Find the slope and slope of the straight lines. The equation of a straight line with a slope: theory, examples, problem solving

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;
• - a 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 slope the tangent will be 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 graph of the function 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) will be obtuse, and the third (α3) zero, since the tangent line is parallel to the x-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.

Useful advice

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 also an exotic option using the definition of cosine and tangent through sharp corners 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.

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 one. 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 solution to such tasks in the exam, you need to remember 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 right decision 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 completing tasks. different levels difficulties. If necessary, the exercise can be saved in the "Favorites" section, so that later you can discuss the decision with the teacher.

Learn to take derivatives of functions. The derivative characterizes the rate of change of a function at a certain point lying on the graph of this function. In this case, the graph can be either a straight line or a curved line. That is, the derivative characterizes the rate of change of the function at a particular point in time. Remember general rules for which derivatives are taken, and only then proceed to the next step.

• Read the article.
• How to take the simplest derivatives, for example, the derivative exponential equation, described . The calculations presented in the following steps will be based on the methods described there.

Learn to distinguish between problems in which the slope needs to be calculated in terms of the derivative of a function. In tasks, it is not always suggested to find the slope or derivative of a function. For example, you may be asked to find the rate of change of a function at point A(x, y). You may also be asked to find the slope of the tangent at point A(x, y). In both cases, it is necessary to take the derivative of the function.

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 relationship with other types of equations. Everything will be discussed on examples of problem solving.

Yandex.RTB R-A-339285-1

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 canonical equation 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 obtained 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

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.

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 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.

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

The slope coefficient is straight. In this article, we will consider tasks related to the coordinate plane included in the exam in mathematics. These are assignments for:

- determination of the slope of a straight line, when two points through which it passes are known;
- determination of the abscissa or ordinate of the point of intersection of two lines on the plane.

What is the abscissa and ordinate of a point was described in this section. In it, we have already considered several problems related to the coordinate plane. What needs to be understood for the type of tasks under consideration? A bit of theory.

The equation of a straight line on the coordinate plane has the form:

where k – this is the slope of the straight line.

Next moment! Slope of a straight line equal to tangent angle of inclination of a straight line. This is the angle between the given line and the axisoh.

It lies between 0 and 180 degrees.

That is, if we reduce the equation of a straight line to the form y = kx + b, then further we can always determine the coefficient k (slope coefficient).

Also, if we can determine the tangent of the slope of the straight line based on the condition, then we will thereby find its slope.

The next theoretical moment!Equation of a straight line passing through two given points.The formula looks like:

Consider problems (similar to those from open bank assignments):

Find the slope of the straight line passing through the points with coordinates (–6; 0) and (0; 6).

In this problem, the most rational way to solve this is to find the tangent of the angle between the x-axis and the given straight line. It is known that it is equal to the angular coefficient. Consider a right triangle formed by a straight line and the x and y axes:

The tangent of an angle in right triangle is the ratio of the opposite leg to the adjacent:

* Both legs are equal to six (these are their lengths).

Of course, this task can be solved using the formula for finding the equation of a straight line passing through two given points. But it will be a longer solution path.

Answer: 1

Find the slope of the straight line passing through the points with coordinates (5;0) and (0;5).

Our points have coordinates (5;0) and (0;5). Means,

Let's bring the formula to the form y = kx + b

We got that the angular coefficient k = – 1.

Answer: -1

Straight a passes through points with coordinates (0;6) and (8;0). Straight b passes through the point with coordinates (0;10) and is parallel to the line a b with axle ox.

In this problem, you can find the equation of a straight line a, determine the slope for it. Straight line b the slope will be the same since they are parallel. Next, you can find the equation of a straight line b. And then, substituting the value y = 0 into it, find the abscissa. BUT!

In this case, it is easier to use the triangle similarity property.

The right triangles formed by the given (parallel) lines of coordinates are similar, which means that the ratios of their respective sides are equal.

The desired abscissa is 40/3.

Answer: 40/3

Straight a passes through points with coordinates (0;8) and (–12;0). Straight b passes through the point with coordinates (0; -12) and is parallel to the line a. Find the abscissa of the point of intersection of the line b with axle ox.

For this problem, the most rational way to solve it is to use the similarity property of triangles. But we will solve it in a different way.

We know the points through which the line passes a. We can write the equation of a straight line. The formula for the equation of a straight line passing through two given points is:

By condition, the points have coordinates (0;8) and (–12;0). Means,

Let's bring to mind y = kx + b:

Got that corner k = 2/3.

*The angular coefficient could be found through the tangent of the angle in a right triangle with legs 8 and 12.

We know that parallel lines have equal slopes. So the equation of a straight line passing through the point (0;-12) has the form:

Find value b we can substitute the abscissa and ordinate into the equation:

So the line looks like:

Now, to find the desired abscissa of the point of intersection of the line with the x-axis, you need to substitute y \u003d 0:

Answer: 18

Find the ordinate of the point of intersection of the axis oy and a straight line passing through point B(10;12) and a parallel line passing through the origin and point A(10;24).

Let's find the equation of a straight line passing through the points with coordinates (0;0) and (10;24).

The formula for the equation of a straight line passing through two given points is:

Our points have coordinates (0;0) and (10;24). Means,

Let's bring to mind y = kx + b

The slopes of the parallel lines are equal. Hence, the equation of a straight line passing through the point B (10; 12) has the form:

Meaning b we find by substituting the coordinates of the point B (10; 12) into this equation:

We got the equation of a straight line:

To find the ordinate of the point of intersection of this line with the axis OU must be substituted into the found equation X= 0:

*Easiest solution. With the help of parallel translation, we shift this line down along the axis OU to the point (10;12). The shift occurs by 12 units, that is, point A(10;24) "passed" to point B(10;12), and point O(0;0) "passed" to point (0;–12). So the resulting line will intersect the axis OU at the point (0;–12).

The desired ordinate is -12.

Answer: -12

Find the ordinate of the point of intersection of the line given by the equation

3x + 2y = 6, with axis Oy.

Coordinate of the point of intersection of the given line with the axis OU has the form (0; at). Substitute the abscissa into the equation X= 0, and find the ordinate:

Ordinate of point of intersection of a line with an axis OU equals 3.

*The system is being solved:

Answer: 3

Find the ordinate of the point of intersection of the lines given by the equations

3x + 2y = 6 and y = - x.

When two lines are given, and the question is about finding the coordinates of the point of intersection of these lines, the system of these equations is solved:

In the first equation, we substitute - X instead of at:

The ordinate is minus six.

Answer: – 6

Find the slope of the straight line passing through the points with coordinates (–2; 0) and (0; 2).

Find the slope of the straight line passing through the points with coordinates (2;0) and (0;2).

The line a passes through the points with coordinates (0;4) and (6;0). Line b passes through the point with coordinates (0;8) and is parallel to line a. Find the abscissa of the point of intersection of line b with the x-axis.

Find the ordinate of the point of intersection of the y-axis and the line passing through point B (6;4) and the parallel line passing through the origin and point A (6;8).

1. It is necessary to clearly understand that the slope of the straight line is equal to the tangent of the slope of the straight line. This will help you in solving many problems of this type.

2. The formula for finding a straight line passing through two given points must be understood. With its help, you can always find the equation of a straight line if the coordinates of two of its points are given.

3. Remember that the slopes of parallel lines are equal.

4. As you understand, in some problems it is convenient to use the sign of similarity of triangles. Problems are solved practically orally.

5. Tasks in which two lines are given and it is required to find the abscissa or ordinate of their intersection point can be solved graphically. That is, build them on the coordinate plane (on a sheet in a cell) and determine the intersection point visually. *But this method is not always applicable.

6. And the last. If a straight line and the coordinates of the points of its intersection with the coordinate axes are given, then in such problems it is convenient to find the slope through finding the tangent of the angle in the formed right triangle. How to "see" this triangle for various arrangements of lines on the plane is schematically shown below:

>> Line inclination angle from 0 to 90 degrees<<

>> Straight line angle from 90 to 180 degrees<<

That's all. Good luck to you!

Sincerely, Alexander.

P.S: I would be grateful if you tell about the site in social networks.