The task is to construct a line passing through the given coordinates of the end of a segment.

We believe that the segment is non-degenerate, i.e. has a length greater than zero (otherwise, of course, there are infinitely many different lines passing through it).

Two-dimensional case

Let a segment be given, i.e. the coordinates of its ends , , , are known.

Required to build equation of a line in a plane, passing through this segment, i.e. find the coefficients , , in the equation of the straight line:

Note that the required triples passing through a given segment are infinitely many: You can multiply all three coefficients by an arbitrary non-zero number and get the same straight line. Therefore, our task is to find one of these triplets.

It is easy to verify (by substituting these expressions and the coordinates of the points and into the equation of the straight line) that the following set of coefficients is suitable:

Integer case

An important advantage of this method of constructing a straight line is that if the coordinates of the ends were integer, then the resulting coefficients will also be integers. In some cases, this allows geometric operations to be performed without resorting to real numbers at all.

However, there is a small drawback: for the same line, different triplets of coefficients can be obtained. To avoid this, but not move away from integer coefficients, you can use the following technique, often called rationing. Let's find the greatest common divisor of the numbers , , , divide all three coefficients by it, and then normalize the sign: if or , then multiply all three coefficients by . As a result, we will come to the conclusion that for identical lines we will obtain identical triplets of coefficients, which will make it easy to check lines for equality.

Real-valued case

When working with real numbers, you should always be aware of errors.

The coefficients we obtain are of the order of the original coordinates, the coefficient is already of the order of the square of them. These can already be quite large numbers, and, for example, when the lines intersect, they will become even larger, which can lead to large rounding errors even with the original coordinates of order .

Therefore, when working with real numbers, it is advisable to perform the so-called normalization direct: namely, to make the coefficients such that . To do this you need to calculate the number:

and divide all three coefficients , , by it.

Thus, the order of the coefficients and will no longer depend on the order of the input coordinates, and the coefficient will be of the same order as the input coordinates. In practice, this leads to a significant improvement in calculation accuracy.

Finally, let's mention comparison straight lines - after all, after such normalization for the same straight line, only two triplets of coefficients can be obtained: up to multiplication by . Accordingly, if we carry out additional normalization taking into account the sign (if or , then multiply by ), then the resulting coefficients will be unique.

We continue to study the section “Equation of a line on a plane” and in this article we will examine the topic “Equation of a line in segments”. We will sequentially consider the form of the equation of a line in segments, the construction of a straight line, which is given by this equation, the transition from the general equation of a line to the equation of a line in segments. All this will be accompanied by examples and analysis of problem solving.

Let there be a rectangular coordinate system O x y on the plane.

A straight line on a plane in the Cartesian coordinate system O x y is given by an equation of the form x a + y b = 1, where a and b are some real numbers, non-zero, the values ​​of which are equal to the lengths of the segments cut off by the straight line on the O x and O y axes. The lengths of the segments are calculated from the origin.

As we know, the coordinates of any of the points belonging to a straight line given by the equation of a straight line satisfy the equation of this straight line. Points a, 0 and 0, b belong to this straight line, since a a + 0 b = 1 ⇔ 1 ≡ 1 and 0 a + b b = 1 ⇔ 1 ≡ 1. Points a, 0 and b, 0 are located on the coordinate axes O x and O y and are removed from the origin by a and b units. The direction in which the length of the segment should be plotted is determined by the sign that appears before the numbers a and b. The “-” sign means that the length of the segment must be plotted in the negative direction of the coordinate axis.

Let us explain all of the above by placing straight lines relative to a fixed Cartesian coordinate system O x y on a schematic drawing. The equation of a straight line in segments x a + y b = 1 is used to construct a straight line in the Cartesian coordinate system O x y. To do this, we need to mark points a, 0 and b, 0 on the axes, and then connect these points with a line using a ruler.

The drawing shows cases when the numbers a and b have different signs, and, therefore, the lengths of the segments are plotted in different directions of the coordinate axes.

Let's look at an example.

Example 1

A straight line is given by the equation of a straight line in segments of the form x 3 + y - 5 2 = 1. It is necessary to construct this line on a plane in the Cartesian coordinate system O x y.

Solution

Using the equation of a straight line in segments, we determine the points through which the straight line passes. This is 3, 0, 0, - 5 2. Let's mark them and draw a line.

Reducing the general equation of a line to the equation of a line in segments

The transition from a given equation of a line to an equation of a line in segments makes it easier for us to solve various problems. Having a complete general equation of a line, we can obtain the equation of a line in segments.

The complete general equation of a straight line on a plane is A x + B y + C = 0, where A, B and C are not equal to zero. We transfer the number C to the right side of the equality, divide both sides of the resulting equality by – C. At the same time, we send the coefficients of x and y to the denominators:

A x + B y + C = 0 ⇔ A x + B y = - C ⇔ ⇔ A - C x + B - C y = 1 ⇔ x - C A + y - C B = 1

To carry out the last transition, we used the equality p q = 1 q p, p ≠ 0, q ≠ 0.

As a result, we have made the transition from the general equation of the straight line A x + B y + C = 0 to the equation of the straight line in the segments x a + y b = 1, where a = - C A, b = - C B.

Let's look at the following example.

Example 2

Let us make the transition to the equation of a straight line in segments, having a general equation of a straight line x - 7 y + 1 2 = 0.

Solution

We move one second to the right side of the equality x - 7 y + 1 2 = 0 ⇔ x - 7 y = - 1 2 .

We divide both sides of the equality by - 1 2: x - 7 y = - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y = 1.

Let us transform the resulting equality to the right type: 1 - 1 2 x - 7 - 1 2 y = 1 ⇔ x - 1 2 + y 1 14 = 1 .

We have obtained the equation of a straight line in segments.

Answer: x - 1 2 + y 1 14 = 1

In cases where a straight line is given by a canonical or parametric equation of a line on a plane, then first we move on to the general equation of the line, and then to the equation of the line in segments.

Moving from the equation of a line in segments to the general equation of a line is simple: we transfer the unit from the right side of the equation of a line in segments of the form x a + y b = 1 to the left side with the opposite sign, selecting the coefficients in front of the unknowns x and y.

x a + y b = 1 ⇔ x a + y b - 1 = 0 ⇔ 1 a x + 1 b y - 1 = 0

We obtain a general equation of a line, from which we can go to any other type of equation of a line on the plane. We discussed the transition process in detail in the topic “Reducing the general equation of a line to other types of equation of a line.”

Example 3

The equation of a straight line in segments has the form x 2 3 + y - 12 = 1. It is necessary to write the general equation of a straight line on a plane.

Solution

It operates according to a previously described algorithm:

x 2 3 + y - 12 = 1 ⇔ 1 2 3 x + 1 - 12 y - 1 = 0 ⇔ ⇔ 3 2 x - 1 12 y - 1 = 0

Answer: 3 2 x - 1 12 y - 1 = 0

If you notice an error in the text, please highlight it and press Ctrl+Enter

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 junior classes, 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 linear function It turned out 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 for independent decision:

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 has rung, the graduation ceremony has ended, 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:

General equation the 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 some 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 general appearance:

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 convert 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 an infinite number of 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, 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 unit 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 reconstruct 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 straight 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 giving practical examples, since we have already actually solved this 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 types of equations of a 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” to zero, 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.

Line equation of the form , where a And b– some real numbers other than zero are called equation of a straight line in segments. This name is not accidental, since the absolute values ​​of numbers A And b equal to the lengths of the segments that the straight line cuts off on the coordinate axes Ox And Oy respectively (segments are counted from the origin). Thus, the equation of a line in segments makes it easy to construct this line in a drawing. To do this, you should mark the points with coordinates and in a rectangular coordinate system on the plane, and use a ruler to connect them with a straight line.

For example, let's construct a straight line given by an equation in segments of the form . Mark the points and connect them.

You can get detailed information about this type of equation of a line on a plane in the article equation of a line in segments.

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Algebra and analytical geometry. The concept of a matrix, operations on matrices and their properties

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Definition of differentiability
The operation of finding the derivative is called differentiation of a function. A function is said to be differentiable at some point if it has a finite derivative at that point, and

Rule of differentiation
Corollary 1. The constant factor can be taken out of the sign of the derivative:

Geometric meaning of derivative. Tangent equation
The angle of inclination of a straight line y = kx+b is the angle measured from the position

Geometric meaning of the derivative of a function at a point
Let us consider the secant AB of the graph of the function y = f(x) such that points A and B have coordinates, respectively

Solution
Function defined for everyone real numbers. Since (-1; -3) is a point of tangency, then

Necessary conditions for an extremum and sufficient conditions for an extremum
Definition of an increasing function. The function y = f(x) increases on the interval X if for any

Sufficient signs of an extremum of a function
To find the maxima and minima of a function, you can use any of three sufficient indications extremum. Although the most common and convenient is the first one.

Basic properties of a definite integral. Property 1. Derivative of definite integral at the upper limit is equal to the integrand into which instead of a variable is integrated

Newton-Leibniz formula (with proof)
Newton-Leibniz formula. Let the function y = f(x) be continuous on an interval and F(x) be one of the antiderivatives of the function on this interval, then the equation

Equation of a line in segments

Let the general equation of a straight line be given:

The equation of a straight line in segments, where are the segments that the straight line cuts off on the corresponding coordinate axes.

Construct a straight line given by the general equation:

From which we can construct an equation of this line in segments:

The relative position of lines on a plane.

Statement 1.

In order for straight lines and given by equations:

Coincidence is necessary and sufficient so that:

Proof: and coincide, their direction vectors and are collinear, i.e.:

Let's take point M 0 with this straight line, then:

Multiplying the first equation by and adding to the second by (2) we get:

So, formulas (2), (3) and (4) are equivalent. Let (2) be satisfied, then the equations of the system (*) are equivalent; the corresponding straight lines coincide.

Statement 2.

The lines and given by equations (*) are parallel and do not coincide if and only if:

Proof:

Even if they don't match:

Inconsistent, i.e., according to the Kronecker-Capelli theorem:

This is only possible if:

That is, when condition (5) is met.

When the first equality (5) is fulfilled, - failure to satisfy the second equality results in the incompatibility of the system (*) the lines are parallel and do not coincide.

Note 1.

Polar coordinate system.

Let us fix a point on the plane and call it a pole. The ray emanating from the pole will be called the polar axis.

Let's choose a scale for measuring the lengths of segments and agree that the rotation around the point counterclockwise will be considered positive. Consider any point on given plane, denote by its distance to the pole and call it the polar radius. The angle by which the polar axis must be rotated so that it coincides with will be denoted by and called the polar angle.

Definition 3.

The polar coordinates of a point are its polar radius and polar angle:

Remark 2. in the pole. The value for points other than a point is determined up to a term.

Consider a Cartesian rectangular coordinate system: the pole coincides with the origin, and the polar axis coincides with the positive semi-axis. Here. Then:

What is the relationship between rectangular Cartesian and polar coordinate systems.

Bernoulli's lemniscate equation. Write it in the polar coordinate system.

Normal equation of a line on a plane. Let the polar axis coincide with, - the axis passing through the origin. Let be:

Let then:

Condition (**) for point:

Equation of a straight line in a polar coordinate system.

Here - the length drawn from the origin to the straight line, - the angle of inclination of the normal to the axis.

Equation (7) can be rewritten:

Normal equation of a line on a plane.