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 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 the 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 a perpendicular from a point A to the horizontal h.
Let's consider more complex example, when the straight line takes 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 locus 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 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
Let two points be given M(X 1 ,U 1) and N(X 2,y 2). Let's find the equation of the line passing through these points.
Since this line passes through the point M, then according to formula (1.13) its equation has the form
U – Y 1 = K(X–x 1),
Where K– unknown angular coefficient.
The value of this coefficient is determined from the condition that the desired straight line passes through the point N, which means its coordinates satisfy equation (1.13)
Y 2 – Y 1 = K(X 2 – X 1),
From here you can find the slope of this line:
,
Or after conversion
(1.14)
Formula (1.14) determines Equation of a line passing through two points M(X 1, Y 1) and N(X 2, Y 2).
In the special case when points M(A, 0), N(0, B), A ¹ 0, B¹ 0, lie on the coordinate axes, equation (1.14) will take a simpler form
Equation (1.15) called Equation of a straight line in segments, Here A And B denote the segments cut off by a straight line on the axes (Figure 1.6).
Figure 1.6
Example 1.10. Write an equation for a line passing through the points M(1, 2) and B(3, –1).
. According to (1.14), the equation of the desired line has the form
2(Y – 2) = -3(X – 1).
Transferring all terms to the left side, we finally obtain the desired equation
3X + 2Y – 7 = 0.
Example 1.11. Write an equation for a line passing through a point M(2, 1) and the point of intersection of the lines X+ Y – 1 = 0, X – y+ 2 = 0.
. We will find the coordinates of the point of intersection of the lines by solving these equations together
If we add these equations term by term, we get 2 X+ 1 = 0, whence . Substituting the found value into any equation, we find the value of the ordinate U:
Now let’s write the equation of the straight line passing through the points (2, 1) and:
or .
Hence or –5( Y – 1) = X – 2.
We finally obtain the equation of the desired line in the form X + 5Y – 7 = 0.
Example 1.12. Find the equation of the line passing through the points M(2.1) and N(2,3).
Using formula (1.14), we obtain the equation
It doesn't make sense since the second denominator is zero. From the conditions of the problem it is clear that the abscissas of both points have the same value. This means that the desired straight line is parallel to the axis OY and its equation is: x = 2.
Comment . If, when writing the equation of a line using formula (1.14), one of the denominators turns out to be equal to zero, then the desired equation can be obtained by equating the corresponding numerator to zero.
Let's consider other ways to define a line on a plane.
1. Let a non-zero vector be perpendicular to the given line L, and point M 0(X 0, Y 0) lies on this line (Figure 1.7).
Figure 1.7
Let's denote M(X, Y) any point on a line L. Vectors and 
Orthogonal. Using the conditions of orthogonality of these vectors, we obtain or A(X – X 0) + B(Y – Y 0) = 0.
We have obtained the equation of a line passing through a point M 0 is perpendicular to the vector. This vector is called Normal vector to a straight line L. The resulting equation can be rewritten as
Oh + Wu + WITH= 0, where WITH = –(AX 0 + By 0), (1.16),
Where A And IN– coordinates of the normal vector.
We obtain the general equation of the line in parametric form.
2. A straight line on a plane can be defined as follows: let a non-zero vector be parallel to the given straight line L and period M 0(X 0, Y 0) lies on this line. Let's take an arbitrary point again M(X, y) on a straight line (Figure 1.8).
Figure 1.8
Vectors and 
collinear.
Let us write down the condition for the collinearity of these vectors: , where T– an arbitrary number called a parameter. Let's write this equality in coordinates:
These equations are called Parametric equations Straight. Let us exclude the parameter from these equations T:
These equations can otherwise be written in the form
. (1.18)
The resulting equation is called The canonical equation of the line. The vector is called The directing vector is straight .
Comment . It is easy to see that if is the normal vector to the line L, then its direction vector can be the vector since , i.e. .
Example 1.13. Write the equation of a line passing through a point M 0(1, 1) parallel to line 3 X + 2U– 8 = 0.
Solution . The vector is the normal vector to the given and desired lines. Let's use the equation of a line passing through a point M 0 with a given normal vector 3( X –1) + 2(U– 1) = 0 or 3 X + 2у– 5 = 0. We obtained the equation of the desired line.
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 relative position two straight lines:
Straight line— algebraic curve of the first order: a straight line in the Cartesian coordinate system
is given on the plane by an equation of the first degree (linear equation).
General equation of a straight line.
Definition. Any straight line on the plane can be specified by a first-order equation
Ax + Wu + C = 0,
and constant A, B are not equal to zero at the same time. This first order equation is called general
equation of a straight line. Depending on the values of the constants A, B And WITH The following special cases are possible:
. C = 0, A ≠0, B ≠ 0- a straight line passes through the origin
. A = 0, B ≠0, C ≠0 (By + C = 0)- straight line parallel to the axis Oh
. B = 0, A ≠0, C ≠ 0 (Ax + C = 0)- straight line parallel to the axis OU
. B = C = 0, A ≠0- the straight line coincides with the axis OU
. A = C = 0, B ≠0- the straight line coincides with the axis Oh
The equation of a straight line can be represented in in various forms depending on any given
initial conditions.
Equation of a straight line from a point and a normal vector.
Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)
perpendicular to the line given by the equation
Ax + Wu + C = 0.
Example. Find the equation of a line passing through a point A(1, 2) perpendicular to the vector (3, -1).
Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C
Let's substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore
C = -1. Total: the required equation: 3x - y - 1 = 0.
Equation of a line passing through two points.
Let two points be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2, z 2), Then equation of a line,
passing through these points:
If any of the denominators is zero, the corresponding numerator should be set equal to zero. On
plane, the equation of the straight line written above is simplified:
If x 1 ≠ x 2 And x = x 1, If x 1 = x 2 .
Fraction = k called slope straight.
Example. Find the equation of the line passing through points A(1, 2) and B(3, 4).
Solution. Applying the formula written above, we get:
Equation of a straight line using a point and slope.
If the general equation of the 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 OU.
Example. The general equation of a straight line is given x - y + 1 = 0. Find the equation of this line in segments.
C = 1, , a = -1, b = 1.
Normal equation of a line.
If both sides of the equation Ax + Wu + C = 0 divide by number 
which is called
normalizing factor, then we get
xcosφ + ysinφ - p = 0 -normal equation of a line.
The sign ± of the normalizing factor must be chosen so that μ*C< 0.
R- the length of the perpendicular dropped from the origin to the straight line,
A φ - the angle formed by this perpendicular with the positive direction of the axis Oh.
Example. The general equation of the line is given 12x - 5y - 65 = 0. Required to write Various types equations
this straight line.
The equation of this line in segments:
The equation of this line with the slope: (divide by 5)
Equation of a line:
cos φ = 12/13; sin φ= -5/13; p = 5.
It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines,
parallel to the axes or passing through the origin.
The angle between straight lines on a plane.
Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2, That sharp corner 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.
Equation of a line passing through two points. In the article" " I promised you to look at the second way to solve the presented problems of finding the derivative, given a graph of a function and a tangent to this graph. We will discuss this method in , do not miss! Why in the next one?
The fact is that the formula for the equation of a straight line will be used there. Of course, we could simply show this formula and advise you to learn it. But it’s better to explain where it comes from (how it is derived). It's necessary! If you forget it, you can quickly restore itwill not be difficult. Everything is outlined below in detail. So, we have two points A on the coordinate plane(x 1;y 1) and B(x 2;y 2), a straight line is drawn through the indicated points: 
Here is the direct formula itself:
*That is, when substituting specific coordinates of points, we get an equation of the form y=kx+b.
**If you simply “memorize” this formula, then there is a high probability of getting confused with the indices when X. In addition, indices can be designated in different ways, for example:
That's why it's important to understand the meaning.
Now the derivation of this formula. Everything is very simple!
Triangles ABE and ACF are similar in acute angle (the first sign of similarity right triangles). It follows from this that the ratios of the corresponding elements are equal, that is:
Now we simply express these segments through the difference in the coordinates of the points:
Of course, there will be no error if you write the relationships of the elements in a different order (the main thing is to maintain consistency):
The result will be the same equation of the line. This is all!
That is, no matter how the points themselves (and their coordinates) are designated, by understanding this formula you will always find the equation of a straight line.
The formula can be derived using the properties of vectors, but the principle of derivation will be the same, since we will be talking about the proportionality of their coordinates. In this case, the same similarity of right triangles works. In my opinion, the conclusion described above is more clear)).
View output using vector coordinates >>>
Let a straight line be constructed on the coordinate plane passing through two given points A(x 1;y 1) and B(x 2;y 2). Let us mark an arbitrary point C on the line with coordinates ( x; y). We also denote two vectors:
It is known that for vectors lying on parallel lines (or on the same line), their corresponding coordinates are proportional, that is:
— we write down the equality of the ratios of the corresponding coordinates:
Let's look at an example:
Find the equation of a straight line passing through two points with coordinates (2;5) and (7:3).
You don’t even have to build the straight line itself. We apply the formula:
It is important that you grasp the correspondence when drawing up the ratio. You can't go wrong if you write:
Answer: y=-2/5x+29/5 go y=-0.4x+5.8
In order to make sure that the resulting equation is found correctly, be sure to check - substitute the coordinates of the data in the condition of the points into it. The equations should be correct.
That's all. I hope the material was useful to you.
Sincerely, Alexander.
P.S: I would be grateful if you tell me about the site on social networks.
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 Ox axis
The equation of a straight line can be presented in different forms depending on any given initial conditions.
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.
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:
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.
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:
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.
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. The general equation of the line 12x – 5y – 65 = 0 is given. It is required to write various types of equations for 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 for 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; y2 = -3.
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.
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:
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 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 = -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.
