Consider the equation of a straight line passing through a point and a normal vector. Let a point and a non-zero vector be given in the coordinate system (Fig. 1).
Definition
As we can see, there is a single straight line that passes through the point perpendicular to the direction of the vector (in this case it is called normal vector straight ).
Rice. 1
Let us prove that the linear equation
this is an equation of a line, that is, the coordinates of each point of the line satisfy equation (1), but the coordinates of a point that does not lie on do not satisfy equation (1).
To prove this, let us note that the scalar product of vectors and = in coordinate form coincides with the left side of equation (1).
Next we use the obvious property of the line: the vectors and are perpendicular if and only if the point lies on . And provided that both vectors are perpendicular, their scalar product (2) turns into for all points that lie on, and only for them. This means (1) is the equation of the straight line.
Definition
Equation (1) is called equation of the line that passes through a given pointwith normal vector = .
Let's transform equation (1)
Denoting = , we get
Thus, a linear equation of the form (3) corresponds to a straight line. On the contrary, using a given equation of the form (3), where at least one of the coefficients is not equal to zero, a straight line can be constructed.
Indeed, let a pair of numbers satisfy equation (3), that is
Subtracting the latter from (3), we obtain the relation that determines the straight line behind the vector and the point.
Study of the general equation of a line
It is useful to know the features of placing a line in certain cases when one or two of the numbers are equal to zero.
1. The general equation looks like this: . The point satisfies it, which means the line passes through the origin. It can be written: = – x (see Fig. 2).
Rice. 2
We believe that:
If we put , then , we get another point (see Fig. 2).
2. , then the equation looks like this, where = –. The normal vector lies on the axis, a straight line. Thus, the straight line is perpendicular at point , or parallel to the axis (see Fig. 3). In particular, if and , then and the equation is the equation of the ordinate axis.
Rice. 3
3. Similarly, when the equation is written, where . The vector belongs to the axis. Straight line at a point (Fig. 4).
If, then the equation of the axis is .
The study can be formulated in this form: the straight line is parallel to the coordinate axis, the change of which is absent in the general equation of the straight line.
For example:
Let's construct a straight line using the general equation, provided that - are not equal to zero. To do this, it is enough to find two points that lie on this line. It is sometimes more convenient to find such points on coordinate axes.
Let us then = –.
When , then = –.
Let us denote – = , – = . Points and were found. Let us plot and draw a straight line on the axes and through them (see Fig. 5).
Rice. 5
From the general, you can move on to an equation that will include the numbers and:
And then it turns out:
Or, according to the notation, we obtain the equation
Which is called equation of a straight line in segments. The numbers and, accurate to the sign, are equal to the segments that are cut off by a straight line on the coordinate axes.
Equation of a straight line with slope
To find out what the equation of a straight line with a slope is, consider equation (1):
Denoting – = , we get
equation of a line that passes through a point in a given direction. The geometric content of the coefficient is clear from Fig. 6.
B = = , where is the smallest angle by which the positive direction of the axis needs to be rotated around the common point until it aligns with the straight line. Obviously, if the angle is acute, then title="Rendered by QuickLaTeX.com" height="17" width="97" style="vertical-align: -4px;">; если же – тупой угол, тогда .!}
Let's open the brackets in (5) and simplify it:
Where . Relationship (6) – equation straight line with slope. When , is a segment that cuts off a straight line on the axis (see Fig. 6).
Note!
To move from a general straight line equation to an equation with a slope coefficient, you must first solve for .
Rice. 6
= – x + – =
where denoted = –, = –. If, then from the study of the general equation it is already known that such a straight line is perpendicular to the axis.
Let's look at the canonical equation of a straight line using an example.
Let a point and a nonzero vector be specified in the coordinate system (Fig. 7).
Rice. 7
It is necessary to create an equation for a straight line that passes through a point parallel to the vector, which is called the direction vector. An arbitrary point belongs to this line if and only if . Since the vector is given, and the vector is , then, according to the parallelism condition, the coordinates of these vectors are proportional, that is:
Definition
Relationship (7) is called the equation of a line that passes through a given point in a given direction or the canonical equation of a line.
Let us note that we can move to an equation of the form (7), for example, from the equation of a pencil of lines (4)
or from the equation of a straight line through a point and a normal vector (1):
It was assumed above that the direction vector is non-zero, but it may happen that one of its coordinates, for example, . Then expression (7) will be formally written:
which doesn't make sense at all. However, we accept and obtain the equation of the straight line perpendicular to the axis. Indeed, from the equation it is clear that the straight line is defined by a point and a direction vector perpendicular to the axis. If we remove the denominator from this equation, then we get:
Or - the equation of a straight line perpendicular to the axis. A similar result would be obtained for the vector .
Parametric equation of a line
To understand what a parametric equation of a line is, you need to return to equation (7) and equate each fraction (7) to a parameter. Since at least one of the denominators in (7) is not equal to zero, and the corresponding numerator can acquire arbitrary values, then the region of parameter change is the entire numerical axis.
Definition
Equation (8) is called the parametric equation of a straight line.
Examples of straight line problems
Of course, it is difficult to solve anything solely based on definitions, because you need to solve at least a few examples or problems on your own that will help consolidate the material you have covered. Therefore, let's analyze the main tasks in a straight line, since similar problems often come across in exams and tests.
Canonical and parametric equation
Example 1
On a straight line given by the equation, find a point that is located at a distance of 10 units from the point of this straight line.
Solution:
Let sought after point of a straight line, then for the distance we write . Given that . Since the point belongs to a line that has a normal vector, then the equation of the line can be written: = = and then it turns out:
Then the distance . Subject to , or . From the parametric equation:
Example 2
Task
The point moves uniformly with speed in the direction of the vector from the starting point. Find the coordinates of the point through from the beginning of the movement.
Solution
First you need to find the unit vector. Its coordinates are direction cosines:
Then the velocity vector:
X = x = .
The canonical equation of the line will now be written:
= = , = – parametric equation. After this, you need to use the parametric equation of the straight line at .
Solution:
The equation of a line that passes through a point is found using the formula for a pencil of lines, where slope for a straight line and = for a straight line.
Considering the figure, where you can see that between straight lines and - there are two angles: one is acute, and the second is obtuse. According to formula (9), this is the angle between the straight lines and by which you need to rotate the straight line counterclockwise relative to their intersection point until it aligns with the straight line .
So, we remembered the formula, we figured out the angles and now we can return to our example. This means, taking into account formula (9), we first find the equations of the leg.
Since rotating the straight line by an angle counterclockwise relative to the point leads to alignment with the straight line, then in formula (9) , a . From the equation:
Using the beam formula, the equation of a straight line will be written:
Similarly we find , and ,
Line equation:
Equation of a line – types of equation of a line: passing through a point, general, canonical, parametric, etc. updated: November 22, 2019 by: Scientific Articles.Ru
Let two points be given M 1 (x 1,y 1) And M 2 (x 2,y 2). Let us write the equation of the straight line in the form (5), where k still unknown coefficient:
Since the point M 2 belongs to a given line, then its coordinates satisfy equation (5): . Expressing from here and substituting it into equation (5), we obtain the required equation:
If 
this equation can be rewritten in a form that is more convenient for memorization:
(6)
Example. Write down the equation of a straight line passing through points M 1 (1,2) and M 2 (-2,3)
Solution. 
. Using the property of proportion and performing the necessary transformations, we obtain the general equation of a straight line:
Angle between two straight lines
Consider two straight lines l 1 And l 2:
l 1: , , And
l 2: , ,
φ is the angle between them (). From Fig. 4 it is clear: .
 |
From here 
, or
Using formula (7) you can determine one of the angles between straight lines. The second angle is equal to .
Example. Two straight lines are given by the equations y=2x+3 and y=-3x+2. find the angle between these lines.
Solution. From the equations it is clear that k 1 =2, and k 2 =-3. Substituting these values into formula (7), we find
. Thus, the angle between these lines is equal to .
Conditions for parallelism and perpendicularity of two straight lines
If straight l 1 And l 2 are parallel, then φ=0 And tgφ=0. from formula (7) it follows that , whence k 2 =k 1. Thus, the condition for parallelism of two lines is the equality of their angular coefficients.
If straight l 1 And l 2 are perpendicular, then φ=π/2, α 2 = π/2+ α 1 . 
. Thus, the condition for the perpendicularity of two straight lines is that their angular coefficients are inverse in magnitude and opposite in sign.
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:
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 tanj= ; j = p/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 height passes through point C, then its coordinates satisfy this equation: whence b = 17. Total: .
Answer: 3x + 2y – 34 = 0.
The distance from a point to a line is determined by the length of the perpendicular drawn from the point to the line.
If the line is parallel to the projection plane (h | | P 1), then in order to determine the distance from the point A to a straight line h it is necessary to lower the perpendicular from the point A to the horizontal h.
Let's consider a more complex example, when the straight line occupies a general position. Let it be necessary to determine the distance from a point M to a straight line A general position.
Determination task distances between parallel lines is solved similarly to the previous one. A point is taken on one line and a perpendicular is dropped from it to another line. The length of a perpendicular is equal to the distance between parallel lines.
Second order curve is a line defined by an equation of the second degree relative to the current Cartesian coordinates. In the general case, Ax 2 + 2Bxy + Su 2 + 2Dx + 2Ey + F = 0,
where A, B, C, D, E, F are real numbers and at least one of the numbers A 2 + B 2 + C 2 ≠0.
Circle
Circle center– this is the geometric locus of points in the plane equidistant from a point in the plane C(a,b).
The circle is given by the following equation:
Where x,y are the coordinates of an arbitrary point on the circle, R is the radius of the circle.
Sign of the equation of a circle
1. The term with x, y is missing
2. The coefficients for x 2 and y 2 are equal
Ellipse
Ellipse is called the geometric locus of points in a plane, the sum of the distances of each of which from two given points of this plane is called foci (a constant value).
The canonical equation of the ellipse:
X and y belong to the ellipse.
a – semimajor axis of the ellipse
b – semi-minor axis of the ellipse
The ellipse has 2 axes of symmetry OX and OU. The axes of symmetry of an ellipse are its axes, the point of their intersection is the center of the ellipse. The axis on which the foci are located is called focal axis. The point of intersection of the ellipse with the axes is the vertex of the ellipse.
Compression (tension) ratio: ε = s/a– eccentricity (characterizes the shape of the ellipse), the smaller it is, the less the ellipse is extended along the focal axis.
If the centers of the ellipse are not at the center C(α, β)
Hyperbola
Hyperbole is called the geometric locus of points in a plane, the absolute value of the difference in distances, each of which from two given points of this plane, called foci, is a constant value different from zero.
Canonical hyperbola equation
A hyperbola has 2 axes of symmetry:
a – real semi-axis of symmetry
b – imaginary semi-axis of symmetry
Asymptotes of a hyperbola:
Parabola
Parabola is the locus of points in the plane equidistant from a given point F, called the focus, and a given line, called the directrix.
The canonical equation of a parabola:
У 2 =2рх, where р is the distance from the focus to the directrix (parabola parameter)
If the vertex of the parabola is C (α, β), then the equation of the parabola (y-β) 2 = 2р(x-α)
If the focal axis is taken as the ordinate axis, then the equation of the parabola will take the form: x 2 =2qу
This article is part of the topic equation of a line in a plane. Here we will look at it from all sides: we will start with the proof of the theorem that specifies the form of the general equation of a line, then we will consider an incomplete general equation of a line, we will give examples of incomplete equations of a line with graphic illustrations, and in conclusion we will dwell on the transition from a general equation of a line to other types of equations of this line and give detailed solutions to typical problems for composing the general equation of a straight line.
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General equation of a straight line - basic information.
Let's analyze this algorithm when solving an example.
Example.
Write parametric equations of a line that is given by the general equation of a line 
.
Solution.
First, we reduce the original general equation of the line to the canonical equation of the line:
Now we take the left and right sides of the resulting equation to be equal to the parameter. We have 
Answer:
From a general equation of a straight line, it is possible to obtain an equation of a straight line with an angle coefficient only when . What do you need to do to make the transition? Firstly, on the left side of the general straight line equation, leave only the term , the remaining terms must be moved to the right side with the opposite sign: 
. Secondly, divide both sides of the resulting equality by the number B, which is non-zero, 
. That's all.
Example.
A straight line in a rectangular coordinate system Oxy is given by the general equation of a straight line. Get the equation of this line with the slope.
Solution.
Let's carry out the necessary actions: .
Answer:
When a line is given by the complete general equation of the line, it is easy to obtain the equation of the line in segments of the form. To do this, we transfer the number C to the right side of the equality with the opposite sign, divide both sides of the resulting equality by –C, and finally transfer the coefficients for the variables x and y to the denominators: 
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).
Dependence between ellipse parameters
And 
is expressed by the ratio:
(4)
Ellipse eccentricitycalled the interfocal distance ratio2sto the major axis2a:
Headmistresses
ellipse are straight lines parallel to the Oy axis, which are located at a distance from this axis. Directrix equations: 
.
If in the equation of the ellipse 
, then the foci of the ellipse are on the Oy axis.
So, 
This article received equation of a line passing through two given points in a rectangular Cartesian coordinate system on a plane, and also derived the equations of a straight line that passes through two given points in a rectangular coordinate system in three-dimensional space. After presenting the theory, solutions to typical examples and problems are shown in which it is necessary to construct equations of a line of various types when the coordinates of two points on this line are known.
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Equation of a line passing through two given points on a plane.
Before obtaining the equation of a straight line passing through two given points in a rectangular coordinate system on a plane, let us recall some facts.
One of the axioms of geometry states that through two divergent points on a plane a single straight line can be drawn. In other words, by specifying two points on a plane, we uniquely define a straight line that passes through these two points (if necessary, refer to the section on methods for specifying a straight line on a plane).
Let Oxy be fixed on the plane. In this coordinate system, any straight line corresponds to some equation of a straight line on the plane. The directing vector of the straight line is inextricably linked with this same straight line. This knowledge is quite enough to create an equation of a straight line passing through two given points.
Let us formulate the condition of the problem: create an equation for the straight line a, which in the rectangular Cartesian coordinate system Oxy passes through two divergent points and.
We will show you the simplest and most universal solution to this problem.
We know that the canonical equation of a line on a plane is of the form 
defines in the rectangular coordinate system Oxy a straight line passing through a point and having a direction vector .
Let's write the canonical equation of a straight line a passing through two given points and .
Obviously, the direction vector of the straight line a, which passes through the points M 1 and M 2, is the vector, it has the coordinates 
(see article if necessary). Thus, we have all the necessary data to write the canonical equation of straight line a - the coordinates of its direction vector 
and the coordinates of the point lying on it (and ). It looks like 
(or 
).
We can also write down the parametric equations of a line on a plane passing through two points and. They look like 
or 
.
Let's look at the solution to the example.
Example.
Write the equation of a line that passes through two given points 
.
Solution.
We found out that the canonical equation of a line passing through two points with coordinates and has the form .
From the problem conditions we have 
. Let's substitute this data into the equation . We get 
.
Answer:
.
If we need not the canonical equation of a line and not the parametric equations of a line passing through two given points, but an equation of a line of a different type, then we can always arrive at it from the canonical equation of a line.
Example.
Write down the general equation of the straight line, which in the rectangular coordinate system Oxy on the plane passes through two points and.
Solution.
First, let's write the canonical equation of a line passing through two given points. It looks like . Now let's bring the resulting equation to the required form: .
Answer:
.
At this point we can finish with the equation of a straight line passing through two given points in a rectangular coordinate system on a plane. But I would like to remind you how we solved such a problem in high school in algebra lessons.
At school we only knew the equation of a straight line with an angular coefficient of the form . Let us find the value of the angular coefficient k and the number b at which the equation defines in the rectangular coordinate system Oxy on the plane a straight line passing through the points and at . (If x 1 =x 2, then the angular coefficient of the line is infinite, and the line M 1 M 2 is determined by the general incomplete equation of the line of the form x-x 1 =0).
Since points M 1 and M 2 lie on a line, the coordinates of these points satisfy the equation of the line, that is, the equalities and are valid. Solving a system of equations of the form 
regarding unknown variables k and b, we find 
or 
. For these values of k and b, the equation of a straight line passing through two points and takes the form 
or 
.
There is no point in memorizing these formulas; when solving examples, it is easier to repeat the indicated actions.
Example.
Write the equation of a line with slope if this line passes through the points and .
Solution.
In the general case, the equation of a straight line with an angle coefficient has the form . Let us find k and b for which the equation corresponds to a line passing through two points and .
Since points M 1 and M 2 lie on a straight line, their coordinates satisfy the equation of the straight line, that is, the equalities are true 
And . The values of k and b are found by solving the system of equations 
(if necessary, refer to the article): 
It remains to substitute the found values into the equation. Thus, the required equation of a line passing through two points and has the form .
Colossal work, isn't it?
It is much easier to write the canonical equation of a line passing through two points and , it has the form 
, and from it go to the equation of a straight line with an angular coefficient: .
Answer:
Equations of a line that passes through two given points in three-dimensional space.
Let a rectangular coordinate system Oxyz be fixed in three-dimensional space, and two divergent points be given 
And 
, through which the straight line M 1 M 2 passes. Let us obtain the equations of this line.
We know that the canonical equations of a straight line in space are of the form 
and parametric equations of a straight line in space of the form 
define a straight line in the rectangular coordinate system Oxyz, which passes through the point with coordinates and has a direction vector 
.
The direction vector of the line M 1 M 2 is the vector, and this line passes through the point (And ), then the canonical equations of this line have the form (or 
), and the parametric equations are 
(or 
).
.
If you need to define a straight line M 1 M 2 using the equations of two intersecting planes, then you must first draw up the canonical equations of a straight line passing through two points And , and from these equations obtain the required plane equations.
Bibliography.
- Atanasyan L.S., Butuzov V.F., Kadomtsev S.B., Poznyak E.G., Yudina I.I. Geometry. Grades 7 – 9: textbook for general education institutions.
- Atanasyan L.S., Butuzov V.F., Kadomtsev S.B., Kiseleva L.S., Poznyak E.G. Geometry. Textbook for 10-11 grades of secondary school.
- Pogorelov A.V., Geometry. Textbook for grades 7-11 in general education institutions.
- Bugrov Ya.S., Nikolsky S.M. Higher mathematics. Volume one: elements of linear algebra and analytical geometry.
- Ilyin V.A., Poznyak E.G. Analytic geometry.