An example solution is to indicate a straight line. Equation of a straight line on a plane
Definition. Any straight line on a plane can be given by a first-order equation
Ax + Wu + C \u003d 0,
and the constants A, B are not equal to zero at the same time. This first-order equation is called the general equation of the straight line.Depending on the values \u200b\u200bof constants A, B and C, the following special cases are possible:
C \u003d 0, A ≠ 0, B ≠ 0 - the line passes through the origin
A \u003d 0, B ≠ 0, C ≠ 0 (By + C \u003d 0) - the straight line is parallel to the Ox axis
B \u003d 0, A ≠ 0, C ≠ 0 (Ax + C \u003d 0) - the straight line is parallel to the Oy axis
B \u003d C \u003d 0, A ≠ 0 - the straight line coincides with the Oy axis
A \u003d C \u003d 0, B ≠ 0 - the straight line coincides with the Ox axis
The equation of the straight line can be presented in different forms depending on any given initial conditions.
Equation of a straight line along a point and a normal vector
Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B) is perpendicular to the straight line given by the equation Ax + Vy + C \u003d 0.
Example... Find the equation of the straight line passing through the point A (1, 2) perpendicular to (3, -1).
Decision... At A \u003d 3 and B \u003d -1, we compose the equation of the straight line: 3x - y + C \u003d 0. To find the coefficient C, we substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C \u003d 0, therefore, C \u003d -1 ... Total: the required equation: 3x - y - 1 \u003d 0.
Equation of a straight 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 straight line passing through these points:
If any of the denominators is zero, the corresponding numerator should be equated to zero. On the plane, the equation of the straight line written above is simplified:
if x 1 ≠ x 2 and x \u003d x 1, if x 1 \u003d x 2.
Fraction \u003d k is called slope straight.
Example... Find the equation of the straight line passing through the points A (1, 2) and B (3, 4).
Decision. Applying the above formula, we get:
Equation of a straight line by point and slope
If the total Ax + Vu + C \u003d 0 bring to the form:
and designate 
, then the resulting equation is called equation of a straight line with slopek.
Equation of a straight line along a point and a direction vector
By analogy with the paragraph considering the equation of a straight line through the normal vector, you can enter the specification of a straight line through a point and a direction vector of a straight line.
Definition. Each nonzero vector (α 1, α 2) whose components satisfy the condition А α 1 + В α 2 \u003d 0 is called the directing vector of the line
Ax + Wu + C \u003d 0.
Example. Find the equation of a straight line with a direction vector (1, -1) and passing through point A (1, 2).
Decision. The equation of the desired straight line will be sought in the form: Ax + By + C \u003d 0. According to the definition, the coefficients must satisfy the conditions:
1 * A + (-1) * B \u003d 0, i.e. A \u003d B.
Then the equation of the straight line has the form: Ax + Ay + C \u003d 0, or x + y + C / A \u003d 0. for x \u003d 1, y \u003d 2 we obtain C / A \u003d -3, i.e. required equation:
Equation of a straight line in segments
If in the general equation of the straight line Ax + Vy + C \u003d 0 C ≠ 0, then, dividing by –C, we get: 
or
The geometric meaning of the coefficients is that the coefficient and is the coordinate of the point of intersection of the straight line with the Ox axis, and b - the coordinate of the point of intersection of the straight line with the Oy axis.
Example. A general equation of the straight line x - y + 1 \u003d 0 is given. Find the equation of this straight line in segments.
C \u003d 1, a \u003d -1, b \u003d 1.
Normal Equation of a Line
If both sides of the equation Ax + Vy + C \u003d 0 multiply by the number 
which is called normalizing factor, then we get
xcosφ + ysinφ - p \u003d 0 -
normal equation of a straight line. The ± sign of the normalizing factor should be chosen so that μ * С< 0. р – длина перпендикуляра, опущенного из начала координат на прямую, а φ - угол, образованный этим перпендикуляром с положительным направлением оси Ох.
Example... A general equation of the straight line 12x - 5y - 65 \u003d 0 is given. It is required to write various types of equations of this straight line.
the equation of this straight line in segments: 
equation of this line with slope: (divide by 5)
; cos φ \u003d 12/13; sin φ \u003d -5/13; p \u003d 5.
It should be noted that not every line can be represented by an equation in segments, for example, straight lines parallel to axes or passing through the origin.
Example... The straight line cuts off equal positive segments on the coordinate axes. Make a straight line equation if the area of \u200b\u200bthe triangle formed by these segments is 8 cm 2.
Decision. The straight line equation has the form:, ab / 2 \u003d 8; ab \u003d 16; a \u003d 4, a \u003d -4. a \u003d -4< 0 не подходит по условию задачи. Итого: или х + у – 4 = 0.
Example... Make the equation of the straight line passing through the point A (-2, -3) and the origin.
Decision.
The straight line equation has the form: 
, where x 1 \u003d y 1 \u003d 0; x 2 \u003d -2; y 2 \u003d -3.
Angle between straight lines on a plane
Definition. If two straight lines y \u003d k 1 x + b 1, y \u003d k 2 x + b 2 are given, then the acute angle between these lines will be defined as
.
Two straight lines are parallel if k 1 \u003d k 2. Two straight lines are perpendicular if k 1 \u003d -1 / k 2.
Theorem.Straight lines Ax + By + C \u003d 0 and A 1 x + B 1 y + C 1 \u003d 0 are parallel when the proportional coefficients A 1 \u003d λA, B 1 \u003d λB. If also С 1 \u003d λС, then the lines coincide. The coordinates of the point of intersection of two straight lines are found as a solution to the system of equations of these straight lines.
Equation of a straight line passing through a given point perpendicular to a given straight line
Definition. The straight line passing through the point M 1 (x 1, y 1) and perpendicular to the straight line y \u003d 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 straight line Ax + Vy + C \u003d 0 is determined as
.
Evidence. Let point M 1 (x 1, y 1) be the base of the perpendicular dropped from point M onto a given line. Then the distance between points M and M 1:
(1)
The coordinates x 1 and y 1 can be found as a solution to the system of equations:
The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicular to a given straight 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 \u003d 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem is proved.
Example... Determine the angle between the straight lines: y \u003d -3 x + 7; y \u003d 2 x + 1.
k 1 \u003d -3; k 2 \u003d 2; tgφ \u003d 
; φ \u003d π / 4.
Example... Show that the straight lines 3x - 5y + 7 \u003d 0 and 10x + 6y - 3 \u003d 0 are perpendicular.
Decision... We find: k 1 \u003d 3/5, k 2 \u003d -5/3, k 1 * k 2 \u003d -1, therefore, the straight lines are perpendicular.
Example... The vertices of the triangle A (0; 1), B (6; 5), C (12; -1) are given. Find the equation for the height drawn from vertex C.
Decision... We find the equation of the side AB: 
; 4 x \u003d 6 y - 6;
2 x - 3 y + 3 \u003d 0;
The required height equation is: Ax + By + C \u003d 0 or y \u003d kx + b. k \u003d. Then y \u003d. Because height passes through point C, then its coordinates satisfy this equation: 
whence b \u003d 17. Total:. 
Answer: 3 x + 2 y - 34 \u003d 0.
Equation of a straight line passing through two points. The article" " I promised you to analyze the second method of solving the presented problems of finding the derivative, for a given graph of a function and a tangent to this graph. We will analyze this method in , do not miss! Why in the next one?
The fact is that the formula for the equation of a straight line will be used there. Of course, one could just show the given formula and advise you to learn it. But it is better to explain - where it comes from (how it is derived). It's necessary! If you forget it, then quickly restore it will not be difficult. Everything is detailed below. So, we have two points A on the coordinate plane(x 1; y 1) and B (x 2; y 2), a straight line is drawn through the indicated points: 
Here is the straight line formula itself:
* That is, when substituting specific coordinates of points, we get an equation of the form y \u003d kx + b.
** If this formula is simply "jagged", then there is a high probability of getting confused with the indices at x... In addition, indices can be denoted in different ways, for example:
That is why it is important to understand the meaning.
Now the conclusion of this formula. Everything is very simple!
Triangles ABE and ACF are similar in acute angle (the first sign of similarity of right-angled triangles). It follows from this that the relations of the respective elements are equal, that is:
Now we simply express these segments through the difference in the coordinates of the points:
Of course, there will be no mistake if you write the relations of the elements in a different order (the main thing is to keep the correspondence):
The result will be the same equation of the straight line. It's all!
That is, no matter how the points themselves (and their coordinates) are designated, understanding this formula you will always find the equation of a straight line.
The formula can be derived using the properties of vectors, but the principle of derivation will be the same, since we will talk about the proportionality of their coordinates. In this case, the same semblance of right-angled triangles works. In my opinion, the output described above is clearer)).
View output through vector coordinates \u003e\u003e\u003e
Let a straight line be constructed on the coordinate plane passing through two given points A (x 1; y 1) and B (x 2; y 2). Let's mark on the straight line an arbitrary point C with coordinates ( x; y). We also denote two vectors:
It is known that for vectors lying on parallel lines (or on one straight line), their corresponding coordinates are proportional, that is:
- we write down the equality of the ratios of the corresponding coordinates:
Let's consider an example:
Find the equation of a straight line passing through two points with coordinates (2; 5) and (7: 3).
You don't even have to build the straight line itself. We apply the formula:
It is important that you catch the correspondence when drawing up the ratio. You can't go wrong if you write down:
Answer: y \u003d -2 / 5x + 29/5 go y \u003d -0.4x + 5.8
In order to make sure that the obtained equation is found correctly, be sure to do a check - substitute the coordinates of the data in the condition of points into it. You should get correct equalities.
That's all. I hope the material was useful to you.
Sincerely, Alexander.
P.S: I would be grateful if you could tell us about the site on social networks.
Properties of a straight line in Euclidean geometry.
You can draw infinitely many straight lines through any point.
A single straight line can be drawn through any two non-coinciding points.
Two mismatched straight lines on the plane either intersect at a single point, or are
parallel (follows from the previous one).
In three-dimensional space, there are three options for the relative position of two straight lines:
- straight lines intersect;
- straight lines are parallel;
- straight lines intersect.
Straight line - algebraic curve of the first order: in a Cartesian coordinate system, a straight line
is given on the plane by an equation of the first degree (linear equation).
General equation of the line.
Definition... Any straight line on a plane can be given by a first-order equation
Ax + Wu + C \u003d 0,
with constant A, B not equal to zero at the same time. This first-order equation is called common
equation of a straight line. Depending on the values \u200b\u200bof the constants A, B and FROM the following special cases are possible:
. C \u003d 0, A ≠ 0, B ≠ 0 - the straight line passes through the origin
. A \u003d 0, B ≠ 0, C ≠ 0 (By + C \u003d 0)- straight line parallel to the axis Oh
. B \u003d 0, A ≠ 0, C ≠ 0 (Ax + C \u003d 0) - straight line parallel to the axis OU
. B \u003d C \u003d 0, A ≠ 0 - the straight line coincides with the axis OU
. A \u003d C \u003d 0, B ≠ 0 - the straight line coincides with the axis Oh
The equation of a straight line can be presented in different forms, depending on any given
initial conditions.
Equation of a straight line along a point and a normal vector.
Definition... In a Cartesian rectangular coordinate system, a vector with components (A, B)
perpendicular to the straight line given by the equation
Ax + Wu + C \u003d 0.
Example... Find the equation of a straight line passing through a point A (1, 2) perpendicular to vector (3, -1).
Decision... At A \u003d 3 and B \u003d -1, we compose the equation of the straight line: 3x - y + C \u003d 0. To find the coefficient C
substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C \u003d 0, therefore
C \u003d -1. Total: the required equation: 3x - y - 1 \u003d 0.
Equation of a straight line passing through two points.
Let two points be given in space M 1 (x 1, y 1, z 1)and M2 (x 2, y 2, z 2), then straight line equation,
passing through these points:
If any of the denominators is zero, the corresponding numerator should be equated to zero. On the
plane, the equation of the straight line written above is simplified:
if a x 1 ≠ x 2 and x \u003d x 1 , if a x 1 \u003d x 2 .
Fraction \u003d k called slope straight.
Example... Find the equation of the straight line passing through the points A (1, 2) and B (3, 4).
Decision... Applying the above formula, we get:
Equation of a straight line by point and slope.
If the general equation of the straight line Ax + Wu + C \u003d 0 lead to the form:
and designate 
, then the resulting equation is called
equation of a straight line with slope k.
Equation of a straight line along a point and a direction vector.
By analogy with the paragraph considering the equation of a straight line through the normal vector, you can enter the task
a straight line through a point and a direction vector of a straight line.
Definition... Every nonzero vector (α 1, α 2)whose components satisfy the condition
Аα 1 + Вα 2 \u003d 0 called directing vector of a straight line.
Ax + Wu + C \u003d 0.
Example... Find the equation of a straight line with a direction vector (1, -1) and passing through point A (1, 2).
Decision... The equation of the desired straight line will be sought in the form: Ax + By + C \u003d 0. According to the definition,
the coefficients must meet the conditions:
1 * A + (-1) * B \u003d 0, i.e. A \u003d B.
Then the equation of the straight line has the form: Ax + Ay + C \u003d 0, or x + y + C / A \u003d 0.
at x \u003d 1, y \u003d 2we get C / A \u003d -3, i.e. required equation:
x + y - 3 \u003d 0
Equation of a straight line in segments.
If in the general equation of the straight line Ax + Vy + C \u003d 0 C ≠ 0, then, dividing by -C, we get:
or where
The geometric meaning of the coefficients is that the coefficient a is the coordinate of the intersection point
straight with axis Oh, and b - the coordinate of the point of intersection of the straight line with the axis OU.
Example... The general equation of the straight line is given x - y + 1 \u003d 0.Find the equation of this straight line in segments.
C \u003d 1, a \u003d -1, b \u003d 1.
Normal equation of a straight line.
If both sides of the equation Ax + Wu + C \u003d 0 divide by number 
which is called
normalizing factor, then we get
xcosφ + ysinφ - p \u003d 0 -normal line equation.
The ± sign of the normalizing factor should be chosen so that μ * C< 0.
r - the length of the perpendicular dropped from the origin to the straight line,
and φ - the angle formed by this perpendicular with the positive direction of the axis Oh.
Example... The general equation of the line is given 12x - 5y - 65 \u003d 0... Required to write different types of equations
this straight line.
The equation of this line in segments:
Equation of this line with slope: (divide by 5)
Straight line equation:
cos φ \u003d 12/13; sin φ \u003d -5/13; p \u003d 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.
The angle between straight lines on the plane.
Definition... If two lines are given y \u003d k 1 x + b 1, y \u003d k 2 x + b 2 , then an acute angle between these lines
will be defined as
Two straight lines are parallel if k 1 \u003d k 2... Two straight lines are perpendicular,
if a k 1 \u003d -1 / k 2 .
Theorem.
Direct Ax + Wu + C \u003d 0and A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients are proportional
А 1 \u003d λА, В 1 \u003d λВ... If also С 1 \u003d λС, then the straight lines coincide. Coordinates of the point of intersection of two lines
are found as a solution to the system of equations of these straight lines.
Equation of a straight line passing through a given point perpendicular to a given straight line.
Definition... Line through point M 1 (x 1, y 1) and perpendicular to the line y \u003d kx + b
represented by the equation:
Distance from point to line.
Theorem... If a point is given M (x 0, y 0), the distance to the straight line Ax + Wu + C \u003d 0defined as:
Evidence... Let the point M 1 (x 1, y 1) - the base of the perpendicular dropped from the point Mfor a given
straight line. Then the distance between the points Mand M 1:
(1)
Coordinates x 1 and at 1 can be found as a solution to the system of equations:
The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicular to
a given straight 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 \u003d 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem is proved.
This article discloses the derivation of the equation of a straight line passing through two given points in a rectangular coordinate system located on a plane. Let us derive the equation of a straight line passing through two given points in a rectangular coordinate system. We will clearly show and solve several examples related to the material covered.
Yandex.RTB R-A-339285-1
Before obtaining the equation of a straight line passing through two given points, it is necessary to pay attention to some facts. There is an axiom that says that it is possible to draw a straight line and only one through two non-coinciding points on the plane. In other words, two given points of the plane are defined by a straight line passing through these points.
If the plane is specified by a rectangular coordinate system Oxy, then any straight line depicted in it will correspond to the equation of a straight line on the plane. There is also a connection with the direction vector of the line. This data is enough to generate the equation of a line passing through two given points.
Let's consider an example of solving a similar problem. It is necessary to draw up an equation of the straight line a passing through two non-coincident points M 1 (x 1, y 1) and M 2 (x 2, y 2), which are in the Cartesian coordinate system.
In the canonical equation of a straight line on a plane, which has the form x - x 1 ax \u003d y - y 1 ay, a rectangular coordinate system O x y with a straight line is specified, which intersects with it at a point with coordinates M 1 (x 1, y 1) with a guide vector a → \u003d (ax, ay).
It is necessary to compose the canonical equation of the straight line a, which passes through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2).
Line a has a direction vector M 1 M 2 → with coordinates (x 2 - x 1, y 2 - y 1), since it intersects points M 1 and M 2. We received the necessary data in order to transform the canonical equation with the coordinates of the direction vector M 1 M 2 → \u003d (x 2 - x 1, y 2 - y 1) and the coordinates of the points M 1 (x 1, y 1) lying on them and M 2 (x 2, y 2). We get an equation of the form x - x 1 x 2 - x 1 \u003d y - y 1 y 2 - y 1 or x - x 2 x 2 - x 1 \u003d y - y 2 y 2 - y 1.
Consider the figure below.
Following the calculations, we write down the parametric equations of a straight line on the plane that passes through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2). We obtain an equation of the form x \u003d x 1 + (x 2 - x 1) λ y \u003d y 1 + (y 2 - y 1) λ or x \u003d x 2 + (x 2 - x 1) λ y \u003d y 2 + (y 2 - y 1) λ.
Let's take a closer look at the solution of several examples.
Example 1
Write down the equation of a straight line passing through 2 given points with coordinates M 1 - 5, 2 3, M 2 1, - 1 6.
Decision
The canonical equation for a straight line intersecting at two points with coordinates x 1, y 1 and x 2, y 2 takes the form x - x 1 x 2 - x 1 \u003d y - y 1 y 2 - y 1. By the condition of the problem, we have that x 1 \u003d - 5, y 1 \u003d 2 3, x 2 \u003d 1, y 2 \u003d - 1 6. Substitute numeric values \u200b\u200binto the equation x - x 1 x 2 - x 1 \u003d y - y 1 y 2 - y 1. From here we get that the canonical equation takes the form x - (- 5) 1 - (- 5) \u003d y - 2 3 - 1 6 - 2 3 ⇔ x + 5 6 \u003d y - 2 3 - 5 6.
Answer: x + 5 6 \u003d y - 2 3 - 5 6.
If you need to solve a problem with a different kind of equation, then first you can go to the canonical, since it is easier to come from it to any other.
Example 2
Draw up the general equation of a straight line passing through points with coordinates M 1 (1, 1) and M 2 (4, 2) in the coordinate system O x y.
Decision
First, you need to write down the canonical equation of a given straight line that passes through two given points. We get an equation of the form x - 1 4 - 1 \u003d y - 1 2 - 1 ⇔ x - 1 3 \u003d y - 1 1.
Let us bring the canonical equation to the required form, then we get:
x - 1 3 \u003d y - 1 1 ⇔ 1 x - 1 \u003d 3 y - 1 ⇔ x - 3 y + 2 \u003d 0
Answer: x - 3 y + 2 \u003d 0.
Examples of such tasks were considered in school textbooks in algebra lessons. School problems differed in that the well-known equation of a straight line with a slope, which has the form y \u003d k x + b. If you need to find the value of the slope k and the number b at which the equation y \u003d kx + b defines a line in the O x y system that passes through the points M 1 (x 1, y 1) and M 2 (x 2, y 2) , where x 1 ≠ x 2. When x 1 \u003d x 2 , then the slope takes on the value of infinity, and the straight line М 1 М 2 is determined by a general incomplete equation of the form x - x 1 \u003d 0 .
Because the points M 1 and M 2are on the straight line, then their coordinates satisfy the equation y 1 \u003d k x 1 + b and y 2 \u003d k x 2 + b. It is necessary to solve the system of equations y 1 \u003d k x 1 + b y 2 \u003d k x 2 + b for k and b.
To do this, find k \u003d y 2 - y 1 x 2 - x 1 b \u003d y 1 - y 2 - y 1 x 2 - x 1 x 1 or k \u003d y 2 - y 1 x 2 - x 1 b \u003d y 2 - y 2 - y 1 x 2 - x 1 x 2.
With such values \u200b\u200bof k and b, the equation of the line passing through the given two points takes the following form y \u003d y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 1 or y \u003d y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 2.
Remembering such a huge number of formulas at once will not work. To do this, you need to increase the number of repetitions in problem solutions.
Example 3
Write the equation of the straight line with the slope passing through the points with coordinates M 2 (2, 1) and y \u003d k x + b.
Decision
To solve the problem, we use the formula with the slope, which has the form y \u003d k x + b. Coefficients k and b should take such a value that this equation corresponds to a straight line passing through two points with coordinates M 1 (- 7, - 5) and M 2 (2, 1).
Points M 1 and M 2 are located on a straight line, then their coordinates should reverse the equation y \u003d k x + b true equality. From this we obtain that - 5 \u003d k (- 7) + b and 1 \u003d k 2 + b. Combine the equation into the system - 5 \u003d k · - 7 + b 1 \u003d k · 2 + b and solve.
Upon substitution, we obtain
5 \u003d k - 7 + b 1 \u003d k 2 + b ⇔ b \u003d - 5 + 7 k 2 k + b \u003d 1 ⇔ b \u003d - 5 + 7 k 2 k - 5 + 7 k \u003d 1 ⇔ ⇔ b \u003d - 5 + 7 kk \u003d 2 3 ⇔ b \u003d - 5 + 7 2 3 k \u003d 2 3 ⇔ b \u003d - 1 3 k \u003d 2 3
Now the values \u200b\u200bk \u003d 2 3 and b \u003d - 1 3 are substituted into the equation y \u003d k x + b. We get that the desired equation passing through the given points will be an equation of the form y \u003d 2 3 x - 1 3.
This way of solving it predetermines the waste of a lot of time. There is a way in which the task is solved literally in two steps.
Let us write the canonical equation of the straight line passing through M 2 (2, 1) and M 1 (- 7, - 5), which has the form x - (- 7) 2 - (- 7) \u003d y - (- 5) 1 - (- 5) ⇔ x + 7 9 \u003d y + 5 6.
We now turn to the equation in the slope. We get that: x + 7 9 \u003d y + 5 6 ⇔ 6 (x + 7) \u003d 9 (y + 5) ⇔ y \u003d 2 3 x - 1 3.
Answer: y \u003d 2 3 x - 1 3.
If in three-dimensional space there is a rectangular coordinate system O x y z with two specified non-coinciding points with coordinates M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), the straight line M 1 M 2, it is necessary to obtain the equation of this straight line.
We have that canonical equations of the form x - x 1 ax \u003d y - y 1 ay \u003d z - z 1 az and parametric equations of the form x \u003d x 1 + ax λ y \u003d y 1 + ay λ z \u003d z 1 + az λ are able to define a line in the coordinate system O x y z passing through points having coordinates (x 1, y 1, z 1) with a direction vector a → \u003d (ax, ay, az).
Straight M 1 M 2 has a direction vector of the form M 1 M 2 → \u003d (x 2 - x 1, y 2 - y 1, z 2 - z 1), where the line passes through the point M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), hence the canonical equation can be of the form x - x 1 x 2 - x 1 \u003d y - y 1 y 2 - y 1 \u003d z - z 1 z 2 - z 1 or x - x 2 x 2 - x 1 \u003d y - y 2 y 2 - y 1 \u003d z - z 2 z 2 - z 1, in turn parametric x \u003d x 1 + (x 2 - x 1) λ y \u003d y 1 + (y 2 - y 1) λ z \u003d z 1 + (z 2 - z 1) λ or x \u003d x 2 + (x 2 - x 1) λ y \u003d y 2 + (y 2 - y 1) Λ z \u003d z 2 + (z 2 - z 1) λ.
Consider a figure that shows 2 given points in space and the equation of a straight line.
Example 4
Write the equation of a straight line defined in a rectangular coordinate system O x y z of three-dimensional space, passing through given two points with coordinates M 1 (2, - 3, 0) and M 2 (1, - 3, - 5).
Decision
It is necessary to find the canonical equation. Since we are talking about a three-dimensional space, it means that when a straight line passes through given points, the desired canonical equation takes the form x - x 1 x 2 - x 1 \u003d y - y 1 y 2 - y 1 \u003d z - z 1 z 2 - z 1 ...
By hypothesis, we have that x 1 \u003d 2, y 1 \u003d - 3, z 1 \u003d 0, x 2 \u003d 1, y 2 \u003d - 3, z 2 \u003d - 5. Hence it follows that the necessary equations can be written as follows:
x - 2 1 - 2 \u003d y - (- 3) - 3 - (- 3) \u003d z - 0 - 5 - 0 ⇔ x - 2 - 1 \u003d y + 3 0 \u003d z - 5
Answer: x - 2 - 1 \u003d y + 3 0 \u003d z - 5.
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Let the line pass through the points M 1 (x 1; y 1) and M 2 (x 2; y 2). The equation of the 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), the coordinates of this point must satisfy the 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 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 ordinate axis. Its equation has the form x \u003d x 1 .
If y 2 \u003d y I, then the equation of the straight line can be written in the form y \u003d y 1, the straight line M 1 M 2 is parallel to the abscissa 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, since the numbers a and b indicate which segments are cut off by a straight line on the coordinate axes.
Equation of a straight 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 nonzero vector n \u003d (A; B).
Take an arbitrary point M (x; y) on a 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 zero: that is
A (x - xo) + B (y - yo) \u003d 0. (10.8)
Equation (10.8) is called the equation of a straight line passing through a given point perpendicular to a given vector .
The vector n \u003d (A; B), perpendicular to the straight line, is called normal the normal vector of this line .
Equation (10.8) can be rewritten as Ax + Wu + C \u003d 0 , (10.9)
where A and B are the coordinates of the normal vector, C \u003d -Aх о - Ву о - free term. Equation (10.9) is the general equation of the straight line (see fig. 2).
Fig. 1 Fig. 2
Canonical equations of the line
,
Where 
- coordinates of the point through which the straight 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.
The canonical equation of a circle of radius
R centered at 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 on a plane, the sum of the distances from each of which to two given points
and 
, which are called foci, are constant 
greater than the distance between the foci 
.
The canonical equation of an ellipse, the foci of which lie on the Ox axis, and the origin of coordinates midway between the foci has the form 
r 
dea the length of the semi-major axis;b - the length of the semi-minor axis (Fig. 2).
