If two straight lines l 1 and l 2 lie on a plane, then three different cases of their relative position are possible: 1) intersect (that is, they have one common point); 2) parallel and do not coincide; 3) match.
Let's find out how to find out which of these cases occurs if these lines are given by their equations in general form:
If the lines l 1 and l 2 intersect at some point M(x,y), then the coordinates of this point must satisfy both equations of system (12).
Therefore, in order to find the coordinates of the intersection point of lines l 1 and l 2, it is necessary to solve the system of equations (12):
1) if system (12) has a unique solution, then the lines l 1 and l 2 intersect;
2) if system (12) has no solution, then the lines l 1 and l 2 are parallel;
3) if system (12) has many solutions, then the lines l 1 and l 2 coincide.
The condition for the coincidence of two straight lines is the proportionality of the corresponding coefficients of their equations.
Example 10. Do the lines 3x+4y-1=0 and 2x+3y-1=0 intersect?
Solution: Let's solve the system of equations: 
the system has a unique solution, therefore the lines intersect. The intersection point of the lines has coordinates (-1;1).
Example 11. Are the lines 2x-y+2=0 and 4x-2y-1=0 parallel?
Solution: Let's solve the system of equations 
This system has no solutions, therefore the lines are parallel.
Example 12. Are the straight lines x+y+1=0 and 3x+3y+3=0 the same?
Solution: They coincide, since the coefficients are proportional.
Example 13. Write an equation for a straight line passing through the point of intersection of the lines x+y-1=0, x-y+2=0 and through the point (2,1).
Solution: Find the coordinates of the point of intersection of two given straight lines. To do this, we solve these equations together. Adding, we find: 2x+1=0, from where
Subtracting the second from the first equation, we get: 2у-3=0, whence . Next, it remains to create an equation of a straight line using two points () and (2;1)
The required equation will be 
, or or from where 
or x+5y-7=0
Angle between two straight lines on the plane is called the angle between their direction vectors. By this definition, we get not one angle, but two adjacent angles that complement each other up to . In elementary geometry, from two adjacent angles, as a rule, the smaller one is chosen, i.e. the magnitude of the angle between two straight lines satisfies the condition.
If 
And 
the direction vectors of the lines and, respectively (Fig. 3.23, a), then the angle between these lines is calculated by the formula:
The angle between lines (3.19) can be calculated as the angle between their normals 
And 
:
| (3.22) |
To obtain the value of an acute angle between straight lines, you need to take the right side in absolute value:
A necessary and sufficient condition for the perpendicularity of lines (3.19) is the condition for the orthogonality of their normals, i.e. equality to zero of the scalar product of their normals:
Using formula (3.22), we obtain an acute angle between straight lines (3.19) if (Fig. 3.23,a), and an obtuse angle otherwise: (Fig. 3.23,6). In other words, using formula (3.22) we find the angle between the lines in which lie the points belonging to opposite half-planes defined by these lines. In Fig. 3.23, positive and negative half-planes are marked with plus “+” or minus “–” signs, respectively.
Chapter V*. Equations of lines and planes in space.
§ 66. Conditions for the coincidence and intersection of planes
If the plane r 1 and r 2 given by the equations
A 1 X+ B 1 y+ C 1 z+ D 1 = 0 and A 2 X+ B 2 y+ C 2 z+ D 2 = 0, (1)
have a common point, then its coordinates satisfy each of equations (1). Therefore, to find the common points of these planes, you need to solve the system of equations
i.e., a system of two equations with three unknowns. When the condition is met
(3)
system (2) has no solutions. In fact, let's assume the opposite.
Let's assume that ( X 0 ; at 0 , z 0) - solution of the system. Then if
then from the second equation of system (2) we obtain
A 2 X 0 + B 2 at 0+C2 z 0 = - D 2 ,
and from the first
k(A 2 X 0 + B 2 at 0+C2 z 0) = - D 1 ,
and, therefore, which contradicts condition (3).
We know that the condition 
there is a condition for the planes to be parallel. Thus, if condition (3) is met, the plane r 1 and r 2 are parallel and do not coincide.
In the case when the coefficients and free terms of system (2) satisfy the condition
(4)
the system looks like
Each of the equations of the system defines the same plane. Thus, condition (4) is a necessary and sufficient condition for the coincidence of planes.
If the plane r 1 and r 2 are not parallel, i.e. if they intersect, then
In this case, equations (2) are equations of the straight line l plane intersections r 1 and r 2. Let us show how you can find the canonical equations of this line. To compose the canonical equations of a line, you need to know the coordinates of a certain point and the coordinates of its direction vector A . Any solution of system (2) can be taken as the coordinates of point M0. As a guide vector A direct l you can take the vector product of vectors n 1 = (A 1 ; B 1 ; C 1) and n 2 = (A 2 ; B 2 ; C 2), i.e. normal vectors of planes r 1 and r 2 .
Indeed (Fig. 203), the vector [ n 1 ; n 2 ] by definition of the vector product is perpendicular to the vectors n 1 and n 2 and therefore parallel to the planes r 1 and r 2 and, therefore, collinear to the line l their intersections.
Problem 1. Compose canonical equations of a line that is the intersection of planes
X - 2at + z+ 1 = 0 and 2 X - at+ 3z - 2 = 0.
Because n 1 = (1; - 2; 1), n 2 = (2; -1; 3), then
To determine the coordinates of any point on a given line, we find any solution to the system of equations
Let's put, for example, z= 0, then we get
where X = 5 / 3 , y= 4 / 3 . Consequently, the original system has a solution (5 / 3; 4 / 3; 0), and therefore this line passes through the point M (5 / 3; 4 / 3; 0).
Knowing the coordinates of a point on a line and the coordinates of its direction vector, we write down the canonical equations of this line
Note that if plane A 1 X+ B 1 y+ C 1 z+ D 1 = 0 and A 2 X+ B 2 y+ C 2 z+ D 2 = 0 intersect, then the equation of any plane passing through the line of their intersection can be written in the form
α (A 1 X+ B 1 y+ C 1 z+ D 1) + β(A 2 X+ B 2 y+ C 2 z+ D 2) = 0,
where α and β are some numbers.
Task 2. Write an equation for a plane passing through the line of intersection of planes 3 x - 2at - z+ 4 = 0 and X - 4at - 3z- 2 = 0 and point M 0 (1; 1; - 2).
Let's create an equation for the planes passing through the line of intersection of these planes:
α(3 x - 2at - z+ 4) + β( X - 4at - 3z - 2) = 0.
Since M 0 belongs to the desired plane, then
α (3 1 - 2 1 + 2 + 4) + β(1- 4 1 + 6 -2) = 0,
and therefore
whence, for example, α = 1, β = -7.
The required equation of the plane will be
3x - 2at - z + 4 - 7 (X - 4at - 3z - 2) = 0,
2x - 13at - 10z- 9 = 0.
Not even a minute had passed before I created a new Verd file and continued such a fascinating topic. You need to capture moments of a working mood, so there will be no lyrical introduction. There will be a prosaic spanking =)
Two straight spaces can:
1) interbreed;
2) intersect at the point ;
3) be parallel;
4) match.
Case No. 1 is fundamentally different from other cases. Two straight lines intersect if they do not lie in the same plane. Raise one arm up and extend the other arm forward - here is an example of crossing lines. In points No. 2-4 the straight lines must lie in one plane.
How to find out the relative positions of lines in space?
Consider two direct spaces:
– a straight line defined by a point and a direction vector;
– a straight line defined by a point and a direction vector.
For a better understanding, let's make a schematic drawing: 
The drawing shows intersecting straight lines as an example.
How to deal with these straight lines?
Since the points are known, it is easy to find the vector.
If straight interbreed, then the vectors not coplanar(see lesson Linear (non) dependence of vectors. Basis of vectors), and, therefore, the determinant composed of their coordinates is non-zero. Or, which is actually the same thing, it will be non-zero: 
.
In cases No. 2-4, our structure “falls” into one plane, and the vectors coplanar, and the mixed product of linearly dependent vectors equals zero: 
.
Let's expand the algorithm further. Let's assume that 
Therefore, the lines either intersect, are parallel, or coincide.
If the direction vectors collinear, then the lines are either parallel or coincident. For the final nail, I propose the following technique: take any point on one line and substitute its coordinates into the equation of the second line; if the coordinates “fit,” then the lines coincide; if they “don’t fit,” then the lines are parallel.
The algorithm is simple, but practical examples still wouldn’t hurt:
Example 11
Find out the relative position of two lines
Solution: as in many geometry problems, it is convenient to formulate the solution point by point:
1) We take out points and direction vectors from the equations:
2) Find the vector:
Thus, the vectors are coplanar, which means that the lines lie in the same plane and can intersect, be parallel, or coincide.
4) Let's check the direction vectors for collinearity.
Let's create a system from the corresponding coordinates of these vectors:
From everyone equations it follows that, therefore, the system is consistent, the corresponding coordinates of the vectors are proportional, and the vectors are collinear.
Conclusion: the lines are parallel or coincide.
5) Find out whether the lines have common points. Let's take a point belonging to the first line and substitute its coordinates into the equations of the line: 
Thus, the lines have no common points, and they have no choice but to be parallel.
Answer:
An interesting example to solve on your own:
Example 12
Find out the relative positions of the lines
This is an example for you to solve on your own. Please note that the second line has the letter as a parameter. Logical. In the general case, these are two different lines, so each line has its own parameter.
And again I urge you not to skip the examples, the tasks I propose are far from random ;-)
Problems with a line in space
In the final part of the lesson, I will try to consider the maximum number of different problems with spatial lines. In this case, the original order of the story will be observed: first we will consider problems with crossing lines, then with intersecting lines, and at the end we will talk about parallel lines in space. However, I must say that some tasks of this lesson can be formulated for several cases of the location of lines at once, and in this regard, the division of the section into paragraphs is somewhat arbitrary. There are simpler examples, there are more complex examples, and hopefully everyone will find what they need.
Crossing lines
Let me remind you that straight lines intersect if there is no plane in which they both lie. When I was thinking through the practice, a monster problem came to mind, and now I’m glad to present to your attention a dragon with four heads:
Example 13
Given straight lines. Required:
a) prove that lines intersect;
b) find the equations of a line passing through a point perpendicular to the given lines;
c) compose equations of a straight line that contains common perpendicular crossing lines;
d) find the distance between the lines.
Solution: The one who walks will master the road:
a) Let us prove that lines intersect. Let's find the points and direction vectors of these lines: 
Let's find the vector:
Let's calculate mixed product of vectors:
Thus, the vectors not coplanar, which means that the lines intersect, which is what needed to be proven.
Probably everyone has long noticed that for crossing lines the verification algorithm is the shortest.
b) Find the equations of the line that passes through the point and is perpendicular to the lines. Let's make a schematic drawing: 
For a change I posted a direct FOR straight, look how it is a little erased at the crossing points. Crossbreeding? Yes, in general, the straight line “de” will be crossed with the original straight lines. Although we are not interested in this moment, we just need to construct a perpendicular line and that’s it.
What is known about the direct “de”? The point belonging to it is known. There is not enough guide vector.
According to the condition, the straight line must be perpendicular to the straight lines, which means that its direction vector will be orthogonal to the direction vectors. Already familiar from Example No. 9, let’s find the vector product:
Let’s compose the equations of the straight line “de” using a point and a direction vector:
Ready. In principle, you can change the signs in the denominators and write the answer in the form 
, but there is no need for this.
To check, you need to substitute the coordinates of the point into the resulting straight line equations, then use scalar product of vectors make sure that the vector is really orthogonal to the direction vectors “pe one” and “pe two”.
How to find the equations of a line containing a common perpendicular?
c) This problem will be more difficult. I recommend skipping this point for dummies, I don’t want to cool your sincere sympathy for analytical geometry =) By the way, it might be better for more prepared readers to hold off too, the fact is that in terms of complexity the example should be placed last in the article, but according to the logic of presentation it should be located here.
So, you need to find the equations of a line that contains a common perpendicular to skew lines.
– this is a segment connecting these lines and perpendicular to these lines: 
Here is our handsome guy: - common perpendicular of intersecting lines. He's the only one. There is no other like it. We need to create equations for the line that contains this segment.
What is known about the direct “um”? Its direction vector is known, found in the previous paragraph. But, unfortunately, we do not know a single point belonging to the straight line “em”, nor do we know the ends of the perpendicular – the points . Where does this perpendicular line intersect the two original lines? In Africa, in Antarctica? From the initial review and analysis of the condition, it is not at all clear how to solve the problem... But there is a tricky trick associated with the use of parametric equations of a straight line.
We will formalize the decision point by point:
1) Let's rewrite the equations of the first line in parametric form: 
Let's consider the point. We don't know the coordinates. BUT. If a point belongs to a given line, then its coordinates correspond to , let us denote it by . Then the coordinates of the point will be written in the form: 
Life is getting better, one unknown is still not three unknowns.
2) The same outrage must be carried out on the second point. Let us rewrite the equations of the second line in parametric form: 
If a point belongs to a given line, then with a very specific meaning its coordinates must satisfy the parametric equations:
Or: 
3) Vector, like the previously found vector, will be the directing vector of the straight line. How to construct a vector from two points was discussed in time immemorial in class Vectors for dummies. Now the difference is that the coordinates of the vectors are written with unknown parameter values. So what? No one forbids subtracting the corresponding coordinates of the beginning of the vector from the coordinates of the end of the vector.
There are two points: 
.
Finding the vector:
4) Since the direction vectors are collinear, one vector is linearly expressed through the other with a certain proportionality coefficient “lambda”:
Or coordinate-by-coordinate: 
It turned out to be the most ordinary system of linear equations with three unknowns, which is standardly solvable, for example, Cramer's method. But here there is an opportunity to get off with little loss, let’s express “lambda” from the third equation and substitute it into the first and second equations:
Thus: 
, and we don’t need “lambda”. The fact that the parameter values turned out to be the same is purely an accident.
5) The sky is completely clearing, let’s substitute the found values 
to our points:
The direction vector is not particularly needed, since its counterpart has already been found.
It's always interesting to check after a long journey.
:
The correct equalities are obtained.
Let's substitute the coordinates of the point into the equations 
:
The correct equalities are obtained.
6) Final chord: let’s create the equations of a straight line using a point (you can take it) and a direction vector:
In principle, you can select a “good” point with intact coordinates, but this is cosmetic.
How to find the distance between intersecting lines?
d) We cut off the fourth head of the dragon.
Method one. Not even a method, but a small special case. The distance between crossing lines is equal to the length of their common perpendicular: 
.
Extreme points of the common perpendicular 
found in the previous paragraph, and the task is elementary:
Method two. In practice, most often the ends of the common perpendicular are unknown, so a different approach is used. Parallel planes can be drawn through two intersecting straight lines, and the distance between these planes is equal to the distance between these straight lines. In particular, a common perpendicular sticks out between these planes.
In the course of analytical geometry, from the above considerations, a formula is derived for finding the distance between intersecting straight lines: 
(instead of our points “um one, two” you can take arbitrary points of lines).
Mixed product of vectors already found in point "a": 
.
Vector product of vectors found in paragraph "be": 
, let's calculate its length:
Thus:
Let's proudly display the trophies in one row:
Answer:
A) 
, which means that straight lines intersect, which is what was required to be proved;
b) 
;
V) 
;
G) 
What else can you tell about crossing lines? There is a defined angle between them. But we will consider the universal angle formula in the next paragraph:
Intersecting straight spaces necessarily lie in the same plane: 
The first thought is to lean on the intersection point with all your might. And I immediately thought, why deny yourself the right desires?! Let's get on top of her right now!
How to find the point of intersection of spatial lines?
Example 14
Find the point of intersection of lines
Solution: Let's rewrite the equations of lines in parametric form: 
This task was discussed in detail in Example No. 7 of this lesson (see. Equations of a line in space). And by the way, I took the straight lines themselves from Example No. 12. I won’t lie, I’m too lazy to come up with new ones.
The solution is standard and has already been encountered when we were trying to figure out the equations for the common perpendicular of intersecting lines.
The point of intersection of the lines belongs to the line, therefore its coordinates satisfy the parametric equations of this line, and they correspond to a very specific parameter value:
But this same point also belongs to the second line, therefore: 
We equate the corresponding equations and carry out simplifications: 
A system of three linear equations with two unknowns is obtained. If the lines intersect (which is proven in Example No. 12), then the system is necessarily consistent and has a unique solution. It can be solved Gaussian method, but let’s not sin with such kindergarten fetishism, let’s do it simpler: from the first equation we express “te zero” and substitute it into the second and third equations: 
The last two equations turned out to be essentially the same, and it follows from them that . Then:
Let's substitute the found value of the parameter into the equations: 
Answer:
To check, we substitute the found value of the parameter into the equations: 
The same coordinates were obtained that needed to be checked. Meticulous readers can substitute the coordinates of the point into the original canonical equations of lines.
By the way, it was possible to do the opposite: find the point through “es zero”, and check it through “te zero”.
A well-known mathematical superstition says: where the intersection of lines is discussed, there is always a smell of perpendiculars.
How to construct a line of space perpendicular to a given one?
(lines intersect)
Example 15
a) Write down the equations of a line passing through a point perpendicular to the line 
(lines intersect).
b) Find the distance from the point to the line.
Note
: clause “lines intersect” – significant. Through the point
you can draw an infinite number of perpendicular lines that will intersect with the straight line “el”. The only solution occurs in the case when a straight line perpendicular to a given point is drawn two given by a straight line (see Example No. 13, point “b”).
A) Solution: We denote the unknown line by . Let's make a schematic drawing:
What is known about the straight line? According to the condition, a point is given. In order to compose the equations of a straight line, it is necessary to find the direction vector. The vector is quite suitable as such a vector, so we will deal with it. More precisely, let's take the unknown end of the vector by the scruff of the neck.
1) Let’s take out its direction vector from the equations of the straight line “el”, and rewrite the equations themselves in parametric form: 
Many guessed that now for the third time during the lesson the magician will pull a white swan out of his hat. Consider a point with unknown coordinates. Since the point is , its coordinates satisfy the parametric equations of the straight line “el” and they correspond to a specific parameter value: 
Or in one line:
2) According to the condition, the lines must be perpendicular, therefore, their direction vectors are orthogonal. And if the vectors are orthogonal, then their dot product equals zero: 
What happened? The simplest linear equation with one unknown: 
3) The value of the parameter is known, let’s find the point:
And the direction vector:
.
4) Let’s compose the equations of a straight line using a point and a direction vector 
:
The denominators of the proportion turned out to be fractional, and this is exactly the case when it is appropriate to get rid of fractions. I'll just multiply them by -2: 
Answer: 
Note
: a more rigorous ending to the solution is formalized as follows: let’s compose the equations of a straight line using a point and a direction vector 
. Indeed, if a vector is a directing vector of a straight line, then the collinear vector , naturally, will also be a directing vector of this straight line.
The verification consists of two stages:
1) check the direction vectors of the lines for orthogonality;
2) we substitute the coordinates of the point into the equations of each line, they should “fit” both there and there.
There was a lot of talk about typical actions, so I checked on a draft.
By the way, I forgot another point - to construct a point “zyu” symmetrical to the point “en” relative to the straight line “el”. However, there is a good “flat analogue”, which can be found in the article The simplest problems with a straight line on a plane. Here the only difference will be in the additional “Z” coordinate.
How to find the distance from a point to a line in space?
b) Solution: Let's find the distance from a point to a line.
Method one. This distance is exactly equal to the length of the perpendicular: . The solution is obvious: if the points are known 
, That:
Method two. In practical problems, the base of the perpendicular is often a sealed secret, so it is more rational to use a ready-made formula.
The distance from a point to a line is expressed by the formula: 
, where is the directing vector of the straight line “el”, and – free a point belonging to a given line.
1) From the equations of the line 
we take out the direction vector and the most accessible point.
2) The point is known from the condition, sharpen the vector:
3) Let's find vector product and calculate its length: 
4) Calculate the length of the guide vector:
5) Thus, the distance from a point to a line:
By the way, the last inequality indicates that their normal vectors are not parallel.
If the lines are parallel, then the system has no solution. Analytically it would look like this:
But if all three fractions are equal, then the lines coincide with each other, and therefore the system has an infinite number of solutions.
Angle between two straight lines can be found using two formulas.
If straight lines are given by general equations, then the angle between them coincides with the angle between their normal vectors. It is calculated using formula (6.9) from the previous lecture. For our case it will look like:
. (7.7)
Condition for parallel lines:
;
Perpendicularity condition:
.
If the lines are given by equations with angular coefficients of the form:
And 
,
then the tangent of the angle between them is determined by the formula:
. (7.8)
Parallel condition:
Perpendicularity condition:
.
Example 7.4. Find the point of intersection of lines 
And 
and the angle between them.
Solutions e. Let’s find the point of intersection of the lines by solving the system of equations using Cramer’s method:
, 
,
,
We define the angle between straight lines as the angle between their normal vectors (2, 5) and (5, –2). According to formula (7.7) we have:
.
What does this answer say? Straight lines are perpendicular, because .
Example 7.5. At what parameter values a And b straight and 
: A) intersect, b) are parallel, V) match?
Solutions e. Two lines intersect if the condition is met. In our case
.
Lines are parallel if 
, i.e.
.
And, lastly, two straight lines coincide provided that 
, i.e. If 
.
Example 7.6. Given a point and a line 
. Write equations of lines L 1 and L 2 passing through a point A, and .
Solutions e. Let's make a schematic drawing.
Rice. 7.6
Slope of the original straight line L equals k= –2. By condition, therefore 
. Using formula (7.4) we find the equation of the straight line L 1:
, or 
.
Since then 
. Then the equation of the line L 2 will look like:
, or 
.
7.4. Definition of a second order curve
Definition 7.1.Second order curve is a line defined by an equation of the second degree relative to the current coordinates. In general, this equation looks like:
where are all the numbers A, IN, WITH, etc. are real numbers, and, in addition, at least one of the numbers A, IN, WITH- different from zero.
Before the introduction of the Cartesian coordinate system, all curves were described verbally, based on the geometric properties of the curve in question. So, the definition of a circle read like this:
Definition 7.2. Circle – this is the locus of points on the plane equidistant from a given point, called the center.
Equation of a circle, centered at point ( A,b) and radius R in the Cartesian coordinate system, what you received at school looks like this:
If we open the parentheses, we obtain an equation similar to equation (7.9), in which there is no term containing the product of the current coordinates, and the coefficients of the higher powers are equal to each other.
The derivation of all second-order equations is similar to the derivation of straight line equations and follows the same algorithm.
Let us derive the equation of a parabola based on its definition.
7.5. Canonical parabola equation
Definition 7.3. Parabola is the locus of points in the plane equidistant from a given point F, called focus, and this line, called headmistress.
Let us denote the distance from the focus to the directrix by p. This quantity is called parameter parabolas.
1. Let's position the x-axis so that it passes through the focus, perpendicular to the directrix and has a positive direction from the directrix to the focus.
2. Place the origin of coordinates in the middle of this perpendicular. Then the coordinates of the point will be F(p/2, 0), and the directrix equation: 
.
3. Take the current point on the parabola M(x, y).
4. By definition of a parabola, the distance MN from point M to the directrix is equal to its distance MF from focus: M.F.= MN. As can be seen from the drawing (Fig. 7.7), the coordinates of the point N(–p/2, y). Let's find these distances using the formula for the distance between two points from point 1 of the previous lecture.
, 
.
Equating the right-hand sides of these expressions and squaring both sides of the equality, we obtain:
,
or after contractions
. (7.11)
Equation (7.11) is called canonical parabola equation. Only points lying on the curve will satisfy it, and the rest will not. Let's study the shape of its graph using the canonical equation.
Since y is included in an even power, then the axis OH will be the axis of symmetry, i.e. one value X will correspond to two values Y– positive and negative. Because the right side is non-negative at, then the left one too. Because r– the distance between the focus and the directrix is always greater than zero, then X. If X=0, then at=0, i.e. the parabola passes through the origin. With unlimited increase x absolute value at will also increase without limit.
The graph of the parabola defined by equation (7.11) is shown in Fig. 7.7.
Rice. 7.7 Fig. 7.8
The axis of symmetry of a parabola is called the focal axis, because the focus is on her. If the focal axis of the parabola is taken as the ordinate axis, then its equation takes the form:
.
Its drawing is shown in Fig. 7.8. In this case, the focus will be at the point F(0, p/2), and the directrix equation will have the form at = –r/2.
Thus, we examined the parabola, found its equation and showed possible locations relative to the origin.
If the vertex of the parabola is displaced to the point 
, then the canonical equation will look like this:
.
We will not deal with the derivation of the remaining second-order curves. Those interested can find all the calculations in the recommended literature.
Let us limit ourselves to their definitions and equations.
If two straight lines l 1 and l 2 lie on a plane, then three different cases of their relative position are possible: 1) intersect (that is, they have one common point); 2) parallel and do not coincide; 3) match.
Let's find out how to find out which of these cases occurs if these lines are given by their equations in general form:
If the lines l 1 and l 2 intersect at some point M(x,y), then the coordinates of this point must satisfy both equations of system (12).
Therefore, in order to find the coordinates of the intersection point of lines l 1 and l 2, it is necessary to solve the system of equations (12):
1) if system (12) has a unique solution, then the lines l 1 and l 2 intersect;
2) if system (12) has no solution, then the lines l 1 and l 2 are parallel;
3) if system (12) has many solutions, then the lines l 1 and l 2 coincide.
The condition for the coincidence of two straight lines is the proportionality of the corresponding coefficients of their equations.
Example 10. Do the lines 3x+4y-1=0 and 2x+3y-1=0 intersect?
Solution: Let's solve the system of equations: 
the system has a unique solution, therefore the lines intersect. The intersection point of the lines has coordinates (-1;1).
Example 11. Are the lines 2x-y+2=0 and 4x-2y-1=0 parallel?
Solution: Let's solve the system of equations 
This system has no solutions, therefore the lines are parallel.
Example 12. Are the straight lines x+y+1=0 and 3x+3y+3=0 the same?
Solution: They coincide, since the coefficients are proportional.
Example 13. Write an equation for a straight line passing through the point of intersection of the lines x+y-1=0, x-y+2=0 and through the point (2,1).
Solution: Find the coordinates of the point of intersection of two given straight lines. To do this, we solve these equations together. Adding, we find: 2x+1=0, from where
Subtracting the second from the first equation, we get: 2у-3=0, whence . Next, it remains to create an equation of a straight line using two points () and (2;1)
The required equation will be 
, or or from where 
or x+5y-7=0
Angle between two straight lines on the plane is called the angle between their direction vectors. By this definition, we get not one angle, but two adjacent angles that complement each other up to . In elementary geometry, from two adjacent angles, as a rule, the smaller one is chosen, i.e. the magnitude of the angle between two straight lines satisfies the condition.
If 
And 
the direction vectors of the lines and, respectively (Fig. 3.23, a), then the angle between these lines is calculated by the formula:
The angle between lines (3.19) can be calculated as the angle between their normals 
And 
:
| (3.22) |
To obtain the value of an acute angle between straight lines, you need to take the right side in absolute value:
A necessary and sufficient condition for the perpendicularity of lines (3.19) is the condition for the orthogonality of their normals, i.e. equality to zero of the scalar product of their normals:
Using formula (3.22), we obtain an acute angle between straight lines (3.19) if (Fig. 3.23,a), and an obtuse angle otherwise: (Fig. 3.23,6). In other words, using formula (3.22) we find the angle between the lines in which lie the points belonging to opposite half-planes defined by these lines. In Fig. 3.23, positive and negative half-planes are marked with plus “+” or minus “–” signs, respectively.
