Let the straight line pass through the points M 1 (x 1; y 1) and M 2 (x 2; y 2). The equation of a straight line passing through the point M 1 has the form y- y 1 \u003d k (x - x 1), (10.6)
where k - still unknown coefficient.
Since the straight line passes through the point M 2 (x 2 y 2), then the coordinates of this point must satisfy equation (10.6): y 2 -y 1 \u003d k (x 2 -x 1).
From here we find Substituting the found value k
into equation (10.6), we obtain the equation of a straight line passing through the points M 1 and M 2: 
It is assumed that in this equation x 1 ≠ x 2, y 1 ≠ y 2
If x 1 \u003d x 2, then the straight line passing through the points M 1 (x 1, y I) and M 2 (x 2, y 2) is parallel to the y-axis. Its equation is x = x 1 .
If y 2 \u003d y I, then the equation of the straight line can be written as y \u003d y 1, the straight line M 1 M 2 is parallel to the x-axis.
Equation of a straight line in segments
Let the straight line intersect the Ox axis at the point M 1 (a; 0), and the Oy axis - at the point M 2 (0; b). The equation will take the form: 
those. 
. This equation is called the equation of a straight line in segments, because the numbers a and b indicate which segments the straight line cuts off on the coordinate axes.
Equation of a straight line passing through a given point perpendicular to a given vector
Let's find the equation of a straight line passing through a given point Mo (x O; y o) perpendicular to a given non-zero vector n = (A; B).
Take an arbitrary point M(x; y) on the straight line and consider the vector M 0 M (x - x 0; y - y o) (see Fig. 1). Since the vectors n and M o M are perpendicular, their scalar product is equal to zero: that is,
A(x - xo) + B(y - yo) = 0. (10.8)
Equation (10.8) is called equation of a straight line passing through a given point perpendicular to a given vector .
The vector n = (A; B) perpendicular to the line is called normal normal vector of this line .
Equation (10.8) can be rewritten as Ah + Wu + C = 0 , (10.9)
where A and B are the coordinates of the normal vector, C \u003d -Ax o - Vu o - free member. Equation (10.9) is the general equation of a straight line(see Fig.2).
Fig.1 Fig.2
Canonical equations of the straight line
,
Where 
are the coordinates of the point through which the line passes, and 
- direction vector.
Curves of the second order Circle
A circle is the set of all points of a plane equidistant from a given point, which is called the center.
Canonical equation of a circle of radius
R centered on a point 
:
In particular, if the center of the stake coincides with the origin, then the equation will look like: 
Ellipse
An ellipse is a set of points in a plane, the sum of the distances from each of them to two given points
and 
, which are called foci, is a constant value 
, greater than the distance between the foci 
.
The canonical equation of an ellipse whose foci lie on the Ox axis and whose origin is in the middle between the foci has the form 
G 
de a the length of the major semiaxis; b is the length of the minor semiaxis (Fig. 2).
Equation of a straight line on a plane.
The direction vector is straight. Normal vector
A straight line on a plane is one of the simplest geometric shapes, familiar to you since elementary grades, and today we will learn how to deal with it using the methods of analytical geometry. To master the material, it is necessary to be able to build a straight line; know which equation defines a straight line, in particular, a straight line passing through the origin and straight lines parallel to the coordinate axes. This information can be found in the manual. Graphs and properties of elementary functions, I created it for matan, but the section on the linear function turned out to be very successful and detailed. Therefore, dear teapots, first warm up there. In addition, you need to have basic knowledge of vectors otherwise the understanding of the material will be incomplete.
In this lesson, we will look at ways in which you can write the equation of a straight line in a plane. I recommend not neglecting practical examples (even if it seems very simple), as I will supply them with elementary and important facts, technical methods that will be required in the future, including in other sections of higher mathematics.
- How to write the equation of a straight line with a slope?
- How ?
- How to find the direction vector by the general equation of a straight line?
- How to write an equation of a straight line given a point and a normal vector?
and we start:
Line Equation with Slope
The well-known "school" form of the equation of a straight line is called equation of a straight line with a slope. For example, if a straight line is given by the equation, then its slope: . Consider the geometric meaning of this coefficient and how its value affects the location of the line: 
In the course of geometry it is proved that the slope of the straight line is tangent of an angle between positive axis directionand given line: , and the corner is “unscrewed” counterclockwise.
In order not to clutter up the drawing, I drew angles for only two straight lines. Consider the "red" straight line and its slope. According to the above: (angle "alpha" is indicated by a green arc). For the "blue" straight line with the slope, equality is true (the angle "beta" is indicated by the brown arc). And if the tangent of the angle is known, then if necessary it is easy to find and the corner using the inverse function - arc tangent. As they say, a trigonometric table or a calculator in hand. Thus, the slope characterizes the degree of inclination of the straight line to the x-axis.
In this case, the following cases are possible:
1) If the slope is negative: , then the line, roughly speaking, goes from top to bottom. Examples are "blue" and "crimson" straight lines in the drawing.
2) If the slope is positive: , then the line goes from bottom to top. Examples are "black" and "red" straight lines in the drawing.
3) If the slope is equal to zero: , then the equation takes the form , and the corresponding line is parallel to the axis. An example is the "yellow" line.
4) For a family of straight lines parallel to the axis (there is no example in the drawing, except for the axis itself), the slope does not exist (tangent of 90 degrees not defined).
The greater the slope modulo, the steeper the line graph goes.
For example, consider two straight lines. Here , so the straight line has a steeper slope. I remind you that the module allows you to ignore the sign, we are only interested in absolute values angular coefficients.
In turn, a straight line is steeper than straight lines. 
.
Vice versa: the smaller the slope modulo, the straight line is flatter.
For straight lines 
the inequality is true, thus, the straight line is more than a canopy. Children's slide, so as not to plant bruises and bumps.
Why is this needed?
Prolong your torment Knowing the above facts allows you to immediately see your mistakes, in particular, errors when plotting graphs - if the drawing turned out “clearly something is wrong”. It is desirable that you straightaway it was clear that, for example, a straight line is very steep and goes from bottom to top, and a straight line is very flat, close to the axis and goes from top to bottom.
In geometric problems, several straight lines often appear, so it is convenient to denote them somehow.
Notation: straight lines are indicated by small Latin letters: . A popular option is the designation of the same letter with natural subscripts. For example, the five lines that we have just considered can be denoted by 
.
Since any straight line is uniquely determined by two points, it can be denoted by these points: 
etc. The notation quite obviously implies that the points belong to the line.
Time to loosen up a bit:
How to write the equation of a straight line with a slope?
If a point is known that belongs to a certain line, and the slope of this line, then the equation of this line is expressed by the formula:
Example 1
Compose the equation of a straight line with a slope if it is known that the point belongs to this straight line.
Decision: We will compose the equation of a straight line according to the formula 
. In this case: 
Answer:
Examination performed elementarily. First, we look at the resulting equation and make sure that our slope is in its place. Second, the coordinates of the point must satisfy the given equation. Let's plug them into the equation:
The correct equality is obtained, which means that the point satisfies the resulting equation.
Conclusion: Equation found correctly.
A more tricky example for a do-it-yourself solution:
Example 2
Write the equation of a straight line if it is known that its angle of inclination to the positive direction of the axis is , and the point belongs to this straight line.
If you have any difficulties, re-read the theoretical material. More precisely, more practical, I miss many proofs.
The last bell rang, the graduation ball died down, and behind the gates of our native school, in fact, analytical geometry is waiting for us. Jokes are over... Maybe it's just getting started =)
Nostalgically we wave the handle to the familiar and get acquainted with the general equation of a straight line. Since in analytic geometry it is precisely this that is in use:
The general equation of a straight line has the form: , where are some numbers. At the same time, the coefficients simultaneously are not equal to zero, since the equation loses its meaning.
Let's dress in a suit and tie an equation with a slope. First, we move all the terms to the left side:
The term with "x" must be put in first place:
In principle, the equation already has the form , but according to the rules of mathematical etiquette, the coefficient of the first term (in this case ) must be positive. Changing signs:
Remember this technical feature! We make the first coefficient (most often ) positive!
In analytic geometry, the equation of a straight line will almost always be given in a general form. Well, if necessary, it is easy to bring it to a “school” form with a slope (with the exception of straight lines parallel to the y-axis).
Let's ask ourselves what enough know to build a straight line? Two points. But about this childhood case later, now sticks with arrows rule. Each straight line has a well-defined slope, to which it is easy to "adapt" vector.
A vector that is parallel to a line is called the direction vector of that line.. Obviously, any straight line has infinitely many direction vectors, and all of them will be collinear (co-directed or not - it does not matter).
I will denote the direction vector as follows: .
But one vector is not enough to build a straight line, the vector is free and is not attached to any point of the plane. Therefore, it is additionally necessary to know some point that belongs to the line.
How to write an equation of a straight line given a point and a direction vector?
If a certain point belonging to the line and the directing vector of this line are known, then the equation of this line can be compiled by the formula:
Sometimes it is called canonical equation of the line .
What to do when one of the coordinates is zero, we will look into practical examples below. By the way, note - both at once coordinates cannot be zero, since the zero vector does not specify a specific direction.
Example 3
Write an equation of a straight line given a point and a direction vector
Decision: We will compose the equation of a straight line according to the formula. In this case:
Using the properties of proportion, we get rid of fractions: 
And we bring the equation to a general form:
Answer:
Drawing in such examples, as a rule, is not necessary, but for the sake of understanding: 
In the drawing, we see the starting point, the original direction vector (it can be postponed from any point on the plane) and the constructed line. By the way, in many cases, the construction of a straight line is most conveniently carried out using the slope equation. Our equation is easy to convert to the form and without any problems pick up one more point to build a straight line.
As noted at the beginning of the section, a line has infinitely many direction vectors, and they are all collinear. For example, I drew three such vectors: 
. Whichever direction vector we choose, the result will always be the same straight line equation.
Let's compose the equation of a straight line by a point and a directing vector:
Breaking down the proportion:
Divide both sides by -2 and get the familiar equation:
Those who wish can similarly test vectors 
or any other collinear vector.
Now let's solve the inverse problem:
How to find the direction vector by the general equation of a straight line?
Very simple:
If a straight line is given by a general equation in a rectangular coordinate system, then the vector is the direction vector of this straight line.
Examples of finding direction vectors of straight lines: 
The statement allows us to find only one direction vector from an infinite set, but we don’t need more. Although in some cases it is advisable to reduce the coordinates of the direction vectors:
So, the equation specifies a straight line that is parallel to the axis, and the coordinates of the resulting steering vector are conveniently divided by -2, getting exactly the basis vector as the steering vector. Logically.
Similarly, the equation defines a straight line parallel to the axis, and dividing the coordinates of the vector by 5, we get the ort as the direction vector.
Now let's execute check example 3. The example went up, so I remind you that in it we made up the equation of a straight line using a point and a direction vector
First of all, according to the equation of a straight line, we restore its directing vector: 
- everything is fine, we got the original vector (in some cases, it can turn out to be collinear to the original vector, and this is usually easy to see by the proportionality of the corresponding coordinates).
Secondly, the coordinates of the point must satisfy the equation . We substitute them into the equation:
The correct equality has been obtained, which we are very pleased with.
Conclusion: Job completed correctly.
Example 4
Write an equation of a straight line given a point and a direction vector
This is a do-it-yourself example. Solution and answer at the end of the lesson. It is highly desirable to make a check according to the algorithm just considered. Try to always (if possible) check on a draft. It is foolish to make mistakes where they can be 100% avoided.
In the event that one of the coordinates of the direction vector is zero, it is very simple to do:
Example 5
Decision: The formula is invalid because the denominator on the right side is zero. There is an exit! Using the properties of proportion, we rewrite the formula in the form , and the rest rolled along a deep rut:
Answer:
Examination:
1) Restore the direction vector of the straight line:
– the resulting vector is collinear to the original direction vector.
2) Substitute the coordinates of the point in the equation: 
The correct equality is obtained
Conclusion: job completed correctly
The question arises, why bother with the formula if there is a universal version that will work anyway? There are two reasons. First, the fractional formula much better to remember. And secondly, the disadvantage of the universal formula is that markedly increased risk of confusion when substituting coordinates.
Example 6
Compose the equation of a straight line given a point and a direction vector.
This is a do-it-yourself example.
Let's return to the ubiquitous two points:
How to write the equation of a straight line given two points?
If two points are known, then the equation of a straight line passing through these points can be compiled using the formula:
In fact, this is a kind of formula, and here's why: if two points are known, then the vector will be the direction vector of this line. On the lesson Vectors for dummies we considered the simplest problem - how to find the coordinates of a vector from two points. According to this problem, the coordinates of the direction vector: 
Note
: points can be "swapped" and use the formula 
. Such a decision would be equal.
Example 7
Write the equation of a straight line from two points 
.
Decision: Use the formula: 
We comb the denominators:
And shuffle the deck: 
It is now convenient to get rid of fractional numbers. In this case, you need to multiply both parts by 6: 
Open the brackets and bring the equation to mind: 
Answer:
Examination is obvious - the coordinates of the initial points must satisfy the resulting equation:
1) Substitute the coordinates of the point: 
True equality.
2) Substitute the coordinates of the point: 
True equality.
Conclusion: the equation of the straight line is correct.
If a at least one of points does not satisfy the equation, look for an error.
It is worth noting that the graphical verification in this case is difficult, because to build a line and see if the points belong to it 
, not so easy.
I will note a couple of technical points of the solution. Perhaps in this problem it is more advantageous to use the mirror formula 
and, for the same points 
make an equation: 
There are fewer fractions. If you want, you can complete the solution to the end, the result should be the same equation.
The second point is to look at the final answer and see if it can be further simplified? For example, if an equation is obtained, then it is advisable to reduce it by two: - the equation will set the same straight line. However, this is already a topic of conversation about mutual arrangement of straight lines.
Having received an answer 
in Example 7, just in case, I checked if ALL coefficients of the equation are divisible by 2, 3 or 7. Although, most often such reductions are made during the solution.
Example 8
Write the equation of a straight line passing through the points 
.
This is an example for an independent solution, which will just allow you to better understand and work out the calculation technique.
Similar to the previous paragraph: if in the formula 
one of the denominators (direction vector coordinate) vanishes, then we rewrite it as . And again, notice how awkward and confused she began to look. I don’t see much point in giving practical examples, since we have already actually solved such a problem (see Nos. 5, 6).
Straight line normal vector (normal vector)
What is normal? In simple terms, a normal is a perpendicular. That is, the normal vector of a line is perpendicular to the given line. It is obvious that any straight line has an infinite number of them (as well as directing vectors), and all the normal vectors of the straight line will be collinear (codirectional or not - it does not matter).
Dealing with them will be even easier than with direction vectors:
If a straight line is given by a general equation in a rectangular coordinate system, then the vector is the normal vector of this straight line.
If the coordinates of the direction vector have to be carefully “pulled out” from the equation, then the coordinates of the normal vector can simply be “removed”.
The normal vector is always orthogonal to the direction vector of the line. We will verify the orthogonality of these vectors using dot product:
I will give examples with the same equations as for the direction vector: 
Is it possible to write an equation of a straight line, knowing one point and a normal vector? It feels like it's possible. If the normal vector is known, then the direction of the straightest line is also uniquely determined - this is a “rigid structure” with an angle of 90 degrees.
How to write an equation of a straight line given a point and a normal vector?
If some point belonging to the line and the normal vector of this line are known, then the equation of this line is expressed by the formula:
Here everything went without fractions and other surprises. Such is our normal vector. Love it. And respect =)
Example 9
Compose the equation of a straight line given a point and a normal vector. Find the direction vector of the straight line.
Decision: Use the formula: 
The general equation of the straight line is obtained, let's check:
1) "Remove" the coordinates of the normal vector from the equation: 
- yes, indeed, the original vector is obtained from the condition (or the vector should be collinear to the original vector).
2) Check if the point satisfies the equation: 
True equality.
After we are convinced that the equation is correct, we will complete the second, easier part of the task. We pull out the direction vector of the straight line: 
Answer: 
In the drawing, the situation is as follows: 
For the purposes of training, a similar task for an independent solution:
Example 10
Compose the equation of a straight line given a point and a normal vector. Find the direction vector of the straight line.
The final section of the lesson will be devoted to less common, but also important types of equations of a straight line in a plane
Equation of a straight line in segments.
Equation of a straight line in parametric form
The equation of a straight line in segments has the form , where are nonzero constants. Some types of equations cannot be represented in this form, for example, direct proportionality (since the free term is zero and there is no way to get one on the right side).
This is, figuratively speaking, a "technical" type of equation. The usual task is to represent the general equation of a straight line as an equation of a straight line in segments. Why is it convenient? The equation of a straight line in segments allows you to quickly find the points of intersection of a straight line with coordinate axes, which is very important in some problems of higher mathematics.
Find the point of intersection of the line with the axis. We reset the “y”, and the equation takes the form . The desired point is obtained automatically: .
Same with axis 
is the point where the line intersects the y-axis.
In this article, we will consider the general equation of a straight line in a plane. Let us give examples of constructing the general equation of a straight line if two points of this straight line are known or if one point and the normal vector of this straight line are known. Let us present methods for transforming an equation in general form into canonical and parametric forms.
Let an arbitrary Cartesian rectangular coordinate system be given Oxy. Consider a first degree equation or a linear equation:
| Ax+By+C=0, | (1) |
where A, B, C are some constants, and at least one of the elements A and B different from zero.
We will show that a linear equation in the plane defines a straight line. Let us prove the following theorem.
Theorem 1. In an arbitrary Cartesian rectangular coordinate system on a plane, each straight line can be given by a linear equation. Conversely, each linear equation (1) in an arbitrary Cartesian rectangular coordinate system on the plane defines a straight line.
Proof. It suffices to prove that the line L is determined by a linear equation for any one Cartesian rectangular coordinate system, since then it will be determined by a linear equation and for any choice of Cartesian rectangular coordinate system.
Let a straight line be given on the plane L. We choose a coordinate system so that the axis Ox aligned with the line L, and the axis Oy was perpendicular to it. Then the equation of the line L will take the following form:
| y=0. | (2) |
All points on a line L will satisfy the linear equation (2), and all points outside this straight line will not satisfy the equation (2). The first part of the theorem is proved.
Let a Cartesian rectangular coordinate system be given and let linear equation (1) be given, where at least one of the elements A and B different from zero. Find the locus of points whose coordinates satisfy equation (1). Since at least one of the coefficients A and B is different from zero, then equation (1) has at least one solution M(x 0 ,y 0). (For example, when A≠0, dot M 0 (−C/A, 0) belongs to the given locus of points). Substituting these coordinates into (1) we obtain the identity
| Ax 0 +By 0 +C=0. | (3) |
Let us subtract identity (3) from (1):
| A(x−x 0)+B(y−y 0)=0. | (4) |
Obviously, equation (4) is equivalent to equation (1). Therefore, it suffices to prove that (4) defines some line.
Since we are considering a Cartesian rectangular coordinate system, it follows from equality (4) that the vector with components ( x−x 0 , y−y 0 ) is orthogonal to the vector n with coordinates ( A,B}.
Consider some line L passing through the point M 0 (x 0 , y 0) and perpendicular to the vector n(Fig.1). Let the point M(x,y) belongs to the line L. Then the vector with coordinates x−x 0 , y−y 0 perpendicular n and equation (4) is satisfied (scalar product of vectors n and equals zero). Conversely, if the point M(x,y) does not lie on a line L, then the vector with coordinates x−x 0 , y−y 0 is not orthogonal to vector n and equation (4) is not satisfied. The theorem has been proven.
Proof. Since lines (5) and (6) define the same line, the normal vectors n 1 ={A 1 ,B 1 ) and n 2 ={A 2 ,B 2) are collinear. Since the vectors n 1 ≠0, n 2 ≠ 0, then there is a number λ , what n 2 =n 1 λ . Hence we have: A 2 =A 1 λ , B 2 =B 1 λ . Let's prove that C 2 =C 1 λ . It is obvious that coinciding lines have a common point M 0 (x 0 , y 0). Multiplying equation (5) by λ and subtracting equation (6) from it we get:
Since the first two equalities from expressions (7) are satisfied, then C 1 λ −C 2=0. Those. C 2 =C 1 λ . The remark has been proven.
Note that equation (4) defines the equation of a straight line passing through the point M 0 (x 0 , y 0) and having a normal vector n={A,B). Therefore, if the normal vector of the line and the point belonging to this line are known, then the general equation of the line can be constructed using equation (4).
Example 1. A line passes through a point M=(4,−1) and has a normal vector n=(3, 5). Construct the general equation of a straight line.
Decision. We have: x 0 =4, y 0 =−1, A=3, B=5. To construct the general equation of a straight line, we substitute these values into equation (4):
Answer:
Vector parallel to line L and hence is perpendicular to the normal vector of the line L. Let's construct a normal line vector L, given that the scalar product of vectors n and is equal to zero. We can write, for example, n={1,−3}.
To construct the general equation of a straight line, we use formula (4). Let us substitute into (4) the coordinates of the point M 1 (we can also take the coordinates of the point M 2) and the normal vector n:
Substituting point coordinates M 1 and M 2 in (9) we can make sure that the straight line given by equation (9) passes through these points.
Answer:
Subtract (10) from (1):
We have obtained the canonical equation of a straight line. Vector q={−B, A) is the direction vector of the straight line (12).
See reverse transformation.
Example 3. A straight line in a plane is represented by the following general equation:
Move the second term to the right and divide both sides of the equation by 2 5.
The line passing through the point K(x 0; y 0) and parallel to the line y = kx + a is found by the formula:
y - y 0 \u003d k (x - x 0) (1)
Where k is the slope of the straight line.
Alternative formula:
The line passing through the point M 1 (x 1 ; y 1) and parallel to the line Ax+By+C=0 is represented by the equation
A(x-x 1)+B(y-y 1)=0 . (2)
Example #1. Compose the equation of a straight line passing through the point M 0 (-2.1) and at the same time:a) parallel to the straight line 2x+3y -7 = 0;
b) perpendicular to the line 2x+3y -7 = 0.
Decision . Let's represent the slope equation as y = kx + a . To do this, we will transfer all values except y to the right side: 3y = -2x + 7 . Then we divide the right side by the coefficient 3 . We get: y = -2/3x + 7/3
Find the equation NK passing through the point K(-2;1) parallel to the straight line y = -2 / 3 x + 7 / 3
Substituting x 0 \u003d -2, k \u003d -2 / 3, y 0 \u003d 1 we get:
y-1 = -2 / 3 (x-(-2))
or
y = -2 / 3 x - 1 / 3 or 3y + 2x +1 = 0
Example #2. Write the equation of a straight line parallel to the straight line 2x + 5y = 0 and forming, together with the coordinate axes, a triangle whose area is 5.
Decision
. Since the lines are parallel, the equation of the desired line is 2x + 5y + C = 0. The area of a right triangle, where a and b are its legs. Find the points of intersection of the desired line with the coordinate axes: 
;
.
So, A(-C/2,0), B(0,-C/5). Substitute in the formula for the area: 
. We get two solutions: 2x + 5y + 10 = 0 and 2x + 5y - 10 = 0 .
Example #3. Write the equation of the line passing through the point (-2; 5) and the parallel line 5x-7y-4=0 .
Decision. This straight line can be represented by the equation y = 5/7 x – 4/7 (here a = 5/7). The equation of the desired line is y - 5 = 5 / 7 (x - (-2)), i.e. 7(y-5)=5(x+2) or 5x-7y+45=0 .
Example #4. Solving example 3 (A=5, B=-7) using formula (2), we find 5(x+2)-7(y-5)=0.
Example number 5. Write the equation of a straight line passing through the point (-2;5) and a parallel straight line 7x+10=0.
Decision. Here A=7, B=0. Formula (2) gives 7(x+2)=0, i.e. x+2=0. Formula (1) is not applicable, since this equation cannot be solved with respect to y (this straight line is parallel to the y-axis).
General equation of a straight line:
Particular cases of the general equation of a straight line:
and if C= 0, equation (2) will have the form
Ax + By = 0,
and the straight line defined by this equation passes through the origin, since the coordinates of the origin x = 0, y= 0 satisfy this equation.
b) If in the general equation of the straight line (2) B= 0, then the equation takes the form
Ax + With= 0, or .
Equation does not contain a variable y, and the straight line defined by this equation is parallel to the axis Oy.
c) If in the general equation of the straight line (2) A= 0, then this equation takes the form
By + With= 0, or ;
the equation does not contain a variable x, and the straight line defined by it is parallel to the axis Ox.
It should be remembered: if a straight line is parallel to any coordinate axis, then its equation does not contain a term containing a coordinate of the same name with this axis.
d) When C= 0 and A= 0 equation (2) takes the form By= 0, or y = 0.
This is the axis equation Ox.
e) When C= 0 and B= 0 equation (2) can be written in the form Ax= 0 or x = 0.
This is the axis equation Oy.
Mutual arrangement of straight lines on a plane. Angle between lines on a plane. Condition of parallel lines. The condition of perpendicularity of lines.
l 1 l 2 l 1: A 1 x + B 1 y + C 1 = 0
l 2: A 2 x + B 2 y + C 2 = 0
S 2 S 1 The vectors S 1 and S 2 are called guides for their lines.
The angle between the lines l 1 and l 2 is determined by the angle between the direction vectors.
Theorem 1: cos angle between l 1 and l 2 \u003d cos (l 1; l 2) \u003d
Theorem 2: In order for 2 lines to be equal, it is necessary and sufficient:
Theorem 3: so that 2 lines are perpendicular is necessary and sufficient:
L 1 l 2 ó A 1 A 2 + B 1 B 2 = 0
General equation of the plane and its special cases. Equation of a plane in segments.
General plane equation:
Ax + By + Cz + D = 0
Special cases:
1. D=0 Ax+By+Cz = 0 - the plane passes through the origin
2. С=0 Ax+By+D = 0 – plane || oz
3. В=0 Ax+Cz+d = 0 – plane || OY
4. A=0 By+Cz+D = 0 – plane || OX
5. A=0 and D=0 By+Cz = 0 - the plane passes through OX
6. B=0 and D=0 Ax+Cz = 0 - the plane passes through OY
7. C=0 and D=0 Ax+By = 0 - the plane passes through OZ
Mutual arrangement of planes and straight lines in space:
1. The angle between lines in space is the angle between their direction vectors.
Cos (l 1 ; l 2) = cos(S 1 ; S 2) = = 
2. The angle between the planes is determined through the angle between their normal vectors.
Cos (l 1 ; l 2) = cos(N 1 ; N 2) = = 
3. The cosine of the angle between a line and a plane can be found through the sin of the angle between the direction vector of the line and the normal vector of the plane.
4. 2 lines || in space when their || vector guides
5. 2 planes || when || normal vectors
6. The concepts of perpendicularity of lines and planes are introduced similarly.
Question #14
Various types of the equation of a straight line on a plane (the equation of a straight line in segments, with a slope, etc.)
Equation of a straight line in segments:
Suppose that in the general equation of a straight line:
1. C \u003d 0 Ah + Wu \u003d 0 - the straight line passes through the origin.
2. a \u003d 0 Wu + C \u003d 0 y \u003d
3. in \u003d 0 Ax + C \u003d 0 x \u003d
4. v \u003d C \u003d 0 Ax \u003d 0 x \u003d 0
5. a \u003d C \u003d 0 Wu \u003d 0 y \u003d 0
The equation of a straight line with a slope:
Any straight line that is not equal to the y-axis (B not = 0) can be written in the following. form:
k = tgα α is the angle between the straight line and the positively directed line ОХ
b - point of intersection of the straight line with the OS axis
Doc-in:
Ax+By+C = 0
Wu \u003d -Ax-C |: B
Equation of a straight line on two points:
Question #16
The finite limit of a function at a point and for x→∞
End limit at point x 0:
The number A is called the limit of the function y \u003d f (x) for x → x 0, if for any E > 0 there is b > 0 such that for x ≠ x 0, satisfying the inequality |x - x 0 |< б, выполняется условие |f(x) - A| < Е
The limit is denoted: = A
End limit at point +∞:
The number A is called the limit of the function y = f(x) for x → + ∞ , if for any E > 0 there exists C > 0 such that for x > C the inequality |f(x) - A|< Е
The limit is denoted: = A
End limit at point -∞:
The number A is called the limit of the function y = f(x) for x→-∞, if for any E< 0 существует С < 0 такое, что при х < -С выполняется неравенство |f(x) - A| < Е
