line in question 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 abscissa axis is 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 angular coefficient of similar straight lines is also equal to infinity, which confirms what was written above.
Tangent slope
A common task often encountered in practice is also to find the slope of a tangent to the graph of a function at a certain point. A 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 derivative. The derivative of any function at some point is a constant, numerically equal to tangent the angle formed between the tangent at a specified point to the graph of this function and the abscissa axis. It turns out that to determine the angular coefficient of the tangent at point x 0, we need to calculate the value of the derivative of the original function at this point k = f"(x 0). Let's look at an example:
Problem: 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 required slope at point x = 0.1 is 4.831
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 defined 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 obtain 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 variations in 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 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
By condition, we have that parallel lines have identical 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 very conveniently written. 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 general equation straight line to equations of another 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.
Need to move from general view equations A x + B y + C = 0, where B ≠ 0, to an equation with a slope. 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 topic “The angular coefficient of a tangent as the tangent of the angle of inclination” in the certification exam is given several tasks at once. Depending on their condition, the graduate may be required to provide either a full answer or a short answer. In preparation for passing the Unified State Exam In mathematics, the student should definitely repeat problems in which it is necessary to calculate the angular coefficient of a tangent.
It will help you do this educational portal"Shkolkovo". Our specialists prepared and presented theoretical and practical material in the most accessible way possible. Having become familiar with it, graduates with any level of training will be able to successfully solve problems related to derivatives in which it is necessary to find the tangent of the tangent angle.
Highlights
To find the correct and rational solution to such tasks in the Unified State Exam, it is necessary to remember the basic definition: the derivative represents the rate of change of a function; it is equal to the tangent of the tangent angle drawn to the graph of the function at a certain point. It is equally important to complete the drawing. It will allow you to find the right decision Unified State Exam problems on the derivative, in which it is necessary to calculate the tangent of the tangent angle. For clarity, it is best to plot the graph on the OXY plane.
If you have already familiarized yourself with the basic material on the topic of derivatives and are ready to begin solving problems on calculating the tangent of the tangent angle, such as Unified State Exam assignments, you can do this online. For each task, for example, problems on the topic “Relationship of a derivative with the speed and acceleration of a body,” we wrote down the correct answer and solution algorithm. At the same time, students can practice performing tasks of varying levels of complexity. If necessary, the exercise can be saved in the “Favorites” section so that you can discuss the solution with the teacher later.