Let the line pass through the points M 1 (x 1; y 1) and M 2 (x 2; y 2). The equation of the straight line passing through the point M 1 has the form y-y 1 = k (x - x 1), (10.6)

where k - still unknown coefficient.

Since the straight line passes through the point M 2 (x 2 y 2), the coordinates of this point must satisfy the equation (10.6): y 2 -y 1 = k (x 2 -x 1).

From here we find Substituting the found value k into equation (10.6), we obtain the equation of a straight line passing through points M 1 and M 2:

It is assumed that in this equation x 1 ≠ x 2, y 1 ≠ y 2

If x 1 = x 2, then the straight line passing through the points M 1 (x 1, y I) and M 2 (x 2, y 2) is parallel to the ordinate axis. Its equation has the form x = x 1 .

If y 2 = y I, then the equation of the straight line can be written as y = y 1, the straight line M 1 M 2 is parallel to the abscissa axis.

Equation of a straight line in segments

Let the straight line intersect the Ox axis at the point M 1 (a; 0), and the Oy axis - at the point M 2 (0; b). The equation will take the form:
those.
... This equation is called the equation of a straight line in segments, since the numbers a and b indicate which segments are cut off by a straight line on the coordinate axes.

Equation of a straight line passing through a given point perpendicular to a given vector

Let us find the equation of a straight line passing through a given point Mo (x O; y o) perpendicular to a given nonzero vector n = (A; B).

Take an arbitrary point M (x; y) on a straight line and consider the vector M 0 M (x - x 0; y - y o) (see Fig. 1). Since the vectors n and M o M are perpendicular, their scalar product is zero: that is,

A (x - xo) + B (y - yo) = 0. (10.8)

Equation (10.8) is called the equation of a straight line passing through a given point perpendicular to a given vector .

The vector n = (A; B), perpendicular to the straight line, is called normal normal vector of this line .

Equation (10.8) can be rewritten as Ax + Wu + C = 0 , (10.9)

where A and B are the coordinates of the normal vector, C = -Aх о - Ву о - free term. Equation (10.9) is the general equation of the straight line(see fig. 2).

Fig. 1 Fig. 2

Canonical equations of the straight line

,

Where
- coordinates of the point through which the straight line passes, and
is the direction vector.

Second-order Curves Circle

A circle is the set of all points of the plane equidistant from a given point, which is called the center.

The canonical equation of a circle of radius R centered at point
:

In particular, if the center of the stake coincides with the origin, then the equation will look like:

Ellipse

An ellipse is a set of points on a plane, the sum of the distances from each of which to two given points and , which are called foci, have a constant
greater than the distance between the foci
.

The canonical equation of an ellipse, the foci of which lie on the Ox axis, and the origin of coordinates in the middle between the foci has the form
G de a the length of the semi-major axis; b - the length of the semi-minor axis (Fig. 2).

Relationship between ellipse parameters
and expressed by the ratio:

(4)

Eccentricity ellipsecalled the ratio of the interfocal distance2cto the major axis2a:

Headmistresses ellipses are called straight lines parallel to the axis Oy, which are at a distance from this axis. Directrix equations:
.

If in the ellipse equation
, then the foci of the ellipse are on the Oy axis.

So,

Given two points M 1 (x 1, y 1) and M 2 (x 2, y 2)... We write the equation of the straight line in the form (5), where k still unknown coefficient:

Since the point M 2 belongs to a given straight line, then its coordinates satisfy equation (5):. Expressing from this and substituting it into equation (5), we obtain the required equation:

If this equation can be rewritten in a form more convenient for memorization:

(6)

Example. Write down the equation of the straight line passing through the 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 the straight line:

Angle between two straight lines

Consider two lines l 1 and l 2:

l 1: , , and

l 2: , ,

φ is the angle between them (). Figure 4 shows:.

From here , or

Using formula (7), one of the angles between the straight lines can be determined. The second angle is.

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 can be seen that k 1 = 2, and k 2 = -3. substituting these values ​​into formula (7), we find

... Thus, the angle between these lines is equal.

Conditions for parallelism and perpendicularity of two straight lines

If straight l 1 and l 2 are parallel, then φ=0 and tgφ = 0... it follows from formula (7) that, whence k 2 = k 1... Thus, the condition for the parallelism of two straight lines is the equality of their slopes.

If straight l 1 and l 2 are perpendicular, then φ = π / 2, α 2 = π / 2 + α 1. ... Thus, the condition of perpendicularity of two straight lines is that their slopes are reciprocal 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 straight line Ax + Vy + C = 0 is determined as

Proof. Let point M 1 (x 1, y 1) be the base of the perpendicular dropped from point M onto a given straight line. Then the distance between points M and M 1:

The coordinates x 1 and y 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 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 is proved.

Example. Determine the angle between the straight lines: y = -3x + 7; y = 2x + 1.

k 1 = -3; k 2 = 2 tgj =; j = p / 4.

Example. Show that the straight 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 straight lines are perpendicular.

Example. The vertices of the triangle A (0; 1), B (6; 5), C (12; -1) are given. Find the equation for the height drawn from vertex C.

We find the equation of the side AB:; 4x = 6y - 6;

2x - 3y + 3 = 0;

The required height equation is: 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 straight line is determined by the length of the perpendicular dropped from a point to a straight 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 straight h it is necessary to lower the perpendicular from the point A on 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 the point M to straight a general position.

The task of determining distance between parallel lines solved similarly to the previous one. A point is taken on one straight line, a perpendicular is lowered from it to another straight line. The length of the perpendicular is equal to the distance between the parallel lines.

Curve of the second order is called a line determined by an equation of the second degree relative to the current Cartesian coordinates. In general, Ax 2 + 2Bxy + Cy 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 locus of points in the plane equidistant from the point of the plane C (a, b).

The circle is given by the following equation:

Where x, y are the coordinates of an arbitrary point of the circle, R is the radius of the circle.

Circumference Equation

1. There is no term with x, y

2. Equal Coefficients at x 2 and y 2

Ellipse

Ellipse is called the 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 (constant value).

Canonical Ellipse Equation:

X and y belong to an ellipse.

a - semi-major axis of the ellipse

b - semi-minor axis of the ellipse

The ellipse has 2 axes of symmetry OX and OY. The axes of symmetry of the ellipse are its axes, the point of their intersection is the center of the ellipse. The axis on which the focuses are located is called focal axis... The point of intersection of the ellipse with the axes is the vertex of the ellipse.

Compression (stretching) ratio: ε = s / a- eccentricity (characterizes the shape of the ellipse), the smaller it is, the less the ellipse will stretch along the focal axis.

If the centers of the ellipse are not in the center of C (α, β)

Hyperbola

Hyperbole is called the locus of points in the 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 other than zero.

Canonical hyperbola equation

The hyperbola has 2 axes of symmetry:

a is the real semiaxis of symmetry

b - imaginary semiaxis of symmetry

Hyperbola assymptotes:

Parabola

Parabola is called the locus of points in the plane equidistant from a given point F, called the focus and a given straight line, called the directrix.

Canonical parabola equation:

Y 2 = 2px, where p is the distance from the focus to the directrix (parabola parameter)

If the vertex of the parabola C (α, β), then the equation of the parabola (y-β) 2 = 2p (x-α)

If the focal axis is taken as the ordinate axis, then the parabola equation will take the form: x 2 = 2qу

Properties of a straight line in Euclidean geometry.

You can draw infinitely many straight lines through any point.

A single straight line can be drawn through any two non-coinciding points.

Two mismatched straight lines on 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 straight lines:

• straight lines intersect;

• straight lines are parallel;

• straight lines intersect.

Straight line- algebraic curve of the first order: in a Cartesian coordinate system, a straight line

is given on the plane by an equation of the first degree (linear equation).

General equation of the straight line.

Definition... Any straight line on a plane can be given by a first-order equation

Ax + Wu + C = 0,

with constant A, B are not equal to zero at the same time. This first-order equation is called common

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- the 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 presented in different forms, depending on any given

initial conditions.

Equation of a straight line along a point and a normal vector.

Definition... In a Cartesian rectangular coordinate system, a vector with components (A, B)

perpendicular to the straight line given by the equation

Ax + Wu + C = 0.

Example... Find the equation of a straight line passing through a point A (1, 2) perpendicular to vector (3, -1).

Solution... At A = 3 and B = -1, we compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C

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 straight 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 straight line,

passing through these points:

If any of the denominators is zero, the corresponding numerator should be equated to zero. On the

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 straight line passing through the points A (1, 2) and B (3, 4).

Solution... Applying the above formula, we get:

Equation of a straight line by point and slope.

If the general equation of the straight line Ax + Wu + C = 0 lead to the form:

and designate , then the resulting equation is called

equation of a straight line with slope k.

Equation of a straight line along a point and a direction vector.

By analogy with the paragraph considering the equation of a straight line through the normal vector, you can enter the task

a straight line through a point and a direction vector of a straight line.

Definition... Every nonzero vector (α 1, α 2) whose components satisfy the condition

Аα 1 + Вα 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 point A (1, 2).

Solution... The equation of the required straight line will be sought in the form: Ax + By + C = 0. According to the definition,

the coefficients must meet 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.

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 Ax + Vy + C = 0 C ≠ 0, then, dividing by -C, 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- the coordinate of the point of intersection of the straight line with the axis OU.

Example... The general equation of the straight line is given x - y + 1 = 0. Find the equation of this straight line in segments.

C = 1, a = -1, b = 1.

Normal equation of a straight 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 the line.

The ± sign of the normalizing factor should 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... A general equation of the straight line is given 12x - 5y - 65 = 0... Required to write different types of equations

this straight line.

The equation of this line in segments:

Equation of this line with slope: (divide by 5)

Equation of a straight 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 the plane.

Definition... If two lines are given y = k 1 x + b 1, y = k 2 x + b 2, then an acute angle between these lines

will be defined as

Two lines are parallel if k 1 = k 2... Two straight 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 are parallel when the coefficients are proportional

А 1 = λА, В 1 = λВ... If also С 1 = λС, then the straight lines coincide. Coordinates of the point of intersection of two lines

are found as a solution to the system of equations of these straight lines.

Equation of a straight line passing through a given point perpendicular to a given straight line.

Definition... Line through point M 1 (x 1, y 1) and perpendicular to the line y = kx + b

is represented by the equation:

Distance from point to line.

Theorem... If a point is given M (x 0, y 0), 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 the perpendicular dropped from the point M for a given

straight line. Then the distance between the 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 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 is proved.

Given two points M(X 1 ,Have 1) and N(X 2,y 2). Let's find the equation of the straight line passing through these points.

Since this line passes through the point M, then according to formula (1.13) its equation has the form

Have – Y 1 = K(X - x 1),

Where K- unknown slope.

The value of this coefficient is determined from the condition that the desired straight line passes through the point N, and hence, its coordinates satisfy the equation (1.13)

Y 2 – Y 1 = K(X 2 – X 1),

From here you can find the slope of this straight line:

,

Or after conversion

(1.14)

Formula (1.14) determines Equation of a straight line passing through two points M(X 1, Y 1) and N(X 2, Y 2).

In the special case, when the points M(A, 0), N(0, B), A ¹ 0, B¹ 0, lie on the coordinate axes, equation (1.14) takes on a simpler form

Equation (1.15) called By the 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. Equate a straight line through points M(1, 2) and B(3, –1).

. According to (1.14), the equation of the sought line has the form

2(Y – 2) = -3(X – 1).

Transferring all terms to the left side, we finally obtain the required equation

3X + 2Y – 7 = 0.

Example 1.11. Equate a straight line through a point M(2, 1) and the point of intersection of the lines X+ Y - 1 = 0, X - y+ 2 = 0.

. We find the coordinates of the point of intersection of the straight lines by solving together the given equations

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 Have:

Now we write the equation of the straight line passing through the points (2, 1) and:

or .

Hence, or –5 ( Y – 1) = X – 2.

Finally, we obtain the equation of the desired straight line in the form X + 5Y – 7 = 0.

Example 1.12. Find the equation of the straight 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. It can be seen from the problem statement that the abscissas of both points have the same value. Hence, the sought line is parallel to the axis OY and its equation is: x = 2.

Comment . If, when writing the equation of a straight line according to 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.

Consider other ways to define a straight line on a plane.

1. Let a nonzero vector be perpendicular to the given line L and point M 0(X 0, Y 0) lies on this straight line (Figure 1.7).

Figure 1.7

We denote M(X, Y) an arbitrary point on the line L... Vectors and Orthogonal. Using the orthogonality conditions for these vectors, we obtain either A(X – X 0) + B(Y – Y 0) = 0.

We got the equation of a straight line passing through a point M 0 perpendicular to the vector. This vector is called The normal vector to straight L... The resulting equation can be rewritten as

Oh + Woo + WITH= 0, where WITH = –(AX 0 + By 0), (1.16),

Where A and V- coordinates of the normal vector.

Let's get the general equation of the straight line in parametric form.

2. A straight line on a plane can be specified as follows: let a nonzero vector be parallel to a given straight line L and point M 0(X 0, Y 0) lies on this straight 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 collinearity condition for these vectors:, where T- an arbitrary number called a parameter. Let's write this equality in coordinates:

These equations are called Parametric equations Straight... We exclude from these equations the parameter T:

These equations can otherwise be written in the form

. (1.18)

The resulting equation is called The canonical equation of the straight line... The vector is called The direction vector of the straight line .

Comment . It is easy to see that if is the normal vector to the line L, then its direction vector can be a vector, since, i.e.

Example 1.13. Write the equation of the straight line passing through the point M 0 (1, 1) parallel to straight line 3 X + 2Have– 8 = 0.

Solution . The vector is the normal vector to the given and desired straight lines. We will use the equation of the straight line passing through the point M 0 with a given normal vector 3 ( X –1) + 2(Have- 1) = 0 or 3 X + 2y- 5 = 0. Received the equation of the desired straight line.

Let's look at how to draw up the equation of a straight line passing through two points, using examples.

Example 1.

Make the equation of the straight line passing through the points A (-3; 9) and B (2; -1).

Method 1 - compose the equation of a straight line with a slope.

The equation of a straight line with a slope has the form. Substituting the coordinates of points A and B into the equation of the straight line (x = -3 and y = 9 in the first case, x = 2 and y = -1 in the second), we obtain a system of equations from which we find the values ​​of k and b:

Adding the 1st and 2nd equations term by term, we get: -10 = 5k, whence k = -2. Substituting k = -2 into the second equation, we find b: -1 = 2 · (-2) + b, b = 3.

Thus, y = -2x + 3 is the desired equation.

Method 2 - compose the general equation of the straight line.

The general equation of the straight line has the form. Substituting the coordinates of points A and B into the equation, we get the system:

Since the number of unknowns is greater than the number of equations, the system is not solvable. But you can express all the variables through one. For example, through b.

Multiplying the first equation of the system by -1 and adding term by term with the second:

we get: 5a-10b = 0. Hence a = 2b.

Substitute the resulting expression into the second equation: 2 · 2b -b + c = 0; 3b + c = 0; c = -3b.
Substitute a = 2b, c = -3b into the equation ax + by + c = 0:

2bx + by-3b = 0. It remains to split both parts by b:

The general equation of a straight line is easily reduced to the equation of a straight line with a slope:

Method 3 - compose the equation of a straight line passing through 2 points.

The equation of a straight line passing through two points has:

Substitute into this equation the coordinates of points A (-3; 9) and B (2; -1)

(that is, x 1 = -3, y 1 = 9, x 2 = 2, y 2 = -1):

and simplify:

whence 2x + y-3 = 0.

In the school course, the equation of a straight line with a slope is most often used. But the easiest way is to derive and use the formula for the equation of a straight line passing through two points.

Comment.

If, when substituting the coordinates of the given points, one of the denominators of the equation

turns out to be equal to zero, then the desired equation is obtained by equating to zero of the corresponding numerator.

Example 2.

Make an equation of a straight line passing through two points C (5; -2) and D (7; -2).

Substitute into the equation a straight line passing through 2 points, the coordinates of points C and D.