Definition. Any straight line on the plane can be specified by a first-order equation
Ax + Wu + C = 0,
Moreover, the constants A and B are not equal to zero at the same time. This first order equation is called general equation of a straight line. Depending on the values constant A, B and C the following special cases are possible:
C = 0, A ≠0, B ≠ 0 – the straight line passes through the origin
A = 0, B ≠0, C ≠0 (By + C = 0) - straight line parallel to the Ox axis
B = 0, A ≠0, C ≠ 0 (Ax + C = 0) – straight line parallel to the Oy axis
B = C = 0, A ≠0 – the straight line coincides with the Oy axis
A = C = 0, B ≠0 – the straight line coincides with the Ox axis
The equation of a straight line can be represented in in various forms depending on any given initial conditions.
Equation of a straight line from a point and normal vector
Definition. In the Cartesian rectangular coordinate system, a vector with components (A, B) is perpendicular to the straight line given by the equation Ax + By + C = 0.
Example. Find the equation of the line passing through the point A(1, 2) perpendicular to (3, -1).
Solution. With A = 3 and B = -1, let us compose the equation of the straight line: 3x – y + C = 0. To find the coefficient C, we substitute the coordinates into the resulting expression given point A. We get: 3 – 2 + C = 0, therefore, C = -1. Total: the required equation: 3x – y – 1 = 0.
Equation of a line passing through two points
Let two points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2) be given in space, then the equation of the line passing through these points is:
If any of the denominators is equal to zero, the corresponding numerator should be equal to zero. On the plane, the equation of the line written above is simplified:
if x 1 ≠ x 2 and x = x 1, if x 1 = x 2.
The fraction = k is called slope straight.
Example. Find the equation of the line passing through points A(1, 2) and B(3, 4).
Solution. Applying the formula written above, we get:
Equation of a straight line from a point and slope
If the total Ax + Bu + C = 0, lead to the form:
and designate 
, then the resulting equation is called equation of a straight line with slopek.
Equation of a straight line from a point and a direction vector
By analogy with the point considering the equation of a straight line through a normal vector, you can enter the definition of a straight line through a point and the directing vector of the straight line.
Definition. Each non-zero vector (α 1, α 2), the components of which satisfy the condition A α 1 + B α 2 = 0 is called a directing vector of the line
Ax + Wu + C = 0.
Example. Find the equation of a straight line with a direction vector (1, -1) and passing through the point A(1, 2).
Solution. We will look for the equation of the desired line in the form: Ax + By + C = 0. In accordance with the definition, the coefficients must satisfy the conditions:
1 * A + (-1) * B = 0, i.e. A = B.
Then the equation of the straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0. for x = 1, y = 2 we obtain C/ A = -3, i.e. required equation:
Equation of a line in segments
If in the general equation of the straight line Ах + Ву + С = 0 С≠0, then, dividing by –С, we get: 
or
Geometric meaning coefficients is that the coefficient A is the coordinate of the point of intersection of the line with the Ox axis, and b– the coordinate of the point of intersection of the straight line with the Oy axis.
Example. The general equation of the line x – y + 1 = 0 is given. Find the equation of this line in segments.
C = 1, , a = -1, b = 1.
Normal equation of a line
If both sides of the equation Ax + By + C = 0 are multiplied by the number 
which is called normalizing factor, then we get
xcosφ + ysinφ - p = 0 –
normal equation of a line. The sign ± of the normalizing factor must be chosen so that μ * C< 0. р – длина перпендикуляра, опущенного из начала координат на прямую, а φ - угол, образованный этим перпендикуляром с положительным направлением оси Ох.
Example. Given the general equation of the straight line 12x – 5y – 65 = 0. You need to write Various types equations of this line.
equation of this line in segments: 
equation of this line with slope: (divide by 5)
; cos φ = 12/13; sin φ= -5/13; p = 5.
It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines parallel to the axes or passing through the origin of coordinates.
Example. The straight line cuts off equal positive segments on the coordinate axes. Write an equation of a straight line if the area of the triangle formed by these segments is 8 cm 2.
Solution. The equation of the straight line has the form: , ab /2 = 8; ab=16; a=4, a=-4. a = -4< 0 не подходит по условию задачи. Итого: или х + у – 4 = 0.
Example. Write an equation for a straight line passing through point A(-2, -3) and the origin.
Solution.
The equation of the straight line is: 
, where x 1 = y 1 = 0; x 2 = -2; y 2 = -3.
Angle between straight lines on a plane
Definition. If two lines are given y = k 1 x + b 1, y = k 2 x + b 2, then the acute angle between these lines will be defined as
.
Two lines are parallel if k 1 = k 2. Two lines are perpendicular if k 1 = -1/ k 2.
Theorem. The lines Ax + Bу + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A 1 = λA, B 1 = λB are proportional. If also C 1 = λC, then the lines coincide. The coordinates of the point of intersection of two lines are found as a solution to the system of equations of these lines.
Equation of a line passing through a given point perpendicular to a given line
Definition. A straight line passing through the point M 1 (x 1, y 1) and perpendicular to the straight line y = kx + b is represented by the equation:
Distance from point to line
Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Bу + C = 0 is determined as
.
Proof. Let point M 1 (x 1, y 1) be the base of the perpendicular dropped from point M to a given straight line. Then the distance between points M and M 1:
(1)
The coordinates x 1 and y 1 can be found by solving the system of equations:
The second equation of the system is the equation of a line passing through a given point M 0 perpendicular to a given line. If we transform the first equation of the system to the form:
A(x – x 0) + B(y – y 0) + Ax 0 + By 0 + C = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem has been proven.
Example. Determine the angle between the lines: y = -3 x + 7; y = 2 x + 1.
k 1 = -3; k 2 = 2; tgφ = 
; φ= π /4.
Example. Show that the lines 3x – 5y + 7 = 0 and 10x + 6y – 3 = 0 are perpendicular.
Solution. We find: k 1 = 3/5, k 2 = -5/3, k 1* k 2 = -1, therefore, the lines are perpendicular.
Example. Given are the vertices of the triangle A(0; 1), B (6; 5), C (12; -1). Find the equation of the height drawn from vertex C.
Solution. We find the equation of side AB: 
; 4 x = 6 y – 6;
2 x – 3 y + 3 = 0;
The required height equation has the form: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: 
from where b = 17. Total: . 
Answer: 3 x + 2 y – 34 = 0.
Canonical equations of a line in space are equations that define a line passing through a given point collinear to the direction vector.
Let a point and a direction vector be given. An arbitrary point lies on a line l only if the vectors and are collinear, i.e., the condition is satisfied for them:
.
The above equations are canonical equations straight.
Numbers m , n And p are projections of the direction vector onto the coordinate axes. Since the vector is non-zero, then all numbers m , n And p cannot simultaneously be equal to zero. But one or two of them may turn out to be zero. In analytical geometry, for example, the following entry is allowed:
,
which means that the projections of the vector on the axis Oy And Oz are equal to zero. Therefore, both the vector and the straight line defined by the canonical equations are perpendicular to the axes Oy And Oz, i.e. planes yOz .
Example 1. Write equations for a line in space perpendicular to a plane 
and passing through the point of intersection of this plane with the axis Oz
.
Solution. Let's find the point of intersection of this plane with the axis Oz. Since any point lying on the axis Oz, has coordinates , then, assuming in the given equation of the plane x = y = 0, we get 4 z- 8 = 0 or z= 2 . Therefore, the point of intersection of this plane with the axis Oz has coordinates (0; 0; 2) . Since the desired line is perpendicular to the plane, it is parallel to its normal vector. Therefore, the directing vector of the straight line can be the normal vector 
given plane.
Now let’s write down the required equations of a straight line passing through a point A= (0; 0; 2) in the direction of the vector:
Equations of a line passing through two given points
A straight line can be defined by two points lying on it 
And 
In this case, the directing vector of the straight line can be the vector . Then the canonical equations of the line take the form
.
The above equations determine a line passing through two given points.
Example 2. Write an equation for a line in space passing through the points and .
Solution. Let us write down the required equations of the straight line in the form given above in the theoretical reference:
.
Since , then the desired straight line is perpendicular to the axis Oy .
Straight as the line of intersection of planes
A straight line in space can be defined as the line of intersection of two non-parallel planes and, i.e., as a set of points satisfying a system of two linear equations
The equations of the system are also called the general equations of a straight line in space.
Example 3. Compose canonical equations of a line in space given by general equations
Solution. To write the canonical equations of a line or, what is the same, the equations of a line passing through two given points, you need to find the coordinates of any two points on the line. They can be the points of intersection of a straight line with any two coordinate planes, for example yOz And xOz .
Point of intersection of a line and a plane yOz has an abscissa x= 0 . Therefore, assuming in this system of equations x= 0, we get a system with two variables:
Her decision y = 2 , z= 6 together with x= 0 defines a point A(0; 2; 6) the desired line. Then assuming in the given system of equations y= 0, we get the system
Her decision x = -2 , z= 0 together with y= 0 defines a point B(-2; 0; 0) intersection of a line with a plane xOz .
Now let's write down the equations of the line passing through the points A(0; 2; 6) and B (-2; 0; 0) :
,
or after dividing the denominators by -2:
,
In this article we will consider the general equation of a straight line on a plane. Let us give examples of constructing a general equation of a line if two points of this line are known or if one point and the normal vector of this line are known. Let us introduce methods for transforming the equation into general view into canonical and parametric views.
Let an arbitrary Cartesian rectangular coordinate system be given Oxy. Consider the equation of the first degree or linear equation:
| Ax+By+C=0, | (1) |
Where A, B, C− some constants, and at least one of the elements A And B different from zero.
We will show that a linear equation on a 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 specified by a linear equation. Conversely, each linear equation (1) in an arbitrary Cartesian rectangular coordinate system on a plane defines a straight line.
Proof. It is enough to prove that the straight 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 for any choice of Cartesian rectangular coordinate system.
Let a straight line be given on the plane L. Let us choose a coordinate system so that the axis Ox coincided with a straight 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 linear equation (2), and all points outside this line will not satisfy equation (2). The first part of the theorem has been proven.
Let a Cartesian rectangular coordinate system be given and let a linear equation (1) be given, where at least one of the elements A And B different from zero. Let us find the geometric 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, point M 0 (−C/A, 0) belongs to the given geometric 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 is enough to prove that (4) defines a certain line.
Since we are considering the Cartesian rectangular system coordinates, then from equality (4) it follows that the vector with components ( x−x 0 , y−y 0 ) orthogonal to the vector n with coordinates ( A,B}.
Let's consider some straight 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 equal to zero). Conversely, if 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 the vector n and equation (4) is not satisfied. The theorem has been proven.
Proof. Since lines (5) and (6) define the same line, then the normal vectors n 1 ={A 1 ,B 1) and n 2 ={A 2 ,B 2) collinear. Since vectors n 1 ≠0, n 2 ≠0, then there is such a number λ , What n 2 =n 1 λ . From here we have: A 2 =A 1 λ , B 2 =B 1 λ . Let's prove that C 2 =C 1 λ . Obviously, 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 the 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 a 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 straight line passes through a point M=(4,−1) and has a normal vector n=(3, 5). Construct the general equation of a line.
Solution. 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:
The vector is parallel to the line L and, therefore, perperdicular to the normal vector of the line L. Let's construct a normal line vector L, taking into account that the scalar product of vectors n and 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 the coordinates of the point into (4) M 1 (we can also take the coordinates of the point M 2) and normal vector n:
Substituting the coordinates of the points 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 the line. Vector q={−B, A) is the direction vector of line (12).
See reverse conversion.
Example 3. A straight line on a plane is represented by the following general equation:
Let's move the second term to the right and divide both sides of the equation by 2·5.
Equation of a line on a plane.
As is known, any point on the plane is determined by two coordinates in some coordinate system. Coordinate systems can be different depending on the choice of basis and origin.
Definition. Line equation is called the relation y = f(x) between the coordinates of the points that make up this line.
Note that the equation of a line can be expressed parametrically, that is, each coordinate of each point is expressed through some independent parameter t.
A typical example is the trajectory of a moving point. In this case, the role of the parameter is played by time.
Equation of a straight line on a plane.
Definition. Any straight line on the plane can be specified by a first-order equation
Ax + Wu + C = 0,
Moreover, the constants A and B are not equal to zero at the same time, i.e. A 2 + B 2 0. This first order equation is called general equation of a straight line.
Depending on the values of constants A, B and C, the following special cases are possible:
C = 0, A 0, B 0 – the straight line passes through the origin
A = 0, B 0, C 0 (By + C = 0) - straight line parallel to the Ox axis
B = 0, A 0, C 0 (Ax + C = 0) – straight line parallel to the Oy axis
B = C = 0, A 0 – the straight line coincides with the Oy axis
A = C = 0, B 0 – the straight line coincides with the Ox axis
The equation of a straight line can be presented in different forms depending on any given initial conditions.
Equation of a straight line from a point and a normal vector.
Definition. In the Cartesian rectangular coordinate system, a vector with components (A, B) is perpendicular to the straight line given by the equation Ax + By + C = 0.
Example. Find the equation of the line passing through the point A(1, 2) perpendicular to the vector 
(3,
-1).
With A = 3 and B = -1, let’s compose the equation of the straight line: 3x – y + C = 0. To find the coefficient C, we substitute the coordinates of the given point A into the resulting expression.
We get: 3 – 2 + C = 0, therefore C = -1.
Total: the required equation: 3x – y – 1 = 0.
Equation of a line passing through two points.
Let two points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2) be given in space, then the equation of the line passing through these points is:
If any of the denominators is zero, the corresponding numerator should be set equal to zero.
On the plane, the equation of the straight line written above is simplified:
if x 1 x 2 and x = x 1, if x 1 = x 2.
Fraction 
=k is called slope straight.
Example. Find the equation of the line passing through points A(1, 2) and B(3, 4).
Applying the formula written above, we get:
Equation of a straight line using a point and slope.
If the general equation of the straight line Ax + By + C = 0 is reduced to the form:
and designate 
, then the resulting equation is called equation of a straight line with slopek.
Equation of a straight line from a point and a direction vector.
By analogy with the point considering the equation of a straight line through a normal vector, you can enter the definition of a straight line through a point and the directing vector of the straight line.
Definition.
Every non-zero vector 
( 1, 2), the components of which satisfy the condition A 1 + B 2 = 0 is called the directing vector of the line
Ax + Wu + C = 0.
Example. Find the equation of a line with a direction vector 
(1, -1) and passing through point A(1, 2).
We will look for the equation of the desired line in the form: Ax + By + C = 0. In accordance with the definition, the coefficients must satisfy the conditions:
1A + (-1)B = 0, i.e. A = B.
Then the equation of the straight line has the form: Ax + Ay + C = 0, or x + y + C/A = 0.
at x = 1, y = 2 we get C/A = -3, i.e. required equation:
Equation of a straight line in segments.
If in the general equation of the straight line Ах + Ву + С = 0 С 0, then, dividing by –С, we get: 
or
, Where
The geometric meaning of the coefficients is that the coefficient A is the coordinate of the point of intersection of the line with the Ox axis, and b– the coordinate of the point of intersection of the straight line with the Oy axis.
Example. The general equation of the line x – y + 1 = 0 is given. Find the equation of this line in segments.
C = 1, 
, a = -1,b = 1.
Normal equation of a line.
If both sides of the equation Ax + By + C = 0 are divided by the number 
which is called normalizing factor, then we get
xcos + ysin - p = 0 –
normal equation of a line.
The sign of the normalizing factor must be chosen so that С< 0.
p is the length of the perpendicular dropped from the origin to the straight line, and is the angle formed by this perpendicular with the positive direction of the Ox axis.
Example. The general equation of the line 12x – 5y – 65 = 0 is given. It is required to write various types of equations for this line.
equation of this line in segments: 
equation of this line with slope: (divide by 5)
normal equation of a line:
; cos = 12/13; sin = -5/13; p = 5.
It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines parallel to the axes or passing through the origin of coordinates.
Example. The straight line cuts off equal positive segments on the coordinate axes. Write an equation of a straight line if the area of the triangle formed by these segments is 8 cm 2.
The equation of the straight line is: 
, a = b = 1; ab/2 = 8; a = 4; -4.
a = -4 is not suitable according to the conditions of the problem.
Total: 
or x + y – 4 = 0.
Example. Write an equation for a straight line passing through point A(-2, -3) and the origin.
The equation of the straight line is: 
, where x 1 = y 1 = 0; x 2 = -2; y 2 = -3.
The angle between straight lines on a plane.
Definition. If two lines are given y = k 1 x + b 1, y = k 2 x + b 2, then the acute angle between these lines will be defined as
.
Two lines are parallel if k 1 = k 2.
Two lines are perpendicular if k 1 = -1/k 2 .
Theorem. Direct lines Ax + Wu + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A are proportional 1 = A, B 1 = B. If also C 1 = C, then the lines coincide.
The coordinates of the point of intersection of two lines are found as a solution to the system of equations of these lines.
Equation of a line passing through a given point
perpendicular to this line.
Definition. A straight line passing through the point M 1 (x 1, y 1) and perpendicular to the straight line y = kx + b is represented by the equation:
Distance from a point to a line.
Theorem. If the point M(x) is given 0 , y 0 ), then the distance to the straight line Ах + Ву + С =0 is defined as
.
Proof. Let point M 1 (x 1, y 1) be the base of the perpendicular dropped from point M to a given straight line. Then the distance between points M and M 1:
The coordinates x 1 and y 1 can be found by solving the system of equations:
The second equation of the system is the equation of a line passing through a given point M 0 perpendicular to a given line.
If we transform the first equation of the system to the form:
A(x – x 0) + B(y – y 0) + Ax 0 + By 0 + C = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
.
The theorem has been proven.
Example. Determine the angle between the lines: y = -3x + 7; y = 2x + 1.
k 1 = -3; k 2 = 2 tg = 
; = /4.
Example. Show that the lines 3x – 5y + 7 = 0 and 10x + 6y – 3 = 0 are perpendicular.
We find: k 1 = 3/5, k 2 = -5/3, k 1 k 2 = -1, therefore, the lines are perpendicular.
Example. Given are the vertices of the triangle A(0; 1), B(6; 5), C(12; -1). Find the equation of the height drawn from vertex C.
We find the equation of side AB: 
; 4x = 6y – 6;
2x – 3y + 3 = 0; 
The required height equation has the form: Ax + By + C = 0 or y = kx + b.
k = 
. Then y = 
. Because the height passes through point C, then its coordinates satisfy this equation: 
whence b = 17. Total: 
.
Answer: 3x + 2y – 34 = 0.
Analytical geometry in space.
Equation of a line in space.
Equation of a line in space given a point and
direction vector.
Let's take an arbitrary line and a vector 
(m, n, p), parallel to the given line. Vector 
called guide vector straight.
On the straight line we take two arbitrary points M 0 (x 0 , y 0 , z 0) and M (x, y, z).
z
M 1
Let us denote the radius vectors of these points as 
And 
, it's obvious that 
-
=
.
Because vectors 
And 
are collinear, then the relation is true 
=
t, where t is some parameter.
In total, we can write: 
=
+
t.
Because this equation is satisfied by the coordinates of any point on the line, then the resulting equation is parametric equation of a line.
This vector equation can be represented in coordinate form:
By transforming this system and equating the values of the parameter t, we obtain the canonical equations of a straight line in space:
.
Definition.
Direction cosines direct are the direction cosines of the vector 
, which can be calculated using the formulas:
;
.
From here we get: m: n: p = cos : cos : cos.
The numbers m, n, p are called angle coefficients straight. Because 
is a non-zero vector, then m, n and p cannot be equal to zero at the same time, but one or two of these numbers can be equal to zero. In this case, in the equation of the line, the corresponding numerators should be set equal to zero.
Equation of a straight line in space passing
through two points.
If on a straight line in space we mark two arbitrary points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), then the coordinates of these points must satisfy the straight line equation obtained above:
.
In addition, for point M 1 we can write:
.
Solving these equations together, we get:
.
This is the equation of a line passing through two points in space.
General equations of a straight line in space.
The equation of a straight line can be considered as the equation of the line of intersection of two planes.
As discussed above, a plane in vector form can be specified by the equation:
+ D = 0, where
- plane normal; 
- radius is the vector of an arbitrary point on the plane.
Let the 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 point M 1 has the form y-y 1 = 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), the coordinates of this point must satisfy equation (10.6): y 2 -y 1 = 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 points M 1 and M 2: 
It is assumed that in this equation x 1 ≠ x 2, y 1 ≠ y 2
If x 1 = 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 ordinate axis. Its equation is x = x 1 .
If y 2 = y I, then the equation of the line can be written as y = y 1, the straight line M 1 M 2 is parallel to the abscissa axis.
Equation of a line in segments
Let the straight line intersect the Ox axis at point M 1 (a;0), and the Oy axis at point M 2 (0;b). The equation will take the form: 
those. 
. This equation is called equation of a straight line in segments, because numbers a and b indicate which segments the line cuts off on the coordinate axes.
Equation of a line passing through a given point perpendicular to a given vector
Let us 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).
Let's take an arbitrary point M(x; y) on the 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 .
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 = -Ax o - Vu o is the free term. Equation (10.9) is the general equation of the line(see Fig. 2).
Fig.1 Fig.2
Canonical equations of the line
,
Where 
- coordinates of the point through which the line passes, and 
- direction vector.
Second order curves Circle
A circle is the set of all points of the plane equidistant from a given point, which is called the center.
Canonical equation of a circle of radius
R centered at a point 
:
In particular, if the center of the stake coincides with the origin of coordinates, then the equation will look like: 
Ellipse
An ellipse is a set of points on a plane, the sum of the distances from each of which to two given points
And 
, which are called foci, is a constant quantity 
, greater than the distance between foci 
.
The canonical equation of an ellipse whose foci lie on the Ox axis, and the origin of coordinates in the middle between the foci has the form 
G 
de a semi-major axis length; b – length of the semi-minor axis (Fig. 2).