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. Let's consider the definitions, get the equation itself, reveal the connection with other types of equations. Everything will be discussed on examples of problem solving.
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 right angle relative to the coordinate system with the coefficient value.
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.
Solution
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.
Solution
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 .
Solution
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 \u003d 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.
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, 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.
Solution
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.
Solution
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.
Solution
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. The transition is made from general equation direct to equations of another kind.
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?
Solution
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.
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 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
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.
Solution
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 ax = y - y 1 ay ⇔ ay (x - x 1) = ax (y - y 1) ⇔ ⇔ ax y = ay x - ay x 1 + ax y 1 ⇔ y = ayax x - ayax 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.
Solution.
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.
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 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, parametric equations of the straight line of the form x \u003d x 1 + a x λ y \u003d y 1 + a y λ should be reduced to the canonical equation of the straight line, only after that you can proceed to the equation with the slope.
Example 12
Find the slope of the straight line if it is given by parametric equations x = λ y = - 1 + 2 · λ .
Solution
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 .
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
Numerically equal to the tangent of the angle (constituting the smallest rotation from the Ox axis to the Oy axis) between the positive direction of the x-axis and the given straight line.
The tangent of an angle can be calculated as the ratio of the opposite leg to the adjacent one. k is always equal to , that is, the derivative of the straight line equation with respect to x.
With positive values of the angular coefficient k and zero value of the shift coefficient b line will lie in the first and third quadrants (in which x And y both positive and negative). At the same time, large values of the angular coefficient k a steeper straight line will correspond, and a smaller one - a flatter one.
Lines and are perpendicular if , and parallel when .
Notes
Wikimedia Foundation. 2010 .
- Ifit (king of Elis)
- List of Decrees of the President of the Russian Federation "On awarding state awards" for 2001
See what the "Line Slope" is in other dictionaries:
slope (straight)- — Topics oil and gas industry EN slope … Technical Translator's Handbook
Slope- (mathematical) number k in the equation of a straight line on the plane y \u003d kx + b (see Analytical geometry), characterizing the slope of the straight line relative to the abscissa axis. In a rectangular coordinate system U. to. k \u003d tg φ, where φ is the angle between ... ... Great Soviet Encyclopedia
Line equations
ANALYTIC GEOMETRY- a branch of geometry that explores the simplest geometric objects by means of elementary algebra based on the method of coordinates. Creation analytical geometry usually attributed to R. Descartes, who outlined its foundations in the last chapter of his ... ... Collier Encyclopedia
Reaction time- The measurement of reaction time (RT) is probably the most venerable subject in empirical psychology. It originated in the field of astronomy, in 1823, with the measurement of individual differences in the speed at which a star was perceived to cross the telescope's line of sight. These … Psychological Encyclopedia
MATHEMATICAL ANALYSIS- a section of mathematics that provides methods for the quantitative study of various processes of change; deals with the study of the rate of change (differential calculus) and the determination of the lengths of curves, areas and volumes of figures bounded by curved contours and ... Collier Encyclopedia
Straight- This term has other meanings, see Direct (meanings). A straight line is one of the basic concepts of geometry, that is, it does not have an exact universal definition. In a systematic presentation of geometry, a straight line is usually taken as one ... ... Wikipedia
Straight line- Image of straight lines in a rectangular coordinate system A straight line is one of the basic concepts of geometry. In a systematic presentation of geometry, a straight line is usually taken as one of the initial concepts, which is only indirectly determined ... ... Wikipedia
Direct- Image of straight lines in a rectangular coordinate system A straight line is one of the basic concepts of geometry. In a systematic presentation of geometry, a straight line is usually taken as one of the initial concepts, which is only indirectly determined ... ... Wikipedia
Minor axis- Not to be confused with the term "Ellipsis". Ellipse and its foci Ellipse (other Greek ἔλλειψις disadvantage, in the sense of lack of eccentricity up to 1) the locus of points M of the Euclidean plane, for which the sum of the distances from two given points F1 ... ... Wikipedia
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 view 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 line is equal to infinity, and the 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 acute angleα 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. The tangent of a 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 the tangent of the angle that forms 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
Tasks for finding the derivative of the tangent are included in the exam in mathematics and are met there annually. At the same time, statistics recent years shows that such tasks cause certain difficulties for graduates. Therefore, if a student expects to get decent scores on the basis of passing the exam, then he should definitely learn how to cope with the tasks from the section "The angular coefficient of the tangent as the value of the derivative at the point of contact", prepared by the specialists of the educational portal "Shkolkovo". Having dealt with the algorithm for solving them, the student will be able to successfully overcome the certification test.
Basic moments
Coming to a decision USE tasks on this topic, it is necessary to recall the basic definition: the derivative of a function at a point is equal to the slope of the tangent to the graph of the function at that point. This is what it consists geometric meaning derivative.
Another important definition needs to be refreshed. It sounds like this: the slope equals the tangent of the angle of inclination of the tangent to the x-axis.
What other important points should be noted in this topic? When solving problems for finding the derivative in the USE, it must be remembered that the angle that the tangent forms can be less, more than 90 degrees, or equal to zero.
How to prepare for the exam?
In order for the tasks in the USE on the topic “The slope of the tangent as the value of the derivative at the point of contact” to be given to you quite easily, use the information on this section on the Shkolkovo educational portal when preparing for the final test. Here you will find the necessary theoretical material, collected and clearly presented by our experts, and you will also be able to practice the exercises.
For each task, for example, tasks on the topic "The angular coefficient of the tangent as the tangent of the angle of inclination", we wrote down the correct answer and the solution algorithm. At the same time, students can perform exercises of various levels of complexity online. If necessary, the task can be saved in the "Favorites" section in order to discuss its solution with the teacher later.
The topic "The angular coefficient of the 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 both a full answer and a short answer. In preparation for passing the exam in mathematics, the student should definitely repeat the tasks in which it is required to calculate the slope of the tangent.
Doing this will help you educational portal"Shkolkovo". Our experts have prepared and presented theoretical and practical material as accessible as possible. Having become acquainted with it, graduates with any level of training will be able to successfully solve problems related to derivatives, in which it is required to find the tangent of the slope of the tangent.
Basic moments
To find the correct and rational decision similar tasks in the exam must be remembered basic definition: the derivative is the rate of change of the function; it is equal to the tangent of the slope of the tangent 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 correct solution to the USE problems on the derivative, in which it is required to calculate the tangent of the slope of the tangent. For clarity, it is best to plot a graph on the OXY plane.
If you have already familiarized yourself with the basic material on the topic of the derivative and are ready to start solving problems for calculating the tangent of the angle of inclination of a tangent, similar to USE assignments you can do it online. For each task, for example, tasks on the topic "Relationship of the derivative with the speed and acceleration of the body", we wrote down the correct answer and the solution algorithm. In this case, students can practice performing tasks of various levels of complexity. If necessary, the exercise can be saved in the "Favorites" section, so that later you can discuss the decision with the teacher.
