Find the equation of a line online using coordinates. Equation of a line passing through two points
This article reveals how to obtain the equation of a straight line passing through two given points in rectangular system coordinates located on the plane. Let us derive the equation of a straight line passing through two given points in a rectangular coordinate system. We will clearly show and solve several examples related to the material covered.
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Before obtaining the equation of a line passing through two given points, it is necessary to pay attention to some facts. There is an axiom that says that through two divergent points on a plane it is possible to draw a straight line and only one. In other words, two given points on a plane are defined by a straight line passing through these points.
If the plane is defined by the rectangular coordinate system Oxy, then any straight line depicted in it will correspond to the equation of a straight line on the plane. There is also a connection with the directing vector of the straight line. This data is sufficient to compile the equation of a straight line passing through two given points.
Let's look at an example of solving a similar problem. It is necessary to create an equation for a straight line a passing through two divergent 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 line on a plane, having the form x - x 1 a x = y - y 1 a y, a rectangular coordinate system O x y is specified with a 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 create a canonical equation of a straight line a, which will pass through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2).
Straight a has a direction 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 obtain 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 down the parametric equations of a line on a plane that passes through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2). We obtain an equation of the form x = x 1 + (x 2 - x 1) · λ y = y 1 + (y 2 - y 1) · λ or x = x 2 + (x 2 - x 1) · λ y = y 2 + (y 2 - y 1) · λ .
Let's take a closer look at solving several examples.
Example 1
Write down 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 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 conditions of the problem, we have that x 1 = - 5, y 1 = 2 3, x 2 = 1, y 2 = - 1 6. It is necessary to substitute numeric 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 takes 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 you need to solve a problem with a different type of equation, then first you can go to the canonical one, since it is easier to come from it to any other one.
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 given two points. We get an equation of the form x - 1 4 - 1 = y - 1 2 - 1 ⇔ x - 1 3 = y - 1 1 .
Let's bring the canonical equation to the desired form, then we get:
x - 1 3 = y - 1 1 ⇔ 1 x - 1 = 3 y - 1 ⇔ x - 3 y + 2 = 0
Answer: x - 3 y + 2 = 0 .
Examples of such tasks were discussed in school textbooks in algebra lessons. School problems differed in that the equation of a straight line with slope, having the form y = k x + b. If you need to find the value of the slope k and the number b for which the equation y = 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 angular coefficient takes on the value of infinity, and the straight line M 1 M 2 is defined by a general incomplete equation of the form x - x 1 = 0 .
Because the points 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 for k and b.
To do this, we find k = y 2 - y 1 x 2 - x 1 b = y 1 - y 2 - y 1 x 2 - x 1 x 1 or k = y 2 - y 1 x 2 - x 1 b = y 2 - y 2 - y 1 x 2 - x 1 x 2 .
With these values of k and b, the equation of a 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 away great amount formulas won't work. To do this, it is necessary to increase the number of repetitions in solving problems.
Example 3
Write down the equation of a straight line with an angular coefficient passing through points with coordinates M 2 (2, 1) and y = k x + b.
Solution
To solve the problem, we use a formula with an angular coefficient of the form y = 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 are located on a straight line, then their coordinates must make the equation y = k x + b a true equality. From this 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 find that the required equation passing through the given points will be an equation of the form y = 2 3 x - 1 3 .
This method of solution predetermines spending large quantity time. There is a way in which the task is solved in literally two steps.
Let us write the canonical equation of the 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), the straight line M passing through them 1 M 2 , it is necessary to obtain the equation of this line.
We have that canonical equations of the form x - x 1 a x = y - y 1 a y = z - z 1 a z and parametric equations of the form x = x 1 + a x · λ y = y 1 + a y · λ z = z 1 + a z · λ are able to define a line in the coordinate system O x y z, passing through points having coordinates (x 1, y 1, z 1) with a direction 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 straight 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 = y - y 2 y 2 - y 1 = z - z 2 z 2 - z 1, in turn parametric x = x 1 + (x 2 - x 1) λ y = 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 = z 2 + (z 2 - z 1) · λ .
Consider a drawing that shows 2 given points in space and the equation of a straight line.
Example 4
Write the equation of a line defined in a rectangular coordinate system O x y z of three-dimensional space, passing through given two points with coordinates M 1 (2, - 3, 0) and M 2 (1, - 3, - 5).
Solution
It is necessary to find the canonical equation. Since we are talking about three-dimensional space, it means that when a line passes through given points, the desired canonical equation will take the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 .
By condition we have that x 1 = 2, y 1 = - 3, z 1 = 0, x 2 = 1, y 2 = - 3, z 2 = - 5. It follows that the necessary equations will be written as follows:
x - 2 1 - 2 = y - (- 3) - 3 - (- 3) = z - 0 - 5 - 0 ⇔ x - 2 - 1 = y + 3 0 = z - 5
Answer: x - 2 - 1 = y + 3 0 = z - 5.
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Definition. Any straight line on the plane can be specified by a first-order equation
Ax + Wu + C = 0,
Moreover, the constants A and 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 constant A, B and C the following special cases are possible:
C = 0, A ≠0, B ≠ 0 – the straight line passes through the origin
A = 0, B ≠0, C ≠0 (By + C = 0) - straight line parallel to the Ox axis
B = 0, A ≠0, C ≠ 0 (Ax + C = 0) – straight line parallel to the Oy axis
B = C = 0, A ≠0 – the straight line coincides with the Oy axis
A = C = 0, B ≠0 – the straight line coincides with the Ox axis
The equation of a straight line can be presented in different forms depending on any given initial conditions.
Equation of a straight line from a point and normal vector
Definition. In the Cartesian rectangular coordinate system, a vector with components (A, B) is perpendicular to the straight line given by the equation Ax + By + C = 0.
Example. Find the equation of the line passing through the point A(1, 2) perpendicular to (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, we 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 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 the line passing through these points is:
If any of the denominators is equal to zero, the corresponding numerator should be equal to zero. On the plane, the equation of the line written above is simplified:
if x 1 ≠ x 2 and x = x 1, if x 1 = x 2.
The fraction = k is 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 from a point and slope
If the total Ax + Bu + C = 0, lead to the form:
and designate 
, then the resulting equation is called equation of a straight line with slopek.
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 a normal vector, you can enter the definition of a straight line through a point and the directing vector of the 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 a directing vector of the 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. In accordance with the definition, the coefficients must satisfy the 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. for x = 1, y = 2 we obtain C/ A = -3, i.e. required equation:
Equation of a 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
The geometric meaning of the coefficients is that the coefficient A is the coordinate of the point of intersection of the line with the Ox axis, and b– the coordinate of the point of intersection of the straight line with the Oy axis.
Example. The general equation of the line x – y + 1 = 0 is given. 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 + By + C = 0 are multiplied by the 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. 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 = 0. You need to write Various types equations of this line.
equation of this line in segments: 
equation of this line with 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 of coordinates.
Example. The straight line cuts off equal positive segments on the coordinate axes. Write an equation of a straight line if the area of the triangle formed by these segments is 8 cm 2.
Solution. The equation of the straight line 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 an equation for a straight line passing through point A(-2, -3) and the origin.
Solution.
The equation of the straight line is: 
, where x 1 = y 1 = 0; x 2 = -2; y 2 = -3.
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 lines are perpendicular if k 1 = -1/ k 2.
Theorem. The lines Ax + Bу + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A 1 = λA, B 1 = λB are proportional. If also C 1 = λC, 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. A straight line passing through the point M 1 (x 1, y 1) and perpendicular to the straight line y = 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 + Bу + C = 0 is determined as
.
Proof. Let point M 1 (x 1, y 1) be the base of the perpendicular dropped from point M to a given straight line. Then the distance between points M and M 1:
(1)
The coordinates x 1 and y 1 can be found by solving the system of equations:
The second equation of the system is the equation of the line passing through given point M 0 is perpendicular to a given straight line. If we transform the first equation of the system to the form:
A(x – x 0) + B(y – y 0) + Ax 0 + By 0 + C = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem has been proven.
Example. Determine the angle between the lines: y = -3 x + 7; y = 2 x + 1.
k 1 = -3; k 2 = 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 = 3/5, k 2 = -5/3, k 1* k 2 = -1, therefore, the lines are perpendicular.
Example. Given are the vertices of the triangle A(0; 1), B (6; 5), C (12; -1). Find the equation of the height drawn from vertex C.
Solution. We find the equation of side AB: 
; 4 x = 6 y – 6;
2 x – 3 y + 3 = 0;
The required height equation has the form: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: 
from where b = 17. Total: . 
Answer: 3 x + 2 y – 34 = 0.
General equation straight:
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 No. 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
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 O 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 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 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, died down prom, 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 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 (co-directional 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 transform 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 infinitely many 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 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 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 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.
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 there is a 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 as a final 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 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 compile 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 the point M 0 (- 3, 4) specified by the problem condition, through which the straight 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 move the term B y to the right side of the general equation. On the left side we take A out of brackets. As a result, we get: A x + C A = - B y.
This equality can be written as a proportion: x + C A - B = y A.
If B ≠ 0, we leave only the term A x on the left side of the general equation, 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 convert the general equation of a straight line into parametric ones, first go to canonical form, and then the transition from the canonical line equation 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 transition to the general form of the equation.
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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