1. Given the vertices of a triangle ABC.A(–9; –2), IN(3; 7), WITH(1; –7).
1) side length AB;
2) equations of the sides AB And AC and their angular coefficients;
3) angle A in radians;
4) height equation WITHD and its length;
5) the equation of a circle for which the height WITHD there is a diameter;
6) a system of linear inequalities defining a triangle ABC.
Solution. Let's make a drawing.
1. Let's find the length of side AB. The distance between two points is determined by the formula
2. Let's find the equations of the sidesAB AndAC and their angular coefficients.
Let's write down the equation of a straight line passing through two points.
This is the general equation of a line. Let's resolve it with respect to y, we get
, the slope of the straight line is equal to 
Similarly for side AC we have.
the slope of the straight line is equal to 
3. We'll findcornerA
in radians.
This is the angle between two vectors 
And 
. Let's write down the coordinates of the vectors. The cosine of the angle between the vectors is equal to
4. We'll findheight equationWITH
D
and its length.
, therefore, their angular coefficients are related by the relation 
.
Let's write the height equation through the angular coefficient
Dot 
belongs to the line CD, therefore its coordinates satisfy the equation of the line, hence we have 
Finally 
or
We calculate the length of the height as the distance from point C to straight line AB
5. Let's find the equation of a circle, for which heightWITH D there is a diameter.
We find the coordinates of point D as the point of intersection of two straight lines AB and CD, the equations of which are known.
Let's find the coordinates of point O - the center of the circle. This is the middle of the CD section.
The radius of the circle is 
Let's write down the equation of a circle.
6) Let's define a triangleABC system of linear inequalities.
Let's find the equation of line CB.
The system of linear inequalities will look like this.
2. Solve this system of equations using Cramer's formulas. Check the resulting solution.
Solution. Let us calculate the determinant of this system:
.
Let's find the determinants 
and solve the system:
Examination:
Answer:
3. Write the system of equations in matrix form and solve it using
inverse matrix. Check the resulting solution
Solution.
Let's find the determinant of matrix A
the matrix is non-singular and has an inverse. Let's find all the algebraic complements and create a union matrix. 
The inverse matrix has the form: 
Let's do the multiplication 
and find the vector of solutions.
Examination
. 
Answer: 
Solution.
N = (2, 1). Draw a level line perpendicular to the normal vector and move it in the direction of the normal,
The objective function reaches its minimum at point A, and its maximum at point B. We find the coordinates of these points by jointly solving the equations of the lines at the intersection of which they are located.
5. A travel company requires no more A three-ton buses and no more V
five-ton buses. The selling price of buses of the first brand is 20,000 USD, of the second brand
40000 USD A travel company can allocate no more than With c.u.
How many buses of each brand should be purchased separately so that their total
(total) load capacity was maximum. Solve the problem graphically.
A= 20 V= 18 With= 1000000
Solution.
Let's create a mathematical model of the problem .
Let us denote by 
- the number of buses of each tonnage that will be purchased. The purpose of procurement is to have the maximum carrying capacity of purchased machines, described by the goal function
The limitations of the task are determined by the number of purchased buses and their cost.
Let's solve the problem graphically. . We construct the region of feasible solutions to the problem and the normal to the level lines N = (3, 5). Draw a level line perpendicular to the normal vector and move it in the direction of the normal.
The goal function reaches its maximum at the point 
, the goal function takes on the value .
Solution. 1. The domain of definition of the function is the entire numerical axis.
2, The function is neither even nor odd.
3. When x=0, y=20
4. We examine the function for monotonicity and extrema.
Let's find the zeros of the derivative
Stationary points of a function.
Let's plot stationary points on the Ox axis and check the signs of the derivative on each section of the axis.
–maximum point 
; 
-minimum point 
5. We examine the graph of the function for convexity and concavity. Let's take the 2nd derivative
The inflection point of a function graph.
At 
- the function is convex; at 
- the function is concave.
The graph of the function looks like
6. Find the largest and smallest value of the function on the interval [-1; 4]
Let's calculate the value of the function at the ends of the segment 
At the minimum point, the function takes on the values , therefore, the smallest value on the segment [-1; 4] the function takes at the minimum point, and the maximum at the left boundary of the interval.
7. Find indefinite integrals and check integration results
differentiation.
Solution.
Examination.
Here the product of cosines has been replaced by a sum, according to trigonometric formulas.
1. Equation of sides AB and BC and their angular coefficients.
The assignment gives the coordinates of the points through which these lines pass, so we will use the equation of a line passing through two given points $$\frac(x-x_1)(x_2-x_1)=\frac(y-y_1)(y_2-y_1)$ $ substitute and get the equations
equation of line AB $$\frac(x+6)(6+6)=\frac(y-8)(-1-8) => y = -\frac(3)(4)x + \frac(7 )(2)$$ the slope of straight line AB is equal to \(k_(AB) = -\frac(3)(4)\)
equation of line BC $$\frac(x-4)(6-4)=\frac(y-13)(-1-13) => y = -7x + 41$$ slope of line BC is equal to \(k_( BC) = -7\)
2. Angle B in radians with an accuracy of two digits
Angle B is the angle between lines AB and BC, which is calculated by the formula $$tg\phi=|\frac(k_2-k_1)(1+k_2*k_1)|$$substitute the values of the angular coefficients of these lines and get $$tg\ phi=|\frac(-7+\frac(3)(4))(1+7*\frac(3)(4))| = 1 => \phi = \frac(\pi)(4) \approx 0.79$$
3.Length of side AB
The length of side AB is calculated as the distance between the points and is equal to \(d = \sqrt((x_2-x_1)^2+(y_2-y_1)^2)\) => $$d_(AB) = \sqrt((6+ 6)^2+(-1-8)^2) = 15$$
4. Equation of CD height and its length.
We will find the height equation using the formula of a straight line passing through a given point C(4;13) in a given direction - perpendicular to the straight line AB using the formula \(y-y_0=k(x-x_0)\). Let's find the angular coefficient of height \(k_(CD)\) using the property of perpendicular lines \(k_1=-\frac(1)(k_2)\) we get $$k_(CD)= -\frac(1)(k_(AB) ) = -\frac(1)(-\frac(3)(4)) = \frac(4)(3)$$ We substitute a straight line into the equation, we get $$y - 13 = \frac(4)(3) (x-4) => y = \frac(4)(3)x+\frac(23)(3)$$ We will look for the length of the height as the distance from point C(4;13) to straight line AB using the formula $$d = \frac(Ax_0+By_0+C)(\sqrt(A^2+B^2))$$ in the numerator is the equation of the straight line AB, let's reduce it to this form \(y = -\frac(3)(4)x + \frac(7)(2) => 4y+3x-14 = 0\) , substitute the resulting equation and the coordinates of the point into the formula $$d = \frac(4*13+3*4-14 )(\sqrt( 4^2+3^2)) = \frac(50)(5) =10$$
5. Equation of the median AE and the coordinates of the point K, the intersection of this median with the height CD.
We will look for the equation of the median as the equation of a straight line passing through two given points A(-6;8) and E, where point E is the midpoint between points B and C and its coordinates are found according to the formula \(E(\frac(x_2+x_1) (2);\frac(y_2+y_1)(2))\) substitute the coordinates of the points \(E(\frac(6+4)(2);\frac(-1+13)(2))\) = > \(E(5; 6)\), then the equation of the median AE will be the following $$\frac(x+6)(5+6)=\frac(y-8)(6-8) => y = - \frac(2)(11)x + \frac(76)(11)$$Let's find the coordinates of the point of intersection of the heights and the median, i.e. let's find their common point. To do this, we will create a system equation $$\begin(cases)y = -\frac(2)(11)x + \frac(76)(11)\\y = \frac(4)(3)x+ \frac(23)(3)\end(cases)=>\begin(cases)11y = -2x +76\\3y = 4x+23\end(cases)=>$$$$\begin(cases)22y = -4x +152\\3y = 4x+23\end(cases)=> \begin(cases)25y =175\\3y = 4x+23\end(cases)=> $$$$\begin(cases) y =7\\ x=-\frac(1)(2)\end(cases)$$ Coordinates of the intersection point \(K(-\frac(1)(2);7)\)
6. Equation of a line that passes through point K parallel to side AB.
If the straight line is parallel, then their angular coefficients are equal, i.e. \(k_(AB)=k_(K) = -\frac(3)(4)\), the coordinates of the point \(K(-\frac(1)(2);7)\) are also known, i.e. . to find the equation of a straight line, we apply the formula for the equation of a straight line passing through a given point in a given direction \(y - y_0=k(x-x_0)\), substitute the data and get $$y - 7= -\frac(3)(4) (x-\frac(1)(2)) => y = -\frac(3)(4)x + \frac(53)(8)$$
8. Coordinates of point M which is symmetrical to point A relative to straight line CD.
Point M lies on line AB, because CD is the height to this side. Let's find the intersection point of CD and AB; to do this, solve the system of equations $$\begin(cases)y = \frac(4)(3)x+\frac(23)(3)\\y = -\frac(3)(4) x + \frac(7)(2)\end(cases) =>\begin(cases)3y = 4x+23\\4y =-3x + 14\end(cases) => $$$$\begin(cases )12y = 16x+92\\12y =-9x + 42\end(cases) =>
\begin(cases)0= 25x+50\\12y =-9x + 42\end(cases) => $$$$\begin(cases)x=-2\\y=5 \end(cases)$$ Coordinates of point D(-2;5). According to the condition AD=DK, this distance between points is found by the Pythagorean formula \(d = \sqrt((x_2-x_1)^2+(y_2-y_1)^2)\), where AD and DK are the hypotenuses of equal right triangles, and \(Δx =x_2-x_1\) and \(Δy=y_2-y_1\) are the legs of these triangles, i.e. let's find the legs and find the coordinates of point M. \(Δx=x_D-x_A = -2+6=4\), and \(Δy=y_D-y_A = 5-8=-3\), then the coordinates of point M will be equal \ (x_M-x_D = Δx => x_D +Δx =-2+4=2 \), and \(y_M-y_D = Δy => y_D +Δy =5-3=2 \), we found that the coordinates of the point \( M(2;2)\)
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 for the height dropped from the vertex A to the side Sun, consider the equation of the side Sun:
. Vector with coordinates 
perpendicular to the side Sun and therefore parallel to the height. Let us write down the equation of a straight line passing through a point A parallel to the vector 
:
This is the equation for the height omitted from t. A to the side Sun.
b) Find the coordinates of the middle of the side Sun according to the formulas: 
Here 
– these are the coordinates of t. IN, A 
– coordinates t. WITH. Let's substitute and get:
The straight line passing through this point and the point A 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
there is a vector that coincides in direction with the bisector of the angle A. Thus, the equation of the desired bisector can be written as:
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 IN. Side AC given by the equation 
So the vector 
perpendicular AC, and thus parallel to the desired height. Then the equation of the line passing through the vertex IN in the direction of the vector 
(i.e. perpendicular AC), 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.
5. Middle AB has coordinates 
. Let us write the equation of the median to the side AB. This line passes through points with coordinates (3, 2) and (3, 6), which means its equation has the form:
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 
And 
, which is equal to the ratio of the scalar product of these vectors to the product of their lengths:
.