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 the elementary grades, and today we will learn how to deal with it using the methods of analytical geometry. To master the material, it is necessary to be able to build a straight line; know which equation defines a straight line, in particular, a straight line passing through the origin 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 turned out to be very successful and detailed. Therefore, dear teapots, first warm up there. In addition, you need to have basic knowledge O vectors otherwise the understanding of the material will be incomplete.
On this lesson we will consider ways in which you can write the equation of a straight line in a plane. I recommend that you do not neglect practical examples (even if it seems very simple), as I will supply them with elementary and important facts, techniques, which will be required in the future, including in other sections of higher mathematics.
- How to write the equation of a straight line with a slope?
- How ?
- How to find the direction vector by 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 start:
Line Equation with Slope
The well-known "school" form of the equation of a straight line is called equation of a straight line with slope factor
. For example, if a straight line is given by the equation, then its slope: . Consider the geometric meaning given coefficient and how its value affects the location of the line: 
In the course of geometry it is proved that the slope of the straight line is tangent of an angle between positive axis directionand given line: , and the corner is “unscrewed” counterclockwise.
In order not to clutter up the drawing, I drew angles for only two straight lines. Consider the "red" straight line and its slope. According to the above: (angle "alpha" is indicated by a green arc). For the "blue" straight line with the slope, equality is true (the angle "beta" is indicated by the brown arc). And if the tangent of the angle is known, then if necessary it is easy to find and the corner by using inverse function- arctangent. As they say, a trigonometric table or a calculator in hand. Thus, the slope characterizes the degree of inclination of the straight line to the x-axis.
At the same time, it is possible following cases:
1) If the slope is negative: , then the line, roughly speaking, goes from top to bottom. Examples are "blue" and "crimson" straight lines in the drawing.
2) If the slope is positive: , then the line goes from bottom to top. Examples are "black" and "red" straight lines in the drawing.
3) If the slope zero: , then the equation takes the form , and the corresponding straight line is parallel to the axis . An example is the "yellow" line.
4) For a family of straight lines parallel to the axis (there is no example in the drawing, except for the axis itself), the slope does not exist (tangent of 90 degrees not defined).
The greater the slope modulo, the steeper the line graph goes.
For example, consider two straight lines. Here , so the straight line has a steeper slope. I 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. 
.
Vice versa: the smaller the slope modulo, the straight line is flatter.
For straight lines 
the inequality is true, thus, the straight line is more than a canopy. Children's slide, so as not to plant bruises and bumps.
Why is this needed?
Prolong your torment Knowing the above facts allows you to immediately see your mistakes, in particular, errors when plotting graphs - if the drawing turned out “clearly something is wrong”. It is desirable that you straightaway it was clear that, for example, a straight line is very steep and goes from bottom to top, and a straight line is very flat, close to the axis and goes from top to bottom.
IN geometric problems often several straight lines appear, so it is convenient to denote them somehow.
Notation: straight lines are indicated by small with Latin letters: . A popular option is the designation of the same letter with natural subscripts. For example, the five lines that we have just considered can be denoted by 
.
Since any straight line is uniquely determined by two points, it can be denoted by these points: 
etc. The notation quite obviously implies that the points belong to the line.
Time to loosen up a bit:
How to write the equation of a straight line with a slope?
If a point is known that belongs to a certain line, and the slope of this line, then the equation of this line is expressed by the formula:
Example 1
Compose the equation of a straight line with a slope if it is known that the point belongs to this straight line.
Solution: We will compose the equation of a straight line according to the formula 
. IN this case:
Answer:
Examination performed elementarily. First, we look at the resulting equation and make sure that our slope is in its place. Second, the coordinates of the point must satisfy the given equation. Let's plug them into the equation:
The correct equality is obtained, which means that the point satisfies the resulting equation.
Conclusion: Equation found correctly.
A trickier example for independent solution:
Example 2
Write the equation of 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're having trouble, re-read theoretical material. More precisely, more practical, I miss many proofs.
rang last call, the prom has died down, and outside the gates home school we are waiting, in fact, analytic geometry. Jokes are over... Maybe it's just getting started =)
Nostalgically we wave the handle to the familiar and get acquainted with the general equation of a straight line. Since in analytic geometry it is precisely this that is in use:
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 an equation with a slope. First, we 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 analytic geometry, the equation of a straight line will almost always be given in general form. Well, if necessary, it is easy to bring it to a “school” form with a slope (with the exception of straight lines parallel to the y-axis).
Let's ask ourselves what enough know to build a straight line? Two points. But about this childhood case later, now sticks with arrows rule. Each straight line has a well-defined slope, to which it is easy to "adapt" vector.
A vector that is parallel to a line is called the direction vector of that line.. Obviously, any straight line has infinitely many direction vectors, and all of them will be collinear (co-directed or not - it does not matter).
I will denote the direction vector as follows: .
But one vector is not enough to build a straight line, the vector is free and is not attached to any point of 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 given a point and a direction vector?
If a certain point belonging to the line and the directing vector of this line are known, then the equation of this line can be compiled by the formula:
Sometimes it is called canonical equation of the line .
What to do when one of the coordinates is zero, we will look into practical examples below. By the way, note - both at once coordinates cannot be zero, because null vector does not provide a specific direction.
Example 3
Write an equation of a straight line given a point and a direction vector
Solution: We will compose the equation of a straight line according to the formula. In this case:
Using the properties of proportion, we get rid of fractions: 
And we bring the equation to general view:
Answer:
Drawing in such examples, as a rule, is not necessary, but for the sake of understanding: 
In the drawing, we see the starting point, the original direction vector (it can be postponed from any point on the plane) and the constructed line. By the way, in many cases, the construction of a straight line is most conveniently carried out using the slope equation. Our equation is easy to convert to the form and without any problems pick up one more point to build a straight line.
As noted at the beginning of the section, a line has infinitely many direction vectors, and they are all collinear. For example, I drew three such vectors: 
. Whichever direction vector we choose, the result will always be the same straight line equation.
Let's compose the equation of a straight line by a point and a directing vector:
Breaking down the proportion:
Divide both sides by -2 and get the familiar equation:
Those who wish can similarly test vectors 
or any other collinear vector.
Now let's decide inverse problem:
How to find the direction vector by the general equation of a straight line?
Very simple:
If the line is given by the general equation in rectangular system coordinates, then the vector is the direction vector of the given line.
Examples of finding direction vectors of straight lines: 
The statement allows us to find only one direction vector from an infinite set, but we don’t need more. Although in some cases it is advisable to reduce the coordinates of the direction vectors:
So, the equation specifies a straight line that is parallel to the axis and the coordinates of the resulting steering vector are conveniently divided by -2, getting exactly the basis vector as the steering vector. Logically.
Similarly, the equation defines a straight line parallel to the axis, and dividing the coordinates of the vector by 5, we get the ort as the direction vector.
Now let's execute check example 3. The example went up, so I remind you that in it we made up the equation of a straight line using a point and a direction vector
Firstly, according to the equation of a straight line, we restore its directing vector: 
- everything is fine, we got the original vector (in some cases, it can turn out to be collinear to the original vector, and this is usually easy to see 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 has been obtained, which we are very pleased with.
Conclusion: Job completed correctly.
Example 4
Write an equation of a straight line given a point and a direction vector
This is a do-it-yourself example. Solution and answer at the end of the lesson. It is highly desirable to make a check according to the algorithm just considered. Try to always (if possible) check on a draft. It is foolish to make mistakes where they can be 100% avoided.
In the event that one of the coordinates of the direction vector is zero, it is very simple to do:
Example 5
Solution: The formula is invalid because 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 direction vector of the straight line:
– the resulting vector is collinear to the original direction vector.
2) Substitute the coordinates of the point in the equation: 
The correct equality is obtained
Conclusion: job completed correctly
The question arises, why bother with the formula if there is a universal version that will work anyway? There are two reasons. First, the fractional formula much better to remember. Second, the disadvantage universal formula is that markedly increased risk of confusion when substituting coordinates.
Example 6
Compose the equation of a straight line given a point and a direction vector.
This is a do-it-yourself example.
Let's return to the ubiquitous two points:
How to write the equation of a straight line given 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 kind of formula, and here's why: if two points are known, then the vector will be the direction vector of this line. At the lesson Vectors for dummies we considered the simplest task– how to find the coordinates of a vector from two points. According to this problem, the coordinates of the direction vector: 
Note
: points can be "swapped" and use the formula 
. Such a decision would be equal.
Example 7
Write the equation of a straight line from two points 
.
Solution: Use the formula: 
We comb the denominators:
And shuffle the deck: 
Now is the time to get rid of fractional numbers. In this case, you need to multiply both parts by 6: 
Open the brackets and bring the equation to mind: 
Answer:
Examination obvious - coordinates starting 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 straight line is correct.
If at least one of points does not satisfy the equation, look for an error.
It is worth noting that the graphical verification in this case is difficult, because to build a line and see if the points belong to it 
, not so easy.
I will note a couple of technical points of the solution. Perhaps in this problem it is more advantageous to use the mirror formula 
and, for the same points 
make an equation: 
There are fewer fractions. If you want, you can complete the solution to the end, the result should be the same equation.
The second point is to look at the final answer and see if it can be further simplified? For example, if an equation is obtained, then it is advisable to reduce it by two: - the equation will set the same straight line. However, this is already a topic of conversation about mutual arrangement of straight lines.
Having received an answer 
in Example 7, just in case, I checked if 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 the equation of a straight line passing through the points 
.
This is an example for an independent solution, which will just allow you to better understand and work out the calculation technique.
Similar to the previous paragraph: if in the formula 
one of the denominators (direction vector coordinate) vanishes, then we rewrite it as . And again, notice how awkward and confused she began to look. I don't see much point in bringing practical examples, since we have already actually solved such a problem (see Nos. 5, 6).
Straight line normal vector (normal vector)
What is normal? In simple words, the normal is the perpendicular. That is, the normal vector of a line is perpendicular to the given line. It is obvious that any straight line has an infinite number of them (as well as directing vectors), and all the normal vectors of the straight line will be collinear (codirectional or not - it does not matter).
Dealing with them will be even easier than with direction vectors:
If a straight line is given by a general equation in a rectangular coordinate system, then the vector is the normal vector of this straight 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 simply be “removed”.
The normal vector is always orthogonal to the direction vector of the line. We will 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 write an equation of a straight line, knowing one point and a normal vector? It feels like it's possible. If the normal vector is known, then the direction of the straightest line is also uniquely determined - 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 some point belonging to the line and the normal vector of this line are known, then the equation of this line is expressed by the formula:
Here everything went without fractions and other surprises. Such is our normal vector. Love it. And respect =)
Example 9
Compose the equation of a straight line given a point and a normal vector. Find the direction vector of the straight line.
Solution: Use the formula: 
The general equation of the straight line is obtained, let's check:
1) "Remove" the coordinates of the normal vector from the equation: 
- yes, indeed, the original vector is obtained from the condition (or the vector should be collinear to the original vector).
2) Check if the point satisfies the equation: 
True equality.
After we have made sure that the equation is correct, we will perform the second, more easy part tasks. We pull out the direction vector of the straight line: 
Answer: 
In the drawing, the situation is as follows: 
For the purposes of training, a similar task for an independent solution:
Example 10
Write an equation of a straight line given a point and normal vector. Find the direction vector of the straight line.
Final section 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 straight 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 zero and there is no way to get one on the right side).
This is, figuratively speaking, a "technical" type of equation. The usual task is to general equation represent a straight line as an equation of a straight line in segments . Why is it convenient? The equation of a straight line in segments allows you to quickly find the points of intersection of a straight line with coordinate axes, which is very important in some problems of higher mathematics.
Find the point of intersection of the line with the axis. We reset the “y”, and the equation takes the form . Desired point obtained automatically: .
Same with axis 
is the point where the line intersects the y-axis.
Definition. Any line in the plane can be given by a first order equation
Ah + Wu + C = 0,
and the constants A, B are not equal to zero at the same time. This first order equation is called the general equation of a straight line. Depending on the values constant A, B and C, the following special cases are possible:
C \u003d 0, A ≠ 0, B ≠ 0 - the line passes through the origin
A \u003d 0, B ≠ 0, C ≠ 0 (By + C \u003d 0) - the line is parallel to the Ox axis
B \u003d 0, A ≠ 0, C ≠ 0 ( Ax + C \u003d 0) - the line is parallel to the Oy axis
B \u003d C \u003d 0, A ≠ 0 - the straight line coincides with the Oy axis
A \u003d C \u003d 0, B ≠ 0 - the line coincides with Ox axis
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) is perpendicular to a line, given by the equation Ah + Wu + C = 0.
Example. Find the equation of a straight line passing through the point A(1, 2) perpendicular to (3, -1).
Solution. At A = 3 and B = -1, we compose the equation of a straight line: 3x - y + C = 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 line passing through two points
Let two points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2) be given in space, then the equation of a straight line 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 straight line equation written above is simplified:
if x 1 ≠ x 2 and x = x 1 if x 1 = x 2.
Fraction = k is 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 from a point and a slope
If the total Ax + Wu + C = 0 lead to the form:
and designate 
, then the resulting equation is called equation of a straight line with a slopek.
Equation of a straight line with a point and direction vector
By analogy with the point considering the equation of a straight line through the normal vector, you can enter the assignment of a straight line through a point and a directing vector of a straight line.
Definition. Each non-zero vector (α 1, α 2), the components of which satisfy the condition A α 1 + B α 2 = 0 is called the directing vector of the 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. In accordance with the definition, the 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. for x = 1, y = 2 we get C / A = -3, i.e. desired equation:
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
geometric sense coefficients in that the coefficient A is the coordinate of the point of intersection of the line with the x-axis, and b- the coordinate of the point of intersection of the straight line with the Oy axis.
Example. Given the general equation of the line x - y + 1 = 0. Find the equation of this line in the segments.
C \u003d 1, , a \u003d -1, b \u003d 1.
Normal equation of a straight line
If both sides of the equation Ax + Vy + C = 0 are multiplied by the 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 * С< 0. р – длина перпендикуляра, опущенного из начала координат на прямую, а φ - угол, образованный этим перпендикуляром с положительным направлением оси Ох.
Example. Given the general equation of the straight line 12x - 5y - 65 \u003d 0. It is required to write Various types equations of this line.
the equation of this straight line in segments: 
the equation of this line with the slope: (divide by 5)
; 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.
Example. Direct cuts off coordinate axes equal positive segments. Write the equation of a straight line if the area of the triangle formed by these segments is 8 cm 2.
Solution. The straight line equation has the form: , ab /2 = 8; ab=16; a=4, a=-4. a = -4< 0 не подходит по условию задачи. Итого: или х + у – 4 = 0.
Example. Write the equation of a straight line passing through the point A (-2, -3) and the origin.
Solution.
The equation of a straight line has the form: 
, where x 1 \u003d y 1 \u003d 0; x 2 \u003d -2; y 2 \u003d -3.
Angle between lines on a plane
Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2 , then sharp corner 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. The straight lines Ax + Vy + C \u003d 0 and A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients A 1 \u003d λA, B 1 \u003d λB are proportional. If also С 1 = λС, then the lines coincide. The coordinates of the point of intersection of two lines are found as a solution to the system of equations of these lines.
Equation of a line passing through a given point perpendicular to a given line
Definition. The line passing through the point M 1 (x 1, y 1) and perpendicular to the line y \u003d kx + b is represented by the equation:
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:
(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 given point M 0 is perpendicular to a 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.
Example. Determine the angle between the lines: y = -3 x + 7; y = 2 x + 1.
k 1 \u003d -3; k2 = 2; tgφ = 
; φ= π /4.
Example. Show that the lines 3x - 5y + 7 = 0 and 10x + 6y - 3 = 0 are perpendicular.
Solution. 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.
Solution. We find the equation of the side AB: 
; 4 x = 6 y - 6;
2x – 3y + 3 = 0;
The desired height equation is: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: 
whence b = 17. Total: . 
Answer: 3x + 2y - 34 = 0.
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 \u003d k (x - x 0) (1)
Where k is the slope of the straight line.
Alternative formula:
The 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 #1. Compose the equation of 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 line 2x+3y -7 = 0.
Solution . Let's represent the slope equation as y = kx + a . To do this, we transfer all values except y to right side: 3y = -2x + 7 . Then we divide the right side by the coefficient 3 . We get: y = -2/3x + 7/3
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 \u003d -2, k \u003d -2 / 3, y 0 \u003d 1 we get:
y-1 = -2 / 3 (x-(-2))
or
y = -2 / 3 x - 1 / 3 or 3y + 2x +1 = 0
Example #2. Write the equation of a straight line parallel to the straight 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 required line is 2x + 5y + C = 0. Area right triangle, where a and b are its legs. Find the points of intersection of the desired line with the coordinate axes: 
;
.
So, A(-C/2,0), B(0,-C/5). Substitute in the formula for the area: 
. We get two solutions: 2x + 5y + 10 = 0 and 2x + 5y - 10 = 0 .
Example #3. Write the equation of the line passing through the point (-2; 5) and the parallel 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 #4. Solving example 3 (A=5, B=-7) using formula (2), we find 5(x+2)-7(y-5)=0.
Example number 5. Write the equation of a straight line passing through the point (-2;5) and a parallel straight 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 this equation cannot be solved with respect to y (this straight line is parallel to the y-axis).
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).
There are three options in 3D space. relative position two straight lines:
- lines intersect;
- straight lines are parallel;
- straight lines intersect.
Straight line is an algebraic curve of the first order: in Cartesian system coordinates straight line
is given on the plane by an equation of the first degree (linear equation).
General equation of a straight line.
Definition. Any line in the plane can be given by a first order equation
Ah + Wu + C = 0,
and constant A, B not equal to zero at the same time. This first order equation is called general
straight line equation. Depending on the values of the constants A, B And WITH The following special cases are possible:
. C = 0, A ≠ 0, B ≠ 0- the line passes through the origin
. A = 0, B ≠0, C ≠0 ( By + C = 0)- straight line parallel to the axis Oh
. B = 0, A ≠ 0, C ≠ 0 ( Ax + C = 0)- straight line parallel to the axis OU
. B = C = 0, A ≠ 0- the line coincides with the axis OU
. A = C = 0, B ≠ 0- the line coincides with the axis Oh
The equation of a straight line can be represented in various forms depending on any given
initial conditions.
Equation of a straight line by a point and a normal vector.
Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)
perpendicular to the line given by the equation
Ah + Wu + C = 0.
Example. Find the equation of a straight line passing through a point A(1, 2) perpendicular to the vector (3, -1).
Solution. Let's compose at A \u003d 3 and B \u003d -1 the equation of the straight line: 3x - y + C \u003d 0. To find the coefficient C
we substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore
C = -1. Total: the desired equation: 3x - y - 1 \u003d 0.
Equation of a straight line passing through two points.
Let two points be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2 , z 2), Then straight line equation,
passing through these points:
If any of the denominators is equal to zero, the corresponding numerator should be set equal to zero. On
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 straight.
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 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.
Yandex.RTB R-A-339285-1
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-coincident 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 draw up canonical equation straight line a that 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 parametric equations a straight line on the 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. Need to substitute numerical values into 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 have been discussed in school textbooks in algebra class. 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 line M 1 M 2 is defined by the 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 the given two points takes next view y = y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 1 or y = y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 2 .
Remember this right now great amount formulas 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 spending a large number 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-coinciding points with coordinates M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), a straight line M 1 M 2 passing through them, you need to get 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. Because we are talking about three-dimensional space, which 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 \u003d y - y 1 y 2 - y 1 \u003d 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. Hence it follows that necessary equations will be written like this:
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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