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 Mathan, but the section about the linear function turned out to be 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 of this 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 using the 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.
The following cases are possible:
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 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 for independent decision:
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 passed, 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 general appearance:
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 straight line is given by a general equation in rectangular system coordinates, 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 mention a couple more technical points solutions. 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 types of equations of a 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.
General equation of a straight line:
Special cases of the general equation of a straight line:
and if C= 0, equation (2) will have the form
Ax + By = 0,
and the straight line defined by this equation passes through the origin, since the coordinates of the origin are x = 0, y= 0 satisfy this equation.
b) If in the general equation of the straight line (2) B= 0, then the equation takes the form
Ax + WITH= 0, or .
The equation does not contain a variable y, and the straight line defined by this equation is parallel to the axis Oy.
c) If in the general equation of the straight line (2) A= 0, then this equation will take the form
By + WITH= 0, or ;
the equation does not contain a variable x, and the straight line it defines is parallel to the axis Ox.
It should be remembered: if a straight line is parallel to some coordinate axis, then in its equation there is no term containing a coordinate of the same name as this axis.
d) When C= 0 and A= 0 equation (2) takes the form By= 0, or y = 0.
This is the equation of the axis Ox.
d) When C= 0 and B= 0 equation (2) will be written in the form Ax= 0 or x = 0.
This is the equation of the axis Oy.
The relative position of lines on a plane. The angle between straight lines on a plane. Condition for parallel lines. The condition of perpendicularity of lines.
l 1 l 2 l 1: A 1 x + B 1 y + C 1 = 0
l 2: A 2 x + B 2 y + C 2 = 0
S 2 S 1 Vectors S 1 and S 2 are called guides for their lines.
The angle between straight lines l 1 and l 2 is determined by the angle between the direction vectors.
Theorem 1: cos of the angle between l 1 and l 2 = cos(l 1 ; l 2) =
Theorem 2: In order for 2 lines to be equal it is necessary and sufficient:
Theorem 3: For 2 straight lines to be perpendicular it is necessary and sufficient:
L 1 l 2 ó A 1 A 2 + B 1 B 2 = 0
General plane equation and its special cases. Equation of a plane in segments.
General plane equation:
Ax + By + Cz + D = 0
Special cases:
1. D=0 Ax+By+Cz = 0 – the plane passes through the origin
2. С=0 Ax+By+D = 0 – plane || OZ
3. B=0 Ax+Cz+d = 0 – plane || OY
4. A=0 By+Cz+D = 0 – plane || OX
5. A=0 and D=0 By+Cz = 0 – the plane passes through OX
6. B=0 and D=0 Ax+Cz = 0 – the plane passes through OY
7. C=0 and D=0 Ax+By = 0 – the plane passes through OZ
The relative position of planes and straight lines in space:
1. The angle between straight lines in space is the angle between their direction vectors.
Cos (l 1 ; l 2) = cos(S 1 ; S 2) = = 
2. The angle between the planes is determined through the angle between their normal vectors.
Cos (l 1 ; l 2) = cos(N 1 ; N 2) = = 
3. The cosine of the angle between the line and the plane can be found through the sin of the angle between the direction vector of the line and the normal vector of the plane.
4. 2 straight || in space when their || vector guides
5. 2 planes || when || normal vectors
6. The concepts of perpendicularity of lines and planes are introduced similarly.
Question No. 14
Different kinds equations of a straight line on a plane (equation of a straight line in segments, with an angle coefficient, etc.)
Equation of a straight line in segments:
Let us assume that in the general equation of the straight line:
1. C = 0 Ах + Ву = 0 – the straight line passes through the origin.
2. a = 0 Vu + C = 0 y =
3. b = 0 Ax + C = 0 x =
4. b=C=0 Ax = 0 x = 0
5. a=C=0 Ву = 0 у = 0
Equation of a straight line with a slope:
Any straight line that is not equal to the op-amp axis (B not = 0) can be written down in the next line. form:
k = tanα α – angle between straight line and positively directed line OX
b – point of intersection of the straight line with the axis of the op-amp
Document:
Ax+By+C = 0
Wu= -Ah-S |:B
Equation of a straight line based on two points:
Question #16
Finite limit of a function at a point and for x→∞
End limit at x0:
The number A is called the limit of the function y = f(x) for x→x 0 if for any E > 0 there exists b > 0 such that for x ≠x 0 satisfying the inequality |x – x 0 |< б, выполняется условие |f(x) - A| < Е
The limit is indicated by: = A
End limit at point +∞:
The number A is called the limit of the function y = f(x) at x → + ∞ , if for any E > 0 there exists C > 0 such that for x > C the inequality |f(x) - A|< Е
The limit is indicated by: = A
End limit at point -∞:
The number A is called the limit of the function y = f(x) for x→-∞, if for any E< 0 существует С < 0 такое, что при х < -С выполняется неравенство |f(x) - A| < Е
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 this equation cannot be resolved with respect to y (this straight line is parallel to the ordinate axis).
The canonical equations of a line in space are the equations that determine the line passing through given point collinear to the direction vector.
Let a point and a direction vector be given. An arbitrary point lies on a line l only if the vectors and are collinear, i.e., the condition is satisfied for them:
.
The above equations are canonical equations straight.
Numbers m , n And p are projections of the direction vector onto the coordinate axes. Since the vector is non-zero, then all numbers m , n And p cannot simultaneously be equal to zero. But one or two of them may turn out to be zero. In analytical geometry, for example, the following entry is allowed:
,
which means that the projections of the vector on the axis Oy And Oz are equal to zero. Therefore, both the vector and the straight line defined by the canonical equations are perpendicular to the axes Oy And Oz, i.e. planes yOz .
Example 1. Write equations for a line in space perpendicular to a plane 
and passing through the point of intersection of this plane with the axis Oz
.
Solution. Let's find the point of intersection of this plane with the axis Oz. Since any point lying on the axis Oz, has coordinates , then, assuming in the given equation of the plane x = y = 0, we get 4 z- 8 = 0 or z= 2 . Therefore, the point of intersection of this plane with the axis Oz has coordinates (0; 0; 2) . Since the desired line is perpendicular to the plane, it is parallel to its normal vector. Therefore, the directing vector of the straight line can be the normal vector 
given plane.
Now let’s write down the required equations of a straight line passing through a point A= (0; 0; 2) in the direction of the vector:
Equations of a line passing through two given points
A straight line can be defined by two points lying on it 
And 
In this case, the directing vector of the straight line can be the vector . Then the canonical equations of the line take the form
.
The above equations determine a line passing through two given points.
Example 2. Write an equation for a line in space passing through the points and .
Solution. Let us write down the required equations of the straight line in the form given above in the theoretical reference:
.
Since , then the desired straight line is perpendicular to the axis Oy .
Straight as the line of intersection of planes
A straight line in space can be defined as the line of intersection of two non-parallel planes and, i.e., as a set of points satisfying a system of two linear equations
The equations of the system are also called general equations straight in space.
Example 3. Compose canonical equations of a line in space given by general equations
Solution. To write the canonical equations of a line or, what is the same, the equations of a line passing through two given points, you need to find the coordinates of any two points on the line. They can be the points of intersection of a straight line with any two coordinate planes, for example yOz And xOz .
Point of intersection of a line and a plane yOz has an abscissa x= 0 . Therefore, assuming in this system of equations x= 0, we get a system with two variables:
Her decision y = 2 , z= 6 together with x= 0 defines a point A(0; 2; 6) the desired line. Then assuming in the given system of equations y= 0, we get the system
Her decision x = -2 , z= 0 together with y= 0 defines a point B(-2; 0; 0) intersection of a line with a plane xOz .
Now let's write down the equations of the line passing through the points A(0; 2; 6) and B (-2; 0; 0) :
,
or after dividing the denominators by -2:
,
The equation of a line passing through a given point in a given direction. Equation of a line passing through two given points. The angle between two straight lines. The condition of parallelism and perpendicularity of two straight lines. Determining the point of intersection of two lines
1. Equation of a line passing through a given point A(x 1 , y 1) in a given direction, determined by the slope k,
y - y 1 = k(x - x 1). (1)
This equation defines a pencil of lines passing through a point A(x 1 , y 1), which is called the beam center.
2. Equation of a line passing through two points: A(x 1 , y 1) and B(x 2 , y 2), written like this:
The angular coefficient of a straight line passing through two given points is determined by the formula
3. Angle between straight lines A And B is the angle by which the first straight line must be rotated A around the point of intersection of these lines counterclockwise until it coincides with the second line B. If two straight lines are given by equations with a slope
y = k 1 x + B 1 ,