Let the straight 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 the point M 1 has the form y- y 1 \u003d k (x - x 1), (10.6)
where k - still unknown coefficient.
Since the straight line passes through the point M 2 (x 2 y 2), then the coordinates of this point must satisfy equation (10.6): y 2 -y 1 \u003d k (x 2 -x 1).
From here we find Substituting the found value k
into equation (10.6), we obtain the equation of a straight line passing through the points M 1 and M 2: 
It is assumed that in this equation x 1 ≠ x 2, y 1 ≠ y 2
If x 1 \u003d x 2, then the straight line passing through the points M 1 (x 1, y I) and M 2 (x 2, y 2) is parallel to the y-axis. Its equation is x = x 1 .
If y 2 \u003d y I, then the equation of the straight line can be written as y \u003d y 1, the straight line M 1 M 2 is parallel to the x-axis.
Equation of a straight line in segments
Let the straight line intersect the Ox axis at the point M 1 (a; 0), and the Oy axis - at the point M 2 (0; b). The equation will take the form: 
those. 
. This equation is called the equation of a straight line in segments, because the numbers a and b indicate which segments the straight line cuts off on the coordinate axes.
Equation of a straight line passing through a given point perpendicular to a given vector
Let's find the equation of a straight line passing through a given point Mo (x O; y o) perpendicular to a given non-zero vector n = (A; B).
Take an arbitrary point M(x; y) on the straight line and consider the vector M 0 M (x - x 0; y - y o) (see Fig. 1). Since the vectors n and M o M are perpendicular, their scalar product is 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 .
The 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 \u003d -Ax o - Vu o - free member. Equation (10.9) is the general equation of a straight line(see Fig.2).
Fig.1 Fig.2
Canonical equations of the straight line
,
Where 
are the coordinates of the point through which the line passes, and 
- direction vector.
Curves of the second order Circle
A circle is the set of all points of a plane equidistant from a given point, which is called the center.
Canonical equation of a circle of radius
R centered on a point 
:
In particular, if the center of the stake coincides with the origin, then the equation will look like: 
Ellipse
An ellipse is a set of points in a plane, the sum of the distances from each of them to two given points
and 
, which are called foci, is a constant value 
, greater than the distance between the foci 
.
The canonical equation of an ellipse whose foci lie on the Ox axis and whose origin is in the middle between the foci has the form 
G 
de a the length of the major semiaxis; b is the length of the minor semiaxis (Fig. 2).
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 FROM 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 the general equation of the straight line Ah + Wu + C = 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 various types of 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: 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 the 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 some 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 - this is 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 a specific example of the 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 a non-zero number λ. 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 a 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 of the straight line. 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 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, 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. It needs to be converted to a 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 a straight line equation with a slope y = 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 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 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, whereas, as a normal vector of the line whose equation needs to be written, we take the directing vector of the 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 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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This article reveals the derivation of the equation of a straight line passing through two given points in a rectangular coordinate system located on a plane. We derive the equation of a straight line passing through two given points in a rectangular coordinate system. We will visually show and solve several examples related to the material covered.
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Before obtaining the equation of a straight line passing through two given points, it is necessary to pay attention to some facts. There is an axiom that says that through two non-coinciding points on a plane it is possible to draw a straight line and only one. In other words, two given points of the plane are determined by a straight line passing through these points.
If the plane is given by the rectangular coordinate system Oxy, then any straight line depicted in it will correspond to the equation of the straight line on the plane. There is also a connection with the directing vector of the straight line. These data are sufficient to draw up the equation of a straight line passing through two given points.
Consider an example of solving a similar problem. It is necessary to compose the equation of a straight line a passing through two mismatched points M 1 (x 1, y 1) and M 2 (x 2, y 2) located in the Cartesian coordinate system.
In the canonical equation of a straight line on a plane, having the form x - x 1 a x \u003d y - y 1 a y , a rectangular coordinate system O x y is specified with a straight line that intersects with it at a point with coordinates M 1 (x 1, y 1) with a guide vector a → = (a x , a y) .
It is necessary to compose the canonical equation of the straight line a, which will pass through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2) .
The straight line a has a directing vector M 1 M 2 → with coordinates (x 2 - x 1, y 2 - y 1), since it intersects the points M 1 and M 2. We have obtained the necessary data in order to transform the canonical equation with the coordinates of the direction vector M 1 M 2 → = (x 2 - x 1, y 2 - y 1) and the coordinates of the points M 1 lying on them (x 1, y 1) and M 2 (x 2 , y 2) . We get an equation of the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 or x - x 2 x 2 - x 1 = y - y 2 y 2 - y 1 .
Consider the figure below.
Following the calculations, we write the parametric equations of a straight line in a plane that passes through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2) . We get an equation of the form x \u003d x 1 + (x 2 - x 1) λ y \u003d y 1 + (y 2 - y 1) λ or x \u003d x 2 + (x 2 - x 1) λ y \u003d y 2 + (y 2 - y 1) λ.
Let's take a closer look at a few examples.
Example 1
Write the equation of a straight line passing through 2 given points with coordinates M 1 - 5 , 2 3 , M 2 1 , - 1 6 .
Solution
The canonical equation for a straight line intersecting at two points with coordinates x 1 , y 1 and x 2 , y 2 takes the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 . According to the condition of the problem, we have that x 1 \u003d - 5, y 1 \u003d 2 3, x 2 \u003d 1, y 2 \u003d - 1 6. It is necessary to substitute numerical values in the equation x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 . From here we get that the canonical equation will take the form x - (- 5) 1 - (- 5) = y - 2 3 - 1 6 - 2 3 ⇔ x + 5 6 = y - 2 3 - 5 6 .
Answer: x + 5 6 = y - 2 3 - 5 6 .
If it is necessary to solve a problem with a different type of equation, then for a start you can go to the canonical one, since it is easier to come to any other from it.
Example 2
Compose the general equation of a straight line passing through points with coordinates M 1 (1, 1) and M 2 (4, 2) in the O x y coordinate system.
Solution
First you need to write down the canonical equation of a given line that passes through the given two points. We get an equation of the form x - 1 4 - 1 = y - 1 2 - 1 ⇔ x - 1 3 = y - 1 1 .
We bring the canonical equation to the desired form, then we get:
x - 1 3 = y - 1 1 ⇔ 1 x - 1 = 3 y - 1 ⇔ x - 3 y + 2 = 0
Answer: x - 3 y + 2 = 0 .
Examples of such tasks were considered in school textbooks at algebra lessons. School tasks differed in that the equation of a straight line with a slope coefficient was known, having the form y \u003d k x + b. If you need to find the value of the slope k and the number b, at which the equation y \u003d k x + b defines a line in the O x y system that passes through the points M 1 (x 1, y 1) and M 2 (x 2, y 2) , where x 1 ≠ x 2 . When x 1 = x 2 , then the slope takes on the value of infinity, and the straight line M 1 M 2 is defined by a general incomplete equation of the form x - x 1 = 0 .
Because the dots M 1 and M 2 are on a straight line, then their coordinates satisfy the equation y 1 = k x 1 + b and y 2 = k x 2 + b. It is necessary to solve the system of equations y 1 = k x 1 + b y 2 = k x 2 + b with respect to k and b.
To do this, we find k \u003d y 2 - y 1 x 2 - x 1 b \u003d y 1 - y 2 - y 1 x 2 - x 1 x 1 or k \u003d y 2 - y 1 x 2 - x 1 b \u003d y 2 - y 2 - y 1 x 2 - x 1 x 2 .
With such values of k and b, the equation of a straight line passing through given two points takes the following form y = y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 1 or y \u003d y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 2.
Memorizing such a huge number of formulas at once will not work. To do this, it is necessary to increase the number of repetitions in solving problems.
Example 3
Write the equation of a straight line with a slope passing through points with coordinates M 2 (2, 1) and y = k x + b.
Solution
To solve the problem, we use a formula with a slope that has the form y \u003d k x + b. The coefficients k and b must take such a value that this equation corresponds to a straight line passing through two points with coordinates M 1 (- 7 , - 5) and M 2 (2 , 1) .
points M 1 and M 2 located on a straight line, then their coordinates should invert the equation y = k x + b the correct equality. From here we get that - 5 = k · (- 7) + b and 1 = k · 2 + b. Let's combine the equation into the system - 5 = k · - 7 + b 1 = k · 2 + b and solve.
Upon substitution, we get that
5 = k - 7 + b 1 = k 2 + b ⇔ b = - 5 + 7 k 2 k + b = 1 ⇔ b = - 5 + 7 k 2 k - 5 + 7 k = 1 ⇔ ⇔ b = - 5 + 7 k k = 2 3 ⇔ b = - 5 + 7 2 3 k = 2 3 ⇔ b = - 1 3 k = 2 3
Now the values k = 2 3 and b = - 1 3 are substituted into the equation y = k x + b . We get that the desired equation passing through the given points will be an equation that has the form y = 2 3 x - 1 3 .
This way of solving predetermines the expenditure of a large amount of time. There is a way in which the task is solved literally in two steps.
We write the canonical equation of a straight line passing through M 2 (2, 1) and M 1 (- 7, - 5) , having the form x - (- 7) 2 - (- 7) = y - (- 5) 1 - (- 5) ⇔ x + 7 9 = y + 5 6 .
Now let's move on to the slope equation. We get that: x + 7 9 = y + 5 6 ⇔ 6 (x + 7) = 9 (y + 5) ⇔ y = 2 3 x - 1 3 .
Answer: y = 2 3 x - 1 3 .
If in three-dimensional space there is a rectangular coordinate system O x y z with two given non-coincident points with coordinates M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), the straight line M passing through them 1 M 2 , it is necessary to obtain the equation of this line.
We have that canonical equations of the form x - x 1 a x = y - y 1 a y = z - z 1 a z and parametric equations of the form x = x 1 + a x λ y = y 1 + a y λ z = z 1 + a z λ are able to set a line in the O x y z coordinate system passing through points having coordinates (x 1, y 1, z 1) with a directing vector a → = (a x, a y, a z) .
Straight M 1 M 2 has a direction vector of the form M 1 M 2 → = (x 2 - x 1 , y 2 - y 1 , z 2 - z 1) , where the line passes through the point M 1 (x 1 , y 1 , z 1) and M 2 (x 2, y 2, z 2), hence the canonical equation can be of the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 or x - x 2 x 2 - x 1 \u003d y - y 2 y 2 - y 1 \u003d z - z 2 z 2 - z 1, in turn, parametric x \u003d x 1 + (x 2 - x 1) λ y \u003d y 1 + (y 2 - y 1) λ z = z 1 + (z 2 - z 1) λ or x = x 2 + (x 2 - x 1) λ y = y 2 + (y 2 - y 1) λ z \u003d z 2 + (z 2 - z 1) λ.
Consider a figure that shows 2 given points in space and the equation of a straight line.
Example 4
Write the equation of a straight line defined in a rectangular coordinate system O x y z of three-dimensional space, passing through the given two points with coordinates M 1 (2, - 3, 0) and M 2 (1, - 3, - 5) .
Solution
We need to find the canonical equation. Since we are talking about three-dimensional space, it means that when a straight line passes through given points, the desired canonical equation will take the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 .
By condition, we have that x 1 = 2, y 1 = - 3, z 1 = 0, x 2 = 1, y 2 = - 3, z 2 = - 5. It follows that the necessary equations can be written as follows:
x - 2 1 - 2 = y - (- 3) - 3 - (- 3) = z - 0 - 5 - 0 ⇔ x - 2 - 1 = y + 3 0 = z - 5
Answer: x - 2 - 1 = y + 3 0 = z - 5.
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Let two points be given M 1 (x 1, y 1) and M 2 (x 2, y 2). We write the equation of a straight line in the form (5), where k as yet unknown coefficient:
Since the point M 2 belongs to a given line, then its coordinates satisfy equation (5): 
. Expressing from here and substituting it into equation (5), we obtain the desired equation:
If a 
This equation can be rewritten in a form that is easier to remember:
(6)
Example. Write the equation of a straight line passing through the points M 1 (1.2) and M 2 (-2.3)
Solution. 
. Using the property of proportion, and performing the necessary transformations, we obtain the general equation of a straight line:
Angle between two lines
Consider two lines l 1 and l 2:
l 1: , , and
l 2: , ,
φ is the angle between them (). Figure 4 shows: .
 |
From here 
, or
Using formula (7), one of the angles between the lines can be determined. The second angle is .
Example. Two straight lines are given by the equations y=2x+3 and y=-3x+2. find the angle between these lines.
Solution. It can be seen from the equations that k 1 \u003d 2, and k 2 \u003d-3. substituting these values into formula (7), we find
. So the angle between these lines is .
Conditions for parallelism and perpendicularity of two lines
If straight l 1 and l 2 are parallel, then φ=0 and tgφ=0. from formula (7) it follows that , whence k 2 \u003d k 1. Thus, the condition for the parallelism of two lines is the equality of their slopes.
If straight l 1 and l 2 perpendicular, then φ=π/2, α 2 = π/2+ α 1 . 
. Thus, the condition for two straight lines to be perpendicular is that their slopes are reciprocal in magnitude and opposite in sign.
Distance from point to line
Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Vy + C \u003d 0 is defined as
Proof. Let the point M 1 (x 1, y 1) be the base of the perpendicular dropped from the point M to the given line. Then the distance between points M and M 1:
The x 1 and y 1 coordinates 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 = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem has been proven.
Example. Determine the angle between the lines: y = -3x + 7; y = 2x + 1.
k 1 \u003d -3; k 2 = 2tgj= ; j = p/4.
Example. Show that the lines 3x - 5y + 7 = 0 and 10x + 6y - 3 = 0 are perpendicular.
We find: k 1 \u003d 3/5, k 2 \u003d -5/3, k 1 k 2 \u003d -1, therefore, the lines are perpendicular.
Example. The vertices of the triangle A(0; 1), B(6; 5), C(12; -1) are given. Find the equation for the height drawn from vertex C.
We find the equation of the side AB: ; 4x = 6y - 6;
2x - 3y + 3 = 0;
The desired height equation is: Ax + By + C = 0 or y = kx + b.
k= . Then y = . Because height passes through point C, then its coordinates satisfy this equation: whence b \u003d 17. Total: .
Answer: 3x + 2y - 34 = 0.
The distance from a point to a line is determined by the length of the perpendicular dropped from the point to the line.
If the line is parallel to the projection plane (h | | P 1), then in order to determine the distance from the point BUT to straight h it is necessary to drop a perpendicular from the point BUT to the horizontal h.
Let us consider a more complicated example, when the line occupies a general position. Let it be necessary to determine the distance from the point M to straight a general position.
Definition task distances between parallel lines solved similarly to the previous one. A point is taken on one line, and a perpendicular is drawn from it to another line. The length of the perpendicular is equal to the distance between the parallel lines.
Curve of the second order is a line defined by an equation of the second degree with respect to the current Cartesian coordinates. In the general case, Ax 2 + 2Bxy + Su 2 + 2Dx + 2Ey + F \u003d 0,
where A, B, C, D, E, F are real numbers and at least one of the numbers A 2 + B 2 + C 2 ≠0.
Circle
Circle center- this is the locus of points in the plane equidistant from the point of the plane C (a, b).
The circle is given by the following equation:
Where x, y are the coordinates of an arbitrary point on the circle, R is the radius of the circle.
Sign of the circle equation
1. There is no term with x, y
2. Coefficients at x 2 and y 2 are equal
Ellipse
Ellipse the locus of points in a plane is called, the sum of the distances of each of which from two given points of this plane is called foci (a constant value).
Canonical equation of an ellipse:
X and y belong to an ellipse.
a is the major semiaxis of the ellipse
b is the minor semiaxis of the ellipse
The ellipse has 2 axes of symmetry OX and OY. The axes of symmetry of the ellipse are its axes, the point of their intersection is the center of the ellipse. The axis on which the foci are located is called focal axis. The point of intersection of the ellipse with the axes is the vertex of the ellipse.
Compression (stretching) ratio: ε = c/a- eccentricity (characterizes the shape of the ellipse), the smaller it is, the less the ellipse is extended along the focal axis.
If the centers of the ellipse are not in the center С(α, β)
Hyperbola
Hyperbole called the locus of points in a plane, the absolute value of the difference in distances, each of which from two given points of this plane, called foci, is a constant value other than zero.
Canonical equation of a hyperbola
A hyperbola has 2 axes of symmetry:
a - real semiaxis of symmetry
b - imaginary semiaxis of symmetry
Asymptotes of a hyperbola:
Parabola
parabola is the locus of points in a plane equidistant from a given point F, called the focus, and a given line, called the directrix.
Canonical parabola equation:
Y 2 \u003d 2px, where p is the distance from the focus to the directrix (parabola parameter)
If the vertex of the parabola is C (α, β), then the equation of the parabola (y-β) 2 \u003d 2p (x-α)
If the focal axis is taken as the y-axis, then the parabola equation will take the form: x 2 \u003d 2qy
