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.

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 this equation, you need to remember its basic formula y \u003d 2 x - 2, which implies that k \u003d 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 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.

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 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 parametric equations x = λ y = - 1 + 2 λ .

Solution

You need to make the transition from parametric view to the angle factor. To do this, we find the canonical equation from the given parametric one:

x = λ y = - 1 + 2 λ ⇔ λ = x λ = y + 1 2 ⇔ x 1 = y + 1 2 .

Now it is necessary to resolve this equality with respect to y in order to obtain the equation of a straight line with a slope. To do this, we write this way:

x 1 = y + 1 2 ⇔ 2 x = 1 (y + 1) ⇔ y = 2 x - 1

It follows that the slope of the straight line is equal to 2. This is written as k = 2 .

Answer: k = 2 .

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In Cartesian coordinates, every straight line is defined by a first degree equation and, conversely, every first degree equation defines a straight line.

Type equation

is called the general equation of a straight line.

The angle defined as shown in Fig. is called the angle of inclination of the straight line to the x-axis. The tangent of the angle of inclination of the straight line to the x-axis is called the slope of the straight line; it is usually denoted by the letter k:

The equation is called the equation of a straight line with a slope; k is the slope, b is the value of the segment that the straight line cuts off on the Oy axis, counting from the origin.

If the straight line is given by the general equation

,

then its slope is determined by the formula

The equation is the equation of a straight line that passes through the point (, ) and has a slope k.

If the line passes through the points (, ), (, ), then its slope is determined by the formula

The equation

is the equation of a straight line passing through two points (, ) and (, ).

If the slope coefficients of two straight lines are known, then one of the angles between these straight lines is determined by the formula

.

A sign of parallelism of two lines is the equality of their angular coefficients:.

A sign of perpendicularity of two lines is the ratio , or .

In other words, the slopes of perpendicular lines are reciprocal in absolute value and opposite in sign.

4. General equation of a straight line

The equation

Ah+Wu+C=0

(Where A, B, C can have any values, as long as the coefficients A, B were not zero both at once) represents straight line. Any straight line can be represented by an equation of this type. Therefore it is called the general equation of a straight line.

If AX, then it represents a line, parallel to the x-axis.

If IN=0, that is, the equation does not contain at, then it represents a line, parallel to the OY axis.

Kogla IN is not equal to zero, then the general equation of a straight line can be resolve relative to ordinateat , then it is converted to the form

(Where a=-A/B; b=-C/B).

Similarly, when A different from zero, the general equation of a straight line can be solved with respect to X.

If WITH=0, that is, the general equation of a straight line does not contain a free term, then it represents a straight line passing through the origin

5. Equation of a straight line passing through a given point with a given slope

Equation of a line passing through a given point A(x 1 , y 1) in a given direction, determined by the slope k,

y - y 1 = k(x - x 1). (1)

This equation defines a pencil of lines passing through a point A(x 1 , y 1), which is called the center of the beam.

6. equation of a straight line passing through two given points.

. Equation of a straight line passing through two points: A(x 1 , y 1) and B(x 2 , y 2) is written like this:

The slope of a straight line passing through two given points is determined by the formula

7. Equation of a straight line in segments

If in the general equation of the line , then dividing (1) by , we obtain the equation of the line in the segments

Where , . The line intersects the axis at the point , the axis at the point .

8. Formula: Angle between lines on a plane

At Goal α between two straight lines given by the equations: y=k 1 x+b 1 (first line) and y=k 2 x+b 2 (second line), can be calculated by the formula (the angle is measured from the 1st line to the 2nd counterclock-wise ):

tg(α)=(k 2 -k 1 )/(1+k 1 k 2 )

9. Mutual arrangement of two straight lines on a plane.

Let both now equations direct lines are written in general view.

Theorem. Let

- are common equations two straight lines coordinate Oxy plane. Then

1) if , then straight and match;

2) if , then the lines and

parallel;

3) if , then straight intersect.

Proof. The condition is equivalent to the collinearity of normal vectors direct data:

Therefore, if , then straight intersect.

If , then , , and the equation straight takes the form:

Or , i.e. straight match up. Note that the coefficient of proportionality , otherwise all the coefficients of the total equations would be zero, which is impossible.

If straight do not coincide and do not intersect, then the case remains, i.e. straight are parallel.

The theorem has been proven.

In the previous chapter, it was shown that, by choosing a certain coordinate system on the plane, we can analytically express the geometric properties characterizing the points of the line under consideration by an equation between the current coordinates. Thus, we get the equation of the line. In this chapter, the equations of straight lines will be considered.

To formulate the equation of a straight line in Cartesian coordinates, you need to somehow set the conditions that determine its position relative to the coordinate axes.

First, we introduce the concept of the slope of a straight line, which is one of the quantities characterizing the position of a straight line on a plane.

Let's call the angle of inclination of the line to the Ox axis the angle by which the Ox axis must be rotated so that it coincides with the given line (or turns out to be parallel to it). As usual, we will consider the angle taking into account the sign (the sign is determined by the direction of rotation: counterclockwise or clockwise). Since an additional rotation of the Ox axis by an angle of 180 ° will again combine it with the straight line, the angle of inclination of the straight line to the axis can be chosen ambiguously (up to a multiple of ).

The tangent of this angle is uniquely determined (since changing the angle to does not change its tangent).

The tangent of the angle of inclination of a straight line to the x-axis is called the slope of the straight line.

The slope characterizes the direction of the straight line (here we do not distinguish between two mutually opposite directions of the straight line). If the slope of the line is zero, then the line is parallel to the x-axis. With a positive slope, the angle of inclination of the straight line to the x-axis will be acute (we consider here the smallest positive value tilt angle) (Fig. 39); in this case, the larger the slope, the greater the angle of its inclination to the Ox axis. If the slope is negative, then the angle of inclination of the straight line to the x-axis will be obtuse (Fig. 40). Note that a straight line perpendicular to the x-axis does not have a slope (the tangent of an angle does not exist).

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 the general rules by 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.

Take the derivative of the given function. You don't need to build a graph here - you only need the equation of the function. In our example, take the derivative of the function f (x) = 2 x 2 + 6 x (\displaystyle f(x)=2x^(2)+6x). Take the derivative according to the methods outlined in the article mentioned above:

Substitute the coordinates of the point given to you into the found derivative to calculate the slope. The derivative of the function is equal to the slope at a certain point. In other words, f "(x) is the slope of the function at any point (x, f (x)). In our example:

If possible, check your answer on a graph. Keep in mind that the slope factor cannot be calculated at every point. Differential calculus considers complex functions and complex graphs, where the slope cannot be calculated at every point, and in some cases the points do not lie on the graphs at all. If possible, use a graphing calculator to check that the slope of the function given to you is correct. Otherwise, draw a tangent to the graph at the given point and consider whether the value of the slope you found corresponds to what you see on the graph.

• The tangent will have the same slope as the function graph at a certain point. To draw a tangent at a given point, move right/left on the x-axis (in our example, 22 values ​​to the right) and then up one on the y-axis. Mark the point and then connect it to the point you've given. In our example, connect the points with coordinates (4,2) and (26,3).

In mathematics, one of the parameters describing the position of a straight line on the Cartesian coordinate plane is the slope of this straight line. This parameter characterizes the slope of the straight line to the x-axis. To understand how to find the slope, first recall the general form of the equation of a straight line in the XY coordinate system.

In general, any line can be represented by the expression ax+by=c, where a, b and c are arbitrary real numbers, but necessarily a 2 + b 2 ≠ 0.

With the help of simple transformations, such an equation can be brought to the form y=kx+d, in which k and d are real numbers. The number k is a slope, and the equation of a straight line of this kind is called an equation with a slope. It turns out that to find the slope, you just need to bring the original equation to the above form. For a better understanding, consider a specific example:

Task: Find the slope of the line given by the equation 36x - 18y = 108

Solution: Let's transform the original equation.

Answer: The desired slope of this line is 2.

If, during the transformation of the equation, we obtained an expression of the type x = const and as a result we cannot represent y as a function of x, then we are dealing with a straight line parallel to the X axis. The slope of such a straight line is equal to infinity.

For lines that are expressed by an equation like y = const, the slope is zero. This is typical for straight lines parallel to the x-axis. For example:

Task: Find the slope of the line given by the equation 24x + 12y - 4(3y + 7) = 4

Solution: We bring the original equation to a general form

24x + 12y - 12y + 28 = 4

It is impossible to express y from the resulting expression, therefore the slope of this 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 BA / AO is the tangent of an acute angle α in a right triangle OAB. 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