What is a function? This is the dependence of one quantity on another. In a mathematical function, there are most often two unknowns: independent and dependent, or x and y, respectively.

What does it mean? This means that x can take on absolutely any value, and y will adapt to it, changing in accordance with the coefficients of the function.

There are situations where a function has multiple variables. Dependent is always 1, but there may be several factors that influence it. It is not always possible to display such a function on a graph. At best, you can graphically display the dependence of y on 2 variables.

What is the easiest way to represent the dependence y(x)?

Yes, very simple. Imagine a spoiled child and a rich, loving mother. They come to the store together and start begging for candy. Who knows how many candies the boy will demand today?

No one, but depending on the number of candies, the amount that mom will pay at the checkout will increase. In this case, the dependent variable is the amount in the check, and the independent variable is the number of sweets the boy wants today.

It is very important to understand that one value of the function y always corresponds to 1 value of the argument x. But, as with the roots of a quadratic equation, these values ​​can coincide.

Equation of a straight line

Why do we need the equation of a straight line if we are talking about the equation of the lengths of the sides of a triangle?

Yes, because each side of the triangle is a segment. A segment is a limited part of a straight line. That is, we can specify equations of straight lines. And at the points of their intersection, limit the lines, thereby cutting off the straight lines and turning them into segments.

The equation of the line looks like this:

$$y_1=a_1x+b_1$$

$$y_2=a_2x+b_2$$

$$y_3=a_3x+b_3$$

Equation of the sides of a triangle

It is necessary to find the equation for the lengths of the sides of a triangle with vertices at points A(3,7); B(5,3); C(12;9)

All coordinates are positive, which means the triangle will be located in 1 coordinate quadrant.

Let's draw up equations for each of the lines of the triangle one by one.

• The first line will be AB. We substitute the coordinates of the points into the equation of the straight line in place of x and y. Thus we get a system of two linear equations. Having solved it, you can find the value of the coefficients for the function:

A(3,7) ; B(5,3):

From the first equation we express b and substitute it into the second.

Let's substitute the value of a and find b.

b=7-3a=7-3*(-2)=7+6=13

Let's create an equation for a straight line.

• Let's create the remaining two equations in the same way.

B(5,3); C(12;9)

9=12a+b=12a+3-5a

$$b=3-5*(6\over7)=-(9\over7)$$

$$y=(6\over7)x-(9\over7)$$

• A(3,7) ; C(12;9)

9=12a+b=12a+7-3a=9a+7

$$b=7-(6\over9)=(57\over9)$$

$$y=(2\over9)x+(57\over9)$$

• Let us write the equation for the lengths of the sides of a triangle:

$$y=(6\over7)x-(9\over7)$$

$$y=(2\over9)x+(57\over9)$$

What have we learned?

We learned what a function is, talked about the function of a straight line and learned to derive the equations of the sides of a triangle from the coordinates of its vertices.

Test on the topic

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Example. The vertices of triangle ABC are given.
Find: 1) the length of side AB; 2) equations of sides AB and AC and their angular coefficients; 3) Internal angle A in radians with an accuracy of 0.01; 4) equation for the height of CD and its length; 5) the equation of a circle for which the height CD is the diameter; 6) a system of linear inequalities defining triangle ABC.

Length of triangle sides:
|AB| = 15
|AC| = 11.18
|BC| = 14.14
Distance d from point M: d = 10
The coordinates of the vertices of the triangle are given: A(-5,2), B(7,-7), C(5,7).
2) Length of the sides of the triangle
The distance d between points M 1 (x 1 ; y 1) and M 2 (x 2 ; y 2) is determined by the formula:

8) Equation of a line
A straight line passing through points A 1 (x 1 ; y 1) and A 2 (x 2 ; y 2) is represented by the equations:

Equation of line AB
or
or y = -3 / 4 x -7 / 4 or 4y + 3x +7 = 0
Equation of line AC
Canonical equation of the line: or
or y = 1 / 2 x + 9 / 2 or 2y -x - 9 = 0
Equation of line BC
Canonical equation of the line: or
or y = -7x + 42 or y + 7x - 42 = 0
3) Angle between straight lines
Equation of straight line AB:y = -3 / 4 x -7 / 4
Line equation AC:y = 1 / 2 x + 9 / 2
The angle φ between two straight lines, given by equations with angular coefficients y = k 1 x + b 1 and y 2 = k 2 x + b 2, is calculated by the formula:

The slopes of these lines are -3/4 and 1/2. Let's use the formula, and take its right-hand side modulo:

tg φ = 2
φ = arctan(2) = 63.44 0 or 1.107 rad.
9) Equation of height through vertex C
The straight line passing through the point N 0 (x 0 ;y 0) and perpendicular to the straight line Ax + By + C = 0 has a direction vector (A;B) and, therefore, is represented by the equations:



This equation can be found in another way. To do this, let's find the slope k 1 of straight line AB.
AB equation: y = -3 / 4 x -7 / 4, i.e. k 1 = -3 / 4
Let's find the angular coefficient k of the perpendicular from the condition of perpendicularity of two straight lines: k 1 *k = -1.
Substituting the slope of this line instead of k 1, we get:
-3 / 4 k = -1, whence k = 4 / 3
Since the perpendicular passes through the point C(5,7) and has k = 4 / 3, we will look for its equation in the form: y-y 0 = k(x-x 0).
Substituting x 0 = 5, k = 4 / 3, y 0 = 7 we get:
y-7 = 4 / 3 (x-5)
or
y = 4 / 3 x + 1 / 3 or 3y -4x - 1 = 0
Let's find the point of intersection with line AB:
We have a system of two equations:
4y + 3x +7 = 0
3y -4x - 1 = 0
From the first equation we express y and substitute it into the second equation.
We get: x = -1; y=-1
D(-1;-1)
9) Length of the altitude of the triangle drawn from vertex C
The distance d from the point M 1 (x 1 ;y 1) to the straight line Ax + By + C = 0 is equal to the absolute value of the quantity:

Find the distance between point C(5;7) and line AB (4y + 3x +7 = 0)


The length of the height can be calculated using another formula, as the distance between point C(5;7) and point D(-1;-1).
The distance between two points is expressed in terms of coordinates by the formula:

5) the equation of a circle for which the height CD is the diameter;
The equation of a circle of radius R with center at point E(a;b) has the form:
(x-a) 2 + (y-b) 2 = R 2
Since CD is the diameter of the desired circle, its center E is the midpoint of the segment CD. Using the formulas for dividing a segment in half, we get:


Therefore, E(2;3) and R = CD / 2 = 5. Using the formula, we obtain the equation of the desired circle: (x-2) 2 + (y-3) 2 = 25

6) a system of linear inequalities defining triangle ABC.
Equation of line AB: y = -3 / 4 x -7 / 4
Equation of line AC: y = 1 / 2 x + 9 / 2
Equation of line BC: y = -7x + 42

How to learn to solve problems in analytical geometry?
Typical problem 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. 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 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 problems, the same flat figures + the planes themselves and common triangular pyramids with parallelepipeds can fly.

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 the 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.” We ask the following 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?” does not 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 are 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 problem 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 create an equation for 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 the general equation of the line 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 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 this case, then: . Let's compose the height equation using a point and an angular coefficient (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: