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 ,
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 represent 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. The area of a 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).
Properties of a straight line in Euclidean geometry.
An infinite number of straight lines can be drawn through any point.
Through any two non-coinciding points a single straight line can be drawn.
Two divergent lines in a plane either intersect at a single point or are
parallel (follows from the previous one).
In three-dimensional space, there are three options for the relative position of two lines:
- lines intersect;
- lines are parallel;
- straight lines intersect.
Straight line— algebraic curve of the first order: a straight line in the Cartesian coordinate system
is given on the plane by an equation of the first degree (linear equation).
General equation of a straight line.
Definition. Any straight line on the plane can be specified by a first-order equation
Ax + Wu + C = 0,
and constant A, B are not equal to zero at the same time. This first order equation is called general
equation of a straight line. Depending on the values of the constants A, B And WITH The following special cases are possible:
. C = 0, A ≠0, B ≠ 0- a straight line passes through the origin
. A = 0, B ≠0, C ≠0 (By + C = 0)- straight line parallel to the axis Oh
. B = 0, A ≠0, C ≠ 0 (Ax + C = 0)- straight line parallel to the axis Oh
. B = C = 0, A ≠0- the straight line coincides with the axis Oh
. A = C = 0, B ≠0- the straight line coincides with the axis Oh
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 a normal vector.
Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)
perpendicular to the line given by the equation
Ax + Wu + C = 0.
Example. Find the equation of a line passing through a point A(1, 2) perpendicular to the vector (3, -1).
Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C
Let's substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore
C = -1. Total: the required equation: 3x - y - 1 = 0.
Equation of a line passing through two points.
Let two points be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2, z 2), Then equation of a line,
passing through these points:
If any of the denominators is zero, the corresponding numerator should be set equal to zero. On
plane, the equation of the straight line written above is simplified:
If x 1 ≠ x 2 And x = x 1, If x 1 = x 2 .
Fraction = k called slope direct.
Example. Find the equation of the line passing through points A(1, 2) and B(3, 4).
Solution. Applying the formula written above, we get:
Equation of a straight line using a point and slope.
If the general equation of a line Ax + Wu + C = 0 lead to:
and designate 
, then the resulting equation is called
equation of a straight line with slope k.
Equation of a straight line from a point and a direction vector.
By analogy with the point considering the equation of a straight line through the normal vector, you can enter the task
a straight line through a point and a directing vector of a straight line.
Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition
Aα 1 + Bα 2 = 0 called directing vector of a straight line.
Ax + Wu + C = 0.
Example. Find the equation of a straight line with a direction vector (1, -1) and passing through the point A(1, 2).
Solution. We will look for the equation of the desired line in the form: Ax + By + C = 0. According to the definition,
coefficients must satisfy the following conditions:
1 * A + (-1) * B = 0, i.e. A = B.
Then the equation of the straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0.
at x = 1, y = 2 we get C/A = -3, i.e. required equation:
x + y - 3 = 0
Equation of a straight line in segments.
If in the general equation of the straight line Ах + Ву + С = 0 С≠0, then, dividing by -С, we get:
or where
The geometric meaning of the coefficients is that the coefficient a is the coordinate of the intersection point
straight with axis Oh, A b- coordinate of the point of intersection of the line with the axis Oh.
Example. The general equation of a straight line is given x - y + 1 = 0. Find the equation of this line in segments.
C = 1, , a = -1, b = 1.
Normal equation of a line.
If both sides of the equation Ax + Wu + C = 0 divide by number 
which is called
normalizing factor, then we get
xcosφ + ysinφ - p = 0 -normal equation of a line.
The sign ± of the normalizing factor must be chosen so that μ*C< 0.
r- the length of the perpendicular dropped from the origin to the straight line,
A φ - the angle formed by this perpendicular with the positive direction of the axis Oh.
Example. The general equation of the line is given 12x - 5y - 65 = 0. Required to write different types of equations
this straight line.
The equation of this line in segments:
The equation of this line with the slope: (divide by 5)
Equation of a line:
cos φ = 12/13; sin φ= -5/13; p = 5.
It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines,
parallel to the axes or passing through the origin.
The angle between straight lines on a plane.
Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2, then the acute angle between these lines
will be defined as
Two lines are parallel if k 1 = k 2. Two lines are perpendicular
If k 1 = -1/ k 2 .
Theorem.
Direct Ax + Wu + C = 0 And A 1 x + B 1 y + C 1 = 0 parallel when the coefficients are proportional
A 1 = λA, B 1 = λB. If also С 1 = λС, then the lines coincide. Coordinates of the point of intersection of two lines
are found as a solution to the system of equations of these lines.
The equation of a line passing through a given point perpendicular to a given line.
Definition. Line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b
represented by the equation:
Distance from a point to a line.
Theorem. If a point is given M(x 0, y 0), then the distance to the straight line Ax + Wu + C = 0 defined as:
Proof. Let the point M 1 (x 1, y 1)- the base of a perpendicular dropped from a point M for a given
direct. Then the distance between points M And M 1:
(1)
Coordinates x 1 And at 1 can be found as a solution to the system of equations:
The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicularly
given straight line. If we transform the first equation of the system to the form:
A(x - x 0) + B(y - y 0) + Ax 0 + By 0 + C = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem has been proven.
This article reveals the derivation of the equation of a straight line passing through two given points in a rectangular coordinate system located on a 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 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 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 the 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 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 during algebra lessons. School problems differed in that the equation of a straight line with an angle coefficient was known, 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 becomes 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.
It is impossible to memorize such a huge number of formulas at once. 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 a slope, which has 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 the waste of a lot of 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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Let two points be given M 1 (x 1,y 1) And M 2 (x 2,y 2). Let us write the equation of the line in the form (5), where k still unknown coefficient:
Since the point M 2 belongs to a given line, then its coordinates satisfy equation (5): 
. Expressing from here and substituting it into equation (5), we obtain the required equation:
If 
this equation can be rewritten in a form that is more convenient for memorization:
(6)
Example. Write down the equation of a straight line passing through points M 1 (1,2) and M 2 (-2,3)
Solution. 
. Using the property of proportion and performing the necessary transformations, we obtain the general equation of a straight line:
Angle between two straight lines
Consider two straight lines l 1 And l 2:
l 1: , , And
l 2: , ,
φ is the angle between them (). From Fig. 4 it is clear: .
 |
From here 
, or
Using formula (7) you can determine one of the angles between straight lines. The second angle is equal to .
Example. Two straight lines are given by the equations y=2x+3 and y=-3x+2. find the angle between these lines.
Solution. From the equations it is clear that k 1 =2, and k 2 =-3. Substituting these values into formula (7), we find
. Thus, the angle between these lines is equal to .
Conditions for parallelism and perpendicularity of two straight lines
If straight l 1 And l 2 are parallel, then φ=0 And tgφ=0. from formula (7) it follows that , whence k 2 =k 1. Thus, the condition for parallelism of two lines is the equality of their angular coefficients.
If straight l 1 And l 2 are perpendicular, then φ=π/2, α 2 = π/2+ α 1 . 
. Thus, the condition for the perpendicularity of two straight lines is that their angular coefficients are inverse in magnitude and opposite in sign.
Distance from point to line
Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Bу + C = 0 is determined as
Proof. Let point M 1 (x 1, y 1) be the base of a perpendicular dropped from point M to a given straight line. Then the distance between points M and M 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 a line passing through a given point M 0 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 = -3x + 7; y = 2x + 1.
k 1 = -3; k 2 = 2 tanj= ; j = p/4.
Example. Show that the lines 3x – 5y + 7 = 0 and 10x + 6y – 3 = 0 are perpendicular.
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.
We find the equation of side AB: ; 4x = 6y – 6;
2x – 3y + 3 = 0;
The required height equation has the form: Ax + By + C = 0 or y = kx + b.
k= . Then y = . Because height passes through point C, then its coordinates satisfy this equation: whence b = 17. Total: .
Answer: 3x + 2y – 34 = 0.
The distance from a point to a line is determined by the length of the perpendicular drawn from the point to the line.
If the line is parallel to the projection plane (h | | P 1), then in order to determine the distance from the point A to a straight line h it is necessary to lower the perpendicular from the point A to the horizontal h.
Let's consider a more complex example, when the straight line occupies a general position. Let it be necessary to determine the distance from a point M to a straight line A general position.
Determination task distances between parallel lines is solved similarly to the previous one. A point is taken on one line and a perpendicular is dropped from it to another line. The length of a perpendicular is equal to the distance between parallel lines.
Second order curve is a line defined by an equation of the second degree relative to the current Cartesian coordinates. In the general case, Ax 2 + 2Bxy + Su 2 + 2Dx + 2Ey + F = 0,
where A, B, C, D, E, F are real numbers and at least one of the numbers A 2 + B 2 + C 2 ≠0.
Circle
Circle center– this is the geometric locus of points in the plane equidistant from a point in the plane C(a,b).
The circle is given by the following equation:
Where x,y are the coordinates of an arbitrary point on the circle, R is the radius of the circle.
Sign of the equation of a circle
1. The term with x, y is missing
2. The coefficients for x 2 and y 2 are equal
Ellipse
Ellipse is called the geometric locus of points in a plane, the sum of the distances of each of which from two given points of this plane is called foci (a constant value).
The canonical equation of the ellipse:
X and y belong to the ellipse.
a – semimajor axis of the ellipse
b – semi-minor axis of the ellipse
The ellipse has 2 axes of symmetry OX and OU. The axes of symmetry of an ellipse are its axes, the point of their intersection is the center of the ellipse. The axis on which the foci are located is called focal axis. The point of intersection of the ellipse with the axes is the vertex of the ellipse.
Compression (tension) ratio: ε = s/a– eccentricity (characterizes the shape of the ellipse), the smaller it is, the less the ellipse is extended along the focal axis.
If the centers of the ellipse are not at the center C(α, β)
Hyperbola
Hyperbole is called the geometric locus of points in a plane, the absolute value of the difference in distances, each of which from two given points of this plane, called foci, is a constant value different from zero.
Canonical hyperbola equation
A hyperbola has 2 axes of symmetry:
a – real semi-axis of symmetry
b – imaginary semi-axis of symmetry
Asymptotes of a hyperbola:
Parabola
Parabola is the locus of points in the plane equidistant from a given point F, called the focus, and a given straight line, called the directrix.
The canonical equation of a parabola:
У 2 =2рх, where р is the distance from the focus to the directrix (parabola parameter)
If the vertex of the parabola is C (α, β), then the equation of the parabola (y-β) 2 = 2р(x-α)
If the focal axis is taken as the ordinate axis, then the equation of the parabola will take the form: x 2 =2qу