How to derive the equation of a straight line. General equation of a straight line on a 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's find the equation of the line passing through given point Mo (x O; y o) is perpendicular to the 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).
The equation of a line passing through a given point in a given direction. Equation of a line passing through two given points. The angle between two straight lines. The condition of parallelism and perpendicularity of two straight lines. Determining the point of intersection of two lines
1. Equation of a line passing through a given point A(x 1 , y 1) in a given direction, determined slope k,
y - y 1 = k(x - x 1). (1)
This equation defines a pencil of lines passing through a point A(x 1 , y 1), which is called the beam center.
2. Equation of a line passing through two points: A(x 1 , y 1) and B(x 2 , y 2), written like this:
The angular coefficient of a straight line passing through two given points is determined by the formula
3. Angle between straight lines A And B is the angle by which the first straight line must be rotated A around the point of intersection of these lines counterclockwise until it coincides with the second line B. If two straight lines are given by equations with a slope
y = k 1 x + B 1 ,
This article continues the topic of the equation of a line on a plane: we will consider this type of equation as the general equation of a line. Let us define the theorem and give its proof; Let's figure out what an incomplete general equation of a line is and how to make transitions from a general equation to other types of equations of a line. We will reinforce the entire theory with illustrations and solutions to practical problems.
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Let a rectangular coordinate system O x y be specified on the plane.
Theorem 1
Any equation of the first degree, having the form A x + B y + C = 0, where A, B, C are some real numbers (A and B are not equal to zero at the same time) defines a straight line in rectangular system coordinates on the plane. In turn, any straight line in a rectangular coordinate system on a plane is determined by an equation that has the form A x + B y + C = 0 for a certain set of values A, B, C.
Proof
This theorem consists of two points; we will prove each of them.
- Let us prove that the equation A x + B y + C = 0 defines a straight line on the plane.
Let there be some point M 0 (x 0 , y 0) whose coordinates correspond to the equation A x + B y + C = 0. Thus: A x 0 + B y 0 + C = 0. Subtract from the left and right sides of the equations A x + B y + C = 0 the left and right sides of the equation A x 0 + B y 0 + C = 0, we obtain a new equation that looks like A (x - x 0) + B (y - y 0) = 0 . It is equivalent to A x + B y + C = 0.
The resulting equation A (x - x 0) + B (y - y 0) = 0 is a necessary and sufficient condition for the perpendicularity of the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0 ) . Thus, the set of points M (x, y) defines a straight line in a rectangular coordinate system perpendicular to the direction of the vector n → = (A, B). We can assume that this is not so, but then the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) would not be perpendicular, and the equality A (x - x 0 ) + B (y - y 0) = 0 would not be true.
Consequently, the equation A (x - x 0) + B (y - y 0) = 0 defines a certain line in a rectangular coordinate system on the plane, and therefore the equivalent equation A x + B y + C = 0 defines the same line. This is how we proved the first part of the theorem.
- Let us provide a proof that any straight line in a rectangular coordinate system on a plane can be specified by an equation of the first degree A x + B y + C = 0.
Let us define a straight line a in a rectangular coordinate system on a plane; the point M 0 (x 0 , y 0) through which this line passes, as well as the normal vector of this line n → = (A, B) .
Let there also be some point M (x, y) - a floating point on a line. In this case, the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) are perpendicular to each other, and their scalar product there is a zero:
n → , M 0 M → = A (x - x 0) + B (y - y 0) = 0
Let's rewrite the equation A x + B y - A x 0 - B y 0 = 0, define C: C = - A x 0 - B y 0 and as a final result we get the equation A x + B y + C = 0.
So, we have proved the second part of the theorem, and we have proved the entire theorem as a whole.
Definition 1
An equation of the form A x + B y + C = 0 - This general equation of a line on a plane in a rectangular coordinate systemOxy.
Based on the proven theorem, we can conclude that a straight line and its general equation defined on a plane in a fixed rectangular coordinate system are inextricably linked. In other words, the original line corresponds to its general equation; the general equation of a line corresponds to a given line.
From the proof of the theorem it also follows that the coefficients A and B for the variables x and y are the coordinates of the normal vector of the line, which is given by the general equation of the line A x + B y + C = 0.
Let's consider specific example general equation of a straight line.
Let the equation 2 x + 3 y - 2 = 0 be given, which corresponds to a straight line in a given rectangular coordinate system. The normal vector of this line is the vector n → = (2 , 3) . Let's draw the given straight line in the drawing.
We can also state the following: the straight line that we see in the drawing is determined by the general equation 2 x + 3 y - 2 = 0, since the coordinates of all points on a given straight line correspond to this equation.
We can obtain the equation λ · A x + λ · B y + λ · C = 0 by multiplying both sides of the general equation of the line by a number λ not equal to zero. The resulting equation is equivalent to the original general equation, therefore, it will describe the same straight line on the plane.
Definition 2Complete general equation of a line– such a general equation of the straight line A x + B y + C = 0, in which the numbers A, B, C are different from zero. Otherwise the equation is incomplete.
Let us analyze all variations of the incomplete general equation of a line.
- When A = 0, B ≠ 0, C ≠ 0, the general equation takes the form B y + C = 0. Such an incomplete general equation defines in a rectangular coordinate system O x y a straight line that is parallel to the O x axis, since for any real value of x the variable y will take the value - C B . In other words, the general equation of the line A x + B y + C = 0, when A = 0, B ≠ 0, specifies the locus of points (x, y), whose coordinates are equal to the same number - C B .
- If A = 0, B ≠ 0, C = 0, the general equation takes the form y = 0. This incomplete equation defines the abscissa axis O x .
- When A ≠ 0, B = 0, C ≠ 0, we obtain an incomplete general equation A x + C = 0, defining a straight line parallel to the ordinate.
- Let A ≠ 0, B = 0, C = 0, then the incomplete general equation will take the form x = 0, and this is the equation of the coordinate line O y.
- Finally, for A ≠ 0, B ≠ 0, C = 0, the incomplete general equation takes the form A x + B y = 0. And this equation describes a straight line that passes through the origin. In fact, the pair of numbers (0, 0) corresponds to the equality A x + B y = 0, since A · 0 + B · 0 = 0.
Let us graphically illustrate all of the above types of incomplete general equation of a straight line.
Example 1
It is known that the given straight line is parallel to the ordinate axis and passes through the point 2 7, - 11. It is necessary to write down the general equation of the given line.
Solution
A straight line parallel to the ordinate axis is given by an equation of the form A x + C = 0, in which A ≠ 0. The condition also specifies the coordinates of the point through which the line passes, and the coordinates of this point meet the conditions of the incomplete general equation A x + C = 0, i.e. the equality is true:
A 2 7 + C = 0
From it it is possible to determine C if we give A some non-zero value, for example, A = 7. In this case, we get: 7 · 2 7 + C = 0 ⇔ C = - 2. We know both coefficients A and C, substitute them into the equation A x + C = 0 and get the required straight line equation: 7 x - 2 = 0
Answer: 7 x - 2 = 0
Example 2
The drawing shows a straight line; you need to write down its equation.
Solution
The given drawing allows us to easily take the initial data to solve the problem. We see in the drawing that the given straight line is parallel to the O x axis and passes through the point (0, 3).
The straight line, which is parallel to the abscissa, is determined by the incomplete general equation B y + C = 0. Let's find the values of B and C. The coordinates of the point (0, 3), since the given line passes through it, will satisfy the equation of the line B y + C = 0, then the equality is valid: B · 3 + C = 0. Let's set B to some value other than zero. Let's say B = 1, in which case from the equality B · 3 + C = 0 we can find C: C = - 3. We use known values B and C, we obtain the required equation of the straight line: y - 3 = 0.
Answer: y - 3 = 0 .
General equation of a line passing through a given point in a plane
Let the given line pass through the point M 0 (x 0 , y 0), then its coordinates correspond to the general equation of the line, i.e. the equality is true: A x 0 + B y 0 + C = 0. Let us subtract the left and right sides of this equation from the left and right sides of the general complete equation straight. We get: A (x - x 0) + B (y - y 0) + C = 0, this equation is equivalent to the original general one, passes through the point M 0 (x 0, y 0) and has a normal vector n → = (A, B) .
The result that we obtained makes it possible to write down the general equation of a line with known coordinates of the normal vector of the line and the coordinates of a certain point of this line.
Example 3
Given a point M 0 (- 3, 4) through which a line passes, and the normal vector of this line n → = (1 , - 2) . It is necessary to write down the equation of the given line.
Solution
The initial conditions allow us to obtain the necessary data to compile the equation: A = 1, B = - 2, x 0 = - 3, y 0 = 4. Then:
A (x - x 0) + B (y - y 0) = 0 ⇔ 1 (x - (- 3)) - 2 y (y - 4) = 0 ⇔ ⇔ x - 2 y + 22 = 0
The problem could have been solved differently. The general equation of a straight line is A x + B y + C = 0. The given normal vector allows us to obtain the values of coefficients A and B, then:
A x + B y + C = 0 ⇔ 1 x - 2 y + C = 0 ⇔ x - 2 y + C = 0
Now let’s find the value of C using the point M 0 (- 3, 4) specified by the problem condition, through which the straight line passes. The coordinates of this point correspond to the equation x - 2 · y + C = 0, i.e. - 3 - 2 4 + C = 0. Hence C = 11. The required straight line equation takes the form: x - 2 · y + 11 = 0.
Answer: x - 2 y + 11 = 0 .
Example 4
Given a line 2 3 x - y - 1 2 = 0 and a point M 0 lying on this line. Only the abscissa of this point is known, and it is equal to - 3. It is necessary to determine the ordinate of a given point.
Solution
Let us designate the coordinates of point M 0 as x 0 and y 0 . The source data indicates that x 0 = - 3. Since the point belongs to a given line, then its coordinates correspond to the general equation of this line. Then the equality will be true:
2 3 x 0 - y 0 - 1 2 = 0
Define y 0: 2 3 · (- 3) - y 0 - 1 2 = 0 ⇔ - 5 2 - y 0 = 0 ⇔ y 0 = - 5 2
Answer: - 5 2
Transition from the general equation of a line to other types of equations of a line and vice versa
As we know, there are several types of equations for the same straight line on a plane. The choice of the type of equation depends on the conditions of the problem; it is possible to choose the one that is more convenient for solving it. The skill of converting an equation of one type into an equation of another type is very useful here.
First, let's consider the transition from the general equation of the form A x + B y + C = 0 to the canonical equation x - x 1 a x = y - y 1 a y.
If A ≠ 0, then we move the term B y to the right side of the general equation. On the left side we take A out of brackets. As a result, we get: A x + C A = - B y.
This equality can be written as a proportion: x + C A - B = y A.
If B ≠ 0, we leave only the term A x on the left side of the general equation, transfer the others to the right side, we get: A x = - B y - C. We take – B out of brackets, then: A x = - B y + C B .
Let's rewrite the equality in the form of a proportion: x - B = y + C B A.
Of course, there is no need to memorize the resulting formulas. It is enough to know the algorithm of actions when moving from a general equation to a canonical one.
Example 5
The general equation of the line 3 y - 4 = 0 is given. It is necessary to transform it into a canonical equation.
Solution
Let's write the original equation as 3 y - 4 = 0. Next, we proceed according to the algorithm: the term 0 x remains on the left side; and on the right side we put - 3 out of brackets; we get: 0 x = - 3 y - 4 3 .
Let's write the resulting equality as a proportion: x - 3 = y - 4 3 0 . Thus, we have obtained an equation of canonical form.
Answer: x - 3 = y - 4 3 0.
To convert the general equation of a straight line into parametric ones, first go to canonical form, and then the transition from the canonical line equation to parametric equations.
Example 6
The straight line is given by the equation 2 x - 5 y - 1 = 0. Write down the parametric equations for this line.
Solution
Let us make the transition from the general equation to the canonical one:
2 x - 5 y - 1 = 0 ⇔ 2 x = 5 y + 1 ⇔ 2 x = 5 y + 1 5 ⇔ x 5 = y + 1 5 2
Now we take both sides of the resulting canonical equation equal to λ, then:
x 5 = λ y + 1 5 2 = λ ⇔ x = 5 λ y = - 1 5 + 2 λ , λ ∈ R
Answer:x = 5 λ y = - 1 5 + 2 λ , λ ∈ R
The general equation can be converted into an equation of a straight line with slope y = k · x + b, but only when B ≠ 0. For the transition, we leave the term B y on the left side, the rest are transferred to the right. We get: B y = - A x - C . Let's divide both sides of the resulting equality by B, different from zero: y = - A B x - C B.
Example 7
The general equation of the line is given: 2 x + 7 y = 0. You need to convert that equation into a slope equation.
Solution
Let's perform the necessary actions according to the algorithm:
2 x + 7 y = 0 ⇔ 7 y - 2 x ⇔ y = - 2 7 x
Answer: y = - 2 7 x .
From the general equation of a line, it is enough to simply obtain an equation in segments of the form x a + y b = 1. To make such a transition, we move the number C to the right side of the equality, divide both sides of the resulting equality by – C and, finally, transfer the coefficients for the variables x and y to the denominators:
A x + B y + C = 0 ⇔ A x + B y = - C ⇔ ⇔ A - C x + B - C y = 1 ⇔ x - C A + y - C B = 1
Example 8
It is necessary to transform the general equation of the line x - 7 y + 1 2 = 0 into the equation of the line in segments.
Solution
Let's move 1 2 to the right side: x - 7 y + 1 2 = 0 ⇔ x - 7 y = - 1 2 .
Let's divide both sides of the equality by -1/2: x - 7 y = - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y = 1 .
Answer: x - 1 2 + y 1 14 = 1 .
In general, the reverse transition is also easy: from other types of equations to the general one.
The equation of a line in segments and an equation with an angular coefficient can be easily converted into a general one by simply collecting all the terms on the left side of the equality:
x a + y b ⇔ 1 a x + 1 b y - 1 = 0 ⇔ A x + B y + C = 0 y = k x + b ⇔ y - k x - b = 0 ⇔ A x + B y + C = 0
The canonical equation is converted to a general one according to the following scheme:
x - x 1 a x = y - y 1 a y ⇔ a y · (x - x 1) = a x (y - y 1) ⇔ ⇔ a y x - a x y - a y x 1 + a x y 1 = 0 ⇔ A x + B y + C =
To move from parametric ones, first move to the canonical one, and then to the general one:
x = x 1 + a x · λ y = y 1 + a y · λ ⇔ x - x 1 a x = y - y 1 a y ⇔ A x + B y + C = 0
Example 9
The parametric equations of the line x = - 1 + 2 · λ y = 4 are given. It is necessary to write down the general equation of this line.
Solution
Let us make the transition from parametric equations to canonical ones:
x = - 1 + 2 · λ y = 4 ⇔ x = - 1 + 2 · λ y = 4 + 0 · λ ⇔ λ = x + 1 2 λ = y - 4 0 ⇔ x + 1 2 = y - 4 0
Let's move from the canonical to the general:
x + 1 2 = y - 4 0 ⇔ 0 · (x + 1) = 2 (y - 4) ⇔ y - 4 = 0
Answer: y - 4 = 0
Example 10
The equation of a straight line in the segments x 3 + y 1 2 = 1 is given. It is necessary to make a transition to general appearance equations
Solution:
We simply rewrite the equation in the required form:
x 3 + y 1 2 = 1 ⇔ 1 3 x + 2 y - 1 = 0
Answer: 1 3 x + 2 y - 1 = 0 .
Drawing up a general equation of a line
We said above that the general equation can be written with known coordinates of the normal vector and the coordinates of the point through which the line passes. Such a straight line is defined by the equation A (x - x 0) + B (y - y 0) = 0. There we also analyzed the corresponding example.
Now let's look at more complex examples, in which first you need to determine the coordinates of the normal vector.
Example 11
Given a line parallel to the line 2 x - 3 y + 3 3 = 0. The point M 0 (4, 1) through which the given line passes is also known. It is necessary to write down the equation of the given line.
Solution
The initial conditions tell us that the lines are parallel, then, as the normal vector of the line, the equation of which needs to be written, we take the direction vector of the line n → = (2, - 3): 2 x - 3 y + 3 3 = 0. Now we know all the necessary data to create the general equation of the line:
A (x - x 0) + B (y - y 0) = 0 ⇔ 2 (x - 4) - 3 (y - 1) = 0 ⇔ 2 x - 3 y - 5 = 0
Answer: 2 x - 3 y - 5 = 0 .
Example 12
The given line passes through the origin perpendicular to the line x - 2 3 = y + 4 5. It is necessary to create a general equation for a given line.
Solution
The normal vector of a given line will be the direction vector of the line x - 2 3 = y + 4 5.
Then n → = (3, 5) . The straight line passes through the origin, i.e. through point O (0, 0). Let's create a general equation for a given line:
A (x - x 0) + B (y - y 0) = 0 ⇔ 3 (x - 0) + 5 (y - 0) = 0 ⇔ 3 x + 5 y = 0
Answer: 3 x + 5 y = 0 .
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General equation of a straight line:
Special cases of the general equation of a straight line:
and if C= 0, equation (2) will have the form
Ax + By = 0,
and the straight line defined by this equation passes through the origin, since the coordinates of the origin are x = 0, y= 0 satisfy this equation.
b) If in the general equation of the straight line (2) B= 0, then the equation takes the form
Ax + WITH= 0, or .
The equation does not contain a variable y, and the straight line defined by this equation is parallel to the axis Oy.
c) If in the general equation of the straight line (2) A= 0, then this equation will take the form
By + WITH= 0, or ;
the equation does not contain a variable x, and the straight line it defines is parallel to the axis Ox.
It should be remembered: if a straight line is parallel to some coordinate axis, then in its equation there is no term containing a coordinate of the same name as this axis.
d) When C= 0 and A= 0 equation (2) takes the form By= 0, or y = 0.
This is the equation of the axis Ox.
d) When C= 0 and B= 0 equation (2) will be written in the form Ax= 0 or x = 0.
This is the equation of the axis Oy.
The relative position of lines on a plane. The angle between straight lines on a plane. Condition for parallel lines. The condition of perpendicularity of lines.
l 1 l 2 l 1: A 1 x + B 1 y + C 1 = 0
l 2: A 2 x + B 2 y + C 2 = 0
S 2 S 1 Vectors S 1 and S 2 are called guides for their lines.
The angle between straight lines l 1 and l 2 is determined by the angle between the direction vectors.
Theorem 1: cos of the angle between l 1 and l 2 = cos(l 1 ; l 2) =
Theorem 2: In order for 2 lines to be equal it is necessary and sufficient:
Theorem 3: For 2 straight lines to be perpendicular it is necessary and sufficient:
L 1 l 2 ó A 1 A 2 + B 1 B 2 = 0
General plane equation and its special cases. Equation of a plane in segments.
General plane equation:
Ax + By + Cz + D = 0
Special cases:
1. D=0 Ax+By+Cz = 0 – the plane passes through the origin
2. С=0 Ax+By+D = 0 – plane || OZ
3. B=0 Ax+Cz+d = 0 – plane || OY
4. A=0 By+Cz+D = 0 – plane || OX
5. A=0 and D=0 By+Cz = 0 – the plane passes through OX
6. B=0 and D=0 Ax+Cz = 0 – the plane passes through OY
7. C=0 and D=0 Ax+By = 0 – the plane passes through OZ
The relative position of planes and straight lines in space:
1. The angle between straight lines in space is the angle between their direction vectors.
Cos (l 1 ; l 2) = cos(S 1 ; S 2) = = 
2. The angle between the planes is determined through the angle between their normal vectors.
Cos (l 1 ; l 2) = cos(N 1 ; N 2) = = 
3. The cosine of the angle between the line and the plane can be found through the sin of the angle between the direction vector of the line and the normal vector of the plane.
4. 2 straight || in space when their || vector guides
5. 2 planes || when || normal vectors
6. The concepts of perpendicularity of lines and planes are introduced similarly.
Question No. 14
Various types of equation of a straight line on a plane (equation of a straight line in segments, with an angle coefficient, etc.)
Equation of a straight line in segments:
Let us assume that in the general equation of the straight line:
1. C = 0 Ах + Ву = 0 – the straight line passes through the origin.
2. a = 0 Vu + C = 0 y =
3. b = 0 Ax + C = 0 x =
4. b=C=0 Ax = 0 x = 0
5. a=C=0 Ву = 0 у = 0
Equation of a straight line with a slope:
Any straight line that is not equal to the op-amp axis (B not = 0) can be written down in the next line. form:
k = tanα α – angle between straight line and positively directed line OX
b – point of intersection of the straight line with the axis of the op-amp
Document:
Ax+By+C = 0
Wu= -Ah-S |:B
Equation of a straight line based on two points:
Question No. 16
Finite limit of a function at a point and for x→∞
End limit at x0:
The number A is called the limit of the function y = f(x) for x→x 0 if for any E > 0 there exists b > 0 such that for x ≠x 0 satisfying the inequality |x – x 0 |< б, выполняется условие |f(x) - A| < Е
The limit is indicated by: = A
End limit at point +∞:
The number A is called the limit of the function y = f(x) at x → + ∞ , if for any E > 0 there exists C > 0 such that for x > C the inequality |f(x) - A|< Е
The limit is indicated by: = A
End limit at point -∞:
The number A is called the limit of the function y = f(x) for x→-∞, if for any E< 0 существует С < 0 такое, что при х < -С выполняется неравенство |f(x) - A| < Е
Properties of a straight line in Euclidean geometry.
An infinite number of straight lines can be drawn through any point.
Through any two non-coinciding points a single straight line can be drawn.
Two divergent lines in a plane either intersect at a single point or are
parallel (follows from the previous one).
In three-dimensional space there are three options relative position two straight lines:
- lines intersect;
- lines are parallel;
- straight lines intersect.
Straight line— algebraic curve of the first order: a straight line in the Cartesian coordinate system
is given on the plane by an equation of the first degree (linear equation).
General equation of a straight line.
Definition. Any straight line on the plane can be specified by a first-order equation
Ax + Wu + C = 0,
and constant A, 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 of the constants A, B And WITH The following special cases are possible:
. C = 0, A ≠0, B ≠ 0- a straight line passes through the origin
. A = 0, B ≠0, C ≠0 (By + C = 0)- straight line parallel to the axis Oh
. B = 0, A ≠0, C ≠ 0 (Ax + C = 0)- straight line parallel to the axis OU
. B = C = 0, A ≠0- the straight line coincides with the axis OU
. A = C = 0, B ≠0- the straight line coincides with the axis Oh
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 a normal vector.
Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)
perpendicular to the line given by the equation
Ax + Wu + C = 0.
Example. Find the equation of a line passing through a point A(1, 2) perpendicular to the vector (3, -1).
Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C
Let's 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 be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2, z 2), Then equation of a line,
passing through these points:
If any of the denominators is zero, the corresponding numerator should be set equal to zero. On
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 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 using a point and slope.
If the general equation of the line Ax + Wu + C = 0 lead to:
and designate 
, then the resulting equation is called
equation of a straight line with slope k.
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 the normal vector, you can enter the task
a straight line through a point and a directing vector of a straight line.
Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition
Aα 1 + Bα 2 = 0 called directing vector of a straight 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. According to the definition,
coefficients must satisfy the following 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.
at x = 1, y = 2 we get C/A = -3, i.e. required equation:
x + y - 3 = 0
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
Geometric meaning coefficients is that coefficient a is the coordinate of the intersection point
straight with axis Oh, A b- coordinate of the point of intersection of the line with the axis OU.
Example. The general equation of a straight line is given x - y + 1 = 0. 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 + Wu + C = 0 divide by 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.
R- the length of the perpendicular dropped from the origin to the straight line,
A 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 = 0. Required to write Various types equations
this straight line.
The equation of this line in segments:
The equation of this line with the slope: (divide by 5)
Equation of a line:
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.
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 straight lines are perpendicular,
If k 1 = -1/ k 2 .
Theorem.
Direct Ax + Wu + C = 0 And A 1 x + B 1 y + C 1 = 0 parallel when the coefficients are proportional
A 1 = λA, B 1 = λB. If also С 1 = λС, then the lines coincide. Coordinates of the point of intersection of two lines
are found as a solution to the system of equations of these lines.
The equation of a line passing through a given point perpendicular to a given line.
Definition. Line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b
represented by the equation:
Distance from a point to a line.
Theorem. If a point is given M(x 0, y 0), then the distance to the straight line Ax + Wu + C = 0 defined as:
Proof. Let the point M 1 (x 1, y 1)- the base of a perpendicular dropped from a point M for a given
direct. Then the distance between points M And 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 perpendicularly
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 = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem has been proven.