In geometry, such a concept is often considered as "the top of a triangle." This is the intersection point of the two sides of this shape. This concept is encountered in almost every problem, so it makes sense to consider it in more detail.

Determining the vertex of a triangle

A triangle has three points of intersection of the sides, forming three corners. They are called vertices, and the sides on which they rest are called the sides of a triangle.

Figure: 1. Vertex in a triangle.

The vertices in the triangles are denoted in capital Latin letters. Therefore, most often in mathematics, the sides are denoted by two capital Latin letters, according to the names of the vertices that enter the sides. For example, side AB is the side of a triangle connecting vertices A and B.

Figure: 2. Designation of vertices in a triangle.

Concept characteristics

If we take a triangle arbitrarily oriented in the plane, then in practice it is very convenient to express its geometric characteristics through the coordinates of the vertices of this figure. So, the vertex A of the triangle can be expressed by a point with certain numerical parameters A (x; y).

Knowing the coordinates of the vertices of the triangle, you can find the points of intersection of the medians, the length of the height dropped to one of the sides of the figure, and the area of \u200b\u200bthe triangle.

For this, the properties of vectors depicted in the Cartesian coordinate system are used, because the length of the side of the triangle is determined through the length of the vector with the points at which the corresponding vertices of this figure are located.

Using the vertex of a triangle

At any vertex of the triangle, you can find an angle that will be adjacent to the inner corner of the figure in question. To do this, you will have to extend one of the sides of the triangle. Since there are two sides at each vertex, there are two outer corners at each vertex. The outer angle is equal to the sum of the two inner angles of the triangle that are not adjacent to it.

Figure: 3. Property of the outer corner of a triangle.

If you build two outer corners at one vertex, then they will be equal, as vertical.

What have we learned?

One of the important concepts of geometry when considering different types of triangles is the vertex. This is the point where the two sides of the corner of this geometric shape intersect. It is designated by one of the large letters of the Latin alphabet. The vertex of a triangle can be expressed in terms of x and y coordinates, which helps define the length of the side of a triangle as the length of a vector.

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SectionV... ANALYTICAL GEOMETRY ON A PLANE

AND IN SPACE

The section includes tasks that are considered in the topic "Analytical geometry on a plane and in space": drawing up various equations of straight lines on a plane and in space; determination of the relative position of straight lines on a plane, straight lines, a straight line and a plane, planes in space; image of curves of the second order. It should be noted that this section presents the problems of economic content, in solving which information from analytical geometry on a plane is used.

When solving problems of analytical geometry, it is advisable to use the tutorials of the following authors: D.V. Kletenik, N. Sh. Kremer, D.T. V.I. Written Malykhin, because this literature covers a wider range of tasks that can be used for self-study on this topic. The application of analytical geometry to solving economic problems is described in the educational publications of M.S. Crassus and V.I. Ermakova.

Task 5.1. The coordinates of the vertices of the triangle are givenABC ... It is necessary

a) write the equations of the sides of the triangle;

b) write the equation for the height of the triangle drawn from the vertexFROM to the sideAB and find its length;

c) write the equation for the median of the triangle drawn from the vertexIN to the sideAS ;

d) find the corners of the triangle and establish its appearance (rectangular, acute-angled, obtuse-angled);

e) find the lengths of the sides of the triangle and determine its type (versatile, isosceles, equilateral);

f) find the coordinates of the center of gravity (the point of intersection of the medians) of the triangleABC ;

g) find the coordinates of the orthocenter (point of intersection of heights) of the triangleABC .

For each of points a) - c) decisions to make drawings in the coordinate system. In the figures, mark the lines and points corresponding to the points of the problem.

Example 5.1

The coordinates of the vertices of the triangle are givenABC : ... It is necessary a) write the equations of the sides of the triangle; b) write the equation for the height of the triangle drawn from the vertex FROM to the sideAB and find its length; c) write the equation for the median of the triangle drawn from the vertexIN to the sideAS ; d) find the lengths of the sides of the triangle and determine its type (versatile, isosceles, equilateral); e) find the corners of the triangle and establish its appearance (rectangular, acute-angled, obtuse-angled); f) find the coordinates of the center of gravity (the point of intersection of the medians) of the triangle ABC ; g) find the coordinates of the orthocenter (point of intersection of heights) of the triangleABC .

Decision

and) For each side of the triangle, the coordinates of two points that lie on the sought lines are known, which means that the equations of the sides of the triangle are equations of lines passing through two given points

,

where
and
corresponding coordinates of points.

Thus, substituting the coordinates of the points corresponding to straight lines into formula (5.1), we obtain

,
,
,

whence, after transformations, we write down the equations of the sides

In fig. 7 draw the corresponding sides of the triangle
straight lines.

Answer:

b) Let be
- height drawn from the top to the side
... Insofar as
goes through the point perpendicular to vector
, then we compose the equation of the straight line according to the following formula

where
- coordinates of the vector perpendicular to the sought line,
- coordinates of a point belonging to this line. Find the coordinates of a vector perpendicular to the straight line
, and substitute in formula (5.2)

,
,

.

Find the length of the height CH as distance from point to straight

,

where
- equation of a straight line
,
- point coordinates .

In the previous paragraph, it was found

Substituting the data into formula (5.3), we obtain

,

In fig. 8 draw a triangle and the found height CH.

Answer: .

R is. 8

in) median
triangle
divides the side
into two equal parts, i.e. point is the midpoint of the segment
... Based on this, you can find the coordinates
points

, ,

where
and
and , substituting them into formulas (5.4), we obtain

;
.

Median equation
triangle
as an equation of the straight line passing through the points
and
by formula (5.1)

,

.

Answer: (fig. 9).

R is. nine

d) We find the lengths of the sides of the triangle as the lengths of the corresponding vectors, i.e.

,
,
.

Parties
and
triangle
are equal, which means that the triangle is isosceles with the base
.

Answer: triangle
isosceles base
;

,
.

e) Angles of a triangle
find as the angles between the vectors outgoing from the corresponding vertices of the given triangle, i.e.

,
,
.

Since the triangle is isosceles with a base
then

,

The angles between the vectors are calculated by formula (4.4), which requires the scalar products of vectors
,
.

Find the coordinates and modules of the vectors required to calculate the angles

,
;

,
,
.

Substituting the found data into formula (4.4), we obtain

,

Since the values \u200b\u200bof the cosines of all the angles found are positive, the triangle
is acute-angled.

Answer: triangle
acute-angled;

,
,
.

e) Let be

, then the coordinates
points
can be found by formulas (5.5)

, ,

where
,
and
- coordinates of points respectively , and , hence,

,
.

Answer:
- the center of gravity of the triangle
.

g)Let be - orthocenter of a triangle
... Find the coordinates of the point as the coordinates of the intersection of the triangle's heights. Height equation
was found in paragraph b)... Find the height equation
:

,
,

.

Insofar as
, then the solution of the system

is the coordinates of the point where we find
.

Answer:
- orthocenter of a triangle
.

Task 5.2. The fixed costs at the enterprise for the release of some products areF V 0 rub. per unit of production, while the revenue isR 0 rub. per unit of manufactured products. Make up a profit functionP (q ) (q

Data for the problem condition corresponding to the options:

Example 5.2

The fixed costs at the enterprise for the release of some products are
rub. per month, variable costs -
rub. per unit of production, while the revenue is
rub. per unit of manufactured products. Make up a profit functionP (q ) (q - the number of products produced); build its graph and determine the break-even point.

Decision

We calculate the total production costs for the release q units of some products

If sold q units of production, then the total income will be

Based on the obtained functions of total income and total costs, we find the profit function

,

.

Break-even point - the point at which profit is zero, or the point at which total cost is equal to total income

,

,

where do we find

- breakeven point.

To build a graph (Fig. 10) of the profit function, we find one more point

Answer: profit function
, breakeven point
.

Task 5.3. The laws of supply and demand for a certain product are respectively determined by the equationsp = p D (q ), p = p S (q ), wherep - the price of the product,q - quantity of goods. It is assumed that demand is determined only by the price of the product on the marketp FROM , and the offer is only at the pricep S received by suppliers. It is necessary

a) determine the point of market equilibrium;

b) the equilibrium point after the introduction of a tax equal tot ... Determine price increases and decreases in equilibrium sales;

c) find a subsidys which will lead to an increase in sales byq 0 units relative to the original (defined in point a));

d) find a new equilibrium point and government revenue with the introduction of a tax proportional to the price and equal toN %;

e) determine how much money will be spent by the government to buy up the surplus when setting the minimum price equal to p 0 .

Draw a drawing in the coordinate system for each decision point. In the figure, mark the lines and points corresponding to the task item.

Data for the problem condition corresponding to the options:

How to learn to solve problems in analytical geometry?
A typical problem with a triangle on a plane

This lesson is created on the approach to the equator between plane geometry and space geometry. At the moment, there is a need to systematize the accumulated information and answer a very important question: how to learn to solve problems in analytical geometry? The difficulty lies in the fact that you can think of an infinite number of problems in geometry, and no textbook will contain the entire set and variety of examples. Is not derivative of a function with five rules of differentiation, a table and several techniques….

There is a solution! I will not say big words that I have developed some kind of grandiose technique, however, in my opinion, there is an effective approach to the problem under consideration, which allows even a full teapot to achieve good and excellent performance. At least, the general algorithm for solving geometric problems is very clearly formed in my head.

WHAT YOU NEED to know and be able to
to successfully solve geometry problems?

There is no getting away from this - in order not to poke buttons at random, you need to master the basics of analytical geometry. Therefore, if you have just started studying geometry or have completely forgotten it, please start with a lesson Vectors for dummies ... In addition to vectors and actions with them, you need to know the basic concepts of plane geometry, in particular, equation of a straight line on a plane and. The geometry of space is presented by articles Plane equation , Equations of a straight line in space , The main tasks on the line and plane and some other lessons. Curved lines and spatial surfaces of the second order are somewhat apart, and there are not so many specific problems with them.

Suppose a student already has elementary knowledge and skills in solving the simplest problems of analytical geometry. But it happens like this: you read the condition of the problem, and ... you want to close the whole thing altogether, throw it into a far corner and forget it, like about a bad dream. Moreover, this does not fundamentally depend on the level of your qualifications; from time to time I myself come across tasks for which the solution is not obvious. What to do in such cases? Do not be afraid of a task that you do not understand!

Firstly, you should install - is it a "flat" or spatial problem? For example, if vectors with two coordinates appear in the condition, then, of course, this is the geometry of the plane. And if the teacher loaded the grateful listener with a pyramid, then this is clearly the geometry of space. The results of the first step are already quite good, because they managed to cut off a huge amount of unnecessary information for this task!

Second... The condition will usually preoccupy you with some geometric shape. Indeed, walk along the corridors of your native university, and you will see a lot of concerned faces.

In "flat" problems, not to mention the taken for granted points and lines, the most popular figure is the triangle. We will analyze it in great detail. Next comes the parallelogram, and much less common are rectangles, squares, rhombuses, circles, and other shapes.

In spatial problems, the same plane figures + the planes themselves and common triangular pyramids with parallelepipeds can fly.

Second question - do you know everything about this figure? Suppose the condition is about an isosceles triangle, and you very vaguely remember what kind of triangle it is. We open a school textbook and read about an isosceles triangle. What to do ... the doctor said a diamond, which means a diamond. Analytic geometry is analytic geometry, but the task will help to solve the geometric properties of the figures themselves, known to us from the school curriculum. If you do not know what the sum of the angles of the triangle is equal to, then you can suffer for a long time.

Third. ALWAYS try to follow the drawing (on a draft / clean copy / mentally), even if this is not required by condition. In "flat" problems, Euclid himself ordered to take a ruler and a pencil in hand - and not only in order to understand the condition, but also for the purpose of self-examination. In this case, the most convenient scale is 1 unit \u003d 1 cm (2 tetrad cells). Let’s not talk about careless students and mathematicians rotating in coffins - in such problems it is almost impossible to make a mistake. For spatial tasks, we perform a schematic drawing, which will also help analyze the condition.

A blueprint or schematic drawing often allows you to immediately see the way to solve a problem. Of course, for this you need to know the foundation of geometry and cut into the properties of geometric shapes (see the previous paragraph).

Fourth. Solution algorithm development... Many geometry problems are multi-pass, so it is very convenient to break the solution and its design into points. Often, the algorithm immediately comes to mind after you have read a condition or completed a drawing. In case of difficulties, we begin with the QUESTION of the problem... For example, according to the condition "it is required to build a straight line ...". Here the most logical question is: "What is enough to know to build this straight line?" Suppose “we know the point, we need to know the direction vector”. We ask the next question: “How to find this direction vector? Where from? " etc.

Sometimes there is a "gag" - the problem is not solved and that's it. The reasons for the stopper can be as follows:

- Serious gap in basic knowledge. In other words, you don’t know or (and) don’t see some very simple thing.

- Lack of knowledge of the properties of geometric shapes.

- The task was difficult. Yes, it happens. It makes no sense to bathe for hours and collect tears in a handkerchief. Seek advice from your teacher, fellow students or ask a question on the forum. Moreover, it is better to make its setting specific - about that part of the solution that you do not understand. A cry in the form "How to solve the problem?" doesn't look very good ... and above all for your own reputation.

Stage five... We decide-check, decide-check, decide-check-give the answer. It is profitable to check each point of the problem immediately after its completion... This will help you spot the error immediately. Naturally, no one forbids to quickly solve the entire problem, but there is a risk of rewriting everything from scratch (often several pages).

These are, perhaps, all the main considerations that are advisable to be guided by when solving problems.

The practical part of the lesson is represented by geometry on a plane. There will be only two examples, but it will not seem few \u003d)

Let's walk along the thread of the algorithm that I just discussed in my little scientific work:

Example 1

Three peaks of a parallelogram are given. Find the top.

We begin to understand:

Step one: it is obvious that we are talking about a "flat" problem.

Step two: the problem is about a parallelogram. Everyone remembers such a parallelogram figure? There is no need to smile, many people receive education at the age of 30-40-50 or more, so even simple facts can be erased from memory. The definition of a parallelogram is found in Example # 3 of the lesson. Linear (non) dependence of vectors. Vector basis .

Step three: Let's make a drawing in which we mark three known peaks. It's funny that it's not difficult to immediately plot the desired point:

Building is, of course, good, but the decision must be formulated analytically.

Step four: Development of a solution algorithm. The first thing that comes to mind is that the point can be found as the intersection of straight lines. We do not know their equations, so we have to deal with this issue:

1) Opposite sides are parallel. By points find the direction vector of these sides. This is the simplest task that was considered in the lesson. Vectors for dummies .

Note: it is more correct to say "the equation of a straight line containing a side", but hereinafter for brevity I will use the phrases "side equation", "direction vector of a side", etc.

3) Opposite sides are parallel. Find the direction vector of these sides by points.

4) Draw up the equation of a straight line along a point and a direction vector

In points 1-2 and 3-4, we actually solved the same problem twice, by the way, it was disassembled in example number 3 of the lesson The simplest problems with a straight line on a plane ... It was possible to go a longer way - first, find the equations of straight lines and only then "pull out" the direction vectors from them.

5) Now the equations of the lines are known. It remains to compose and solve the corresponding system of linear equations (see examples No. 4, 5 of the same lesson The simplest problems with a straight line on a plane ).

Point found.

The task is quite simple and its solution is obvious, but there is a shorter way!

Second solution:

The diagonals of the parallelogram are halved by their intersection point. I marked the point, but in order not to clutter the drawing, I did not draw the diagonals themselves.

Let's make the equation of the side by points:

To check, you should mentally or on a draft, substitute the coordinates of each point into the resulting equation. Now let's find the slope. To do this, we rewrite the general equation as an equation with a slope:

So the slope is:

Similarly, we find the equations of the sides. I don't see much point in describing the same thing, so I will immediately give the finished result:

2) Find the length of the side. This is the simplest task discussed in the lesson. Vectors for dummies ... For points we use the formula:

Using the same formula, it is easy to find the lengths of the other sides. The check can be done very quickly with a regular ruler.

We use the formula .

Find vectors:

Thus:

By the way, along the way, we found the lengths of the sides.

As a result:

Well, it seems like the truth, for convincingness, you can attach a protractor to the corner.

Attention! Do not confuse the angle of a triangle with an angle between straight lines. The angle of a triangle may be obtuse, but the angle between straight lines may not (see the last paragraph of the article The simplest problems with a straight line on a plane ). However, to find the angle of a triangle, you can use the formulas from the above lesson, but the roughness is that those formulas always give an acute angle. With their help, I solved this problem on a draft and got the result. And on the clean copy would have to write down additional excuses that.

4) Make an equation of a straight line passing through a point parallel to a straight line.

A standard task, discussed in detail in example number 2 of the lesson The simplest problems with a straight line on a plane ... From the general equation of the straight line pull out the direction vector. Let's compose the equation of a straight line along a point and a direction vector:

How do I find the height of a triangle?

5) Let's make the height equation and find its length.

There is no escape from strict definitions, so you have to steal from a school textbook:

The height of the triangle is called the perpendicular drawn from the vertex of the triangle to the line containing the opposite side.

That is, it is necessary to draw up the equation of the perpendicular drawn from the vertex to the side. This task is considered in examples No. 6, 7 of the lesson The simplest problems with a straight line on a plane ... From the equation remove the normal vector. Let us compose the equation of height by point and direction vector:

Please note that we do not know the coordinates of the point.

Sometimes the equation of height is found from the ratio of the slopes of perpendicular lines:. In this case, then:. We will compose the height equation by point and slope (see the beginning of the lesson Equation of a straight line on a plane ):

The height length can be found in two ways.

There is a roundabout way:

a) we find - the point of intersection of the height and the side;
b) find the length of the segment by two known points.

But in the lesson The simplest problems with a straight line on a plane a convenient formula for the distance from a point to a straight line was considered. The point is known:, the equation of the line is also known: , Thus:

6) Calculate the area of \u200b\u200bthe triangle. In space, the area of \u200b\u200ba triangle is traditionally calculated using vector product of vectors , but here is a triangle on the plane. We use the school formula:
- the area of \u200b\u200ba triangle is equal to half the product of its base and height.

In this case:

How do I find the median of a triangle?

7) Let's compose the median equation.

Median triangle is called the segment connecting the vertex of the triangle with the middle of the opposite side.

a) Find a point - the middle of the side. We use midpoint coordinate formulas ... The coordinates of the ends of the segment are known: , then the coordinates of the middle are:

Thus:

We compose the median equation by points :

To check the equation, you need to substitute the coordinates of the points into it.

8) Find the point of intersection of the height and the median. I think everyone has already learned how to perform this element of figure skating without falling: