How to learn to solve problems in analytical geometry?
Typical task with a triangle on a plane

This lesson is created on the approach to the equator between the geometry of the plane and the geometry of space. IN this moment there is a need to systematize the accumulated information and respond to very important question: how to learn to solve problems in analytical geometry? The difficulty is that you can come up with an infinite number of problems in geometry, and no textbook will contain all the multitude 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 speak loudly about the fact 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 complete dummy to achieve good and excellent results. At least, the general algorithm for solving geometric problems took shape very clearly in my head.

WHAT YOU NEED TO KNOW AND BE ABLE TO DO
for successfully solving geometry problems?

There is no escape from this - in order not to randomly poke the buttons with your nose, 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 the 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 line in a plane And . The geometry of space is presented in articles Plane equation, Equations of a line in space, Basic problems on a straight line and a plane and some other lessons. Curved lines and spatial surfaces of the second order stand somewhat apart, and there are not so many specific problems with them.

Let's assume that the student already has basic knowledge and skills in solving the simplest problems of analytical geometry. But it happens like this: you read the statement of the problem, and... you want to close the whole thing altogether, throw it in the far corner and forget it, like a bad dream. Moreover, this fundamentally does not 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? There is no need to be afraid of a task that you don’t understand!

Firstly, should be installed - Is this a “flat” or spatial problem? For example, if the condition includes vectors with two coordinates, then, of course, this is the geometry of a plane. And if the teacher loaded the grateful listener with a pyramid, then there is clearly the geometry of space. The results of the first step are already quite good, because we managed to cut off a huge amount of information unnecessary for this task!

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

In “flat” problems, not to mention the obvious points and lines, the most popular figure is a triangle. We will analyze it in great detail. Next comes the parallelogram, and much less common are the rectangle, square, rhombus, circle, and other shapes.

In spatial tasks the same ones can fly flat figures+ the planes themselves and common triangular pyramids with parallelepipeds.

Question two - Do you know everything about this figure? Suppose the condition talks 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 rhombus, that means a rhombus. Analytical geometry is analytical geometry, but the problem will be solved by the geometric properties of the figures themselves, known to us from school curriculum. If you don’t know what the sum of the angles of a triangle is, you can suffer for a long time.

Third. ALWAYS try to follow the drawing(on a draft/finish copy/mentally), even if this is not required by the condition. In “flat” problems, Euclid himself ordered to pick up a ruler and a pencil - and not only in order to understand the condition, but also for the purpose of self-test. In this case, the most convenient scale is 1 unit = 1 cm (2 notebook cells). Let's not talk about careless students and mathematicians spinning in their graves - it is almost impossible to make a mistake in such problems. For spatial tasks, we perform a schematic drawing, which will also help analyze the condition.

A drawing 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 understand the properties of geometric shapes (see the previous paragraph).

Fourth. Development of a solution algorithm. Many geometry problems are multi-step, so the solution and its design is very convenient to break down into points. Often the algorithm immediately comes to mind after you read the condition or complete the drawing. In case of difficulties, we start with the QUESTION of the task. For example, according to the condition “you need to construct a straight line...”. Here the most logical question is: “What is enough to know to construct this straight line?” Suppose, “we know the point, we need to know the direction vector.” Let's ask next question: “How to find this direction vector? Where?" etc.

Sometimes there is a “bug” - the problem is not solved and that’s it. The reasons for the stop may be the following:

– Serious gap in basic knowledge. In other words, you do not know and/or do not see some very simple thing.

– Ignorance of the properties of geometric figures.

- The task was difficult. Yes, it happens. There is no point in steaming for hours and collecting 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 statement concrete - about that part of the solution that you do not understand. A cry in the form of “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 an answer. It is beneficial to check each point of the task immediately after it is completed. This will help you spot the error immediately. Naturally, no one forbids quickly solving the entire problem, but there is a risk of rewriting everything again (often several pages).

These are, perhaps, all the main considerations that should be followed when solving problems.

The practical part of the lesson is presented in plane geometry. There will be only two examples, but it won’t seem enough =)

Let's go through the thread of the algorithm that I just looked at in my little scientific work:

Example 1

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

Let's start to understand:

Step one: It is obvious that we are talking about a “flat” problem.

Step two: The problem deals with a parallelogram. Does everyone remember this parallelogram figure? There is no need to smile, many people receive their education at 30-40-50 or more years of age, so even simple facts can be erased from memory. The definition of a parallelogram is found in Example No. 3 of the lesson Linear (non) dependence of vectors. Basis of vectors.

Step three: Let's make a drawing on which we mark three known vertices. It’s funny that it’s not difficult to immediately construct the desired point:

Constructing it is, of course, good, but the solution must be formulated analytically.

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

1) Opposite sides parallel. By points Let's find the direction vector of these sides. This is the simplest problem that was discussed in class. Vectors for dummies.

Note: it is more correct to say “the equation of a line containing a side,” but here and further for brevity I will use the phrases “equation of a side,” “direction vector of a side,” etc.

3) Opposite sides are parallel. Using the points, we find the direction vector of these sides.

4) Let’s create an equation of a straight line using a point and a direction vector

In paragraphs 1-2 and 3-4, we actually solved the same problem twice; by the way, it was discussed in example No. 3 of the lesson The simplest problems with a straight line on a plane. It was possible to take a longer route - first find the equations of the lines and only then “pull out” the direction vectors from them.

5) Now the equations of the lines are known. All that remains is 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).

The point has been found.

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

Second solution:

The diagonals of a parallelogram are bisected by their point of intersection. I marked the point, but so as not to clutter the drawing, I did not draw the diagonals themselves.

Let's compose the equation of the side point by point:

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 in the form of an equation with a slope coefficient:

Thus, the slope is:

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

2) Find the length of the side. This is the simplest problem covered in class. Vectors for dummies. For points we use the formula:

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

We use the formula .

Let's find the vectors:

Thus:

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

As a result:

Well, it seems to be true; to be convincing, you can attach a protractor to the corner.

Attention! Do not confuse the angle of a triangle with the angle between straight lines. The angle of a triangle can be obtuse, but the angle between straight lines cannot (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 also 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 in draft and got the result. And on the final copy I would have to write down additional excuses, that .

4) Write an equation for a line passing through a point parallel to the line.

Standard task, discussed in detail in example No. 2 of the lesson The simplest problems with a straight line on a plane. From general equation straight Let's take out the guide vector. Let's create an equation of a straight line using a point and a direction vector:

How to find the height of a triangle?

5) Let’s create an equation for the height and find its length.

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

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

That is, it is necessary to create an equation for a perpendicular drawn from the vertex to the side. This task is discussed in examples No. 6, 7 of lesson The simplest problems with a straight line on a plane. From Eq. remove the normal vector. Let's compose the height equation using a point and a direction vector:

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

Sometimes the height equation is found from the ratio of the angular coefficients of perpendicular lines: . IN in this case, Then: . Let's compose the equation of height using a 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) find – the point of intersection of height and side;
b) find the length of the segment using two known points.

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

6) Calculate the area of ​​the triangle. In space, the area of ​​a triangle is traditionally calculated using vector product of vectors, but here we are given a triangle on a plane. We use the school formula:
– The area of ​​a triangle is equal to half the product of its base and its height.

In this case:

How to find the median of a triangle?

7) Let's create an equation for the median.

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

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

Thus:

Let's compose the median equation point by point :

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:

ChapterV. ANALYTICAL GEOMETRY ON THE PLANE

AND IN SPACE

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

When solving problems of analytical geometry, it is advisable to use textbooks from the following authors: D.V. Kletenika, N. Sh. Kremer, D.T. Written by V.I. Malykhina, 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 presented in educational publications by M.S. Krass and V.I. Ermakova.

Problem 5.1. Given the coordinates of the vertices of the triangleABC . Necessary

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

b) write the equation of the altitude of a triangle drawn from the vertexWITH to the sideAB and find its length;

c) write the equation of the median of a triangle drawn from the vertexIN to the sideAC ;

d) find the angles of the triangle and establish its type (rectangular, acute, obtuse);

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

e) 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 (the point of intersection of altitudes) of the triangleABC .

For each of points a) – c) of the solution, make drawings in a coordinate system. In the pictures, mark the lines and points corresponding to the points of the task.

Example 5.1

Given the coordinates of the vertices of the triangleABC : . It is necessary to a) write the equations of the sides of the triangle; b) write the equation of the altitude of a triangle drawn from the vertex WITH to the sideAB and find its length; c) write the equation of the median of a triangle drawn from the vertexIN to the sideAC ; d) find the lengths of the sides of the triangle and determine its type (scalene, isosceles, equilateral); e) find the angles of the triangle and establish its type (rectangular, acute, obtuse); e) 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 (the point of intersection of altitudes) of the triangleABC .

Solution

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

,

Where
And
the corresponding coordinates of the points.

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

,
,
,

from where, after transformations, we write down the equations of the sides

In Fig. 7 we depict the corresponding sides of the triangle
straight.

Answer:

b) Let
– height drawn from the vertex to the side
. Because the
passes through a point perpendicular to the vector
, then we will compose the equation of the straight line using the following formula

Where
– coordinates of the vector perpendicular to the desired line,
– coordinates of a point belonging to this line. Find the coordinates of the vector perpendicular to the line
, and substitute into formula (5.2)

,
,

.

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

,

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

, ,

Where
And
And points

;
.

, substituting which into formulas (5.4), we obtain
median
Median equation
And
Let's write it as an equation of a line passing through the points

,

.

Answer: according to formula (5.1)

R (Fig. 9).

is. 9

,
,
.

G)
And
median
We find the lengths of the sides of the triangle as the lengths of the corresponding vectors, i.e.
.

Answer: Parties
are equal, which means the triangle is isosceles with the base
;

,
.

triangle isosceles with base
d)

,
,
.

Angles of a triangle
let's find the angles between the vectors emanating from the corresponding vertices of a given triangle, i.e.

,

Since the triangle is isosceles with a base
,
.

, That

,
;

,
,
.

We calculate the angles between the vectors using formula (4.4), which requires scalar products of vectors

,

Let's find the coordinates and magnitudes of the vectors necessary to calculate the angles
Substituting the found data into formula (4.4), we obtain

Answer: Parties
Since the cosines of all angles found are positive, then the triangle

,
,
.

is acute-angled. Let

acute-angled;
. Based on this, you can find the coordinates
e)

, ,

Where
,
And
, then the coordinates , And can be found using formulas (5.5)

,
.

Answer:
– coordinates of the points respectively
.

, hence,– center of gravity of the triangle and)
Let – orthocenter of the triangle
. Find the coordinates of the point b) as the coordinates of the point of intersection of the altitudes of the triangle. Height Equation
:

,
,

.

was found at
. Let's find the height equation

Because the , then the solution of the system
.

Answer:
and)
.

Problem 5.2. Fixed costs at an enterprise when producing some products areF V 0 rub. per unit of production, with revenue amounting toR 0 rub. per unit of manufactured product. Create a profit functionP (q ) (q

Data for the problem condition corresponding to the options:

Example 5.2

Fixed costs at an enterprise when producing some products are
rub. per month, variable costs –
rub. per unit of production, with revenue amounting to
rub. per unit of manufactured product. Create a profit functionP (q ) (q – quantity of products produced); build its graph and determine the break-even point.

Solution

Let's calculate the total production costs upon release q units of some products

If sold q units of production, then 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 costs equal total revenue

,

,

where do we find it from?

- break even.

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

Answer: profit function
, break even
.

Problem 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 – 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 WITH , and the offer is only by pricep S received by suppliers. Necessary

a) determine the market equilibrium point;

b) the equilibrium point after the introduction of a tax equal tot . Determine the increase in price and decrease in equilibrium sales volume;

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

d) find a new equilibrium point and government income when introducing a tax proportional to the price and equalN %;

e) determine how much money the government will spend on buying up the surplus when setting a minimum price equal to p 0 .

For each solution point, make a drawing in the coordinate system. In the figure, mark the lines and points corresponding to the task point.

Data for the problem condition corresponding to the options:

In geometry, the concept of “vertex of a triangle” is often considered. This is the point of intersection of two sides of a given figure. This concept appears in almost every problem, so it makes sense to consider it in more detail.

Determining the vertex of a triangle

In a triangle, there are three points where the sides intersect, forming three angles. They are called vertices, and the sides on which they rest are called sides of the triangle.

Rice. 1. Vertex in a triangle.

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

Rice. 2. Designation of vertices in a triangle.

Characteristics of the concept

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

Knowing the coordinates of the vertices of the triangle, you can find the intersection points of the medians, the length of the height lowered to one of the sides of the figure, and the area of ​​the triangle.

To do this, the properties of vectors depicted in the Cartesian coordinate system are used, because the length of the side of a 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

For any vertex of a triangle, you can find an angle that will be adjacent to the internal angle 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 external angles at each vertex. The external angle is equal to the sum of two internal corners triangle not adjacent to it.

Rice. 3. Property of the external angle of a triangle.

If you construct two external angles at one vertex, they will be equal, like vertical ones.

What have we learned?

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

Test on the topic

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