Desires equal heights and know their dowry. The meaning of obscurity and the area of geometric figures

Example of the most active tasks from the standard work “Analytical geometry on a plane”

Tribute to the peaks
,
tricutaneum ABC. Know:

equalization of all sides of the trikutnik;

The system of linear irregularities that signifies the tricutaneous ABC;

Level of height, median and bisection of the tricutaneous line drawn from the top A;

I will cross the point of the height of the tricut;

I will cross the point of the median of the tricutaneous;

Dovzhin height, lowered to the side AB;

Kut A;

Zrobiti armchair.

Let the tops of the trikutnik move in coordinates: A (1; 4), U (5; 3), Z(3; 6). Let’s immediately draw the chair:

1. To write down the alignment of all sides of the tricubitule, we draw straight lines that pass through two given points with coordinates ( x 0 , y 0 ) that ( x 1 , y 1 ):

In this manner, presenting vengeance ( x 0 , y 0 ) coordinates of the point A, and instead of ( x 1 , y 1 ) coordinates of the point U, we reject the straight line AB:

The comparison will be straight AB, Let's write it down in the official form. Similarly, we know the straight line AC:

And so the jealousy itself is direct ND:

2. Respectfully, the impersonal point of the tricutaneous ABC is a webbing of three surfaces, and the skin surface can be adjusted for additional linear unevenness. How do we take jealousy from both sides ∆ ABC, for example AB then there is unevenness

set points that lie along different sides in a straight line AB. We need to select the surface where point C lies. Let's substitute these coordinates in the opposite direction:

If the other inequality is correct, then the necessary points are indicated by the inequality

The same applies to the direct aircraft, and the plane
. Let me try the vikory point A (1, 1):

Well, there is a need for uneasiness to appear:

If we check the direct AC (test point), then we reject:

Well, you need to be nervous about your mother's view

The remaining system of irregularities can be eliminated:

The signs “≤”, “≥” mean that the points that lie on the sides of the tricubitule are also included in the faceless point that form the tricubitule ABC.

3. a) In order to know the level of height lowered from the top A on b_k ND, let's look at the side ND:
. Vector with coordinates
perpendicular to the side ND And, therefore, parallel to the height. Let's write a straight line to pass through a point A parallel to the vector
:

Ceremony of height, omitted i.z. A on b_k ND.

b) We know the coordinates of the middle of the side ND for the formulas:

Here
- Tse coordinate t. U, A
- Coordinates t. Z. Presentable and removable:

Straight, what to pass through qiu point that point Aє with a random median:

c) The bisector is linear, which comes from the fact that in the isosfemoral tricumus the height, the median, and the bisector are lowered from one vertex to the base of the tricuput, the plane. We know two vectors
і
ta їх dozhini:

Todi vector
May be the same direct vector
, and yogo dozhina
So just a single vector
go straight to the vector
Sum of vectors

is a vector that runs directly from the bisector A. In this manner, the zeal for the desired bisector can be written down in the form:

4) The heights and the heights have already been forgotten. We will achieve yet another height, for example, from the top U. Side AC asks one's peers
So, vector
perpendicular AC, I, thus, parallel to the height. Then there is a straight line to go through the top U directly from the vector
(i.e. perpendicular AC), it looks like:

It appears that the heights of the tricubitus change at one point. Zokrema, this point is the crossbar of the known heights, then. decisions of the peer system:

- Coordinates of this point.

5. Middle AB May coordinates
. Let's write down the median to the side AB. This straight line passes through the points with coordinates (3, 2) and (3, 6), so the plane looks like this:

Dear, the zero at the sign of the fraction and the straight line means that the straight line runs parallel to the ordinate axis.

To find the cross-point of the medians, it is enough to calculate the system of levels:

The cross point of the tricutaneous median has coordinates
.

6. Dovzhina height, lowered to the side AB, the road to the point Z to straight line AB with equals
and is behind the formula:

7. Cosine of Kuta A can be found using the cosine formula between vectors і what is the relationship between the scalar creation of these vectors and the creation of their dovzhins:

In problems 1 - 20, the tricutaneous apex ABC is given.
Know: 1) dovzhin side AB; 2) the level of sides AB and AC and their respective coefficients; 3) Internal cut A in radians with an accuracy of up to 0.01; 4) equalize the height of the CD and do so; 5) level of stake, for which CD height equals diameter; 6) a system of linear irregularities, which is signified by the tricutaneous ABC.

Dovzhina sides of the trikutnik:
|AB| = 15
|AC| = 11.18
|BC| = 14.14
Stand d in front of point M: d = 10
The coordinates of the tricutaneous vertices are given: A(-5,2), B(7,-7), C(5,7).
2) Dovzhina sides of the trikutnik
The distance d between points M 1 (x 1 ; y 1) and M 2 (x 2 ; y 2) is calculated using the formula:

8) Straight line
The straight line passing through the points A 1 (x 1 ; y 1) and A 2 (x 2 ; y 2) is represented by lines:

Rivnyanna straight AB

or

or
y = -3 / 4 x -7 / 4 or 4y + 3x +7 = 0
Rivnyannya direct AC
Canonical straight line:

or

or
y = 1 / 2 x + 9 / 2 or 2y -x - 9 = 0
Rivnyannya direct BC
Canonical straight line:

or

or
y = -7x + 42 or y + 7x - 42 = 0
3) Cut between straight lines
Direct alignment AB:y = -3/4 x -7/4
Level of direct AC: y = 1/2 x + 9/2
Cut φ between two straight lines, given equalities with cut coefficients y = k 1 x + b 1 and y 2 = k 2 x + b 2 is calculated using the formula:

The cut-out coefficients for these direct lines are -3/4 and 1/2. The formula is quick, and we take the right part per module:

tg φ = 2
φ = arctan(2) = 63.44 0 or 1.107 rad.
9) Level of height through vertex C
The straight line that passes through the point N 0 (x 0 ; y 0) and is perpendicular to the straight line Ax + By + C = 0 is the direct vector (A; B) and, therefore, is represented by the lines:

Truth can be found in another way. For which we know the cut coefficient k1 straight AB.
Rivnyannya AB: y = -3/4 x -7/4, then. k 1 = -3/4
We know the coefficient k of the perpendicular from the perpendicularity of the two straight lines: k 1 *k = -1.
Substituting substitution k 1 cut coefficient given directly, can be removed:
-3/4 k = -1, stars k = 4/3
So, since the perpendicular passes through the point C (5.7) and has k = 4 / 3, we will look at its alignment in the form: y-y 0 = k (x-x 0).
Substitutes x 0 = 5, k = 4/3, y 0 = 7 can be removed:
y-7 = 4/3 (x-5)
or
y = 4 / 3 x + 1 / 3 or 3y -4x - 1 = 0
We know the point along the line AB:
The system consists of two levels:
4y + 3x +7 = 0
3y -4x - 1 = 0
The first level is expressed by and is comparable to the other.
Ignorable:
x = -1
y=-1
D(-1;-1)
9) Dovzhin of the height of the tricubitule, drawn from vertex C
Raise d from point M 1 (x 1; y 1) to the straight line Ax + By + C = 0 equal to the absolute value of the quantity:

We know where to stand between point C(5;7) and straight line AB (4y + 3x +7 = 0)

The maximum height can be calculated using another formula, as if you stand between point C(5;7) and point D(-1;-1).
The distance between two points is expressed through coordinates by the formula:

5) level of stake, for which CD height equals diameter;
A line of radius R centered at point E(a;b) looks like:
(x-a) 2 + (y-b) 2 = R 2
The CD fragments are the diameter of a stake, and their center is the middle of the CD slice. Having quickly scrambled through the formulas below, we can eliminate:

Otzhe, E(2;3) і R = CD / 2 = 5. Vicor formula, subtracting the level of the shukana stake: (x-2) 2 + (y-3) 2 = 25

6) a system of linear irregularities, which signifies the ABC tricuputin.
Straight line AB: y = -3/4 x -7/4
Level of direct AC: y = 1/2 x + 9/2
Direct alignment BC: y = -7x + 42

1. The equal sides of AB and BC and their respective coefficients.
Given the coordinates of the point through which the straight lines pass, it is possible for the straight lines to pass through the two given points $$\frac(x-x_1)(x_2-x_1)=\frac(y-y_1)(y_2-y_1) $ $ presentable and derivable equal
level of straight line AB $$\frac(x+6)(6+6)=\frac(y-8)(-1-8) => y = -\frac(3)(4)x + \frac(7 )(2)$$ cut coefficient direct AB more expensive $k_(AB) = -\frac(3)(4)$
straight line BC $$\frac(x-4)(6-4)=\frac(y-13)(-1-13) => y = -7x + 41$$ cut coefficient straight BC dear $k_( BC) = -7$

2. Kut V in radians to an accuracy of two digits
Cut B - cut between straight lines AB and BC, which is insured using the formula $$tg\phi=|\frac(k_2-k_1)(1+k_2*k_1)|$$is represented by the values of cut coefficients of these straight lines and subtracted by $$tg\ phi=|\frac(-7+\frac(3)(4))(1+7*\frac(3)(4))| = 1 => \phi = \frac(\pi)(4) \approx 0.79$$
3.Dovzhinu side AB
The dovzhina of side AB is covered as it rises between the points and is adjacent $d = \sqrt((x_2-x_1)^2+(y_2-y_1)^2)$ => $$d_(AB) = \sqrt((6+ 6)^2 + (-1-8)^2) = 15$$
4. The height of the CD will be adjusted to the same level.
The height level is determined by the straight line formula, which means to pass through a given point C(4;13) in a given straight line - perpendicular to the straight line AB by the formula $y-y_0=k(x-x_0)$. We know the cutoff coefficient of height $k_(CD)$, which is accelerated by the power of perpendicular straight lines $k_1=-\frac(1)(k_2)$ is removed $$k_(CD)= -\frac(1)(k_(AB) ) = -\frac(1)(-\frac(3)(4)) = \frac(4)(3)$$ Inserted into a straight line, subtracted $$y - 13 = \frac(4)(3) (x-4) => y = \frac(4)(3)x+\frac(23)(3)$$ The dove of the height can be found as you rise from point C(4;13) to straight line AB using the formula $$d = \frac(Ax_0+By_0+C)(\sqrt(A^2+B^2))$$ for a number cruncher, the line AB is equal to this form $y = -\frac(3)(4)x + \frac(7)(2) => 4y+3x-14 = 0$ , the coordinates of the point y are represented by the formula $$d = \frac(4*13+3*4-14 )(\sqrt( 4 ^2+3^2)) = \frac(50)(5) =10$$

5. The alignment of the median AE and the coordinates of the point Before the alignment of the median with the height CD.
The alignment of the median will be considered as the alignment of a straight line that passes through two given points A(-6;8) and E, where point E is the midpoint between points B and C and these coordinates are determined by the formula $E(\frac(x_2+x_1 )) (2);\frac(y_2+y_1)(2))$ is represented by the coordinates of the point $E(\frac(6+4)(2);\frac(-1+13)(2))\ ) => \(E(5; 6)$, then the level of median AE will come $$\frac(x+6)(5+6)=\frac(y-8)(6-8) => y = - \frac(2)(11)x + \frac(76)(11)$$We know the coordinates of the point of height and median, then. Let's find the starting point for which the leveling system is foldable $$\begin(cases)y = -\frac(2)(11)x + \frac(76)(11)\y = \frac(4)(3)x+ \ frac(23)(3)\end(cases)=>\begin(cases)11y = -2x +76\\3y = 4x+23\end(cases)=>$$$$\begin(cases)22y = -4x +152\3y = 4x+23\end(cases)=> \begin(cases)25y =175\\3y = 4x+23\end(cases)=> $$$$\begin(cases) y = 7\\ x=-\frac(1)(2)\end(cases)$$ Coordinates of the crossbar point $K(-\frac(1)(2);7)$

6. A straight line passes through point Do parallel to side AB.
As they are directly parallel, their specific coefficients are equal. $k_(AB)=k_(K) = -\frac(3)(4)$, also depending on the coordinates of the point $K(-\frac(1)(2);7)$, then . To find a straight line, we simply use the formula for a straight line to pass through a given point in a given straight line $y - y_0=k(x-x_0)$, which can be represented by the given data and subtracted from $$y - 7= -\frac(3)(4) ) (x-\frac(1)(2)) => y = -\frac(3)(4)x + \frac(53)(8)$$

8. The coordinates of point M are symmetrical to point A and straight CD.
The speck M lies on straight AB, because CD – height to this side. We know the point of the crossbar CD and AB for which we connect the system of levels $$\begin(cases)y = \frac(4)(3)x+\frac(23)(3)\y = -\frac(3)(4 ) x + \frac(7)(2)\end(cases) =>\begin(cases)3y = 4x+23\\4y =-3x + 14\end(cases) => $$$$\begin( cases ) 12y = 16x + 92 \ 12y = -9x + 42 \ end (cases) =>
\begin(cases)0= 25x+50\\12y =-9x + 42\end(cases) => $$$$\begin(cases)x=-2\y=5 \end(cases)$$ Coordinates points D(-2; 5). Behind the mind AD=DK, where the distance between the points is based on the Pythagorean formula $d = \sqrt((x_2-x_1)^2+(y_2-y_1)^2)$, where AD and DK are the hypotenuses of the equal recticutaneous tricutaneous, and (Δx = x_2-x_1) і (Δy = y_2-y_1) - cathetes of these tricutaneous bodies, then. we know the coordinates of the point M. $Δx=x_D-x_A = -2+6=4$, and $Δy=y_D-y_A = 5-8=-3$, then the coordinates of the point M are more $ x_M-x_D = Δx => x_D +Δx =-2+4=2 $, and $y_M-y_D = Δy => y_D +Δy =5-3=2 $, we took the coordinates of the point $ M (2;2)$

Head 1. The coordinates of the tricutaneous vertices ABC are given: A(4; 3), B(16;-6), C(20; 16). Know: 1) dovzhinu side AB; 2) the level of sides AB and BC and their respective coefficients; 3) kut U radians to an accuracy of two digits; 4) leveling the height of CD and dovzhin; 5) the level of the median AE and the coordinates of the point Before the crossbar of the median with the height CD; 6) a straight line that passes through point Do parallel to side AB; 7) coordinates of point M, moved symmetrically to point A and straight line CD.

Decision:

1. Position d between points A(x 1 ,y 1) and B(x 2 ,y 2) is determined by the formula

Zastosovuychi (1), we know the dovzhin side AB:

2. The straight line that passes through the points A(x 1 ,y 1) and B(x 2 ,y 2) looks like

(2)

Substituting (2) the coordinates of points A and B, the alignment of side AB is taken away:

Having revealed the remaining alignment, we know that the alignment of side AB appears to be aligned directly with the cut coefficient:

stars

By substituting the point B and C into (2) coordinates, we find the alignment of straight line BC:

3. It is clear that the tangent is between two straight lines, the coefficients of which are similar and are calculated using the formula

Shukany kut In creations of direct AB and PS, some coefficients are found: Zastosovuchi (3), can be rejected

Or radium.

4. The straight line that passes through a given point in a given direction can be seen

(4)

The height of CD is perpendicular to side AB. To know the cutoff coefficient of the CD height, the speed of the mental perpendicularity of the straight lines. Bo those By substituting the coordinates of point Z and the cutoff coefficient of height into (4), we can remove

To know the height of the CD, it is important to know the coordinates of the point D-point and the span of the straight lines AB and CD. Virus the sleeping system:

we know that. D(8;0).

Using formula (1) we know the value of the height CD:

5. To know the level of the median AE, especially the coordinates of point E, which is the middle of the side BC, the simple formula for dividing the cut into two equal parts:

Otzhe,

By substituting point A and E into (2) coordinates, we find the level of the median:

To know the coordinates of the cross point of height CD and median AE, it is important to know the alignment system

We know.

6. The fragments are directly parallel to side AB, then their cut coefficient is similar to the cut coefficient of straight AB. By substituting in (4) the coordinates of the found point K and the cut coefficient, we can extract

3x + 4y - 49 = 0 (KF)

7. If the fragments of the straight line AB are perpendicular to the straight line CD, then point M is found, expanded symmetrically to the point A to the straight line CD, lie on the straight line AB. In addition, point D is the middle of section AM. Using Zastos formula (5), we know the coordinates of the Shukan point M:

Tricutnik ABC, height CD, median AE, straight line KF and point M are determined by the xOy coordinate system in Fig. 1.

Task 2. The slope is equal to the geometric location of the point, the relationship of the distances to this point A (4; 0) and to this straight line x = 1 to 2.

Decision:

The xOy coordinate system will have a point A(4;0) and a straight line x = 1. Let M(x;y) be a sufficient point of the chosen geometric location of points. Let us drop the perpendicular MB to the given line x = 1 and the coordinates of point B are significant. If the point lies on the given line, then its abscissa is relative to 1. The ordinate of point B is relative to the ordinate of point M. Otzhe, B(1;y) (Fig. 2).

Behind the think tank | MA |: | MV | = 2. Vіdstan |MA| ta |MB| is known from formula (1) of problem 1:

Having placed the left and right parts into a square, we remove them

The equation is removed by a hyperbole, which has a coefficient of a = 2, and is obvious –

The focus of hyperbole is significant. For hyperbole, the jealousy of Ozhe is determined, and – the focus of hyperbole. Apparently, point A (4; 0) is given as the right focus of hyperbole.

The eccentricity of the removed hyperpain is significant:

Rivnyana asymptotes of hyperbole loom in sight. Well, or else - asymptotes of hyperbole. First, there will be a hyperbole, and there will be asymptotes.

Zavdannya 3. The slope of the alignment of the geometric location is the point that is exactly distant from the point A (4; 3) and straight line y = 1. Reduce the alignment to its simplest appearance.

Decision: Let M(x; y) be one of the points of the identified geometric point. Let us drop the perpendicular MB from the point M to the straight line y = 1 (Fig. 3). The coordinates of point B are significant. Obviously, the abscis of point B is prior to the abscis of point M, and the ordinate of point B is prior to 1, so B (x; 1). Behind the think tank | MA | = | MV |. Therefore, for any point M(x;y) that belongs to the identified geometric place of points, the following is true:

If the equation is removed, it means a parabola with a vertex at the point. To reduce the equation of the parabola to its simplest form, we put y + 2 = Y, then the equation of the parabola will look like this:

How can you learn to understand the problems of analytical geometry?
Typical planting with trikutnik on the square

This lesson was 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 received information and information on important nutrition: How do you learn to understand the problems of analytical geometry? The complexity lies in the fact that problems with geometry can be thought of infinitely richly, and no tool can accommodate the vast variety of butts. Tse not Similar functions with five rules of differentiation, a table and many technical techniques.

It’s a decision! I don’t say bad things about the fact that I have developed such a grand method, but, in my opinion, there is a truly effective approach to this problem that allows you to achieve good and outstanding results for everyone. Zreshta, zagalny algorithm The crown of geometric designs has already clearly taken shape in my head.

WHAT YOU NEED TO KNOW AND FOLLOW
for successful completion of geometry problems?

Because you can’t go anywhere - in order not to tempt the buttons with your nose, you need to master the basics of analytical geometry. Therefore, you just started learning geometry, but you completely forgot, please, start with the lesson Vectors for dummies. In order to work with vectors, you need to know the basic concepts of the geometry of area, area, straight line on the plain ta . The geometry of space is represented by articles Rivnyany Square, Rivnyannya right next to the open space, The main lessons are directly related to and other lessons. Curved lines and spacious surfaces stand quite apart, and there are not so many specific tasks behind them.

Let’s assume that the student already has basic knowledge and skills of the most basic tasks of analytical geometry. But it goes like this: you read the memory of your mind, and... you want to immediately close everything on the right, throw it into a distant corner and forget about it, like a bad dream. Moreover, it does not lie with the level of your qualifications, sometimes I am faced with tasks for which the solution is not obvious. How do you fix such situations? There is no need to be afraid of the unknown, lest you understand!

According to Pershe, insert trace – is it “flat” or is it a spacious place? For example, since in our minds figure vectors with two coordinates, then, obviously, there is the geometry of the plane. And if the bookkeeper has enticed the listening ear with a pyramid, then there is space in the geometry. The results of the first cycle are no longer spoiled, and even a great amount of unnecessary information has been removed!

friend. Umova, as a rule, surprises you with some kind of geometric figure. In fact, walk through the corridors of the local VNZ, and you will see a lot of exciting features.

In “flat” buildings, without even talking about points and straight lines, the most popular figure is the knitted figure. We'll sort it out in detail. Then we go to the parallelograms, and it is much more common to see the rectum, square, rhombus, colo, etc. figures.

In spacious rooms, the very flat figures + the very planes and wide three-piece pyramids with parallelepipeds can fly.

Food for a friend - Do you know everything about this figure? Let’s say that your mind thinks about the equilateral tricubitus, but you almost certainly remember what kind of tricuputin it is. We open the school handbag and read about the isosfemoral tricutule. What to do... the doctor said rhombus, then, rhombus. Analytical geometry and analytical geometry, aka The task will help to enhance the geometric power of the figures themselves, we know about the school program. If you don’t know what the bag of tunics of the tricutaneous tree is like, then you can suffer for a long time.

Third. Get yourself started on your armchair(on black/chistovik/podumki), navіt yakshto no need for the mind. In the “flat” factories, Euclid himself ordered to pick up a ruler with an olive - and not only in order to clear the mind, but as a method of self-verification. In this case, the largest scale is 1 unit = 1 cm (2 scales). Let’s no longer sweat about the poor students and mathematicians who are in trouble - it’s practically impossible to make a deal with such people. For those in the open space, we have created a schematic little thing that will also help analyze the mind.

An armchair or a schematic armchair often immediately allows you to complete your career. Of course, for whom it is necessary to know the foundation of geometry and rubles in power geometric shapes(Div. front point).

Fourth. Discord with the algorithm. There is a lot of detail on the geometry of the many flow paths, so the solution and its design can easily be broken down into points. Often the algorithm immediately falls on a thought, after you have read the mind or the conclusion of the chair. In times of guilt, difficulties begin with a NUTRITION task.. For example, behind the brain "you need to stay straight...". Here the most logical answer is: “What is it enough for the nobility to do this right?” Let’s say, “the speck is visible to us, we need to know the direct vector.” Defined foot food: “How to know this direct vector? Stars? etc.

Sometimes the “zatik” disappears – there is no problem here either. The reasons for the stopper may be as follows:

- A serious clearing in elementary knowledge. In other words, you don’t know what you’re saying even in simple words.

- Ignorance of the powers of geometric figures.

- The manager was very important. Yes, so be it. It’s impossible for me to spend hours sweating and collecting my tears in hustka. Return for advice before payment, classmates or ask on the forum. Moreover, it’s better to make a specific statement - about that little decision, which you didn’t realize. I call from the viewer, “How can I find the secret?” doesn’t look much better... and, first of all, for your powerful reputation.

Fifth stage. Verified-verified, verified-verified, verified-verified-provided evidence. It is important to check the leather item immediately after my martyrdom. This will help you to make peace. Of course, no one wants to review the work as a whole, otherwise there is a risk of having to rewrite everything again (often a few sides).

The axis, perhaps, is all the main problems that are completely cured at the hour of the highest order.

The practical part of the lesson is represented by geometry on a plane. There will be only two applications, otherwise you won’t succeed =)

Let's go through the algorithm, which I carefully looked at with my little one scientific practice:

Butt 1

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

Let's start figuring it out:

Croc first: It’s obvious that we’re talking about a “flat” plant

Croc is different: The manager is talking about the parallelogram. Do everyone remember such a parallelogram figure? There is no need to laugh, many people find light in their 30s, 40s, 50s and more fatefully, so simple facts can be erased from memory. The meaning of the parallelogram is shown in Appendix No. 3 to the lesson Linear (not) location of vectors. Vector basis.

Krok third: There is a chair on which there are significantly three views of the top of Tsikavo, which is difficult to immediately find a point:

Be sure to be kind, but the decision must be formulated analytically.

Croc quarters: Exploration of the decision algorithm. First, what falls on the thought - the point can be known as the span of straight lines. Their jealousy is unknown to us, so he will have to deal with these meals:

1) The opposite sides are parallel. Behind the points we know the direct vector of these sides. This is the simplest thing, as seen in class Vectors for dummies.

Note: It’s more correct to say “the straight line, to move the side”, and here and there for the sake of consistency, the expression “the straight side”, “the straight vector of the side”, etc.

3) The opposite sides are parallel. From the points we know the direct vector of these sides.

4) Composite alignment of a straight line along a point and a direct vector

In paragraphs 1-2 and 3-4, the two actually had the same idea, which, before speaking, was taken from application No. 3 of the lesson The simplest plant directly on the plain. It was possible to follow the current path - to immediately know the alignment of straight lines and then “pull” straight vectors from them.

5) Now the competition is direct. Lost flexibility and balance in the telephone system linear levels(div. apply No. 4, 5 of the same lesson The simplest plant directly on the plain).

The point has been found.

The task is simple and the solution is obvious, but it’s a shorter route!

Another way to grow:

The diagonals of the parallelogram are divided by their cross point. I said a point, so as not to harrass the chairs themselves, the diagonals themselves are not tested.

Let's put together the sides behind the points :

To verify the trace of the thought or on the blackboard, insert the coordinates of the skin point in the straight line. Now we know the cut-throat factor. For which we will rewrite the hidden relationship in the appearance of the relationship with the cut coefficient:

In this manner, the captain of the court:

The alignment of the sides is similarly known. I don’t want to paint a special sense of the same, so I’ll just give you a ready-made result:

2) We know the difference between sides. This is the simplest thing, look at the lesson Vectors for dummies. For points Vikorist formula:

It is easy to understand many other aspects behind this formula. The check is quickly completed with a straight line.

Vikorist's formula .

We know the vectors:

In this manner:

Before the speech, we knew a good deal of the sides.

As a result:

Well, it seems to be true; to make it more flexible, you can put a protractor to the corner.

Respect! Don't confuse the straight lines. The cut of the tricut can be blunt, but the cut between the straight lines is not (div. the remaining point of the statistic The simplest plant directly on the plain). However, to find the trikutnik kut, you can use the formulas of the well-known lesson, but the shortness lies in the fact that the formulas always give the gostry kut. I will help them by relying on the problem and removing the result. And on the clean sheet it would have been possible to write down additional confirmations, so.

4) The slopes are aligned with a straight line that passes through a point parallel to the straight line.

Standard instructions, look in detail at lesson No. 2 The simplest plant directly on the plain. Z Zagalnogo Rivnyannya straight vityagnemo direct vector. Composite alignment of a straight line by point and direct vector:

How to know the height of the tricutaneous?

5) The exact level of height will be known until the end.

If you can't get anywhere, you'll have to steal from your school teacher:

The height of a tricut called a perpendicular, drawing from the top of the tricubitule to the straight line, which is located on the proximal side.

Then it is necessary to bend the line of the perpendicular drawn from the vertex to the side. This information is discussed in lessons No. 6, 7 The simplest plant directly on the plain. Rivnyanya the normal vector is determined. Level of height according to point and direct vector:

Please note that the coordinates of the points are not known to us.

Another way to determine the level of height is from the relationship of cut coefficients perpendicular to straight lines: . At the same time, then: . The level of height is folded behind the point and the cutoff coefficient (marvelous cob to the lesson Rivnyanna straight on the plain):

You can find out your height in two ways.

There is a sign:

a) known - the point is the height of the crossbar and the sides;
b) we know the duration of the cut at two points.

Ale in class The simplest plant directly on the plain I could see a manual formula for the distance from a point to a straight line. The speck is visible: , The straight line is also visible: , In this order:

6) The area of the tricutaneous plant is computable. In the expanse of the trikutnik square it is traditional to seek help vector art vectors but here there is a trikutnik on the square. Vikorist's school formula:
- The area of the tricut is equal to half the height of its base.

In this section:

How to find the median of the tricutaneous plant?

7) The level of the median is folded.

Median tricutaneum This is called the cut that connects the top of the tricube from the middle of the prolong side.

a) We know the point - the middle of the side. Vikoristovuyemo formulas for the coordinates of the middle of the cut. View the coordinates of the ends of the section: , Todi coordinates of the middle: