Instructions
You are given three points. Let us denote them as (x1, y1), (x2, y2), (x3, y3). It is assumed that these points are the vertices of some triangle. The task is to create equations of its sides - more precisely, equations of those lines on which these sides lie. These equations should look like:
y = k1*x + b1;
y = k2*x + b2;
y = k3*x + b3. Thus, you have to find the angular values k1, k2, k3 and the displacements b1, b2, b3.
Find a line passing through the points (x1, y1), (x2, y2). If x1 = x2, then the desired line is vertical and its equation is x = x1. If y1 = y2, then the line is horizontal and its equation is y = y1. In general, these coordinates will not correspond to each other.
Substituting the coordinates (x1, y1), (x2, y2) into the general equation of the straight line, you get a system of two linear equations: k1*x1 + b1 = y1;
k1*x2 + b1 = y2. Subtract one equation from the other and solve the resulting equation for k1: k1*(x2 - x1) = y2 - y1, therefore k1 = (y2 - y1)/(x2 - x1).
Substituting what you found into any of the original equations, find the expression for b1:((y2 - y1)/(x2 - x1))*x1 + b1 = y1;
b1 = y1 - ((y2 - y1)/(x2 - x1))*x1. Since we already know that x2 ≠ x1, we can simplify the expression by multiplying y1 by (x2 - x1)/(x2 - x1). Then for b1 you will get the following expression: b1 = (x1*y2 - x2*y1)/(x2 - x1).
Check if the third of the given points is on the found line. To do this, substitute (x3, y3) into the resulting equation and see if the equality holds. If it is observed, therefore, all three points lie on the same line, and the triangle degenerates into a segment.
In the same way as described above, derive equations for the lines passing through the points (x2, y2), (x3, y3) and (x1, y1), (x3, y3).
The final form of the equations for the sides of a triangle given by the coordinates of the vertices is: (1) y = ((y2 - y1)*x + (x1*y2 - x2*y1))/(x2 - x1);
(2) y = ((y3 - y2)*x + (x2*y3 - x3*y2))/(x3 - x2);
(3) y = ((y3 - y1)*x + (x1*y3 - x3*y1))/(x3 - x1).
To find equations parties triangle, first of all, we must try to solve the question of how to find the equation of a line on a plane if its direction vector s(m, n) and some point M0(x0, y0) belonging to the line are known.
Instructions
Take an arbitrary (variable, floating) point М(x, y) and construct a vector М0M =(x-x0, y-y0) (write also М0M(x-x0, y-y0)), which will obviously be collinear (parallel ) by k s. Then, we can conclude that the coordinates of these vectors are proportional, so we can create a canonical straight line: (x-x0)/m = (y-y0)/n. It is this ratio that will be used in solving the problem.
All further actions are determined based on the method .1st method. A triangle is given by the coordinates of its three vertices, which in school geometry is given by the lengths of its three parties(see Fig. 1). That is, the condition contains points M1(x1, y1), M2(x2, y2), M3(x3, y3). They correspond to their radius vectors) OM1, 0M2 and OM3 with the same coordinates as the points. For getting equations parties s M1M2 requires its direction vector M1M2=OM2 – OM1=M1M2(x2-x1, y2-y1) and any of the points M1 or M2 (here the point with the lower index is taken).
So for parties y M1M2 canonical equation of the line (x-x1)/(x2-x1)=(y-y1)/(y2-y1). Acting purely inductively, we can write equations the rest parties.For parties s М2М3: (x-x2)/(x3-x2)=(y-y2)/(y3-y2). For parties s М1М3: (x-x1)/(x3-x1)=(y-y1)/(y3-y1).
2nd method. The triangle is defined by two points (the same as before M1(x1, y1) and M2(x2, y2)), as well as the unit vectors of the directions of the other two parties. For parties s М2М3: p^0(m1, n1). For M1M3: q^0(m2, n2). Therefore for parties s M1M2 will be the same as in the first method: (x-x1)/(x2-x1)=(y-y1)/(y2-y1).
For parties s М2М3 as a point (x0, y0) of the canonical equations(x1, y1), and the direction vector is p^0(m1, n1). For parties s M1M3, (x2, y2) is taken as the point (x0, y0), the direction vector is q^0(m2, n2). Thus, for M2M3: equation (x-x1)/m1=(y-y1)/n1. For M1M3: (x-x2)/m2=(y-y2)/n2.
Video on the topic
Tip 3: How to find the height of a triangle if the coordinates of the points are given
The height is the straight line segment connecting the top of the figure with the opposite side. This segment must be perpendicular to the side, so only one can be drawn from each vertex height. Since there are three vertices in this figure, there are the same number of heights. If a triangle is given by the coordinates of its vertices, the length of each of the heights can be calculated, for example, by using the formula for finding the area and calculating the lengths of the sides.
Instructions
Start by calculating the lengths of the sides triangle. Designate coordinates figures like this: A(X₁,Y₁,Z₁), B(X₂,Y₂,Z₂) and C(X₃,Y₃,Z₃). Then you can calculate the length of side AB using the formula AB = √((X₁-X₂)² + (Y₁-Y₂)² + (Z₁-Z₂)²). For the other two sides these will look like this: BC = √((X₂-X₃)² + (Y₂-Y₃)² + (Z₂-Z₃)²) and AC = √((X₁-X₃)² + (Y₁-Y₃ )² + (Z₁-Z₃)²). For example, for triangle with coordinates A(3,5,7), B(16,14,19) and C(1,2,13) the length of side AB will be √((3-16)² + (5-14)² + (7 -19)²) = √(-13² + (-9²) + (-12²)) = √(169 + 81 + 144) = √394 ≈ 19.85. The lengths of the sides BC and AC, calculated in the same way, will be √(15² + 12² + 6²) = √405 ≈ 20.12 and √(2² + 3² + (-6²)) = √49 = 7.
Knowing the lengths of the three sides obtained in the previous step is enough to calculate the area triangle(S) according to Heron's formula: S = ¼ * √((AB+BC+CA) * (BC+CA-AB) * (AB+CA-BC) * (AB+BC-CA)). For example, substituting into this formula the values obtained from the coordinates triangle-sample from the previous step, this will give the value: S = ¼*√((19.85+20.12+7) * (20.12+7-19.85) * (19.85+7-20.12 ) * (19.85+20.12-7)) = ¼*√(46.97 * 7.27 * 6.73 * 32.97) ≈ ¼*√75768.55 ≈ ¼*275.26 = 68.815 .
Based on area triangle, calculated in the previous step, and the lengths of the sides obtained in the second step, calculate the heights for each of the sides. Since the area is equal to half the product of the height and the length of the side to which it is drawn, to find the height, divide the doubled area by the length of the desired side: H = 2*S/a. For the example used above, the height lowered to side AB will be 2*68.815/16.09 ≈ 8.55, the height to side BC will have a length of 2*68.815/20.12 ≈ 6.84, and for side AC this value will be equal to 2 *68.815/7 ≈ 19.66.
Sources:
- given points find the area of the triangle
Tip 4: How to use the coordinates of the vertices of a triangle to find the equations of its sides
In analytical geometry, a triangle on a plane can be defined in a Cartesian coordinate system. Knowing the coordinates of the vertices, you can create equations for the sides of the triangle. These will be the equations of three straight lines, which, intersecting, form a figure.
In problems 1 - 20 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.
Triangle side length:
|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
Problem 1. The coordinates of the vertices of triangle ABC are given: A(4; 3), B(16;-6), C(20; 16). Find: 1) the length of side AB; 2) equations of sides AB and BC and their angular coefficients; 3) angle B in radians with an accuracy of two digits; 4) equation of height CD and its length; 5) the equation of the median AE and the coordinates of the point K of the intersection of this median with the height CD; 6) the equation of a straight line passing through point K parallel to side AB; 7) coordinates of point M, located symmetrically to point A relative to straight line CD.
Solution:
1. The distance d between points A(x 1 ,y 1) and B(x 2 ,y 2) is determined by the formula
Applying (1), we find the length of side AB:
2. The equation of the line passing through the points A(x 1 ,y 1) and B(x 2 ,y 2) has the form
(2)
Substituting the coordinates of points A and B into (2), we obtain the equation of side AB:
Having solved the last equation for y, we find the equation of side AB in the form of a straight line equation with an angular coefficient:
where
Substituting the coordinates of points B and C into (2), we obtain the equation of straight line BC:
Or 
3. It is known that the tangent of the angle between two straight lines, the angular coefficients of which are respectively equal, is calculated by the formula
(3)
The desired angle B is formed by straight lines AB and BC, the angular coefficients of which are found: Applying (3), we obtain
Or glad.
4. The equation of a straight line passing through a given point in a given direction has the form
(4)
The height CD is perpendicular to side AB. To find the slope of the height CD, we use the condition of perpendicularity of the lines. Since then 
Substituting into (4) the coordinates of point C and the found angular coefficient of height, we obtain
To find the length of the height CD, we first determine the coordinates of point D - the point of intersection of straight lines AB and CD. Solving the system together:
we find 
those. D(8;0).
Using formula (1) we find the length of the height CD:
5. To find the equation of the median AE, we first determine the coordinates of point E, which is the middle of side BC, using the formulas for dividing a segment into two equal parts:
(5)
Hence,
Substituting the coordinates of points A and E into (2), we find the equation for the median:
To find the coordinates of the point of intersection of the height CD and the median AE, we solve together the system of equations
We find.
6. Since the desired straight line is parallel to side AB, its angular coefficient will be equal to the angular coefficient of straight line AB. Substituting into (4) the coordinates of the found point K and the angular coefficient we obtain 
3x + 4y – 49 = 0 (KF)
7. Since the straight line AB is perpendicular to the straight line CD, the desired point M, located symmetrically to the point A relative to the straight line CD, lies on the straight line AB. In addition, point D is the midpoint of segment AM. Using formulas (5), we find the coordinates of the desired point M:
Triangle ABC, height CD, median AE, straight line KF and point M are constructed in the xOy coordinate system in Fig. 1.
Task 2.
Create an equation for the locus of points whose distances to a given point A(4; 0) and to a given straight line x=1 are equal to 2. 
Solution:
In the xOy coordinate system, we construct the point A(4;0) and the straight line x = 1. Let M(x;y) be an arbitrary point of the desired geometric location of points. Let us lower the perpendicular MB to the given line x = 1 and determine the coordinates of point B. Since point B lies on the given line, its abscissa is equal to 1. The ordinate of point B is equal to the ordinate of point M. Therefore, B(1;y) (Fig. 2 ).
According to the conditions of the problem |MA|: |MV| = 2. Distances |MA| and |MB| we find from formula (1) of problem 1:
Squaring the left and right sides, we get
The resulting equation is a hyperbola in which the real semi-axis is a = 2, and the imaginary half-axis is
Let's define the foci of a hyperbola. For a hyperbola, the equality is satisfied. Therefore, and 
– hyperbole tricks. As you can see, the given point A(4;0) is the right focus of the hyperbola.
Let us determine the eccentricity of the resulting hyperbola:
The equations of the asymptotes of the hyperbola have the form and . Therefore, or and are asymptotes of a hyperbola. Before constructing a hyperbola, we construct its asymptotes.
Problem 3. Create an equation for the locus of points equidistant from the point A(4; 3) and the straight line y = 1. Reduce the resulting equation to its simplest form.
Solution: Let M(x; y) be one of the points of the desired geometric locus of points. Let us drop the perpendicular MB from point M to this straight line y = 1 (Fig. 3). Let us determine the coordinates of point B. Obviously, the abscissa of point B is equal to the abscissa of point M, and the ordinate of point B is equal to 1, i.e. B(x; 1). According to the conditions of the problem |MA|=|MV|. Consequently, for any point M(x;y) belonging to the desired geometric locus of points, the following equality is true:
The resulting equation defines a parabola with its vertex at the point. To bring the parabola equation to its simplest form, let us set and y + 2 = Y, then the parabola equation takes the form: 
Exercise 1
57. The vertices of triangle ABC are given. Find
) length of side AB;
) equations of sides AB and AC and their angular coefficients;
) internal angle A;
) equation of the median drawn from vertex B;
) equation of height CD and its length;
) the equation of a circle for which the height CD is the diameter and the points of intersection of this circle with the side AC;
) equation of the bisector of the internal angle A;
) area of triangle ABC;
) a system of linear inequalities defining triangle ABC.
Make a drawing.
A(7, 9); B(-2, -3); C(-7, 7)
Solution:
1)
Let's find the length of the vector
= (x b -x a )2+ (y b -y a )2 = ((-2)-7)2 + (-3 - 9)2 = 92 + 122 = 225
= = 15 - length of side AB 2)
Let's find the equation of side AB
Equation of a line passing through points Oh A ; at V ) and B(x A ; at V ) in general Let's substitute the coordinates of points A and B into this equation of the straight line =
=
=
S AB = (- 3, - 4) is called the direction vector of straight line AB. This vector is parallel to line AB. 4(x - 7) = - 3(y - 9) 4x + 28 = - 3y + 27 4x + 3y + 1 = 0 - equation of line AB If the equation is written in the form: y = X - then we can isolate its angular coefficient: k 1 =4/3
Vector N AB = (-4, 3) is called the normal vector of the line AB. Vector N AB = (-4, 3) is perpendicular to line AB. Similarly, we find the equation of the side AC =
=
=
S AC = (- 7, - 1) - direction vector of the AC side (x - 7) = - 7(y - 9) x + 7 = - 7y + 63 x + 7y - 56 = 0 - equation of side AC y = = x + 8 whence the slope k 2 = 1/7
Vector N A.C. = (- 1, 7) - normal vector of line AC. Vector N A.C. = (- 1, 7) is perpendicular to line AC. 3)
Let's find angle A
Let's write down the formula for the scalar product of vectors And * = *cos ∟A To find angle A, it is enough to find the cosine of this angle. From the previous formula we write the expression for the cosine of angle A cos ∟A = Finding the scalar product of vectors And = (x V - X A ; at V - y A ) = (- 2 - 7; - 3 - 9) = (-9, -12)
= (x With - X A ; at With - y A ) = (- 7 - 7; 7 - 9) = (-14; -2)
9*(-14) + (-12)*(-2) = 150
Vector length = 15 (found earlier) Let's find the length of the vector = (x WITH -x A )2+ (y With -y a )2 = (-14)2 + (-2)2 = 200
= = 14.14 - side length AC Then cos ∟A = = 0,7072
∟A = 45 0
4)
Let us find the equation of the median BE drawn from point B to side AC
The median equation in general form Now we need to find the direction vector of straight line BE. Let's build triangle ABC to parallelogram ABCD, so that side AC is its diagonal. The diagonals in a parallelogram are divided in half, i.e. AE = EC. Therefore, point E lies on line BF. The vector BE can be taken as the direction vector of straight line BE , which we will find. = +
= (x c - X b ; at c - y b ) = (- 7- (-2); 7 - (-3)) = (-5. 10)
= + = (-5 + 9; 10 + 12) = (4; 22)
Let's substitute into the equation Let's substitute the coordinates of point C (-7; 7) (x + 7) = 2(y - 7) x + 77 = 2y - 14 x - 2y + 91 = 0 - equation of median BE Since point E is the middle of side AC, its coordinates X e = (x A + x With )/2 = (7 - 7)/2 = 0
at e = (y A + y With )/2 = (9 + 7)/2 = 8
Coordinates of point E (0; 8) 5)
Let's find the equation for the height CD and its length
General equation It is necessary to find the direction vector of the straight line CD Line CD is perpendicular to line AB, therefore, the direction vector of line CD is parallel to the normal vector of line AB CD ‖AB That is, the normal vector of straight line AB can be taken as the directing vector of straight line CD Vector AB found earlier: AB (-4, 3)
Let's substitute the coordinates of point C, (- 7; 7) (x + 7) = - 4(y - 7) x + 21 = - 4y + 28 x + 4y - 7 = 0 - equation of height C D Point D coordinates: Point D belongs to line AB, therefore, the coordinates of point D(x d . y d ) must satisfy the equation of straight line AB found earlier Point D belongs to the line CD, therefore, the coordinates of point D(x d . y d ) must satisfy the equation of the straight line CD, Let's create a system of equations based on this Coordinates D(1; 1) Find the length of straight line CD = (x d -x c )2+ (y d -y c )2 = (1 + 7)2 + (1 - 7)2 = 64 +36 = 100
= = 10 - length of straight line CD 6)
Find the equation of a circle with diameter CD
It is obvious that straight line CD passes through the origin of coordinates since its equation is -3x - 4y = 0, therefore, the equation of a circle can be written in the form (x - a) 2 + (y - b) 2= R 2- equation of a circle with center at point (a; b) Here R = СD/2 = 10 /2 = 5 (x - a) 2 + (y - b) 2 = 25
The center of the circle O (a; b) lies in the middle of the segment CD. Let's find its coordinates: X 0= a = = = - 3;
y 0= b = = = 4
Circle equation: (x + 3) 2 + (y - 4) 2 = 25
Let's find the intersection of this circle with side AC: point K belongs to both the circle and the line AC x + 7y - 56 = 0 - the equation of the straight line AC found earlier. Let's create a system Thus, we get the quadratic equation at 2- 750у +2800 = 0 at 2- 15у + 56 = 0 =
at 1 = 8
at 2= 7 - point corresponding to point C therefore the coordinates of point H: x = 7*8 - 56 = 0
An example of solving some tasks from the standard work “Analytical geometry on a plane”
The vertices are given, 
,
triangle ABC. Find:
Equations of all sides of a triangle;
System of linear inequalities defining a triangle ABC;
Equations of altitude, median and bisector of a triangle drawn from the vertex A;
The intersection point of the triangle's altitudes;
The intersection point of the triangle's medians;
Length of the height lowered to the side AB;
Corner A;
Make a drawing.
Let the vertices of the triangle have coordinates: A (1; 4), IN (5; 3), WITH(3; 6). Let's draw a drawing right away:
1. To write down the equations of all sides of a triangle, we use the equation of a straight line passing through two given points with coordinates ( x 0 , y 0 ) And ( x 1 , y 1 ):
=
Thus, substituting instead of ( x 0 , y 0 ) point coordinates A, and instead of ( x 1 , y 1 ) point coordinates IN, we get the equation of the line AB:
The resulting equation will be the equation of the straight line AB, written in general form. Similarly, we find the equation of the straight line AC:
And also the equation of the straight line Sun:
2. Note that the set of points of the triangle ABC represents the intersection of three half-planes, and each half-plane can be defined using a linear inequality. If we take the equation of either side ∆ ABC, For example AB, then the inequalities
And 
define points lying on opposite sides of a line AB. We need to choose the half-plane where point C lies. Let’s substitute its coordinates into both inequalities:
The second inequality will be correct, which means that the required points are determined by the inequality
.
We do the same with straight line BC, its equation 
. We use point A (1, 1) as a test point:
This means that the required inequality has the form:
.
If we check straight line AC (test point B), we get:
This means that the required inequality will have the form
We finally obtain a system of inequalities:
The signs “≤”, “≥” mean that points lying on the sides of the triangle are also included in the set of points that make up the triangle ABC.
3. a) In order to find the equation of the height dropped from the vertex A to the side Sun, consider the equation of the side Sun:
. 
Vector with coordinates Sun perpendicular to the side A and therefore parallel to the height. Let's write down the equation of a straight line passing through a point 
:
parallel to the vector A This is the equation for the height omitted from t. Sun.
to the side Sun b) Find the coordinates of the middle of the side 
according to the formulas: 
Here IN– these are the coordinates of t. 
, A WITH– coordinates t.
. Let's substitute and get: A The straight line passing through this point and the point
is the required median: 
c) We will look for the equation of the bisector based on the fact that in an isosceles triangle the height, median and bisector descended from one vertex to the base of the triangle are equal. Let's find two vectors 
And
and their lengths: 
Then the vector 
has the same direction as the vector 
, and its length 
Likewise, the unit vector 
coincides in direction with the vector
Vector sum A there is a vector that coincides in direction with the bisector of the angle
. Thus, the equation of the desired bisector can be written as: IN 4) We have already constructed the equation for one of the heights. Let's construct an equation for another height, for example, from the vertex AC. Side 
given by the equation 
So the vector AC perpendicular IN, and thus parallel to the desired height. Then the equation of the line passing through the vertex 
in the direction of the vector AC(i.e. perpendicular
), has the form:
It is known that the altitudes of a triangle intersect at one point. In particular, this point is the intersection of the found heights, i.e. solving the system of equations:
- coordinates of this point. AB 5. Middle 
has coordinates . Let us write the equation of the median to the side AB.
Note that a zero in the denominator of a fraction in the equation of a straight line means that this straight line runs parallel to the ordinate axis.
To find the intersection point of the medians, it is enough to solve the system of equations:
The intersection point of the medians of a triangle has coordinates 
.
6. Length of height lowered to the side AB, equal to the distance from the point WITH to a straight line AB with equation 
and is found by the formula:
7. Cosine of angle A can be found using the formula for the cosine of the angle between vectors 
c) We will look for the equation of the bisector based on the fact that in an isosceles triangle the height, median and bisector descended from one vertex to the base of the triangle are equal. Let's find two vectors 
, which is equal to the ratio of the scalar product of these vectors to the product of their lengths:
.
