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

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

First, we will introduce the concept of the angular coefficient of a line, which is one of the quantities characterizing the position of a line on a plane.

Let's call the angle of inclination of the straight line to the Ox axis the angle by which the Ox axis needs to be rotated so that it coincides with the given line (or is parallel to it). As usual, we will consider the angle taking into account the sign (the sign is determined by the direction of rotation: counterclockwise or clockwise). Since an additional rotation of the Ox axis through an angle of 180° will again align it with the straight line, the angle of inclination of the straight line to the axis can not be chosen unambiguously (to within a term, a multiple of ).

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

The tangent of the angle of inclination of the straight line to the Ox axis is called the angular coefficient of the straight line.

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

Continuation of the topic, the equation of a line on a plane is based on the study of a straight line from algebra lessons. This article provides general information on the topic of equation of a straight line with a slope. Let's consider the definitions, get the equation itself, and identify the connection with other types of equations. Everything will be discussed using examples of problem solving.

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Before writing such an equation, it is necessary to define the angle of inclination of the straight line to the O x axis with their angular coefficient. Let us assume that a Cartesian coordinate system O x on the plane is given.

Definition 1

The angle of inclination of the straight line to the O x axis, located in the Cartesian coordinate system O x y on the plane, this is the angle that is measured from the positive direction O x to the straight line counterclockwise.

When the line is parallel to O x or coincides in it, the angle of inclination is 0. Then the angle of inclination of the given straight line α is 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 picture shows various variations location right angle relative to the coordinate system with the coefficient value.

To find given 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, it is necessary 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 this 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 obtain the canonical equation of a line on a plane using the equation of a line with an angle coefficient. We get x - x 1 a x = y - y 1 a y . It is necessary to move the term b to 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 coefficient has become canonical equation this line.

Example 7

Bring the equation of a straight line with an angular coefficient y = - 3 x + 12 to canonical form.

Solution

Let us calculate and present it in the form of a canonical equation of a straight line. We get an equation of the form:

y = - 3 x + 12 ⇔ - 3 x = y - 12 ⇔ - 3 x - 3 = y - 12 - 3 ⇔ x 1 = y - 12 - 3

Answer: x 1 = y - 12 - 3.

The general equation of a straight line is easiest to obtain from y = k · x + b, but for this it is necessary to make transformations: y = k · x + b ⇔ k · x - y + b = 0. A transition is made from the general equation of the line to equations of 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.

It is necessary to move from the general form of the equation A x + B y + C = 0, where B ≠ 0, to an equation with an angular coefficient. To do this, we solve the equation for y. We get A x + B y + C = 0 ⇔ - A B · x - C B .

The result is an equation with a slope equal to - A B .

Example 9

A straight line equation of the form 2 3 x - 4 y + 1 = 0 is given. Obtain the equation of a given line with an angular coefficient.

Solution

Based on the condition, it is necessary to solve for y, then we obtain an equation of the form:

2 3 x - 4 y + 1 = 0 ⇔ 4 y = 2 3 x + 1 ⇔ y = 1 4 2 3 x + 1 ⇔ y = 1 6 x + 1 4 .

Answer: y = 1 6 x + 1 4 .

An equation of the form x a + y b = 1 is solved in a similar way, which is called the equation of a straight line in segments, or canonical of the form 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. 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. 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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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 or curved line. That is, the derivative characterizes the rate of change of a function at a specific point in time. Remember 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 of an exponential equation, is described. The calculations presented in the following steps will be based on the methods described therein.

Learn to distinguish problems in which the slope coefficient needs to be calculated through the derivative of a function. Problems do not always ask you 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 function given to you. There is no need to build a graph here - you only need the equation of the function. In our example, take the derivative of the function. Take the derivative according to the methods outlined in the article mentioned above:

• Derivative:

Substitute the coordinates of the point given to you into the found derivative to calculate the slope. The derivative of a 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:

• Find the slope of the function f (x) = 2 x 2 + 6 x (\displaystyle f(x)=2x^(2)+6x) at point A(4,2).
• Derivative of a function:
  • f ′ (x) = 4 x + 6 (\displaystyle f"(x)=4x+6)
• Substitute the value of the “x” coordinate of this point:
  • f ′ (x) = 4 (4) + 6 (\displaystyle f"(x)=4(4)+6)
• Find the slope:
• Slope function f (x) = 2 x 2 + 6 x (\displaystyle f(x)=2x^(2)+6x) at point A(4,2) is equal to 22.

If possible, check your answer on a graph. Remember that the slope cannot be calculated at every point. Differential calculus is considering 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 you are given is correct. Otherwise, draw a tangent to the graph at the point given to you and think about whether the slope value you found matches what you see on the graph.

• The tangent will have the same slope as the graph of the function at a certain point. To draw a tangent at a given point, move left/right 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 given to you. In our example, connect the points with coordinates (4,2) and (26,3).

Equation of a straight line on a plane.
The direction vector is straight. Normal vector

A straight line on a plane is one of the simplest geometric shapes, familiar to you since elementary school, and today we will learn how to deal with it using the methods of analytical geometry. To master the material, you must be able to build a straight line; know what equation defines a straight line, in particular, a straight line passing through the origin of coordinates and straight lines parallel to the coordinate axes. This information can be found in the manual Graphs and properties of elementary functions, I created it for Matan, but the section about the linear function turned out to be very successful and detailed. Therefore, dear teapots, warm up there first. In addition, you need to have basic knowledge about vectors, otherwise the understanding of the material will be incomplete.

In this lesson we will look at ways in which you can create an equation of a straight line on a plane. I recommend not to neglect practical examples (even if it seems very simple), since I will provide them with elementary and important facts, technical techniques that will be required in the future, including in other sections of higher mathematics.

• How to write an equation of a straight line with an angle coefficient?
• How ?
• How to find a direction vector using the general equation of a straight line?
• How to write an equation of a straight line given a point and a normal vector?

and we begin:

Equation of a straight line with slope

The well-known “school” form of a straight line equation is called equation of a straight line with slope. For example, if a straight line is given by the equation, then its slope is: . Let's consider the geometric meaning of this coefficient and how its value affects the location of the line:

In a geometry course it is proven that the slope of the straight line is equal to tangent of the angle between positive axis directionand this line: , and the angle “unscrews” counterclockwise.

In order not to clutter the drawing, I drew angles only for two straight lines. Let's consider the “red” line and its slope. According to the above: (the “alpha” angle is indicated by a green arc). For the “blue” straight line with the angle coefficient, the equality is true (the “beta” angle is indicated by a brown arc). And if the tangent of the angle is known, then if necessary it is easy to find and the corner itself by using inverse function– arctangent. As they say, a trigonometric table or a microcalculator in your hands. Thus, the angular coefficient characterizes the degree of inclination of the straight line to the abscissa axis.

The following cases are possible:

1) If the slope is negative: then the line, roughly speaking, goes from top to bottom. Examples are the “blue” and “raspberry” straight lines in the drawing.

2) If the slope is positive: then the line goes from bottom to top. Examples - “black” and “red” straight lines in the drawing.

3) If the slope is zero: , then the equation takes the form , and the corresponding straight line is parallel to the axis. An example is the “yellow” straight line.

4) For a family of lines parallel to an axis (there is no example in the drawing, except for the axis itself), the angular coefficient does not exist (tangent of 90 degrees is not defined).

The greater the slope coefficient in absolute value, the steeper the straight line graph goes..

For example, consider two straight lines. Here, therefore, the straight line has a steeper slope. Let me remind you that the module allows you to ignore the sign, we are only interested in absolute values angular coefficients.

In turn, a straight line is steeper than straight lines .

Conversely: the smaller the slope coefficient in absolute value, the flatter the straight line.

For straight lines the inequality is true, thus the straight line is flatter. Children's slide, so as not to give yourself bruises and bumps.

Why is this necessary?

Prolong your torment Knowledge of the above facts allows you to immediately see your mistakes, in particular, errors when constructing graphs - if the drawing turns out to be “obviously something wrong.” It is advisable that you straightaway it was clear that, for example, the straight line is very steep and goes from bottom to top, and the straight line is very flat, pressed close to the axis and goes from top to bottom.

In geometric problems, several straight lines often appear, so it is convenient to designate them somehow.

Designations: straight lines are designated in small Latin letters: . A popular option is to designate them using the same letter with natural subscripts. For example, the five lines we just looked at can be denoted by .

Since any straight line is uniquely determined by two points, it can be denoted by these points: etc. The designation clearly implies that the points belong to the line.

It's time to warm up a little:

How to write an equation of a straight line with an angle coefficient?

If a point belonging to a certain line and the angular coefficient of this line are known, then the equation of this line is expressed by the formula:

Example 1

Write an equation of a straight line with an angular coefficient if it is known that the point belongs to this straight line.

Solution: Let's compose the equation of the straight line using the formula . In this case:

Answer:

Examination is done simply. First, we look at the resulting equation and make sure that our slope is in place. Secondly, the coordinates of the point must satisfy this equation. Let's plug them into the equation:

The correct equality is obtained, which means that the point satisfies the resulting equation.

Conclusion: The equation was found correctly.

A more tricky example to solve on your own:

Example 2

Write an equation for a straight line if it is known that its angle of inclination to the positive direction of the axis is , and the point belongs to this straight line.

If you have any difficulties, re-read theoretical material. More precisely, more practical, I skip a lot of evidence.

The last bell rang and died down prom, and outside the gates of our native school, analytical geometry itself awaits us. The jokes are over... Or maybe they are just beginning =)

We nostalgically wave our pen to the familiar and get acquainted with the general equation of a straight line. Because in analytical geometry this is exactly what is used:

The general equation of a straight line has the form: , where are some numbers. At the same time, the coefficients simultaneously are not equal to zero, since the equation loses its meaning.

Let's dress in a suit and tie the equation with the slope coefficient. First, let's move all the terms to the left side:

The term with “X” must be put in first place:

In principle, the equation already has the form , but according to the rules of mathematical etiquette, the coefficient of the first term (in this case) must be positive. Changing signs:

Remember this technical feature! We make the first coefficient (most often) positive!

In analytical geometry, the equation of a straight line will almost always be given in general form. Well, if necessary, it can be easily reduced to the “school” form with an angular coefficient (with the exception of straight lines parallel to the ordinate axis).

Let's ask ourselves what enough know to construct a straight line? Two points. But more about this childhood incident, now sticks with arrows rule. Each straight line has a very specific slope, which is easy to “adapt” to. vector.

A vector that is parallel to a line is called the direction vector of that line. It is obvious that any straight line has an infinite number of direction vectors, and all of them will be collinear (codirectional or not - it doesn’t matter).

I will denote the direction vector as follows: .

But one vector is not enough to construct a straight line; the vector is free and not tied to any point on the plane. Therefore, it is additionally necessary to know some point that belongs to the line.

How to write an equation of a straight line using a point and a direction vector?

If a certain point belonging to a line and the direction vector of this line are known, then the equation of this line can be compiled using the formula:

Sometimes it is called canonical equation of the line .

What to do when one of the coordinates is equal to zero, we will understand in practical examples below. By the way, please note - both at once coordinates cannot be equal to zero, since the zero vector does not specify a specific direction.

Example 3

Write an equation for a straight line using a point and a direction vector

Solution: Let's compose the equation of a straight line using the formula. In this case:

Using the properties of proportion we get rid of fractions:

And we bring the equation to its general form:

Answer:

As a rule, there is no need to make a drawing in such examples, but for the sake of understanding:

In the drawing we see the starting point, the original direction vector (it can be plotted from any point on the plane) and the constructed straight line. By the way, in many cases it is most convenient to construct a straight line using an equation with an angular coefficient. It’s easy to transform our equation into form and easily select another point to construct a straight line.

As noted at the beginning of the paragraph, a straight line has infinitely many direction vectors, and all of them are collinear. For example, I drew three such vectors: . Whatever direction vector we choose, the result will always be the same straight line equation.

Let's create an equation of a straight line using a point and a direction vector:

Resolving the proportion:

Divide both sides by –2 and get the familiar equation:

Those interested can test vectors in the same way or any other collinear vector.

Now let's solve the inverse problem:

How to find a direction vector using the general equation of a straight line?

Very simple:

If a line is given by a general equation in a rectangular coordinate system, then the vector is the direction vector of this line.

Examples of finding direction vectors of straight lines:

The statement allows us to find only one direction vector out of an infinite number, but we don’t need more. Although in some cases it is advisable to reduce the coordinates of the direction vectors:

Thus, the equation specifies a straight line that is parallel to the axis and the coordinates of the resulting direction vector are conveniently divided by –2, obtaining exactly the basis vector as the direction vector. Logical.

Similarly, the equation specifies a straight line parallel to the axis, and by dividing the coordinates of the vector by 5, we obtain the ort vector as the direction vector.

Now let's do it checking Example 3. The example went up, so I remind you that in it we compiled the equation of a straight line using a point and a direction vector

Firstly, using the equation of the straight line we restore its direction vector: – everything is fine, we have received the original vector (in some cases the result may be a collinear vector to the original one, and this is usually easy to notice by the proportionality of the corresponding coordinates).

Secondly, the coordinates of the point must satisfy the equation. We substitute them into the equation:

The correct equality was obtained, which we are very happy about.

Conclusion: The task was completed correctly.

Example 4

Write an equation for a straight line using a point and a direction vector

This is an example for you to solve on your own. The solution and answer are at the end of the lesson. It is highly advisable to check using the algorithm just discussed. Try to always (if possible) check on a draft. It’s stupid to make mistakes where they can be 100% avoided.

In the event that one of the coordinates of the direction vector is zero, proceed very simply:

Example 5

Solution: The formula is not suitable since the denominator on the right side is zero. There is an exit! Using the properties of proportion, we rewrite the formula in the form, and the rest rolled along a deep rut:

Answer:

Examination:

1) Restore the directing vector of the line:
– the resulting vector is collinear to the original direction vector.

2) Substitute the coordinates of the point into the equation:

The correct equality is obtained

Conclusion: task completed correctly

The question arises, why bother with the formula if there is a universal version that will work in any case? There are two reasons. First, the formula is in the form of a fraction much better remembered. And secondly, the disadvantage of the universal formula is that the risk of getting confused increases significantly when substituting coordinates.

Example 6

Write an equation for a straight line using a point and a direction vector.

This is an example for you to solve on your own.

Let's return to the ubiquitous two points:

How to write an equation of a straight line using two points?

If two points are known, then the equation of a straight line passing through these points can be compiled using the formula:

In fact, this is a type of formula and here's why: if two points are known, then the vector will be the direction vector of the given line. At the lesson Vectors for dummies we considered the simplest problem - how to find the coordinates of a vector from two points. According to this problem, the coordinates of the direction vector are:

Note : the points can be “swapped” and the formula can be used . Such a solution will be equivalent.

Example 7

Write an equation of a straight line using two points .

Solution: We use the formula:

Combing the denominators:

And shuffle the deck:

Now is the time to get rid of fractional numbers. In this case, you need to multiply both sides by 6:

Open the brackets and bring the equation to mind:

Answer:

Examination is obvious - the coordinates of the initial points must satisfy the resulting equation:

1) Substitute the coordinates of the point:

True equality.

2) Substitute the coordinates of the point:

True equality.

Conclusion: The equation of the line is written correctly.

If at least one of the points does not satisfy the equation, look for an error.

It is worth noting that graphical verification in this case is difficult, since construct a straight line and see whether the points belong to it , not so simple.

I’ll note a couple more technical aspects of the solution. Perhaps in this problem it is more profitable to use the mirror formula and, at the same points make an equation:

Fewer fractions. If you want, you can carry out the solution to the end, the result should be the same equation.

The second point is to look at the final answer and figure out whether it can be simplified further? For example, if you get the equation , then it is advisable to reduce it by two: – the equation will define the same straight line. However, this is already a topic of conversation about relative position of lines.

Having received the answer in Example 7, just in case, I checked whether ALL coefficients of the equation are divisible by 2, 3 or 7. Although, most often such reductions are made during the solution.

Example 8

Write an equation for a line passing through the points .

This is an example for an independent solution, which will allow you to better understand and practice calculation techniques.

Similar to the previous paragraph: if in the formula one of the denominators (the coordinate of the direction vector) becomes zero, then we rewrite it in the form . Again, notice how awkward and confused she looks. I don’t see much point in bringing practical examples, since we have already actually solved such a problem (see No. 5, 6).

Direct normal vector (normal vector)

What is normal? In simple words, normal is perpendicular. That is, the normal vector of a line is perpendicular to a given line. Obviously, any straight line has an infinite number of them (as well as direction vectors), and all the normal vectors of the straight line will be collinear (codirectional or not, it makes no difference).

Dealing with them will be even easier than with guide vectors:

If a line is given by a general equation in a rectangular coordinate system, then the vector is the normal vector of this line.

If the coordinates of the direction vector have to be carefully “pulled out” from the equation, then the coordinates of the normal vector can be simply “removed”.

The normal vector is always orthogonal to the direction vector of the line. Let us verify the orthogonality of these vectors using dot product:

I will give examples with the same equations as for the direction vector:

Is it possible to construct an equation of a straight line given one point and a normal vector? I feel it in my gut, it’s possible. If the normal vector is known, then the direction of the straight line itself is clearly defined - this is a “rigid structure” with an angle of 90 degrees.

How to write an equation of a straight line given a point and a normal vector?

If a certain point belonging to a line and the normal vector of this line are known, then the equation of this line is expressed by the formula:

Here everything worked out without fractions and other surprises. This is our normal vector. Love him. And respect =)

Example 9

Write an equation of a straight line given a point and a normal vector. Find the direction vector of the line.

Solution: We use the formula:

The general equation of the straight line has been obtained, let’s check:

1) “Remove” the coordinates of the normal vector from the equation: – yes, indeed, the original vector was obtained from the condition (or a collinear vector should be obtained).

2) Let's check whether the point satisfies the equation:

True equality.

After we are convinced that the equation is composed correctly, we will complete the second, easier part of the task. We take out the directing vector of the straight line:

Answer:

In the drawing the situation looks like this:

For training purposes, a similar task for solving independently:

Example 10

Write an equation of a straight line given a point and a normal vector. Find the direction vector of the line.

The final section of the lesson will be devoted to less common, but also important species equations of a straight line on a plane

Equation of a straight line in segments.
Equation of a line in parametric form

The equation of a straight line in segments has the form , where are nonzero constants. Some types of equations cannot be represented in this form, for example, direct proportionality (since the free term is equal to zero and there is no way to get one on the right side).

This is, figuratively speaking, a “technical” type of equation. A common task is to represent the general equation of a line as an equation of a line in segments. How is it convenient? The equation of a line in segments allows you to quickly find the points of intersection of a line with coordinate axes, which can be very important in some problems of higher mathematics.

Let's find the point of intersection of the line with the axis. We reset the “y” and the equation takes the form . The desired point is obtained automatically: .

Same with the axis – the point at which the straight line intersects the ordinate axis.

In mathematics, one of the parameters that describes the position of a line on the Cartesian coordinate plane is the angular coefficient of this 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 view any 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 of the acute angle α 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. The tangent of a 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 a certain point is a constant numerically equal to the tangent of the angle that is formed between the tangent at the 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 the 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 the 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