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 \u003d k (x - x 0) (1)

Where k is the slope of the straight line.

Alternative formula:
The 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 #1. Compose the equation of 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 line 2x+3y -7 = 0.
Solution . Let's represent the slope equation as y = kx + a . To do this, we transfer all values ​​except y to right side: 3y = -2x + 7 . Then we divide the right side by the coefficient 3 . We get: y = -2/3x + 7/3
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 \u003d -2, k \u003d -2 / 3, y 0 \u003d 1 we get:
y-1 = -2 / 3 (x-(-2))
or
y = -2 / 3 x - 1 / 3 or 3y + 2x +1 = 0

Example #2. Write the equation of a straight line parallel to the straight 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. Find the points of intersection of the desired line with the coordinate axes:
;
.
So, A(-C/2,0), B(0,-C/5). Substitute in the formula for the area: . We get two solutions: 2x + 5y + 10 = 0 and 2x + 5y - 10 = 0 .

Example #3. Write the equation of the line passing through the point (-2; 5) and the parallel 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 #4. Solving example 3 (A=5, B=-7) using formula (2), we find 5(x+2)-7(y-5)=0.

Example number 5. Write the equation of a straight line passing through the point (-2;5) and a parallel straight 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 solved with respect to y (this straight line is parallel to the y-axis).

Properties of a straight line in Euclidean geometry.

There are infinitely many lines that can be drawn through any point.

Through any two non-coinciding points, there is only one straight line.

Two non-coincident lines in the 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;

• straight lines are parallel;

• straight lines intersect.

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

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

General equation of a straight line.

Definition. Any line in the plane can be given by a first order equation

Ah + Wu + C = 0,

and constant A, B not equal to zero at the same time. This first order equation is called general

straight line equation. Depending on the values ​​of the constants A, B and FROM The following special cases are possible:

. C = 0, A ≠ 0, B ≠ 0- the 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 line coincides with the axis OU

. A = C = 0, B ≠ 0- the line coincides with the axis Oh

The equation of a straight line can be represented in various forms depending on any given

initial conditions.

Equation of a straight line by 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

Ah + Wu + C = 0.

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

Solution. Let's compose at A \u003d 3 and B \u003d -1 the equation of the straight line: 3x - y + C \u003d 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 desired equation: 3x - y - 1 \u003d 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 straight line equation,

passing through these points:

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

plane, the equation of a straight line written above is simplified:

if x 1 ≠ x 2 and x = x 1, if x 1 = x 2 .

Fraction = k called slope factor straight.

Example. Find the equation of a 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 a point and a slope.

If the general equation of a straight line Ah + Wu + C = 0 bring to the form:

and designate , then the resulting equation is called

equation of a straight line with slope k.

The equation of a straight line on a point and a directing 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 direction 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 direction vector of the straight line.

Ah + Wu + C = 0.

Example. Find the equation of a straight line with direction vector (1, -1) and passing through point A(1, 2).

Solution. We will look for the equation of the desired straight line in the form: Ax + By + C = 0. According to the definition,

coefficients must satisfy the conditions:

1 * A + (-1) * B = 0, i.e. A = B.

Then the equation of a 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. desired equation:

x + y - 3 = 0

Equation of a straight line in segments.

If in the general equation of the straight line Ah + Wu + 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 axle Oh, a b- the 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 straight line in segments.

C \u003d 1, , a \u003d -1, b \u003d 1.

Normal equation of a straight line.

If both sides of the equation Ah + Wu + C = 0 divide by number , which is called

normalizing factor, then we get

xcosφ + ysinφ - p = 0 -normal equation of a straight 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 line,

a φ - the angle formed by this perpendicular with the positive direction of the axis Oh.

Example. Given the general equation of a straight line 12x - 5y - 65 = 0. Required to write different types equations

this straight line.

The equation of this straight line in segments:

The 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.

Angle between lines on a plane.

Definition. If two lines are given y \u003d k 1 x + b 1, y \u003d 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 \u003d -1 / k 2 .

Theorem.

Direct Ah + Wu + C = 0 and A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients are proportional

A 1 \u003d λA, B 1 \u003d λB. If also С 1 \u003d 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 is perpendicular to a given line.

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

represented by the equation:

The distance from a point to a line.

Theorem. If a point is given M(x 0, y 0), then the distance to the line Ah + 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

direct. Then the distance between the points M and M 1:

(1)

Coordinates x 1 and 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 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 a given point in this direction. Equation of a straight line passing through two given points. Angle between two lines. Condition of parallelism and perpendicularity of two lines. Determining the point of intersection of two lines

Examples of problems with solutions

Find the equation of a straight line passing through two points: (-1, 2) and (2, 1).

Solution.

According to the equation

believing in it x 1 = -1, y 1 = 2, x 2 = 2, y 2 \u003d 1 (no matter which point is considered the first, which - the second), we get

after simplifications, we finally obtain the desired equation in the form

x + 3y - 5 = 0.

The sides of the triangle are given by the equations: (AB ) 2 x + 4 y + 1 = 0, (AC ) x - y + 2 = 0, (BC ) 3 x + 4 y -12 = 0. Find the coordinates of the vertices of the triangle.

Solution.

Vertex coordinates A find by solving a system composed of side equations AB and AC:

system of two linear equations with two unknowns we solve by methods known from elementary algebra, and we obtain

Vertex A has coordinates

Vertex coordinates B find by solving a system of equations of the sides AB and BC:

we get .

Vertex coordinates C we obtain by solving the system from the equations of the sides BC and AC:

Vertex C has coordinates.

A (2, 5) parallel to line 3x - 4 y + 15 = 0.

Solution.

Let us prove that if two lines are parallel, then their equations can always be represented in such a way that they differ only in free terms. Indeed, from the condition of parallelism of two lines it follows that .

Denote by t the total value of these relationships. Then

and hence it follows that

A 1 = A 2 t, B 1 = B 2 t. (1)

If two lines

A 1 x + B 1 y + C 1 = 0 and

A 2 x + B 2 y + C 2 = 0

are parallel, conditions (1) are satisfied, and, replacing in the first of these equations A 1 and B 1 by formulas (1), we will have

A 2 tx + B 2 ty + C 1 = 0,

or, dividing both sides of the equation by , we get

Comparing the resulting equation with the equation of the second line A 2 x + B 2 y + C 2 = 0, we note that these equations differ only in the free term; thus, we have proved the claim. Now let's start solving the problem. We write the equation of the desired straight line in such a way that it will differ from the equation of this straight line only by the free term: the first two terms in the desired equation will be taken from this equation, and its free term will be denoted by C. Then the desired equation can be written in the form

3x - 4y + C = 0, (3)

and to be determined C.

Giving in equation (3) to the value C all possible real values, we get a set of lines parallel to the given one. Thus, equation (3) is not an equation of one line, but of a whole family of lines parallel to this line 3 x - 4y+ 15 = 0. From this family of lines, one should single out the one that passes through the point A(2, 5).

If a line passes through a point, then the coordinates of that point must satisfy the equation of the line. And so we define C, if in (3) we substitute instead of the current coordinates x and y point coordinates A, i.e. x = 2, y= 5. We get and C = 14.

Found value C we substitute into (3), and the desired equation will be written as follows:

3x - 4y + 14 = 0.

The same problem can be solved in another way. Since the slopes of parallel lines are equal to each other, and for a given line 3 x - 4y+ 15 = 0 slope, then the slope of the desired line is also equal to .

Now we use the equation y - y 1 = k(x - x 1) a bundle of straight lines. Dot A(2, 5), through which the straight line passes, is known to us, and therefore, substituting into the equation of the pencil of straight lines y - y 1 = k(x - x 1) values ​​, we get

or after simplifications 3 x - 4y+ 14 = 0, i.e. the same as before.

Find equations of lines passing through a pointA (3, 4) at 60 degrees to line 2x + 3 y + 6 = 0.

Solution.

To solve the problem, we should determine the slopes of lines I and II (see figure). Let us denote these coefficients, respectively, by k 1 and k 2 , and the slope of this straight line - through k. It's obvious that .

Based on the definition of the angle between two straight lines, when determining the angle between a given straight line and a straight line, I follows in the numerator of a fraction in the formula

subtract the slope of the given line, since it needs to be rotated counterclockwise around the point C until it coincides with line I.

Considering that , we get

When determining the angle between line II and a given line, one should subtract the slope of line II in the numerator of the same fraction, i.e. k 2 , since line II should be rotated counterclockwise around the point B until it coincides with this line:

Find the equation of a straight line passing through a pointA (5, -1) perpendicular to line 3x - 7 y + 14 = 0.

Solution.

If two lines

A 1 x + B 1 y + C 1 = 0, A 2 x + B 2 y + C 2 = 0

are perpendicular, then the equality

A 1 A 2 + B 1 B 2 = 0,

or, which is the same,

A 1 A 2 = -B 1 B 2 ,

and hence it follows that

The general meaning of these expressions will be denoted by t.

Then , whence it follows that

A 2 = B 1 t, B 2 = -A 1 t.

Substituting these values A 2 and B 2 and the equation of the second straight line, we get

B 1 tx - A 1 ty + C 2 = 0.

or dividing by t both sides of the equality, we will have

Comparing the resulting equation with the equation of the first straight line

A 1 x + B 1 y + C 1 = 0,

note that they have coefficients at x and y changed places, and the sign between the first and second terms changed to the opposite, while the free terms are different.

Let's now start solving the problem. Wishing to write the equation of a straight line perpendicular to a straight line 3 x - 7y+ 14 = 0, based on the conclusion made above, we proceed as follows: we swap the coefficients at x and y, and the minus sign between them is replaced by a plus sign, the free term is denoted by the letter C. Let's get 7 x + 3y + C= 0. This equation is the equation of a family of lines perpendicular to the line 3 x - 7y+ 14 = 0. Define C from the condition that the desired line passes through the point A(5, -1). It is known that if a line passes through a point, then the coordinates of this point must satisfy the equation of the line. Substituting in the last equation 5 instead of x and -1 instead y, we get

This value C Substitute in the last equation and get

7x + 3y - 32 = 0.

We solve the same problem in a different way, using the equation of a pencil of lines

y - y 1 = k(x - x 1).

The slope of this straight line 3 x - 7y + 14 = 0

then the slope of the line perpendicular to it,

Substituting into the equation of a pencil of lines , and instead of x 1 and y 1 coordinates of the given point A(5, -1), find , or 3 y + 3 = -7x+ 35, and finally 7 x + 3y- 32 = 0, i.e. the same as before.

Equations curves are abundant when reading economic literature. Let us point out some of these curves.

indifference curve - a curve showing various combinations of two products that have the same consumer value, or utility, for the consumer.

Consumer Budget Curve is a curve showing the different combinations of quantities of two goods that a consumer can buy at a given level of his money income.

Production Possibility Curve - a curve showing the various combinations of two goods or services that can be produced at full employment and full output in an economy with constant stocks of resources and unchanged technology.

Investment demand curve - a curve showing the dynamics of the interest rate and the volume of investments at different interest rates.

Phillips curve- a curve showing the existence of a stable relationship between the unemployment rate and the inflation rate.

Laffer curve- a curve showing the relationship between tax rates and tax revenues, revealing such a tax rate at which tax revenues reach a maximum.

Even a simple enumeration of terms shows how important it is for economists to be able to build graphs and analyze the equations of curves, which are straight lines and second-order curves - a circle, an ellipse, a hyperbola, a parabola. In addition, when solving a large class of problems, it is required to select an area on the plane bounded by some curves whose equations are given. Most often, these problems are formulated as follows: find the best production plan for given resources. The assignment of resources usually takes the form of inequalities, the equations of which are given. Therefore, one has to look for the largest or smallest values ​​taken by some function in the region specified by the equations of the system of inequalities.

In analytic geometry line on the plane is defined as the set of points whose coordinates satisfy the equation F(x,y)=0. In this case, restrictions must be imposed on the function F so that, on the one hand, this equation has an infinite set of solutions and, on the other hand, so that this set of solutions does not fill a “piece of the plane”. An important class of lines are those for which the function F(x,y) is a polynomial in two variables, in which case the line defined by the equation F(x,y)=0 is called algebraic. The algebraic lines given by the equation of the first degree are straight lines. An equation of the second degree, which has an infinite number of solutions, defines an ellipse, hyperbola, parabola, or a line splitting into two straight lines.

Let a rectangular Cartesian coordinate system be given on the plane. A straight line on a plane can be given by one of the equations:

ten . General equation of a straight line

Ax + By + C = 0. (2.1)

Vector n(А,В) is orthogonal to a straight line, the numbers A and B are not equal to zero at the same time.

twenty . Line Equation with Slope

y - y o = k (x - x o), (2.2)

where k is the slope of the straight line, i.e. k = tg a , where a - the value of the angle formed by the straight line with the axis Оx, M (x o , y o) - some point belonging to the straight line.

Equation (2.2) takes the form y = kx + b if M (0, b) is the point of intersection of the line with the Oy axis.

thirty . Equation of a straight line in segments

x/a + y/b = 1, (2.3)

where a and b are the values ​​of the segments cut off by a straight line on the coordinate axes.

40 . The equation of a straight line passing through two given points is A(x 1 , y 1) and B(x 2 , y 2):

. (2.4)

fifty . Equation of a straight line passing through a given point A(x 1 , y 1) parallel to a given vector a(m, n)

. (2.5)

60 . Normal equation of a straight line

rn o - p = 0, (2.6)

where r is the radius of an arbitrary point M(x, y) of this line, n o is a unit vector orthogonal to this line and directed from the origin to the line; p is the distance from the origin to the straight line.

Normal in coordinate form has the form:

x cos a + y sin a - p \u003d 0,

where a - the value of the angle formed by a straight line with the x-axis.

The equation of a pencil of lines centered at the point A (x 1, y 1) has the form:

y-y 1 = l (x-x 1),

where l is the beam parameter. If the beam is given by two intersecting lines A 1 x + B 1 y + C 1 = 0, A 2 x + B 2 y + C 2 = 0, then its equation has the form:

l (A 1 x + B 1 y + C 1) + m (A 2 x + B 2 y + C 2)=0,

where l and m are the beam parameters that do not turn to 0 at the same time.

The angle between the lines y \u003d kx + b and y \u003d k 1 x + b 1 is given by the formula:

tg j = .

The equality 1 + k 1 k = 0 is a necessary and sufficient condition for the lines to be perpendicular.

To make two equations

A 1 x + B 1 y + C 1 = 0, (2.7)

A 2 x + B 2 y + C 2 = 0, (2.8)

set the same straight line, it is necessary and sufficient that their coefficients are proportional:

A 1 / A 2 = B 1 / B 2 = C 1 / C 2.

Equations (2.7), (2.8) define two different parallel lines if A 1 /A 2 = B 1 /B 2 and B 1 /B 2¹ C 1 /C 2; lines intersect if A 1 /A 2¹B1/B2.

The distance d from the point M o (x o, y o) to the straight line is the length of the perpendicular drawn from the point M o to the straight line. If the line is given by a normal equation, then d =ê r about n o - r ê , where r o is the radius vector of the point M o or, in coordinate form, d =ê x o cos a + y o sin a - r ê .

The general equation of the second-order curve has the form

a 11 x 2 + 2a 12 xy + a 22 y 2 + 2a 1 x +2a 2 y + a = 0.

It is assumed that among the coefficients of the equation a 11 , a 12 , a 22 there are other than zero.

The equation of a circle centered at the point C(a, b) and with a radius equal to R:

(x - a) 2 + (y - b) 2 = R 2 . (2.9)

Ellipsethe locus of points is called, the sum of the distances of which from two given points F 1 and F 2 (foci) is a constant value equal to 2a.

Canonical (simplest) equation of an ellipse

x 2 /a 2 + y 2 /a 2 = 1. (2.10)

The ellipse given by equation (2.10) is symmetrical with respect to the coordinate axes. Options a and b called axle shafts ellipse.

Let a>b, then the foci F 1 and F 2 are on the Ox axis at a distance
c= from the origin. Ratio c/a = e < 1 называется eccentricity ellipse. The distances from the point M(x, y) of the ellipse to its foci (focal radius vectors) are determined by the formulas:

r 1 \u003d a - e x, r 2 \u003d a + e x.

If a< b, то фокусы находятся на оси Оy, c= , e = c/b,
r 1 \u003d b + e x, r 2 \u003d b - e x.

If a = b, then the ellipse is a circle centered at the origin of the radius a.

Hyperbolethe locus of points is called, the difference of the distances of which from two given points F 1 and F 2 (foci) is equal in absolute value to the given number 2a.

Canonical equation of a hyperbola

x 2 /a 2 - y 2 /b 2 = 1. (2.11)

The hyperbola given by equation (2.11) is symmetric with respect to the coordinate axes. It intersects the Ox axis at points A (a,0) and A (-a,0) - the vertices of the hyperbola and does not intersect the Oy axis. Parameter a called real semiaxis, b -imaginary axis. The parameter c= is the distance from the focus to the origin. Ratio c/a = e >1 is called eccentricity hyperbole. Straight lines whose equations y =± b/a x are called asymptotes hyperbole. The distances from the point M(x,y) of the hyperbola to its foci (focal radius vectors) are determined by the formulas:

r 1 = ê e x - a ê , r 2 = ê e x + a ê .

A hyperbola with a = b is called equilateral, its equation x 2 - y 2 \u003d a 2, and the equation of asymptotes y \u003d± x. Hyperbolas x 2 /a 2 - y 2 /b 2 = 1 and
y 2 /b 2 - x 2 /a 2 = 1 are called conjugated.

parabolais the locus of points equidistant from a given point (focus) and a given line (directrix).

The canonical equation of a parabola has two forms:

1) y 2 \u003d 2px - the parabola is symmetrical about the Ox axis.

2) x 2 \u003d 2py - the parabola is symmetrical about the Oy axis.

In both cases, p>0 and the vertex of the parabola, that is, the point lying on the axis of symmetry, is located at the origin.

A parabola whose equation y 2 = 2рx has focus F(р/2,0) and directrix x = - р/2, focal radius vector of point M(x, y) on it r = x+ р/2.

The parabola whose equation x 2 =2py has focus F(0, p/2) and directrix y = - p/2; the focal radius vector of the point M(x, y) of the parabola is r = y + p/2.

The equation F(x, y) = 0 defines a line that divides the plane into two or more parts. In one of these parts, the inequality F(x, y)<0, а в других - неравенство F(x, y)>0. In other words, the line
F(x, y)=0 separates the part of the plane where F(x, y)>0 from the part of the plane where F(x, y)<0.

The straight line, whose equation is Ax+By+C = 0, divides the plane into two half-planes. In practice, to find out in which half-plane we have Ax + By + C<0, а в какой Ax+By+C>0, apply the breakpoint method. To do this, take a control point (of course, not lying on a straight line, the equation of which is Ax + By + C = 0) and check what sign the expression Ax + By + C has at this point. The same sign has the indicated expression in the entire half-plane where the control point lies. In the second half-plane Ax+By+C has the opposite sign.

Nonlinear inequalities with two unknowns are solved in the same way.

For example, let's solve the inequality x 2 -4x+y 2 +6y-12 > 0. It can be rewritten as (x-2) 2 + (y+3) 2 - 25 > 0.

The equation (x-2) 2 + (y+3) 2 - 25 = 0 defines a circle with a center at point C(2,-3) and a radius of 5. The circle divides the plane into two parts - inner and outer. To find out in which of them this inequality takes place, we take a control point in the inner region, for example, the center C(2,-3) of our circle. Substituting the coordinates of point C into the left side of the inequality, we get a negative number -25. Hence, at all points lying inside the circle, the inequality
x 2 -4x+y 2 +6y-12< 0. Отсюда следует, что данное неравенство имеет место во внешней для окружности области.

Example 1.5.Compose the equations of the lines passing through the point A(3,1) and inclined to the line 2x+3y-1 = 0 at an angle of 45 o .

Solution.We will search in the form y=kx+b. Since the line passes through the point A, its coordinates satisfy the equation of the line, i.e. 1=3k+b,Þ b=1-3k. Angle between lines
y= k 1 x+b 1 and y= kx+b is defined by the formula tg j = . Since the slope k 1 of the original line 2x+3y-1=0 is - 2/3, and the angle j = 45 o , then we have an equation for determining k:

(2/3 + k)/(1 - 2/3k) = 1 or (2/3 + k)/(1 - 2/3k) = -1.

We have two values ​​of k: k 1 = 1/5, k 2 = -5. Finding the corresponding values ​​of b by the formula b=1-3k, we get two desired lines, the equations of which are: x - 5y + 2 = 0 and
5x + y - 16 = 0.

Example 1.6. At what value of the parameter t lines whose equations 3tx-8y+1 = 0 and (1+t)x-2ty = 0 are parallel?

Solution.Straight lines given by general equations are parallel if the coefficients at x and y proportional, i.e. 3t/(1+t) = -8/(-2t). Solving the resulting equation, we find t: t 1 \u003d 2, t 2 \u003d -2/3.

Example 1.7. Find the equation of the common chord of two circles:
x 2 +y 2 =10 and x 2 +y 2 -10x-10y+30=0.

Solution.Find the intersection points of the circles, for this we solve the system of equations:

.

Solving the first equation, we find the values ​​x 1 \u003d 3, x 2 \u003d 1. From the second equation - the corresponding values y: y 1 \u003d 1, y 2 \u003d 3. Now we get the equation of a common chord, knowing two points A (3,1) and B (1,3) belonging to this line: (y-1) / (3-1) \u003d (x-3)/(1-3), or y+ x - 4 = 0.

Example 1.8. How are the points located on the plane, the coordinates of which satisfy the conditions (x-3) 2 + (y-3) 2< 8, x >y?

Solution.The first inequality of the system defines the interior of the circle, not including the boundary, i.e. circle with center at point (3,3) and radius . The second inequality defines a half-plane defined by a straight line whose equation is x = y, and, since the inequality is strict, the points of the straight line itself do not belong to the half-plane, and all points below this straight line belong to the half-plane. Since we are looking for points that satisfy both inequalities, then the desired area is the interior of the semicircle.

Example 1.9.Calculate the length of the side of a square inscribed in an ellipse whose equation is x 2 / a 2 + y 2 / b 2 \u003d 1.

Solution.Let M(s, s)- the vertex of the square, lying in the first quarter. Then the side of the square will be 2 With. Because dot M belongs to the ellipse, its coordinates satisfy the equation of the ellipse c 2 /a 2 + c 2 /b 2 = 1, whence
c = ab/ ; so the side of the square is 2ab/ .

Example 1.10.Knowing the equation of asymptotes of the hyperbola y =± 0.5 x and one of its points M (12, 3), draw up the equation of a hyperbola.

Solution.We write the canonical equation of the hyperbola: x 2 /a 2 - y 2 /b 2 = 1. The asymptotes of the hyperbola are given by the equations y =± 0.5 x, so b/a = 1/2, hence a=2b. Because the M- point of the hyperbola, then its coordinates satisfy the equation of the hyperbola, i.e. 144/a 2 - 27/b 2 = 1. Given that a = 2b, we find b: b 2 =9Þ b=3 and a=6. Then the equation of the hyperbola is x 2 /36 - y 2 /9 = 1.

Example 1.11.Calculate the side length of a regular triangle ABC inscribed in a parabola with parameter R, assuming that point A coincides with the vertex of the parabola.

Solution.The canonical equation of a parabola with a parameter R has the form y 2 = 2рx, its vertex coincides with the origin, and the parabola is symmetrical about the x-axis. Since the line AB forms an angle of 30 o with the axis Ox, the equation of the line is: y = x. lots of charts

Therefore, we can find the coordinates of point B by solving the system of equations y 2 =2px, y = x, whence x = 6p, y = 2p. Hence, the distance between the points A(0,0) and B(6p,2p) is 4p.