Find points of a straight line according to the equation. Equation of a line passing through a point, equation of a line passing through two points, angle between two lines, slope of a line
Properties of a straight line in Euclidean geometry.
There are infinitely many lines that can be drawn through any point.
Through any two non-coinciding points, there is only one straight line.
Two non-coincident lines in 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 lines:
- lines intersect;
- straight lines are parallel;
- straight lines intersect.
Straight line- algebraic curve of the first order: in the Cartesian coordinate system, a straight line
is given on the plane by an equation of the first degree (linear equation).
General equation of a straight line.
Definition. Any line in the plane can be given by a first order equation
Ah + Wu + C = 0,
and constant A, B not equal to zero at the same time. This first order equation is called general
straight line equation. Depending on the values of the constants A, B and WITH The following special cases are possible:
. C = 0, A ≠ 0, B ≠ 0- the 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 line coincides with the axis OU
. A = C = 0, B ≠ 0- the line coincides with the axis Oh
The equation of a straight line can be represented in various forms depending on any given
initial conditions.
Equation of a straight line by 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
Ah + Wu + C = 0.
Example. Find the equation of a straight line passing through a point A(1, 2) perpendicular to the vector (3, -1).
Solution. Let's compose at A \u003d 3 and B \u003d -1 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 = 0, therefore
C = -1. Total: the desired 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 equal to zero, the corresponding numerator should be set equal to zero. On the
plane, the equation of a straight line written above is simplified:
if x 1 ≠ x 2 and x = x 1, if x 1 = x 2 .
Fraction = k called slope factor straight.
Example. Find the equation of a straight line passing through the points A(1, 2) and B(3, 4).
Solution. Applying the above formula, we get:
Equation of a straight line by a point and a slope.
If the general equation of a straight line Ah + Wu + C = 0 bring to the form:
and designate 
, then the resulting equation is called
equation of a straight line with slope k.
The equation of a straight line on a point and a directing 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 direction 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 direction vector of the straight line.
Ah + Wu + C = 0.
Example. Find the equation of a straight line with direction vector (1, -1) and passing through point A(1, 2).
Solution. We will look for the equation of the desired straight line in the form: Ax + By + C = 0. According to the definition,
coefficients must satisfy the conditions:
1 * A + (-1) * B = 0, i.e. A = B.
Then the equation of a 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. desired equation:
x + y - 3 = 0
Equation of a straight line in segments.
If in general equation line Ax + Wu + 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 axle Oh, a b- the 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 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 Ah + Wu + C = 0 divide by number 
, which is called
normalizing factor, then we get
xcosφ + ysinφ - p = 0 -normal equation of a straight 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 line,
a φ - the angle formed by this perpendicular with the positive direction of the axis Oh.
Example. Given the general equation of a straight line 12x - 5y - 65 = 0. Required to write different types equations
this straight line.
The equation of this straight line in segments:
The equation of this line with slope: (divide by 5)
Equation of a straight 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.
Angle between lines on a plane.
Definition. If two lines are given y \u003d k 1 x + b 1, y \u003d 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 \u003d -1 / k 2 .
Theorem.
Direct Ah + Wu + C = 0 and A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients are proportional
A 1 \u003d λA, B 1 \u003d λB. If also С 1 \u003d 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 is perpendicular to a given line.
Definition. A line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b
represented by the equation:
The distance from a point to a line.
Theorem. If a point is given M(x 0, y 0), then the distance to the line Ah + Wu + C = 0 defined as:
Proof. Let the point M 1 (x 1, y 1)- the base of the perpendicular dropped from the point M for a given
direct. Then the distance between the points M and M 1:
(1)
Coordinates x 1 and 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 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.
This article continues the topic of the equation of a straight line on a plane: we will consider such a type of equation as the general equation of a straight line. Let's define a theorem and give its proof; Let's figure out what an incomplete general equation of a straight line is and how to make transitions from a general equation to other types of equations of a straight line. We will consolidate the whole theory with illustrations and solving practical problems.
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Let a rectangular coordinate system O x y be given on the plane.
Theorem 1
Any equation of the first degree, having the form A x + B y + C \u003d 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 a rectangular coordinate system on a plane. In turn, any line in a rectangular coordinate system on the 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 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 \u003d 0 the left and right sides of the equation A x 0 + B y 0 + C \u003d 0, we get 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 in a rectangular coordinate system a straight line 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.
Therefore, the equation A (x - x 0) + B (y - y 0) \u003d 0 defines a certain line in a rectangular coordinate system on the plane, and therefore the equivalent equation A x + B y + C \u003d 0 defines the same line. Thus we have proved the first part of the theorem.
- Let us prove that any straight line in a rectangular coordinate system on a plane can be given by an equation of the first degree A x + B y + C = 0 .
Let's set a straight line a in a rectangular coordinate system on the plane; 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 exist some point M (x , y) - a floating point of the 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 is 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 finally get the equation A x + B y + C = 0 .
So, we have proved the second part of the theorem, and we have proved the whole theorem as a whole.
Definition 1
An equation that looks like A x + B y + C = 0 - it general equation of a straight line on a plane in a rectangular coordinate systemO x y .
Based on the proved theorem, we can conclude that a straight line given on a plane in a fixed rectangular coordinate system and its general equation are inextricably linked. In other words, the original line corresponds to its general equation; the general equation of a straight line corresponds to a given straight line.
It also follows from the proof of the theorem that the coefficients A and B for the variables x and y are the coordinates of the normal vector of the straight line, which is given by the general equation of the straight line A x + B y + C = 0 .
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) . Draw a given straight line in the drawing.
The following can also be argued: 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 of a given straight line correspond to this equation.
We can get the equation λ A x + λ B y + λ C = 0 by multiplying both sides of the general straight line equation by the number λ, not zero. The resulting equation is equivalent to the original general equation, therefore, it will describe the same line in the plane.
Definition 2Complete general equation of a straight line- such a general equation of the line A x + B y + C \u003d 0, in which the numbers A, B, C are non-zero. Otherwise, the equation is incomplete.
Let us analyze all variations of the incomplete general equation of the straight line.
- When A \u003d 0, B ≠ 0, C ≠ 0, the general equation becomes B y + C \u003d 0. Such an incomplete general equation defines a straight line in a rectangular coordinate system O x y that is parallel to the O x axis, since for any real value of x, the variable y will take on the value - C B . In other words, the general equation of the line A x + B y + C \u003d 0, when A \u003d 0, B ≠ 0, defines the locus of points (x, y) whose coordinates are equal to the same number - C B .
- If A \u003d 0, B ≠ 0, C \u003d 0, the general equation becomes y \u003d 0. Such an incomplete equation defines the x-axis O x .
- When A ≠ 0, B \u003d 0, C ≠ 0, we get an incomplete general equation A x + C \u003d 0, defining a straight line parallel to the y-axis.
- Let A ≠ 0, B \u003d 0, C \u003d 0, then the incomplete general equation will take the form x \u003d 0, and this is the equation of the coordinate line O y.
- Finally, when A ≠ 0, B ≠ 0, C \u003d 0, the incomplete general equation takes the form A x + B y \u003d 0. And this equation describes a straight line that passes through the origin. Indeed, 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 the above types of the incomplete general equation of a straight line.
Example 1
It is known that the given straight line is parallel to the y-axis and passes through the point 2 7 , - 11 . It is necessary to write down the general equation of a given straight line.
Solution
A straight line parallel to the y-axis is given by an equation of the form A x + C \u003d 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 correspond to the conditions of the incomplete general equation A x + C = 0 , i.e. equality is correct:
A 2 7 + C = 0
It is possible to determine C from it by giving A some non-zero value, for example, A = 7 . In this case, we get: 7 2 7 + C \u003d 0 ⇔ C \u003d - 2. We know both coefficients A and C, substitute them into the equation A x + C = 0 and get the required equation of the line: 7 x - 2 = 0
Answer: 7 x - 2 = 0
Example 2
The drawing shows a straight line, it is necessary to write down its equation.
Solution
The given drawing allows us to easily take the initial data for solving the problem. We see in the drawing that the given 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 + С = 0. Find the values of B and C . The coordinates of the point (0, 3), since the given straight line passes through it, will satisfy the equation of the straight line B y + С = 0, then the equality is valid: В · 3 + С = 0. Let's set B to some value other than zero. Let's say B \u003d 1, in this case, from the equality B · 3 + C \u003d 0 we can find C: C \u003d - 3. Using the known values of B and C, we obtain the required equation of the straight line: y - 3 = 0.
Answer: y - 3 = 0 .
General equation of a straight line passing through a given point of the 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 . 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 \u003d 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 → \u003d (A, B) .
The result that we have obtained makes it possible to write the general equation of a straight line for known coordinates of the normal vector of the straight line and the coordinates of a certain point of this straight line.
Example 3
Given a point M 0 (- 3, 4) through which the line passes, and the normal vector of this line n → = (1 , - 2) . It is necessary to write down the equation of a given straight line.
Solution
The initial conditions allow us to obtain the necessary data for compiling the equation: A \u003d 1, B \u003d - 2, x 0 \u003d - 3, y 0 \u003d 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 has the form A x + B y + C = 0 . The given normal vector allows you to get the values of the 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) given by the condition of the problem, through which the line passes. The coordinates of this point correspond to the equation x - 2 · y + C = 0 , i.e. - 3 - 2 4 + C \u003d 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 the given point.
Solution
Let's set the designation of the coordinates of the point M 0 as x 0 and y 0 . The initial data indicates that x 0 \u003d - 3. Since the point belongs to a given line, then its coordinates correspond to the general equation of this line. Then the following 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 straight line to other types of equations of a straight line and vice versa
As we know, there are several types of the equation of the same straight line in the 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 its solution. This is where the skill of converting an equation of one kind into an equation of another kind comes in very handy.
First, 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 transfer the term B y to right side 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, we transfer the others to the right side, we get: A x \u003d - B y - C. We take out - B out of brackets, then: A x \u003d - B y + C B.
Let's rewrite the equality as 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 during the transition from the general equation to the canonical one.
Example 5
The general equation of the line 3 y - 4 = 0 is given. You need to convert it to canonical equation.
Solution
We write the original equation as 3 y - 4 = 0 . Next, we act according to the algorithm: the term 0 x remains on the left side; and on the right side we take out - 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 the canonical form.
Answer: x - 3 = y - 4 3 0.
To transform the general equation of a straight line into parametric ones, first, the transition to the canonical form is carried out, and then the transition from the canonical equation of the straight line to parametric equations.
Example 6
The straight line is given by the equation 2 x - 5 y - 1 = 0 . Write down the parametric equations of this line.
Solution
Let's 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 let's take both parts 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 to the equation of a straight line with slope y \u003d k x + b, but only when B ≠ 0. For the transition on the left side, we leave the term B y , the rest are transferred to the right. We get: B y = - A x - C . Let's divide both parts of the resulting equality by B , which is different from zero: y = - A B x - C B .
Example 7
The general equation of a straight line is given: 2 x + 7 y = 0 . You need to convert that equation to 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 straight line, it is enough to simply obtain an equation in segments of the form x a + y b \u003d 1. To make such a transition, we transfer the number C to the right side of the equality, divide both parts of the resulting equality by - С 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 convert the general equation of the straight line x - 7 y + 1 2 = 0 into the equation of a straight line in segments.
Solution
Let's move 1 2 to the right side: x - 7 y + 1 2 = 0 ⇔ x - 7 y = - 1 2 .
Divide by -1/2 both sides of the equation: 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 straight line in segments and the equation with a slope can be easily converted into a general one by simply collecting all the terms on the left side of the equation:
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 the general one according to the following scheme:
x - x 1 ax = y - y 1 ay ⇔ ay (x - x 1) = ax (y - y 1) ⇔ ⇔ ayx - axy - ayx 1 + axy 1 = 0 ⇔ A x + B y + C = 0
To pass from the parametric, first the transition to the canonical is carried out, 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 straight line x = - 1 + 2 · λ y = 4 are given. It is necessary to write down the general equation of this line.
Solution
Let's make the transition from parametric equations to canonical:
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 canonical to 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 segments x 3 + y 1 2 = 1 is given. It is necessary to carry out the transition to the general form of the equation.
Solution:
Let's just 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 straight line
Above, we said 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 . In the same place we analyzed the corresponding example.
Now let's look at more complex examples in which, first, it is necessary to determine the coordinates of the normal vector.
Example 11
Given a line parallel to the line 2 x - 3 y + 3 3 = 0 . Also known is the point M 0 (4 , 1) through which the given line passes. It is necessary to write down the equation of a given straight line.
Solution
The initial conditions tell us that the lines are parallel, then, as the normal vector of the line whose equation needs to be written, we take the directing vector of the line n → = (2, - 3) : 2 x - 3 y + 3 3 = 0. Now we know all the necessary data to compose the general equation of a straight 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 write the general equation of a given straight line.
Solution
The normal vector of the given line will be the directing vector of the line x - 2 3 = y + 4 5 .
Then n → = (3 , 5) . The straight line passes through the origin, i.e. through the point O (0, 0) . Let's compose the general equation of a given straight 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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In this article, we will consider the general equation of a straight line in a plane. Let us give examples of constructing the general equation of a straight line if two points of this straight line are known or if one point and the normal vector of this straight line are known. Let us present methods for converting an equation into general view into canonical and parametric forms.
Let an arbitrary Cartesian rectangular coordinate system be given Oxy. Consider a first degree equation or linear equation:
| Ax+By+C=0, | (1) |
where A, B, C are some constants, and at least one of the elements A and B different from zero.
We will show that a linear equation in the 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 given by a linear equation. Conversely, each linear equation (1) in an arbitrary Cartesian rectangular coordinate system on the plane defines a straight line.
Proof. It suffices to prove that the 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 and for any choice of Cartesian rectangular coordinate system.
Let a straight line be given on the plane L. We choose a coordinate system so that the axis Ox aligned with the 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 the linear equation (2), and all points outside this straight line will not satisfy the equation (2). The first part of the theorem is proved.
Let a Cartesian rectangular coordinate system be given and let linear equation (1) be given, where at least one of the elements A and B different from zero. Find the 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, dot M 0 (−C/A, 0) belongs to the given 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 suffices to prove that (4) defines some line.
Since we are considering a Cartesian rectangular coordinate system, it follows from equality (4) that the vector with components ( x−x 0 , y−y 0 ) is orthogonal to the vector n with coordinates ( A,B}.
Consider some 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 equals zero). Conversely, if the 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 vector n and equation (4) is not satisfied. The theorem has been proven.
Proof. Since lines (5) and (6) define the same line, the normal vectors n 1 ={A 1 ,B 1 ) and n 2 ={A 2 ,B 2) are collinear. Since the vectors n 1 ≠0, n 2 ≠ 0, then there is a number λ , what n 2 =n 1 λ . Hence we have: A 2 =A 1 λ , B 2 =B 1 λ . Let's prove that C 2 =C 1 λ . It is obvious that the coinciding lines have 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 a 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 the 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 line passes through a point M=(4,−1) and has a normal vector n=(3, 5). Construct the general equation of a straight 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:
Vector parallel to line L and hence is perpendicular to the normal vector of the line L. Let's construct a normal line vector L, given that the scalar product of vectors n and is 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 into (4) the coordinates of the point M 1 (we can also take the coordinates of the point M 2) and the normal vector n:
Substituting point coordinates 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 a straight line. Vector q={−B, A) is the direction vector of the straight line (12).
See reverse transformation.
Example 3. A straight line in a plane is represented by the following general equation:
Move the second term to the right and divide both sides of the equation by 2 5.
General equation of a straight line:
Particular 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 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 .
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 takes the form
By + WITH= 0, or ;
the equation does not contain a variable x, and the straight line defined by it is parallel to the axis Ox.
It should be remembered: if a straight line is parallel to any coordinate axis, then its equation does not contain a term containing a coordinate of the same name with this axis.
d) When C= 0 and A= 0 equation (2) takes the form By= 0, or y = 0.
This is the axis equation Ox.
e) When C= 0 and B= 0 equation (2) can be written in the form Ax= 0 or x = 0.
This is the axis equation Oy.
Mutual arrangement of straight lines on a plane. Angle between lines on a plane. Condition of 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 The vectors S 1 and S 2 are called guides for their lines.
The angle between the lines l 1 and l 2 is determined by the angle between the direction vectors.
Theorem 1: cos angle between l 1 and l 2 \u003d cos (l 1; l 2) \u003d
Theorem 2: In order for 2 lines to be equal, it is necessary and sufficient:
Theorem 3: so that 2 lines are perpendicular is necessary and sufficient:
L 1 l 2 ó A 1 A 2 + B 1 B 2 = 0
General equation of the plane 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. В=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
Mutual arrangement of planes and straight lines in space:
1. The angle between 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 a line and a 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 lines || 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 #14
Different kinds equations of a straight line on a plane (the equation of a straight line in segments, with a slope, etc.)
Equation of a straight line in segments:
Suppose that in the general equation of a straight line:
1. C \u003d 0 Ah + Wu \u003d 0 - the straight line passes through the origin.
2. a \u003d 0 Wu + C \u003d 0 y \u003d
3. in \u003d 0 Ax + C \u003d 0 x \u003d
4. v \u003d C \u003d 0 Ax \u003d 0 x \u003d 0
5. a \u003d C \u003d 0 Wu \u003d 0 y \u003d 0
The equation of a straight line with a slope:
Any straight line that is not equal to the y-axis (B not = 0) can be written in the following. form:
k = tgα α is the angle between the straight line and the positively directed line ОХ
b - point of intersection of the straight line with the OS axis
Doc-in:
Ax+By+C = 0
Wu \u003d -Ax-C |: B
Equation of a straight line on two points:
Question #16
The finite limit of a function at a point and for x→∞
End limit at point x 0:
The number A is called the limit of the function y \u003d f (x) for x → x 0, if for any E > 0 there is b > 0 such that for x ≠ x 0, satisfying the inequality |x - x 0 |< б, выполняется условие |f(x) - A| < Е
The limit is denoted: = 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 there exists C > 0 such that for x > C the inequality |f(x) - A|< Е
The limit is denoted: = 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
Equation of a straight line passing through two points. In the article" " I promised you to analyze the second way to solve the presented problems for finding the derivative, with a given function graph and a tangent to this graph. We will explore this method in , do not miss! Why next?
The fact is that the formula of the equation of a straight line will be used there. Of course, one could simply show this 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 itwill 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 direct formula:
*That is, when substituting the specific coordinates of the points, we get an equation of the form y=kx+b.
** If this formula is simply “memorized”, then there is a high probability of getting confused with indices when X. In addition, indexes can be denoted in different ways, for example:
That is why it is important to understand the meaning.
Now the derivation of this formula. Everything is very simple!
Triangles ABE and ACF are similar in sharp corner(the first sign of similarity right triangles). It follows from this that the ratios of the corresponding elements are equal, that is:
Now we simply express these segments in terms of the difference in the coordinates of the points:
Of course, there will be no error if you write the relationships of the elements in a different order (the main thing is to keep the correspondence):
The result is the same equation of a 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 deduced 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 similarity of right triangles works. In my opinion, the conclusion described above is more understandable)).
View output via vector coordinates >>>
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 us mark an arbitrary point C on the line with coordinates ( x; y). We also denote two vectors:
It is known that for vectors lying on parallel lines (or on one line), their corresponding coordinates are proportional, that is:
- we write the equality of the ratios of the corresponding coordinates:
Consider an example:
Find the equation of a straight line passing through two points with coordinates (2;5) and (7:3).
You can not even build the 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:
Answer: y=-2/5x+29/5 go y=-0.4x+5.8
In order to make sure that the resulting equation is found correctly, be sure to check it - substitute the data coordinates into it in the condition of the points. You should get correct equalities.
That's all. I hope the material was useful to you.
Sincerely, Alexander.
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