Equation of a segment. Straight line. Equation of a straight line. What will we do with the received material?
Let some affine coordinate system OXY be given.
Theorem 2.1. Any straight line l coordinate system OX is given by a linear equation of the form
A x+B y+ C = O, (1)
where A, B, C R and A 2 + B 2 0. Conversely, any equation of the form (1) defines a straight line.
Equation like (1) - general equation of a line .
Let all coefficients A, B and C in equation (1) be different from zero. Then
Ah-By=-C, and .
Let's denote -C/A=a, -C/B=b. We get
-equation in segments .
Indeed, the numbers |a| and |b| indicate the size of the segments cut off by a straight line l on the OX and OY axes, respectively.
Let it be straight l is given by the general equation (1) in a rectangular coordinate system and let the points M 1 (x 1,y 1) and M 2 (x 2,y 2) belong to l. Then
A x 1 + V at 1 + C = A X 2 + V at 2 + C, that is, A( x 1 -x 2) + B( at 1 -at 2) = 0.
The last equality means that the vector =(A,B) is orthogonal to the vector =(x 1 -x 2,y 1 -y 2). those. Vector (A,B) is called normal vector of the line l.
Consider the vector =(-B,A). Then
A(-B)+BA=0. those. ^.
Therefore, the vector =(-B,A) is the direction vector of the spicy l.
Parametric and canonical equations of the line
Equation of a line passing through two given points
Let a straight line be given in the affine coordinate system (0, X, Y) l, its direction vector = (m,n) and point M 0 ( x 0 ,y 0) owned l. Then for an arbitrary point M ( x,at) of this line we have
and since then 
.
If we denote and
Radius vectors of points M and M 0 respectively, then
- equation of a line in vector form.
Since =( X,at), =(X 0 ,at 0), then
x= x 0 + mt,
y= y 0 + nt
- parametric equation of a line .
It follows that
- canonical equation of the line .
Finally, if on a straight line l given two points M 1 ( X 1 ,at 1) and
M2( x 2 ,at 2), then vector =( X 2 -X 1 ,y 2 -at 1) is guides vector of a straight line l. Then
- equation of a line passing through two given points.
Mutual arrangement two straight lines.
Let straight l 1 and l 2 are given by their general equations
l 1: A 1 X+ B 1 at+ C 1 = 0, (1)
l 2: A 2 X+ B 2 at+ C 2 = 0.
Theorem. Let straight l 1 and l 2 are given by equations (1). Then and only then:
1) lines intersect when there is no number λ such that
A 1 =λA 2, B 1 =λB 2;
2) the lines coincide when there is a number λ such that
A 1 =λA 2, B 1 =λB 2, C 1 =λC 2;
3) the lines are distinct and parallel when there is a number λ such that
A 1 =λA 2, B 1 =λB 2, C 1 λC 2.
Bunch of straight lines
A bunch of straight lines is the set of all lines in the plane passing through a certain point called center beam.
To specify the beam equation, it is enough to know any two straight lines l 1 and l 2 passing through the center of the beam.
Let the straight lines in the affine coordinate system l 1 and l 2 are given by the equations
l 1: A 1 x+ B 1 y+ C 1 = 0,
l 2: A 2 x+ B 2 y+ C 2 = 0.
The equation:
A 1 x+ B 1 y+ C + λ (A 2 X+ B 2 y+ C) = 0
- equation of a pencil of lines defined by the equations l 1 and l 2.
In the future, by a coordinate system we will understand a rectangular coordinate system .
Conditions for parallelism and perpendicularity of two straight lines
Let the lines be given l 1 and l 2. their general equations; = (A 1 ,B 1), = (A 2 ,B 2) – normal vectors of these lines; k 1 = tgα 1, k 2 = tanα 2 – angular coefficients; = ( m 1 ,n 1), (m 2 ,n 2) – direction vectors. Then, straight l 1 and l 2 are parallel if and only if one of the following conditions is true:
or either k 1 =k 2, or .
Let it be straight now l 1 and l 2 are perpendicular. Then, obviously, that is, A 1 A 2 + B 1 B 2 = 0.
If straight l 1 and l 2 are given respectively by the equations
l 1: at=k 1 x+ b 1 ,
l 2: at=k 2 x+ b 2 ,
then tanα 2 = tan(90º+α) = 
.
It follows that
Finally, if and the direction vectors are straight, then ^, that is
m 1 m 2 + n 1 n 2 = 0
The last relationship expresses the necessary and sufficient condition for the perpendicularity of two planes.
Angle between two straight lines
At an angle φ between two straight lines l 1 and l 2 we will understand the smallest angle by which one straight line must be rotated so that it becomes parallel to or coincides with another straight line, that is, 0 £ φ £
Let the lines be given by general equations. It's obvious that
cosφ= 
Let it be straight now l 1 and l 2 is given by equations with slope coefficients k 1 in k 2 respectively. Then
It is obvious that, that is ( X-X 0) + B( at-at 0) + C( z-z 0) = 0
Let's open the brackets and denote D= -A x 0 - V at 0 - C z 0 . We get
A x+B y+ C z+ D = 0 (*)
- plane equation in general form or general plane equation.
Theorem 3.1 Linear equation(*) (A 2 +B 2 +C 2 ≠ 0) is an equation of the plane and vice versa, any equation of the plane is linear.
1) D = 0, then the plane passes through the origin.
2) A = 0, then the plane is parallel to the OX axis
3) A = 0, B = 0, then the plane is parallel to the OXY plane.
Let all coefficients in the equation be different from zero.
- plane equation in segments. Numbers |a|, |b|, |c| indicate the values of the segments cut off by the plane at coordinate axes.
And we'll analyze it in detail special kind equations of the line - . Let's start with the form of the equation of a straight line in segments and give an example. After this, we will focus on constructing a straight line, which is given by the equation of a straight line in segments. In conclusion, we will show how the transition from the complete general equation of a line to the equation of a line in segments is carried out.
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Equation of a line in segments - description and example.
Let Oxy be fixed on the plane.
Equation of a line in segments on a plane in a rectangular coordinate system, Oxy has the form , where a and b are some nonzero real numbers.
It is no coincidence that the equation of a line in segments received this name - the absolute values of the numbers a and b are equal to the lengths of the segments that the line cuts off on the coordinate axes Ox and Oy, counting from the origin.
Let's clarify this point. We know that the coordinates of any point on a line satisfy the equation of that line. Then it is clearly visible that the line defined by the equation of the line in segments passes through the points and , since 
And 
. And the points and are precisely located on the coordinate axes Ox and Oy, respectively, and are distant from the origin of coordinates by a and b units. The signs of the numbers a and b indicate the direction in which the segments should be laid. The “+” sign means that the segment is plotted in the positive direction of the coordinate axis, the “-” sign means the opposite.
Let us depict a schematic drawing explaining all of the above. It shows the location of the lines relative to the fixed rectangular coordinate system Oxy, depending on the values of the numbers a and b in the equation of the line in segments.
Now it has become clear that the equation of a straight line in segments makes it easy to construct this straight line in the rectangular coordinate system Oxy. To construct a straight line, which is given by the equation of a straight line in segments of the form , you should mark the points and in a rectangular coordinate system on the plane, and then connect them with a straight line using a ruler.
Let's give an example.
Example.
Construct a straight line given by the equation of a line in segments of the form.
Solution.
Based on the given equation of a line in segments, it can be seen that the line passes through the points 
. We mark them and connect them with a straight line.
Reducing the general equation of a line to the equation of a line in segments.
When solving some problems related to a line on a plane, it is convenient to work with the equation of a line in segments. However, there are other types of equations that define a line on a plane. Therefore, it is necessary to carry out the transition from a given equation of a line to the equation of this line in segments.
In this paragraph we will show how to obtain the equation of a line in segments if the complete general equation of the line is given.
Let us know the complete general equation of a line on a plane 
. Since A, B and C are not equal to zero, you can move the number C to the right side of the equality, divide both sides of the resulting equality by –C, and send the coefficients for x and y to the denominators: 
.
(In the last transition we used the equality 
).
So we from the general equation of the straight line passed to the equation of a straight line in segments, where 
.
Example.
The straight line in the rectangular coordinate system Oxy is given by the equation 
. Write the equation of this line in segments.
Solution.
Let's move one second to the right side of the given equality: 
. Now let’s divide the resulting equality into both sides: 
. It remains to transform the resulting equality to the desired form: 
. This is how we obtained the required equation of the line in segments.
Answer:
If a straight line defines
Equation of a line in segments
Let the general equation of a straight line be given:
The equation of a straight line in segments, where are the segments that the straight line cuts off on the corresponding coordinate axes.
Construct a straight line given by the general equation:
From which we can construct an equation of this line in segments:
The relative position of lines on a plane.
Statement 1.
In order for straight lines and given by equations:
Coincidence is necessary and sufficient so that:
Proof: and coincide, their direction vectors and are collinear, i.e.:
Let's take point M 0 with this straight line, then:
Multiplying the first equation by and adding to the second by (2) we get:
So, formulas (2), (3) and (4) are equivalent. Let (2) be satisfied, then the equations of the system (*) are equivalent; the corresponding straight lines coincide.
Statement 2.
The lines and given by equations (*) are parallel and do not coincide if and only if:
Proof:
Even if they don't match:
Inconsistent, i.e., according to the Kronecker-Capelli theorem:
This is only possible if:
That is, when condition (5) is met.
When the first equality (5) is fulfilled, - failure to satisfy the second equality results in the incompatibility of the system (*) the lines are parallel and do not coincide.
Note 1.
Polar coordinate system.
Let us fix a point on the plane and call it a pole. The ray emanating from the pole will be called the polar axis.
Let's choose a scale for measuring the lengths of segments and agree that the rotation around the point counterclockwise will be considered positive. Let's consider any point on a given plane, denote it by its distance to the pole and call it the polar radius. The angle by which the polar axis must be rotated so that it coincides with will be denoted by and called the polar angle.
Definition 3.
The polar coordinates of a point are its polar radius and polar angle:
Remark 2. in the pole. The value for points other than a point is determined up to a term.
Consider a Cartesian rectangular coordinate system: the pole coincides with the origin, and the polar axis coincides with the positive semi-axis. Here. Then:
What is the relationship between rectangular Cartesian and polar coordinate systems.
Bernoulli's lemniscate equation. Write it in the polar coordinate system.
Normal equation of a line on a plane. Let the polar axis coincide with, - the axis passing through the origin. Let be:
Let then:
Condition (**) for point:
Equation of a straight line in a polar coordinate system.
Here - the length drawn from the origin to the straight line, - the angle of inclination of the normal to the axis.
Equation (7) can be rewritten:
Normal equation of a line on a plane.
If in the general equation of the line Ах + Ву + С = 0 С ¹ 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 divided by a number called normalizing factor, then we get
Xcosj + ysinj - p = 0 –
normal equation of a line.
The sign ± of the normalizing factor must be chosen so that m×С< 0.
p is the length of the perpendicular dropped from the origin to the straight line, and j is the angle formed by this perpendicular with the positive direction of the Ox axis.
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)
normal equation of a line:
; cosj = 12/13; sinj = -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.
The equation of the straight line has the form: , a = b = 1; ab/2 = 8; a = 4; -4.
a = -4 is not suitable according to the conditions of the problem.
Total: or x + y – 4 = 0.
Example. Write an equation for a straight line passing through point A(-2, -3) and the origin.
The straight line equation has the form: , where x 1 = y 1 = 0; x 2 = -2; y 2 = -3.
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:
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. The straight lines Ax + Bу + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A 1 = lA, B 1 = lB are proportional. If also С 1 = lС, 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.
Distance from a point to a 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 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 = -3x + 7; y = 2x + 1.
k 1 = -3; k 2 = 2 tgj = ; 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.
Second order curves.
A second order curve can be given by the equation
Ax 2 + 2Bhu + Su 2 + 2Dx + 2Ey + F = 0.
There is a coordinate system (not necessarily Cartesian rectangular) in which this equation can be represented in one of the forms given below.
1) - equation of the ellipse.
2) - equation of an “imaginary” ellipse.
3) - hyperbola equation.
4) a 2 x 2 – c 2 y 2 = 0 – equation of two intersecting lines.
5) y 2 = 2px – equation of a parabola.
6) y 2 – a 2 = 0 – equation of two parallel lines.
7) y 2 + a 2 = 0 – equation of two “imaginary” parallel lines.
8) y 2 = 0 – a pair of coinciding lines.
9) (x – a) 2 + (y – b) 2 = R 2 – equation of a circle.
Circle.
In a circle (x – a) 2 + (y – b) 2 = R 2 the center has coordinates (a; b).
Example. Find the coordinates of the center and the radius of the circle if its equation is given in the form:
2x 2 + 2y 2 – 8x + 5y – 4 = 0.
To find the coordinates of the center and radius of the circle, this equation must be brought to the form indicated above in paragraph 9. To do this, select complete squares:
x 2 + y 2 – 4x + 2.5y – 2 = 0
x 2 – 4x + 4 –4 + y 2 + 2.5y + 25/16 – 25/16 – 2 = 0
(x – 2) 2 + (y + 5/4) 2 – 25/16 – 6 = 0
(x – 2) 2 + (y + 5/4) 2 = 121/16
From here we find O(2; -5/4); R = 11/4.
Ellipse.
Definition. Ellipse is called the curve given by the equation.
Definition. Focuses are called such two points, the sum of the distances from which to any point of the ellipse is a constant value.
F 1, F 2 – focuses. F 1 = (c; 0); F 2 (-c; 0)
c – half the distance between focuses;
a – semimajor axis;
b – semi-minor axis.
Theorem. The focal length and semi-axes of the ellipse are related by the relation:
a 2 = b 2 + c 2 .
Proof: If point M is at the intersection of the ellipse with the vertical axis, r 1 + r 2= 2 (according to the Pythagorean theorem). If point M is at the intersection of the ellipse with the horizontal axis, r 1 + r 2 = a – c + a + c. Because by definition the amount r 1 + r 2 is a constant value, then, equating, we get:
a 2 = b 2 + c 2
r 1 + r 2 = 2a.
Definition. The shape of the ellipse is determined by the characteristic, which is the ratio of the focal length to the major axis and is called eccentricity.
Because With< a, то е < 1.
Definition. The quantity k = b/a is called compression ratio ellipse, and the quantity 1 – k = (a – b)/a is called compression ellipse.
The compression ratio and eccentricity are related by the relation: k 2 = 1 – e 2 .
If a = b (c = 0, e = 0, the foci merge), then the ellipse turns into a circle.
If the condition is satisfied for the point M(x 1, y 1): then it is located inside the ellipse, and if , then the point is outside the ellipse.
Theorem. For an arbitrary point M(x, y) belonging to an ellipse, the following relations are true::
R 1 = a – ex, r 2 = a + ex.
Proof. It was shown above that r 1 + r 2 = 2a. In addition, from geometric considerations we can write:
After squaring and bringing similar terms:
It is proved in a similar way that r 2 = a + ex. The theorem has been proven.
An ellipse is connected to two straight lines called headmistresses. Their equations are:
X = a/e; x = -a/e.
Theorem. In order for a point to lie on an ellipse, it is necessary and sufficient that the ratio of the distance to the focus to the distance to the corresponding directrix is equal to the eccentricity e.
Example. Write an equation for the line passing through the left focus and the lower vertex of the ellipse given by the equation:
1) Coordinates of the bottom vertex: x = 0; y2 = 16; y = -4.
2) Coordinates of the left focus: c 2 = a 2 – b 2 = 25 – 16 = 9; c = 3; F 2 (-3; 0).
3) Equation of a line passing through two points:
Example. Write an equation for an ellipse if its foci are F 1 (0; 0), F 2 (1; 1), and the major axis is 2.
The equation of the ellipse has the form: . Focus distance:
2c = thus a 2 – b 2 = c 2 = ½
by condition 2a = 2, therefore a = 1, b =
Hyperbola.
Definition. Hyperbole is the set of points of the plane for which the modulus of the difference in distances from two given points, called tricks is a constant value less than the distance between the foci.
By definition ïr 1 – r 2 ï= 2a. F 1, F 2 – focuses of the hyperbola. F 1 F 2 = 2c.
Let us choose an arbitrary point M(x, y) on the hyperbola. Then:
let us denote c 2 – a 2 = b 2 (geometrically, this quantity is the minor semi-axis)
We obtained the canonical equation of the hyperbola.
The hyperbola is symmetrical about the middle of the segment connecting the foci and about the coordinate axes.
Axis 2a is called the real axis of the hyperbola.
Axis 2b is called the imaginary axis of the hyperbola.
A hyperbola has two asymptotes, the equations of which are
Definition. The relationship is called eccentricity hyperbolas, where c is half the distance between the foci, and is the real semi-axis.
Taking into account the fact that c 2 – a 2 = b 2:
If a = b, e = , then the hyperbola is called equilateral (equilateral).
Definition. Two straight lines perpendicular to the real axis of the hyperbola and located symmetrically relative to the center at a distance a/e from it are called headmistresses hyperbole. Their equations are: .
Theorem. If r is the distance from an arbitrary point M of the hyperbola to any focus, d is the distance from the same point to the directrix corresponding to this focus, then the ratio r/d is a constant value equal to the eccentricity.
Proof. Let us schematically depict a hyperbole.
From the obvious geometric relationships we can write:
a/e + d = x, therefore d = x – a/e.
(x – c) 2 + y 2 = r 2
From the canonical equation: , taking into account b 2 = c 2 – a 2:
Then because с/a = e, then r = ex – a.
For the left branch of the hyperbola the proof is similar. The theorem has been proven.
Example. Find the equation of a hyperbola whose vertices and foci are at the corresponding vertices and foci of the ellipse.
For an ellipse: c 2 = a 2 – b 2.
For a hyperbola: c 2 = a 2 + b 2.
Hyperbola equation: .
Example. Write an equation for a hyperbola if its eccentricity is 2 and its foci coincide with the foci of the ellipse with the equation being the parameter of the parabola. Let us derive the canonical equation of the parabola.
From geometric relationships: AM = MF; AM = x + p/2;
MF 2 = y 2 + (x – p/2) 2
(x + p/2) 2 = y 2 + (x – p/2) 2
x 2 +xp + p 2 /4 = y 2 + x 2 – xp + p 2 /4
Directrix equation: x = -p/2.
Example . On the parabola y 2 = 8x, find a point whose distance from the directrix is 4.
From the parabola equation we find that p = 4.
r = x + p/2 = 4; hence:
x = 2; y2 = 16; y = ±4. Searched points: M 1 (2; 4), M 2 (2; -4).
Example. The equation of a curve in a polar coordinate system has the form:
Find the equation of a curve in a Cartesian rectangular coordinate system, determine the type of curve, find foci and eccentricity. Schematically draw the curve.
Let's use the connection between the Cartesian rectangular and polar coordinate systems: ;
We obtained the canonical equation of the hyperbola. From the equation it is clear that the hyperbola is shifted along the Ox axis by 5 to the left, the major semi-axis a is equal to 4, the minor semi-axis b is equal to 3, whence we obtain c 2 = a 2 + b 2 ; c = 5; e = c/a = 5/4.
Focuses F 1 (-10; 0), F 2 (0; 0).
Let's construct a graph of this hyperbola.
The task is to construct a line passing through the given coordinates of the end of a segment.
We believe that the segment is non-degenerate, i.e. has a length greater than zero (otherwise, of course, there are infinitely many different lines passing through it).
Two-dimensional case
Let a segment be given, i.e. the coordinates of its ends , , , are known.
Required to build equation of a line in a plane, passing through this segment, i.e. find the coefficients , , in the equation of the straight line:
Note that the required triples passing through a given segment are infinitely many: You can multiply all three coefficients by an arbitrary non-zero number and get the same straight line. Therefore, our task is to find one of these triplets.
It is easy to verify (by substituting these expressions and the coordinates of the points and into the equation of the straight line) that the following set of coefficients is suitable:
Integer case
An important advantage of this method of constructing a straight line is that if the coordinates of the ends were integer, then the resulting coefficients will also be integers. In some cases, this allows geometric operations to be performed without resorting to real numbers at all.
However, there is a small drawback: for the same line, different triplets of coefficients can be obtained. To avoid this, but not move away from integer coefficients, you can use the following technique, often called rationing. Let's find the greatest common divisor of the numbers , , , divide all three coefficients by it, and then normalize the sign: if or , then multiply all three coefficients by . As a result, we will come to the conclusion that for identical lines we will obtain identical triplets of coefficients, which will make it easy to check lines for equality.
Real-valued case
When working with real numbers, you should always be aware of errors.
The coefficients we obtain are of the order of the original coordinates, the coefficient is already of the order of the square of them. These can already be quite large numbers, and, for example, when the lines intersect, they will become even larger, which can lead to large rounding errors even with the original coordinates of order .
Therefore, when working with real numbers, it is advisable to perform the so-called normalization direct: namely, to make the coefficients such that 
. To do this you need to calculate the number:
and divide all three coefficients , , by it.
Thus, the order of the coefficients and will no longer depend on the order of the input coordinates, and the coefficient will be of the same order as the input coordinates. In practice, this leads to a significant improvement in calculation accuracy.
Finally, let's mention comparison straight lines - after all, after such normalization for the same straight line, only two triplets of coefficients can be obtained: up to multiplication by . Accordingly, if we carry out additional normalization taking into account the sign (if or , then multiply by ), then the resulting coefficients will be unique.
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