How to find the coordinates of points from the equation of a straight line. Equation of a line that passes through two given points: examples, solutions
Properties of a straight line in Euclidean geometry.
An infinite number of straight lines can be drawn through any point.
Through any two non-coinciding points a single straight line can be drawn.
Two divergent lines in a plane either intersect at a single point or are
parallel (follows from the previous one).
In three-dimensional space, there are three options for the relative position of two lines:
- lines intersect;
- lines are parallel;
- straight lines intersect.
Straight line— algebraic curve of the first order: a straight line in the Cartesian coordinate system
is given on the plane by an equation of the first degree (linear equation).
General equation of a straight line.
Definition. Any straight line on the plane can be specified by a first-order equation
Ax + Wu + C = 0,
and constant A, B are not equal to zero at the same time. This first order equation is called general
equation of a straight line. Depending on the values of the constants A, B And WITH The following special cases are possible:
. C = 0, A ≠0, B ≠ 0- a straight line passes through the origin
. A = 0, B ≠0, C ≠0 (By + C = 0)- straight line parallel to the axis Oh
. B = 0, A ≠0, C ≠ 0 (Ax + C = 0)- straight line parallel to the axis OU
. B = C = 0, A ≠0- the straight line coincides with the axis OU
. A = C = 0, B ≠0- the straight line coincides with the axis Oh
The equation of a straight line can be represented in in various forms depending on any given
initial conditions.
Equation of a straight line from a point and a normal vector.
Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)
perpendicular to the line given by the equation
Ax + Wu + C = 0.
Example. Find the equation of a line passing through a point A(1, 2) perpendicular to the vector (3, -1).
Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C
Let's substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore
C = -1. Total: the required equation: 3x - y - 1 = 0.
Equation of a line passing through two points.
Let two points be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2, z 2), Then equation of a line,
passing through these points:
If any of the denominators is zero, the corresponding numerator should be set equal to zero. On
plane, the equation of the straight line written above is simplified:
If x 1 ≠ x 2 And x = x 1, If x 1 = x 2 .
Fraction = k called slope straight.
Example. Find the equation of the line passing through points A(1, 2) and B(3, 4).
Solution. Applying the formula written above, we get:
Equation of a straight line using a point and slope.
If general equation straight Ax + Wu + C = 0 lead to:
and designate 
, then the resulting equation is called
equation of a straight line with slope k.
Equation of a straight line from a point and a direction vector.
By analogy with the point considering the equation of a straight line through the normal vector, you can enter the task
a straight line through a point and a directing vector of a straight line.
Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition
Aα 1 + Bα 2 = 0 called directing vector of a straight line.
Ax + Wu + C = 0.
Example. Find the equation of a straight line with a direction vector (1, -1) and passing through the point A(1, 2).
Solution. We will look for the equation of the desired line in the form: Ax + By + C = 0. According to the definition,
coefficients must satisfy the following conditions:
1 * A + (-1) * B = 0, i.e. A = B.
Then the equation of the straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0.
at x = 1, y = 2 we get C/A = -3, i.e. required equation:
x + y - 3 = 0
Equation of a straight line in segments.
If in the general equation of the straight line Ах + Ву + С = 0 С≠0, then, dividing by -С, we get:
or where
The geometric meaning of the coefficients is that the coefficient a is the coordinate of the intersection point
straight with axis Oh, A b- coordinate of the point of intersection of the line with the axis OU.
Example. The general equation of a straight line is given x - y + 1 = 0. Find the equation of this line in segments.
C = 1, , a = -1, b = 1.
Normal equation of a line.
If both sides of the equation Ax + Wu + C = 0 divide by number 
which is called
normalizing factor, then we get
xcosφ + ysinφ - p = 0 -normal equation of a line.
The sign ± of the normalizing factor must be chosen so that μ*C< 0.
R- the length of the perpendicular dropped from the origin to the straight line,
A φ - the angle formed by this perpendicular with the positive direction of the axis Oh.
Example. The general equation of the line is given 12x - 5y - 65 = 0. Required to write Various types equations
this straight line.
The equation of this line in segments:
The equation of this line with the slope: (divide by 5)
Equation of a line:
cos φ = 12/13; sin φ= -5/13; p = 5.
It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines,
parallel to the axes or passing through the origin.
The angle between straight lines on a plane.
Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2, then the acute angle between these lines
will be defined as
Two lines are parallel if k 1 = k 2. Two straight lines are perpendicular,
If k 1 = -1/ k 2 .
Theorem.
Direct Ax + Wu + C = 0 And A 1 x + B 1 y + C 1 = 0 parallel when the coefficients are proportional
A 1 = λA, B 1 = λB. If also С 1 = λС, then the lines coincide. Coordinates of the point of intersection of two lines
are found as a solution to the system of equations of these lines.
The equation of a line passing through a given point perpendicular to a given line.
Definition. Line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b
represented by the equation:
Distance from a point to a line.
Theorem. If a point is given M(x 0, y 0), then the distance to the straight line Ax + Wu + C = 0 defined as:
Proof. Let the point M 1 (x 1, y 1)- the base of a perpendicular dropped from a point M for a given
direct. Then the distance between points M And M 1:
(1)
Coordinates x 1 And at 1 can be found as a solution to the system of equations:
The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicularly
given straight line. If we transform the first equation of the system to the form:
A(x - x 0) + B(y - y 0) + Ax 0 + By 0 + C = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem has been proven.
This article continues the topic of the equation of a line on a plane: we will consider this type of equation as the general equation of a line. Let us define the theorem and give its proof; Let's figure out what an incomplete general equation of a line is and how to make transitions from a general equation to other types of equations of a line. We will reinforce the entire theory with illustrations and solutions to practical problems.
Yandex.RTB R-A-339285-1
Let a rectangular coordinate system O x y be specified on the plane.
Theorem 1
Any equation of the first degree, having the form A x + B y + C = 0, where A, B, C are some real numbers(A and B are not equal to zero at the same time) defines a straight line in 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 = 0 for a certain set of values A, B, C.
Proof
This theorem consists of two points; we will prove each of them.
- Let us prove that the equation A x + B y + C = 0 defines a straight line on the plane.
Let there be some point M 0 (x 0 , y 0) whose coordinates correspond to the equation A x + B y + C = 0. Thus: A x 0 + B y 0 + C = 0. Subtract from the left and right sides of the equations A x + B y + C = 0 the left and right sides of the equation A x 0 + B y 0 + C = 0, we obtain a new equation that looks like A (x - x 0) + B (y - y 0) = 0 . It is equivalent to A x + B y + C = 0.
The resulting equation A (x - x 0) + B (y - y 0) = 0 is a necessary and sufficient condition for the perpendicularity of the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0 ) . Thus, the set of points M (x, y) defines a straight line in a rectangular coordinate system perpendicular to the direction of the vector n → = (A, B). We can assume that this is not so, but then the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) would not be perpendicular, and the equality A (x - x 0 ) + B (y - y 0) = 0 would not be true.
Consequently, the equation A (x - x 0) + B (y - y 0) = 0 defines a certain line in a rectangular coordinate system on the plane, and therefore the equivalent equation A x + B y + C = 0 defines the same line. This is how we proved the first part of the theorem.
- Let us provide a proof that any straight line in a rectangular coordinate system on a plane can be specified by an equation of the first degree A x + B y + C = 0.
Let us define a straight line a in a rectangular coordinate system on a plane; the point M 0 (x 0 , y 0) through which this line passes, as well as the normal vector of this line n → = (A, B) .
Let there also be some point M (x, y) - a floating point on a line. In this case, the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) are perpendicular to each other, and their scalar product 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 in end result we get the equation A x + B y + C = 0.
So, we have proved the second part of the theorem, and we have proved the entire theorem as a whole.
Definition 1
An equation of the form A x + B y + C = 0 - This general equation of a line on a plane in a rectangular coordinate systemOxy.
Based on the proven theorem, we can conclude that a straight line and its general equation defined on a plane in a fixed rectangular coordinate system are inextricably linked. In other words, the original line corresponds to its general equation; the general equation of a line corresponds to a given line.
From the proof of the theorem it also follows that the coefficients A and B for the variables x and y are the coordinates of the normal vector of the line, which is given by the general equation of the line A x + B y + C = 0.
Let's consider specific example general equation of a straight line.
Let the equation 2 x + 3 y - 2 = 0 be given, which corresponds to a straight line in a given rectangular coordinate system. The normal vector of this line is the vector n → = (2 , 3) . Let's draw the given straight line in the drawing.
We can also state the following: the straight line that we see in the drawing is determined by the general equation 2 x + 3 y - 2 = 0, since the coordinates of all points on a given straight line correspond to this equation.
We can obtain the equation λ · A x + λ · B y + λ · C = 0 by multiplying both sides of the general equation of the line by a number λ not equal to zero. The resulting equation is equivalent to the original general equation, therefore, it will describe the same straight line on the plane.
Definition 2Complete general equation of a line– such a general equation of the straight line A x + B y + C = 0, in which the numbers A, B, C are different from zero. Otherwise the equation is incomplete.
Let us analyze all variations of the incomplete general equation of a line.
- When A = 0, B ≠ 0, C ≠ 0, the general equation takes the form B y + C = 0. Such an incomplete general equation defines in a rectangular coordinate system O x y a straight line that is parallel to the O x axis, since for any real value of x the variable y will take the value - C B . In other words, the general equation of the straight line A x + B y + C = 0, when A = 0, B ≠ 0, specifies the locus of points (x, y), whose coordinates are equal to the same number - C B .
- If A = 0, B ≠ 0, C = 0, the general equation takes the form y = 0. This incomplete equation defines the abscissa axis O x .
- When A ≠ 0, B = 0, C ≠ 0, we obtain an incomplete general equation A x + C = 0, defining a straight line parallel to the ordinate.
- Let A ≠ 0, B = 0, C = 0, then the incomplete general equation will take the form x = 0, and this is the equation of the coordinate line O y.
- Finally, for A ≠ 0, B ≠ 0, C = 0, the incomplete general equation takes the form A x + B y = 0. And this equation describes a straight line that passes through the origin. In fact, the pair of numbers (0, 0) corresponds to the equality A x + B y = 0, since A · 0 + B · 0 = 0.
Let us graphically illustrate all of the above types of incomplete general equation of a straight line.
Example 1
It is known that the given straight line is parallel to the ordinate axis and passes through the point 2 7, - 11. It is necessary to write down the general equation of the given line.
Solution
A straight line parallel to the ordinate axis is given by an equation of the form A x + C = 0, in which A ≠ 0. The condition also specifies the coordinates of the point through which the line passes, and the coordinates of this point meet the conditions of the incomplete general equation A x + C = 0, i.e. the equality is true:
A 2 7 + C = 0
From it it is possible to determine C if we give A some non-zero value, for example, A = 7. In this case, we get: 7 · 2 7 + C = 0 ⇔ C = - 2. We know both coefficients A and C, substitute them into the equation A x + C = 0 and get the required straight line equation: 7 x - 2 = 0
Answer: 7 x - 2 = 0
Example 2
The drawing shows a straight line; you need to write down its equation.
Solution
The given drawing allows us to easily take the initial data to solve the problem. We see in the drawing that the given straight line is parallel to the O x axis and passes through the point (0, 3).
The straight line, which is parallel to the abscissa, is determined by the incomplete general equation B y + C = 0. Let's find the values of B and C. The coordinates of the point (0, 3), since the given line passes through it, will satisfy the equation of the line B y + C = 0, then the equality is valid: B · 3 + C = 0. Let's set B to some value other than zero. Let's say B = 1, in which case from the equality B · 3 + C = 0 we can find C: C = - 3. We use known values B and C, we obtain the required equation of the straight line: y - 3 = 0.
Answer: y - 3 = 0 .
General equation of a line passing through a given point in a plane
Let the given line pass through the point M 0 (x 0 , y 0), then its coordinates correspond to the general equation of the line, i.e. the equality is true: A x 0 + B y 0 + C = 0. Let us subtract the left and right sides of this equation from the left and right sides of the general complete equation straight. We get: A (x - x 0) + B (y - y 0) + C = 0, this equation is equivalent to the original general one, passes through the point M 0 (x 0, y 0) and has a normal vector n → = (A, B) .
The result that we obtained makes it possible to write down the general equation of a line with known coordinates of the normal vector of the line and the coordinates of a certain point of this line.
Example 3
Given a point M 0 (- 3, 4) through which a line passes, and the normal vector of this line n → = (1 , - 2) . It is necessary to write down the equation of the given line.
Solution
The initial conditions allow us to obtain the necessary data to compose the equation: A = 1, B = - 2, x 0 = - 3, y 0 = 4. Then:
A (x - x 0) + B (y - y 0) = 0 ⇔ 1 (x - (- 3)) - 2 y (y - 4) = 0 ⇔ ⇔ x - 2 y + 22 = 0
The problem could have been solved differently. The general equation of a straight line is A x + B y + C = 0. The given normal vector allows us to obtain the values of coefficients A and B, then:
A x + B y + C = 0 ⇔ 1 x - 2 y + C = 0 ⇔ x - 2 y + C = 0
Now let's find the value of C using given by the condition problem point M 0 (- 3, 4) 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 = 0. Hence C = 11. The required straight line equation takes the form: x - 2 · y + 11 = 0.
Answer: x - 2 y + 11 = 0 .
Example 4
Given a line 2 3 x - y - 1 2 = 0 and a point M 0 lying on this line. Only the abscissa of this point is known, and it is equal to - 3. It is necessary to determine the ordinate of a given point.
Solution
Let us designate the coordinates of point M 0 as x 0 and y 0 . The source data indicates that x 0 = - 3. Since the point belongs to a given line, then its coordinates correspond to the general equation of this line. Then the equality will be true:
2 3 x 0 - y 0 - 1 2 = 0
Define y 0: 2 3 · (- 3) - y 0 - 1 2 = 0 ⇔ - 5 2 - y 0 = 0 ⇔ y 0 = - 5 2
Answer: - 5 2
Transition from the general equation of a line to other types of equations of a line and vice versa
As we know, there are several types of equations for the same straight line on a plane. The choice of the type of equation depends on the conditions of the problem; it is possible to choose the one that is more convenient for solving it. The skill of converting an equation of one type into an equation of another type is very useful here.
First, let's consider the transition from the general equation of the form A x + B y + C = 0 to the canonical equation x - x 1 a x = y - y 1 a y.
If A ≠ 0, then we 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, transfer the others to the right side, we get: A x = - B y - C. We take – B out of brackets, then: A x = - B y + C B .
Let's rewrite the equality in the form of a proportion: x - B = y + C B A.
Of course, there is no need to memorize the resulting formulas. It is enough to know the algorithm of actions when moving from a general equation to a canonical one.
Example 5
The general equation of the line 3 y - 4 = 0 is given. It is necessary to transform it into a canonical equation.
Solution
Let's write the original equation as 3 y - 4 = 0. Next, we proceed according to the algorithm: the term 0 x remains on the left side; and on the right side we put - 3 out of brackets; we get: 0 x = - 3 y - 4 3 .
Let's write the resulting equality as a proportion: x - 3 = y - 4 3 0 . Thus, we have obtained an equation of canonical form.
Answer: x - 3 = y - 4 3 0.
To transform the general equation of a line into parametric ones, first a transition is made to the canonical form, and then a transition from the canonical equation of a 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 for this line.
Solution
Let us make the transition from the general equation to the canonical one:
2 x - 5 y - 1 = 0 ⇔ 2 x = 5 y + 1 ⇔ 2 x = 5 y + 1 5 ⇔ x 5 = y + 1 5 2
Now we take both sides of the resulting canonical equation equal to λ, then:
x 5 = λ y + 1 5 2 = λ ⇔ x = 5 λ y = - 1 5 + 2 λ , λ ∈ R
Answer:x = 5 λ y = - 1 5 + 2 λ , λ ∈ R
The general equation can be converted into an equation of a straight line with slope y = k · x + b, but only when B ≠ 0. For the transition, we leave the term B y on the left side, the rest are transferred to the right. We get: B y = - A x - C . Let's divide both sides of the resulting equality by B, different from zero: y = - A B x - C B.
Example 7
The general equation of the line is given: 2 x + 7 y = 0. You need to convert that equation into a slope equation.
Solution
Let's perform the necessary actions according to the algorithm:
2 x + 7 y = 0 ⇔ 7 y - 2 x ⇔ y = - 2 7 x
Answer: y = - 2 7 x .
From the general equation of a line, it is enough to simply obtain an equation in segments of the form x a + y b = 1. To make such a transition, we move the number C to the right side of the equality, divide both sides of the resulting equality by – C and, finally, transfer the coefficients for the variables x and y to the denominators:
A x + B y + C = 0 ⇔ A x + B y = - C ⇔ ⇔ A - C x + B - C y = 1 ⇔ x - C A + y - C B = 1
Example 8
It is necessary to transform the general equation of the line x - 7 y + 1 2 = 0 into the equation of the line in segments.
Solution
Let's move 1 2 to the right side: x - 7 y + 1 2 = 0 ⇔ x - 7 y = - 1 2 .
Let's divide both sides of the equality by -1/2: x - 7 y = - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y = 1 .
Answer: x - 1 2 + y 1 14 = 1 .
In general, the reverse transition is also easy: from other types of equations to the general one.
The equation of a line in segments and an equation with an angular coefficient can be easily converted into a general one by simply collecting all the terms on the left side of the equality:
x a + y b ⇔ 1 a x + 1 b y - 1 = 0 ⇔ A x + B y + C = 0 y = k x + b ⇔ y - k x - b = 0 ⇔ A x + B y + C = 0
The canonical equation is converted to a general one according to the following scheme:
x - x 1 a x = y - y 1 a y ⇔ a y · (x - x 1) = a x (y - y 1) ⇔ ⇔ a y x - a x y - a y x 1 + a x y 1 = 0 ⇔ A x + B y + C = 0
To move from parametric ones, first move to the canonical one, and then to the general one:
x = x 1 + a x · λ y = y 1 + a y · λ ⇔ x - x 1 a x = y - y 1 a y ⇔ A x + B y + C = 0
Example 9
The parametric equations of the line x = - 1 + 2 · λ y = 4 are given. It is necessary to write down the general equation of this line.
Solution
Let us make the transition from parametric equations to canonical ones:
x = - 1 + 2 · λ y = 4 ⇔ x = - 1 + 2 · λ y = 4 + 0 · λ ⇔ λ = x + 1 2 λ = y - 4 0 ⇔ x + 1 2 = y - 4 0
Let's move from the canonical to the general:
x + 1 2 = y - 4 0 ⇔ 0 · (x + 1) = 2 (y - 4) ⇔ y - 4 = 0
Answer: y - 4 = 0
Example 10
The equation of a straight line in the segments x 3 + y 1 2 = 1 is given. It is necessary to make a transition to general appearance equations
Solution:
We simply rewrite the equation in the required form:
x 3 + y 1 2 = 1 ⇔ 1 3 x + 2 y - 1 = 0
Answer: 1 3 x + 2 y - 1 = 0 .
Drawing up a general equation of a line
We said above that the general equation can be written with known coordinates of the normal vector and the coordinates of the point through which the line passes. Such a straight line is defined by the equation A (x - x 0) + B (y - y 0) = 0. There we also analyzed the corresponding example.
Now let's look at more complex examples, in which first you need to determine the coordinates of the normal vector.
Example 11
Given a line parallel to the line 2 x - 3 y + 3 3 = 0. The point M 0 (4, 1) through which the given line passes is also known. It is necessary to write down the equation of the given line.
Solution
The initial conditions tell us that the lines are parallel, then, as the normal vector of the line, the equation of which needs to be written, we take the direction vector of the line n → = (2, - 3): 2 x - 3 y + 3 3 = 0. Now we know all the necessary data to create the general equation of the line:
A (x - x 0) + B (y - y 0) = 0 ⇔ 2 (x - 4) - 3 (y - 1) = 0 ⇔ 2 x - 3 y - 5 = 0
Answer: 2 x - 3 y - 5 = 0 .
Example 12
The given line passes through the origin perpendicular to the line x - 2 3 = y + 4 5. It is necessary to create a general equation for a given line.
Solution
The normal vector of a given line will be the direction vector of the line x - 2 3 = y + 4 5.
Then n → = (3, 5) . The straight line passes through the origin, i.e. through point O (0, 0). Let's create a general equation for a given line:
A (x - x 0) + B (y - y 0) = 0 ⇔ 3 (x - 0) + 5 (y - 0) = 0 ⇔ 3 x + 5 y = 0
Answer: 3 x + 5 y = 0 .
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Lesson from the series “Geometric algorithms”
Hello dear reader!
Today we will start learning algorithms related to geometry. The fact is that there are quite a lot of Olympiad problems in computer science related to computational geometry, and solving such problems often causes difficulties.
Over the course of several lessons, we will consider a number of elementary subtasks on which the solution of most problems in computational geometry is based.
In this lesson we will create a program for finding the equation of a line, passing through given two points. To solve geometric problems, we need some knowledge of computational geometry. We will devote part of the lesson to getting to know them.
Insights from Computational Geometry
Computational geometry is a branch of computer science that studies algorithms for solving geometric problems.
The initial data for such problems can be a set of points on a plane, a set of segments, a polygon (specified, for example, by a list of its vertices in clockwise order), etc.
The result can be either an answer to some question (such as does a point belong to a segment, do two segments intersect, ...), or some geometric object (for example, the smallest convex polygon connecting given points, the area of a polygon, etc.) .
We will consider problems of computational geometry only on the plane and only in the Cartesian coordinate system.
Vectors and coordinates
To apply the methods of computational geometry, it is necessary to translate geometric images into the language of numbers. We will assume that the plane is given a Cartesian coordinate system, in which the direction of rotation counterclockwise is called positive.
Now geometric objects receive an analytical expression. So, to specify a point, it is enough to indicate its coordinates: a pair of numbers (x; y). A segment can be specified by specifying the coordinates of its ends; a straight line can be specified by specifying the coordinates of a pair of its points.
But our main tool for solving problems will be vectors. Let me therefore recall some information about them.
Line segment AB, which has a point A is considered the beginning (point of application), and the point IN– end, called a vector AB and denote either , or bold lowercase letter, For example A .
To denote the length of a vector (that is, the length of the corresponding segment), we will use the modulus symbol (for example, ).
An arbitrary vector will have coordinates equal to the difference between the corresponding coordinates of its end and beginning:
,
here are the points A And B
have coordinates 
respectively.
For calculations we will use the concept oriented angle, that is, an angle that takes into account the relative position of the vectors.
Oriented angle between vectors a And b positive if the rotation is from the vector a to vector b is performed in a positive direction (counterclockwise) and negative in the other case. See Fig.1a, Fig.1b. It is also said that a pair of vectors a And b positively (negatively) oriented.
Thus, the value of the oriented angle depends on the order in which the vectors are listed and can take values in the interval.
Many problems in computational geometry use the concept of vector (skew or pseudoscalar) products of vectors.
The vector product of vectors a and b is the product of the lengths of these vectors and the sine of the angle between them:
.
Cross product of vectors in coordinates:
The expression on the right is a second-order determinant:
Unlike the definition given in analytical geometry, it is a scalar.
The sign of the vector product determines the position of the vectors relative to each other:
a And b positively oriented.
If the value is , then a pair of vectors a And b negatively oriented.
The cross product of nonzero vectors is zero if and only if they are collinear ( 
). This means that they lie on the same line or on parallel lines.
Let's look at a few simple problems that are necessary when solving more complex ones.
Let's determine the equation of a straight line from the coordinates of two points.
Equation of a line passing through two different points specified by their coordinates.
Let two non-coinciding points be given on a straight line: with coordinates (x1; y1) and with coordinates (x2; y2). Accordingly, a vector with a start at a point and an end at a point has coordinates (x2-x1, y2-y1). If P(x, y) is an arbitrary point on our line, then the coordinates of the vector are equal to (x-x1, y – y1).
Using the vector product, the condition for collinearity of vectors and can be written as follows:
Those. (x-x1)(y2-y1)-(y-y1)(x2-x1)=0
(y2-y1)x + (x1-x2)y + x1(y1-y2) + y1(x2-x1) = 0
We rewrite the last equation as follows:
ax + by + c = 0, (1)
c = x1(y1-y2) + y1(x2-x1)
So, the straight line can be specified by an equation of the form (1).
Problem 1. The coordinates of two points are given. Find its representation in the form ax + by + c = 0.
In this lesson we learned some information about computational geometry. We solved the problem of finding the equation of a line from the coordinates of two points.
In the next lesson, we will create a program to find the intersection point of two lines given by our equations.
Equation of a straight line on a plane.
The direction vector is straight. Normal vector
A straight line on a plane is one of the simplest geometric shapes, familiar to you since elementary school, and today we will learn how to deal with it using the methods of analytical geometry. To master the material, you must be able to build a straight line; know what equation defines a straight line, in particular, a straight line passing through the origin of coordinates and straight lines parallel to the coordinate axes. This information can be found in the manual Graphs and properties of elementary functions, I created it for matan, but the section about linear function It turned out very successful and detailed. Therefore, dear teapots, warm up there first. In addition, you need to have basic knowledge about vectors, otherwise the understanding of the material will be incomplete.
In this lesson we will look at ways in which you can create an equation of a straight line on a plane. I recommend not to neglect practical examples (even if it seems very simple), since I will provide them with elementary and important facts, technical techniques that will be required in the future, including in other sections of higher mathematics.
- How to write an equation of a straight line with an angle coefficient?
- How ?
- How to find a direction vector using the general equation of a straight line?
- How to write an equation of a straight line given a point and a normal vector?
and we begin:
Equation of a straight line with slope
The well-known “school” form of a straight line equation is called equation of a straight line with slope. For example, if a straight line is given by the equation, then its slope is: . Let's consider geometric meaning given coefficient and how its value affects the location of the line: 
In a geometry course it is proven that the slope of the straight line is equal to tangent of the angle between positive axis directionand this line: , and the angle “unscrews” counterclockwise.
In order not to clutter the drawing, I drew angles only for two straight lines. Let's consider the “red” line and its slope. According to the above: (the “alpha” angle is indicated by a green arc). For the “blue” straight line with the angle coefficient, the equality is true (the “beta” angle is indicated by a brown arc). And if the tangent of the angle is known, then if necessary it is easy to find and the corner itself by using inverse function– arctangent. As they say, a trigonometric table or a microcalculator in your hands. Thus, the angular coefficient characterizes the degree of inclination of the straight line to the abscissa axis.
In this case, it is possible following cases:
1) If the slope is negative: then the line, roughly speaking, goes from top to bottom. Examples are the “blue” and “raspberry” straight lines in the drawing.
2) If the slope is positive: then the line goes from bottom to top. Examples - “black” and “red” straight lines in the drawing.
3) If the slope is zero: , then the equation takes the form , and the corresponding straight line is parallel to the axis. An example is the “yellow” straight line.
4) For a family of lines parallel to an axis (there is no example in the drawing, except for the axis itself), the angular coefficient does not exist (tangent of 90 degrees is not defined).
The greater the slope coefficient in absolute value, the steeper the straight line graph goes..
For example, consider two straight lines. Here, therefore, the straight line has a steeper slope. Let me remind you that the module allows you to ignore the sign, we are only interested in absolute values angular coefficients.
In turn, a straight line is steeper than straight lines 
.
Conversely: the smaller the slope coefficient in absolute value, the flatter the straight line.
For straight lines 
the inequality is true, thus the straight line is flatter. Children's slide, so as not to give yourself bruises and bumps.
Why is this necessary?
Prolong your torment Knowledge of the above facts allows you to immediately see your mistakes, in particular, errors when constructing graphs - if the drawing turns out to be “obviously something wrong.” It is advisable that you straightaway it was clear that, for example, the straight line is very steep and goes from bottom to top, and the straight line is very flat, pressed close to the axis and goes from top to bottom.
In geometric problems, several straight lines often appear, so it is convenient to designate them somehow.
Designations: straight lines are designated in small Latin letters: . A popular option is to designate them using the same letter with natural subscripts. For example, the five lines we just looked at can be denoted by 
.
Since any straight line is uniquely determined by two points, it can be denoted by these points: 
etc. The designation clearly implies that the points belong to the line.
It's time to warm up a little:
How to write an equation of a straight line with an angle coefficient?
If a point belonging to a certain line and the angular coefficient of this line are known, then the equation of this line is expressed by the formula:
Example 1
Write an equation of a straight line with an angular coefficient if it is known that the point belongs to this straight line.
Solution: Let's compose the equation of the straight line using the formula 
. IN in this case:
Answer:
Examination is done simply. First, we look at the resulting equation and make sure that our slope is in place. Secondly, the coordinates of the point must satisfy this equation. Let's plug them into the equation:
The correct equality is obtained, which means that the point satisfies the resulting equation.
Conclusion: The equation was found correctly.
A more tricky example to solve on your own:
Example 2
Write an equation for a straight line if it is known that its angle of inclination to the positive direction of the axis is , and the point belongs to this straight line.
If you have any difficulties, re-read the theoretical material. More precisely, more practical, I skip a lot of evidence.
It rang last call, the graduation party has died down, and outside the gates of our native school, analytical geometry itself awaits us. The jokes are over... Or maybe they are just beginning =)
We nostalgically wave our pen to the familiar and get acquainted with the general equation of a straight line. Because in analytical geometry this is exactly what is used:
The general equation of a straight line has the form: , where are some numbers. At the same time, the coefficients simultaneously are not equal to zero, since the equation loses its meaning.
Let's dress in a suit and tie the equation with the slope coefficient. First, let's move all the terms to the left side:
The term with “X” must be put in first place:
In principle, the equation already has the form , but according to the rules of mathematical etiquette, the coefficient of the first term (in this case) must be positive. Changing signs:
Remember this technical feature! We make the first coefficient (most often) positive!
In analytical geometry, the equation of a straight line will almost always be given in general form. Well, if necessary, it can be easily reduced to the “school” form with an angular coefficient (with the exception of straight lines parallel to the ordinate axis).
Let's ask ourselves what enough know to construct a straight line? Two points. But more about this childhood incident, now sticks with arrows rule. Each straight line has a very specific slope, which is easy to “adapt” to. vector.
A vector that is parallel to a line is called the direction vector of that line. It is obvious that any straight line has an infinite number of direction vectors, and all of them will be collinear (codirectional or not - it doesn’t matter).
I will denote the direction vector as follows: .
But one vector is not enough to construct a straight line; the vector is free and not tied to any point on the plane. Therefore, it is additionally necessary to know some point that belongs to the line.
How to write an equation of a straight line using a point and a direction vector?
If a certain point belonging to a line and the direction vector of this line are known, then the equation of this line can be compiled using the formula:
Sometimes it is called canonical equation of the line .
What to do when one of the coordinates is equal to zero, we will understand in practical examples below. By the way, please note - both at once coordinates cannot be equal to zero, since the zero vector does not specify a specific direction.
Example 3
Write an equation for a straight line using a point and a direction vector
Solution: Let's compose the equation of a straight line using the formula. In this case:
Using the properties of proportion we get rid of fractions: 
And we bring the equation to its general form:
Answer:
As a rule, there is no need to make a drawing in such examples, but for the sake of understanding: 
In the drawing we see the starting point, the original direction vector (it can be plotted from any point on the plane) and the constructed straight line. By the way, in many cases it is most convenient to construct a straight line using an equation with an angular coefficient. It’s easy to convert our equation into form and easily select another point to construct a straight line.
As noted at the beginning of the paragraph, a straight line has an infinite number of direction vectors, and all of them are collinear. For example, I drew three such vectors: 
. Whatever direction vector we choose, the result will always be the same straight line equation.
Let's create an equation of a straight line using a point and a direction vector:
Resolving the proportion:
Divide both sides by –2 and get the familiar equation:
Those interested can test vectors in the same way 
or any other collinear vector.
Now let's solve the inverse problem:
How to find a direction vector using the general equation of a straight line?
Very simple:
If a line is given by a general equation in a rectangular coordinate system, then the vector is the direction vector of this line.
Examples of finding direction vectors of straight lines: 
The statement allows us to find only one direction vector out of an infinite number, but we don’t need more. Although in some cases it is advisable to reduce the coordinates of the direction vectors:
Thus, the equation specifies a straight line that is parallel to the axis and the coordinates of the resulting direction vector are conveniently divided by –2, obtaining exactly the basis vector as the direction vector. Logical.
Similarly, the equation specifies a straight line parallel to the axis, and by dividing the coordinates of the vector by 5, we obtain the unit vector as the direction vector.
Now let's do it checking Example 3. The example went up, so I remind you that in it we compiled the equation of a straight line using a point and a direction vector
Firstly, using the equation of the straight line we reconstruct its direction vector: 
– everything is fine, we have received the original vector (in some cases the result may be a collinear vector to the original one, and this is usually easy to notice by the proportionality of the corresponding coordinates).
Secondly, the coordinates of the point must satisfy the equation. We substitute them into the equation:
The correct equality was obtained, which we are very happy about.
Conclusion: The task was completed correctly.
Example 4
Write an equation for a straight line using a point and a direction vector
This is an example for you to solve on your own. The solution and answer are at the end of the lesson. It is highly advisable to check using the algorithm just discussed. Try to always (if possible) check on a draft. It’s stupid to make mistakes where they can be 100% avoided.
In the event that one of the coordinates of the direction vector is zero, proceed very simply:
Example 5
Solution: The formula is not suitable since the denominator on the right side is zero. There is an exit! Using the properties of proportion, we rewrite the formula in the form, and the rest rolled along a deep rut:
Answer:
Examination:
1) Restore the directing vector of the straight line:
– the resulting vector is collinear to the original direction vector.
2) Substitute the coordinates of the point into the equation: 
The correct equality is obtained
Conclusion: task completed correctly
The question arises, why bother with the formula if there is a universal version that will work in any case? There are two reasons. First, the formula is in the form of a fraction much better remembered. And secondly, the disadvantage of the universal formula is that the risk of getting confused increases significantly when substituting coordinates.
Example 6
Write an equation for a straight line using a point and a direction vector.
This is an example for you to solve on your own.
Let's return to the ubiquitous two points:
How to write an equation of a straight line using two points?
If two points are known, then the equation of a straight line passing through these points can be compiled using the formula:
In fact, this is a type of formula and here's why: if two points are known, then the vector will be the direction vector of the given line. At the lesson Vectors for dummies we considered simplest task– how to find the coordinates of a vector from two points. According to this problem, the coordinates of the direction vector are: 
Note
: the points can be “swapped” and the formula can be used 
. Such a solution will be equivalent.
Example 7
Write an equation of a straight line using two points 
.
Solution: We use the formula: 
Combing the denominators:
And shuffle the deck: 
Now is the time to get rid of fractional numbers. In this case, you need to multiply both sides by 6: 
Open the brackets and bring the equation to mind: 
Answer:
Examination is obvious - the coordinates of the initial points must satisfy the resulting equation:
1) Substitute the coordinates of the point: 
True equality.
2) Substitute the coordinates of the point: 
True equality.
Conclusion: The equation of the line is written correctly.
If at least one of the points does not satisfy the equation, look for an error.
It is worth noting that graphical verification in this case is difficult, since construct a straight line and see whether the points belong to it 
, not so simple.
I’ll note a couple more technical aspects of the solution. Perhaps in this problem it is more profitable to use the mirror formula 
and, at the same points 
make an equation: 
Fewer fractions. If you want, you can carry out the solution to the end, the result should be the same equation.
The second point is to look at the final answer and figure out whether it can be simplified further? For example, if you get the equation , then it is advisable to reduce it by two: – the equation will define the same straight line. However, this is already a topic of conversation about relative position of lines.
Having received the answer 
in Example 7, just in case, I checked whether ALL coefficients of the equation are divisible by 2, 3 or 7. Although, most often such reductions are made during the solution.
Example 8
Write an equation for a line passing through the points 
.
This is an example for an independent solution, which will allow you to better understand and practice calculation techniques.
Similar to the previous paragraph: if in the formula 
one of the denominators (the coordinate of the direction vector) becomes zero, then we rewrite it in the form . Again, notice how awkward and confused she looks. I don’t see much point in bringing practical examples, since we have already actually solved such a problem (see No. 5, 6).
Direct normal vector (normal vector)
What is normal? In simple words, normal is perpendicular. That is, the normal vector of a line is perpendicular to a given line. Obviously, any straight line has an infinite number of them (as well as direction vectors), and all the normal vectors of the straight line will be collinear (codirectional or not, it makes no difference).
Dealing with them will be even easier than with guide vectors:
If a line is given by a general equation in a rectangular coordinate system, then the vector is the normal vector of this line.
If the coordinates of the direction vector have to be carefully “pulled out” from the equation, then the coordinates of the normal vector can be simply “removed”.
The normal vector is always orthogonal to the direction vector of the line. Let us verify the orthogonality of these vectors using dot product:
I will give examples with the same equations as for the direction vector: 
Is it possible to construct an equation of a straight line given one point and a normal vector? I feel it in my gut, it’s possible. If the normal vector is known, then the direction of the straight line itself is clearly defined - this is a “rigid structure” with an angle of 90 degrees.
How to write an equation of a straight line given a point and a normal vector?
If a certain point belonging to a line and the normal vector of this line are known, then the equation of this line is expressed by the formula:
Here everything worked out without fractions and other surprises. This is our normal vector. Love him. And respect =)
Example 9
Write an equation of a straight line given a point and a normal vector. Find the direction vector of the line.
Solution: We use the formula: 
The general equation of the straight line has been obtained, let’s check:
1) “Remove” the coordinates of the normal vector from the equation: 
– yes, indeed, the original vector was obtained from the condition (or a collinear vector should be obtained).
2) Let's check whether the point satisfies the equation: 
True equality.
After we are convinced that the equation is composed correctly, we will complete the second, easier part of the task. We take out the directing vector of the straight line: 
Answer: 
In the drawing the situation looks like this: 
For training purposes, a similar task for solving independently:
Example 10
Write an equation of a straight line given a point and a normal vector. Find the direction vector of the line.
The final section of the lesson will be devoted to less common, but also important species equations of a straight line on a plane
Equation of a straight line in segments.
Equation of a line in parametric form
The equation of a straight line in segments has the form , where are nonzero constants. Some types of equations cannot be represented in this form, for example, direct proportionality (since the free term is equal to zero and there is no way to get one on the right side).
This is, figuratively speaking, a “technical” type of equation. A common task is to represent the general equation of a line as an equation of a line in segments. How is it convenient? The equation of a line in segments allows you to quickly find the points of intersection of a line with coordinate axes, which can be very important in some problems of higher mathematics.
Let's find the point of intersection of the line with the axis. We reset the “y” to zero, and the equation takes the form . The desired point is obtained automatically: .
Same with the axis 
– the point at which the straight line intersects the ordinate axis.
The line passing through the point K(x 0 ; y 0) and parallel to the line y = kx + a is found by the formula:
y - y 0 = k(x - x 0) (1)
Where k is the slope of the line.
Alternative formula:
A line passing through the point M 1 (x 1 ; y 1) and parallel to the line Ax+By+C=0 is represented by the equation
A(x-x 1)+B(y-y 1)=0 . (2)
Example No. 1. Write an equation for a straight line passing through the point M 0 (-2,1) and at the same time:a) parallel to the straight line 2x+3y -7 = 0;
b) perpendicular to the straight line 2x+3y -7 = 0.
Solution . Let's imagine the equation with the slope in the form y = kx + a. To do this, move all values except y to the right side: 3y = -2x + 7 . Then divide the right-hand side by a factor of 3. We get: y = -2/3x + 7/3
Let's find the equation NK passing through the point K(-2;1), parallel to the straight line y = -2 / 3 x + 7 / 3
Substituting x 0 = -2, k = -2 / 3, y 0 = 1 we get:
y-1 = -2 / 3 (x-(-2))
or
y = -2 / 3 x - 1 / 3 or 3y + 2x +1 = 0
Example No. 2. Write the equation of a line parallel to the line 2x + 5y = 0 and forming, together with the coordinate axes, a triangle whose area is 5.
Solution
. Since the lines are parallel, the equation of the desired line is 2x + 5y + C = 0. Area right triangle, where a and b are its legs. Let's find the intersection points of the desired line with the coordinate axes: 
;
.
So, A(-C/2,0), B(0,-C/5). Let's substitute it into the formula for area: 
. We get two solutions: 2x + 5y + 10 = 0 and 2x + 5y – 10 = 0.
Example No. 3. Write an equation for a line passing through the point (-2; 5) and parallel to the line 5x-7y-4=0.
Solution. This straight line can be represented by the equation y = 5 / 7 x – 4 / 7 (here a = 5 / 7). The equation of the desired line is y – 5 = 5 / 7 (x – (-2)), i.e. 7(y-5)=5(x+2) or 5x-7y+45=0 .
Example No. 4. Having solved example 3 (A=5, B=-7) using formula (2), we find 5(x+2)-7(y-5)=0.
Example No. 5. Write an equation for a line passing through the point (-2;5) and parallel to the line 7x+10=0.
Solution. Here A=7, B=0. Formula (2) gives 7(x+2)=0, i.e. x+2=0. Formula (1) is not applicable, since given equation cannot be resolved relative to y (this line is parallel to the ordinate).
