Find points of a straight line by equation. Equation of a straight line passing through a point, equation of a straight line passing through two points, the angle between two straight lines, the slope of a straight line
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 represented 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 in the resulting expression the coordinates of the given point A. 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
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 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 continues the topic of the equation of a straight line on a plane: consider such a form of equation as the general equation of a straight line. Let us 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 straight line in a rectangular coordinate system on a plane is determined by an equation that has the form A x + B y + C \u003d 0 for a certain set of values \u200b\u200bA, B, C.
Evidence
this theorem consists of two points, we will prove each of them.
- Let us prove that the equation A x + B y + C \u003d 0 defines a straight line on the plane.
Let there exist some point М 0 (x 0, y 0), the coordinates of which correspond to the equation A x + B y + C \u003d 0. Thus: A x 0 + B y 0 + C \u003d 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 obtain a new equation that has the form A (x - x 0) + B (y - y 0) \u003d 0. It is equivalent to A x + B y + C \u003d 0.
The resulting equation A (x - x 0) + B (y - y 0) \u003d 0 is a necessary and sufficient condition for the vectors n → \u003d (A, B) and M 0 M → \u003d (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 → \u003d (A, B). We can assume that this is not so, but then the vectors n → \u003d (A, B) and M 0 M → \u003d (x - x 0, y - y 0) would not be perpendicular, and the equality A (x - x 0 ) + B (y - y 0) \u003d 0 would not be true.
Therefore, the equation A (x - x 0) + B (y - y 0) \u003d 0 defines some straight line in a rectangular coordinate system on the plane, and hence the equivalent equation A x + B y + C \u003d 0 defines the same straight line. This is how we proved the first part of the theorem.
- Let us give a proof that any straight line in a rectangular coordinate system on a plane can be defined by an equation of the first degree A x + B y + C \u003d 0.
Let us set the 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 → \u003d (A, B).
Let there also be some point M (x, y) - a floating point of a straight line. In this case, the vectors n → \u003d (A, B) and M 0 M → \u003d (x - x 0, y - y 0) are perpendicular to each other, and their scalar product is zero:
n →, M 0 M → \u003d A (x - x 0) + B (y - y 0) \u003d 0
Rewrite the equation A x + B y - A x 0 - B y 0 \u003d 0, define C: C \u003d - A x 0 - B y 0 and in the end result we get the equation A x + B y + C \u003d 0.
Thus, we have proved the second part of the theorem and proved the whole theorem as a whole.
Definition 1
An equation of the form A x + B y + C \u003d 0 - this is general equation of the line on a plane in a rectangular coordinate system O x y.
Based on the proven theorem, we can conclude that a straight line and its general equation given on a plane in a fixed rectangular coordinate system are inextricably linked. In other words, the initial straight 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 line, which is given by the general equation of the line A x + B y + C \u003d 0.
Consider a specific example of a general equation of a straight line.
Let the equation 2 x + 3 y - 2 \u003d 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 → \u003d (2, 3). Draw a given straight line in the drawing.
You can also say the following: the straight line that we see in the drawing is determined by the general equation 2 x + 3 y - 2 \u003d 0, since the coordinates of all points of a given straight line correspond to this equation.
We can obtain the equation λ · A x + λ · B y + λ · C \u003d 0 by multiplying both sides of the general equation of the line by a number λ that is 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 the line - such a general equation of the straight line A x + B y + C \u003d 0, in which the numbers A, B, C are different from zero. Otherwise the equation is incomplete.
Let us examine all the variations of the incomplete general equation of the 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 the value - C B. In other words, the general equation of the straight line A x + B y + C \u003d 0, when A \u003d 0, B ≠ 0, specifies 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 takes the form y \u003d 0. This incomplete equation defines the abscissa axis O x.
- When A ≠ 0, B \u003d 0, C ≠ 0, we obtain an incomplete general equation A x + C \u003d 0, defining a straight line parallel to the ordinate 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, for 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 \u003d 0, since A · 0 + B · 0 \u003d 0.
Let us graphically illustrate all the above types of the incomplete general equation of a straight line.
Example 1
It is known that a 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 a given straight line.
Decision
A straight line parallel to the ordinate axis is given by an equation of the form A x + C \u003d 0, in which A ≠ 0. Also, the condition 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 \u003d 0, i.e. the equality is true:
A · 2 7 + C \u003d 0
It is possible to determine C from it by giving A some non-zero value, for example, A \u003d 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 \u003d 0 and obtain the required equation of the straight line: 7 x - 2 \u003d 0
Answer: 7 x - 2 \u003d 0
Example 2
The drawing shows a straight line, it is necessary to write down its equation.
Decision
The above 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 eyes of the abscissas, determines the incomplete general equation B y + C \u003d 0. Let's find the values \u200b\u200bof B and C. The coordinates of the point (0, 3), since a given straight line passes through it, will satisfy the equation of the straight line B y + C \u003d 0, then the equality is valid: B · 3 + C \u003d 0. Let's set for B some value other than zero. Suppose B \u003d 1, in this case, from the equality B 3 + C \u003d 0 we can find C: C \u003d - 3. We use the known values \u200b\u200bof B and C, we obtain the required equation of the straight line: y - 3 \u003d 0.
Answer: y - 3 \u003d 0.
General equation of a straight line passing through a given point of the plane
Let the given line pass through the point М 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 \u003d 0. We subtract the left and right sides of this equation from the left and right sides of the general complete equation of the line. We get: A (x - x 0) + B (y - y 0) + C \u003d 0, this equation is equivalent to the original general, passes through the point М 0 (x 0, y 0) and has a normal vector n → \u003d (A, B).
The result we have obtained makes it possible to write down the general equation of a straight line with the 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 М 0 (- 3, 4), through which a straight line passes, and a normal vector of this straight line n → \u003d (1, - 2). It is necessary to write down the equation of a given straight line.
Decision
The initial conditions allow us to obtain the necessary data for drawing up 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) \u003d 0 ⇔ 1 (x - (- 3)) - 2 y (y - 4) \u003d 0 ⇔ ⇔ x - 2 y + 22 \u003d 0
The problem could have been solved differently. The general equation of the line is A x + B y + C \u003d 0. A given normal vector allows you to get the values \u200b\u200bof the coefficients A and B, then:
A x + B y + C \u003d 0 ⇔ 1 x - 2 y + C \u003d 0 ⇔ x - 2 y + C \u003d 0
Now we find the value of C using the point M 0 (- 3, 4) specified by the condition of the problem, through which the straight line passes. The coordinates of this point correspond to the equation x - 2 y + C \u003d 0, i.e. - 3 - 2 4 + C \u003d 0. Hence C \u003d 11. The required equation of the straight line takes the form: x - 2 y + 11 \u003d 0.
Answer: x - 2 y + 11 \u003d 0.
Example 4
A straight line 2 3 x - y - 1 2 \u003d 0 and a point М 0 lying on this straight line are given. 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.
Decision
Let's set the designation of the coordinates of the point М 0 as x 0 and y 0. The initial data indicates that x 0 \u003d - 3. Since a point belongs to a given straight line, then its coordinates correspond to the general equation of this straight line. Then the equality will be true:
2 3 x 0 - y 0 - 1 2 \u003d 0
Determine y 0: 2 3 (- 3) - y 0 - 1 2 \u003d 0 ⇔ - 5 2 - y 0 \u003d 0 ⇔ y 0 \u003d - 5 2
Answer: - 5 2
The 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 equations for the same straight line on 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 solving it. This is where the skill of converting an equation of one kind into an equation of another kind comes in handy.
To begin with, consider the transition from the general equation of the form A x + B y + C \u003d 0 to the canonical equation x - x 1 a x \u003d y - y 1 a y.
If А ≠ 0, then we transfer the term B y to the right-hand side of the general equation. On the left side, place A outside the brackets. As a result, we get: A x + C A \u003d - B y.
This equality can be written as a proportion: x + C A - B \u003d y A.
If В ≠ 0, we leave only the term A x on the left side of the general equation, transfer the others to the right side, we obtain: A x \u003d - B y - C. We take out - B outside the brackets, then: A x \u003d - B y + C B.
Let's rewrite equality as a proportion: x - B \u003d y + C B A.
Of course, there is no need to memorize the resulting formulas. It is enough to know the algorithm of actions in the transition from the general equation to the canonical one.
Example 5
The general equation of the straight line is given: 3 y - 4 \u003d 0. It is necessary to transform it into a canonical equation.
Decision
Rewrite the original equation as 3 y - 4 \u003d 0. Next, we act according to the algorithm: on the left side there remains the term 0 x; and on the right side we take out - 3 outside the brackets; we get: 0 x \u003d - 3 y - 4 3.
Let's write the resulting equality as a proportion: x - 3 \u003d y - 4 3 0. So, we got an equation of the canonical form.
Answer: x - 3 \u003d y - 4 3 0.
To transform the general equation of the straight line into parametric ones, one first makes the transition to the canonical form, and then the transition from the canonical equation of the straight line to the parametric equations.
Example 6
The straight line is given by the equation 2 x - 5 y - 1 \u003d 0. Write down the parametric equations of this straight line.
Decision
Let's make the transition from the general equation to the canonical one:
2 x - 5 y - 1 \u003d 0 ⇔ 2 x \u003d 5 y + 1 ⇔ 2 x \u003d 5 y + 1 5 ⇔ x 5 \u003d y + 1 5 2
Now we take both sides of the resulting canonical equation equal to λ, then:
x 5 \u003d λ y + 1 5 2 \u003d λ ⇔ x \u003d 5 λ y \u003d - 1 5 + 2 λ, λ ∈ R
Answer: x \u003d 5 λ y \u003d - 1 5 + 2 λ, λ ∈ R
The general equation can be transformed into an equation of a straight line with a slope y \u003d k x + b, but only if B ≠ 0. For the transition on the left, we leave the term B y, the rest are transferred to the right. We get: B y \u003d - A x - C. Divide both sides of the resulting equality by B, different from zero: y \u003d - A B x - C B.
Example 7
The general equation of the straight line is given: 2 x + 7 y \u003d 0. You must convert that equation to a slope equation.
Decision
Let's perform the necessary actions according to the algorithm:
2 x + 7 y \u003d 0 ⇔ 7 y - 2 x ⇔ y \u003d - 2 7 x
Answer: y \u003d - 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-hand side of the equality, divide both sides 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 \u003d 0 ⇔ A x + B y \u003d - C ⇔ ⇔ A - C x + B - C y \u003d 1 ⇔ x - C A + y - C B \u003d 1
Example 8
It is necessary to transform the general equation of the line x - 7 y + 1 2 \u003d 0 into the equation of the line in segments.
Decision
Move 1 2 to the right side: x - 7 y + 1 2 \u003d 0 ⇔ x - 7 y \u003d - 1 2.
Divide both sides of the equality by -1/2: x - 7 y \u003d - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y \u003d 1.
Answer: x - 1 2 + y 1 14 \u003d 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 an equation with a slope coefficient can be easily transformed into a general one, simply by collecting all the terms on the left side of the equality:
x a + y b ⇔ 1 a x + 1 b y - 1 \u003d 0 ⇔ A x + B y + C \u003d 0 y \u003d k x + b ⇔ y - k x - b \u003d 0 ⇔ A x + B y + C \u003d 0
The canonical equation is transformed to the general one as follows:
x - x 1 ax \u003d y - y 1 ay ⇔ ay (x - x 1) \u003d ax (y - y 1) ⇔ ⇔ ayx - axy - ayx 1 + axy 1 \u003d 0 ⇔ A x + B y + C \u003d 0
To move from parametric, first, the transition to the canonical is carried out, and then to the general:
x \u003d x 1 + a x λ y \u003d y 1 + a y λ ⇔ x - x 1 a x \u003d y - y 1 a y ⇔ A x + B y + C \u003d 0
Example 9
Parametric equations of the straight line x \u003d - 1 + 2 · λ y \u003d 4 are given. It is necessary to write down the general equation of this straight line.
Decision
Let's make the transition from parametric equations to canonical:
x \u003d - 1 + 2 λ y \u003d 4 ⇔ x \u003d - 1 + 2 λ y \u003d 4 + 0 λ ⇔ λ \u003d x + 1 2 λ \u003d y - 4 0 ⇔ x + 1 2 \u003d y - 4 0
Let's move from the canonical to the general:
x + 1 2 \u003d y - 4 0 ⇔ 0 (x + 1) \u003d 2 (y - 4) ⇔ y - 4 \u003d 0
Answer: y - 4 \u003d 0
Example 10
The equation of a straight line in segments x 3 + y 1 2 \u003d 1 is given. It is necessary to make a transition to the general form of the equation.
Decision:
Let's just rewrite the equation as needed:
x 3 + y 1 2 \u003d 1 ⇔ 1 3 x + 2 y - 1 \u003d 0
Answer: 1 3 x + 2 y - 1 \u003d 0.
Drawing up the general equation of a straight line
Above, we said that the general equation can be written with the known coordinates of the normal vector and the coordinates of the point through which the straight line passes. Such a straight line is determined by the equation A (x - x 0) + B (y - y 0) \u003d 0. We also analyzed the corresponding example there.
Now we will consider more complex examples, in which first it is necessary to determine the coordinates of the normal vector.
Example 11
A straight line parallel to the straight line 2 x - 3 y + 3 3 \u003d 0 is given. 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.
Decision
The initial conditions tell us that the straight lines are parallel, then, as the normal vector of the straight line, the equation of which is to be written, we take the directing vector of the straight line n → \u003d (2, - 3): 2 x - 3 y + 3 3 \u003d 0. Now we know all the necessary data to compose the general equation of the line:
A (x - x 0) + B (y - y 0) \u003d 0 ⇔ 2 (x - 4) - 3 (y - 1) \u003d 0 ⇔ 2 x - 3 y - 5 \u003d 0
Answer: 2 x - 3 y - 5 \u003d 0.
Example 12
The specified line passes through the origin perpendicular to the line x - 2 3 \u003d y + 4 5. It is necessary to draw up a general equation for a given straight line.
Decision
The normal vector of the given line will be the direction vector of the line x - 2 3 \u003d y + 4 5.
Then n → \u003d (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) \u003d 0 ⇔ 3 (x - 0) + 5 (y - 0) \u003d 0 ⇔ 3 x + 5 y \u003d 0
Answer: 3 x + 5 y \u003d 0.
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In this article, we will consider the general equation of a straight line on a plane. Let us give examples of constructing a general equation of a 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 the methods for transforming the equation in general form into canonical and parametric forms.
Let an arbitrary Cartesian rectangular coordinate system be given Oxy... Consider an equation of the first degree or a linear equation:
| Ax + By + C=0, | (1) |
where A, B, C - some constants, and at least one of the elements A and B nonzero.
We will show that a linear equation in a plane defines a line. Let us prove the following theorem.
Theorem 1. In an arbitrary Cartesian rectangular coordinate system on a plane, each straight line can be specified by a linear equation. Conversely, each linear equation (1) in an arbitrary Cartesian rectangular coordinate system on a plane defines a straight line.
Evidence. It is enough 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 a Cartesian rectangular coordinate system.
Let a straight line be given on the plane L... Let us choose a coordinate system so that the axis Ox coincided with a straight line Land the axis Oy was perpendicular to it. Then the equation of the line L will take the following form:
| y \u003d 0. | (2) |
All points on a straight line L will satisfy linear equation (2), and all points outside this straight line will not satisfy equation (2). The first part of the theorem is proved.
Let a Cartesian rectangular coordinate system be given and let a linear equation (1) be given, where at least one of the elements A and B nonzero. Let us find the locus of points whose coordinates satisfy equation (1). Since at least one of the coefficients A and B is nonzero, then equation (1) has at least one solution M(x 0 ,y 0). (For example, for A≠ 0, point M 0 (−C / A, 0) belongs to the given locus of points). Substituting these coordinates in (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 a vector with components ( x − x 0 , y − y 0) is orthogonal to the vector n with coordinates ( A, B}.
Consider some straight line Lpassing 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 straight line L... Then the vector with coordinates x − x 0 , y − y 0 is perpendicular n and equation (4) is satisfied (scalar product of vectors n and is equal to zero). Back if point M(x, y) does not lie on the straight 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 is proved.
Evidence. 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 vectors n 1 ≠0, n 2 ≠ 0, then there exists a number λ , what n 2 =n 1 λ ... Hence we have: A 2 =A 1 λ , B 2 =B 1 λ ... Let us prove that C 2 =C 1 λ ... Obviously, the coinciding lines have a common point M 0 (x 0 , y 0). Multiplying equation (5) by λ and subtracting from it equation (6) we get:
Since the first two equalities from expressions (7) hold, then C 1 λ −C 2 \u003d 0. Those. C 2 =C 1 λ ... The remark is proven.
Note that equation (4) defines the equation of the straight line passing through the point M 0 (x 0 , y 0) and having a normal vector n={A, B). Therefore, if the normal vector of a straight line and a point belonging to this straight line are known, then the general equation of the straight line can be constructed using equation (4).
Example 1. A straight line passes through a point M\u003d (4, −1) and has a normal vector n\u003d (3, 5). Construct the general equation of the line.
Decision. We have: x 0 =4, y 0 =−1, A=3, B\u003d 5. To construct a general equation of a straight line, we substitute these values \u200b\u200binto equation (4):
Answer:
Vector parallel to the straight line L and, therefore, perdicular to the normal vector of the straight line L... We construct a normal vector of a straight line L, taking into account that the scalar product of vectors n and is equal to zero. We can write down, for example, n={1,−3}.
To construct the general equation of a straight line, we will use formula (4). Substitute in (4) the coordinates of the point M 1 (we can also take the coordinates of the point M 2) and normal vector n:
Substituting the coordinates of the points M 1 and M 2 in (9), we can make sure that the straight line given by equation (9) passes through these points.
Answer:
Subtract (10) from (1):
We got the canonical equation of the line. Vector q={−B, A) is the directing vector of the straight line (12).
See reverse transformation.
Example 3. A straight line on 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.
The general equation of the straight line:
Particular cases of the general equation of the straight line:
what if C \u003d 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 \u003d 0 satisfy this equation.
b) If in the general equation of the straight line (2) B \u003d 0, then the equation takes the form
Ax + FROM \u003d 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 \u003d 0, then this equation takes the form
By + FROM \u003d 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 any coordinate axis, then in its equation there is no term containing the coordinate of the same name with this axis.
d) When C \u003d 0 and A \u003d 0 equation (2) takes the form By \u003d 0, or y = 0.
This is the axis equation Ox.
e) When C \u003d 0 and B \u003d 0, equation (2) can be written as Ax \u003d 0 or x = 0.
This is the axis equation Oy.
Mutual arrangement of straight lines on a plane. The angle between straight lines on the plane. The condition for parallelism of lines. Condition of perpendicularity of straight lines.
l 1 l 2 l 1: A 1 x + B 1 y + C 1 \u003d 0
l 2: A 2 x + B 2 y + C 2 \u003d 0
S 2 S 1 Vectors S 1 and S 2 are called guides for their lines.
The angle between the straight 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 straight lines to be equal, it is necessary and sufficient:
Theorem 3:so that 2 straight lines are perpendicular it is necessary and sufficient:
L 1 l 2 ó A 1 A 2 + B 1 B 2 \u003d 0
General equation of the plane and its special cases. Equation of the plane in segments.
General plane equation:
Ax + By + Cz + D \u003d 0
Special cases:
1.D \u003d 0 Ax + By + Cz \u003d 0 - the plane passes through the origin
2.C \u003d 0 Ax + By + D \u003d 0 - plane || OZ
3. В \u003d 0 Ax + Cz + d \u003d 0 - plane || OY
4. A \u003d 0 By + Cz + D \u003d 0 - plane || OX
5.A \u003d 0 and D \u003d 0 By + Cz \u003d 0 - the plane passes through OX
6.B \u003d 0 and D \u003d 0 Ax + Cz \u003d 0 - the plane passes through OY
7.C \u003d 0 and D \u003d 0 Ax + By \u003d 0 - the plane passes through OZ
Mutual arrangement 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) \u003d cos (S 1; S 2) \u003d \u003d 
2. The angle between the planes is defined through the angle between their normal vectors.
Cos (l 1; l 2) \u003d cos (N 1; N 2) \u003d \u003d 
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 lines || in space when their || vector guides
5. 2 planes || when || normal vectors
6. The notions of perpendicularity of lines and planes are introduced in a similar way.
Question number 14
Various types of equation of a straight line on a plane (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 the straight line:
1.C \u003d 0 Ax + Vy \u003d 0 - the straight line passes through the origin.
2.a \u003d 0 Vy + C \u003d 0 y \u003d
3.b \u003d 0 Ax + C \u003d 0 x \u003d
4.b \u003d C \u003d 0 Ax \u003d 0 x \u003d 0
5.a \u003d C \u003d 0 Vy \u003d 0 y \u003d 0
Equation of a straight line with a slope:
Any straight line that is not equal to the OA axis (B not \u003d 0) can be written in the next. form:
k \u003d tgα α - the angle between a straight line and a positively directed line OX
b - the point of intersection of the straight line with the OY axis
Doc:
Ax + Wu + C \u003d 0
Wu \u003d -Ah-C |: B
Equation of a straight line at two points:
Question number 16
The finite limit of the function at the point and as x → ∞
Final limit at point x 0:
The number A is called the limit of the function y \u003d f (x) as x → x 0 if for any E\u003e 0 there exists b\u003e 0 such that for x ≠ x 0 satisfying the inequality | x - x 0 |< б, выполняется условие |f(x) - A| < Е
The limit is denoted: \u003d A
End limit at point + ∞:
The number A is called the limit of the function y \u003d f (x) at x → + ∞ if for any E\u003e 0 there exists C\u003e 0 such that for x\u003e C the inequality | f (x) - A |< Е
The limit is denoted: \u003d A
End limit at -∞:
The number A is called the limit of the function y \u003d 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. 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's 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 the given formula is simply "jagged", then there is a high probability of confusion 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 in terms of 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 inference 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 us 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.
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