Equation of a straight line by a given system of equations. General equation of the straight line: description, examples, problem solving

Equation of a line on a plane.

As you know, any point on a plane is determined by two coordinates in any coordinate system. Coordinate systems can be different depending on the choice of datum and origin.

Definition. Equation line called the ratio y = f (x) between the coordinates of the points that make up this line.

Note that the equation of the line can be expressed parametrically, that is, each coordinate of each point is expressed in terms of some independent parameter t.

A typical example is the trajectory of a moving point. In this case, time plays the role of a parameter.

Equation of a straight line on a plane.

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

Ax + Wu + C = 0,

moreover, constants A, B are not equal to zero at the same time, i.e. А 2 + В 2  0. This first-order equation is called the general equation of the straight line.

Depending on the values of constants A, B and C, 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) - the straight line is parallel to the Ox axis

B = 0, A  0, C  0 (Ax + C = 0) - the straight line is 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 the Ox axis

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) is perpendicular to the straight line given by the equation Ax + Vy + C = 0.

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

At A = 3 and B = -1, we compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C, we substitute the coordinates of a 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 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 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 is called slope straight.

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

Applying the above formula, we get:

Equation of a straight line by point and slope.

If the general equation of the straight line Ax + Vy + C = 0 is reduced 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 specification of a straight line through a point and a direction vector of a straight line.

Definition. Every nonzero vector ( 1,  2), the components of which satisfy the condition А 1 + В 2 = 0 is called the 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 point A (1, 2).

The equation of the desired straight line will be sought in the form: Ax + By + C = 0. According to the definition, the coefficients must satisfy the conditions:

1A + (-1) B = 0, i.e. A = B.

Then the equation of the 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:

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 point of intersection of the straight 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 straight line x - y + 1 = 0 is given. 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 + Vy + C = 0 are divided by the number
which is called normalizing factor, then we get

xcos + ysin - p = 0 -

normal equation of a straight line.

The sign  of the normalizing factor should be chosen so that С< 0.

p is the length of the perpendicular dropped from the origin to the straight line, and  is the angle formed by this perpendicular with the positive direction of the Ox axis.

Example. A general equation of the straight line 12x - 5y - 65 = 0 is given. It is required to write various types of equations of this straight line.

the equation of this straight line in segments:

equation of this straight line with slope: (divide by 5)

normal equation of the line:

; cos = 12/13; sin = -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.

Example. The straight line cuts off equal positive segments on the coordinate axes. Make a straight line equation if the area of the triangle formed by these segments is 8 cm 2.

The straight line equation has the form:
, a = b = 1; ab / 2 = 8; a = 4; -4.

a = -4 does not match the problem statement.

Total:
or x + y - 4 = 0.

Example. Draw up the equation of the 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.

The angle between straight lines on the plane.

Definition. If two straight lines are given y = k 1 x + b 1, y = k 2 x + b 2, then sharp corner between these lines will be defined as

Two straight lines are parallel if k 1 = k 2.

Two straight lines are perpendicular if k 1 = -1 / k 2.

Theorem. Straight lines Ax + Vu + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A are proportional 1 =  A, B 1 =  B. If also C 1 =  C, then the straight lines coincide.

The coordinates of the point of intersection of two straight 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 this line.

Definition. The 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:

Distance from point to line.

Theorem. If point M (x 0 , at 0 ), then the distance to the straight line Ax + Vy + C = 0 is defined 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 the straight line passing through set 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 tg =
;  =  / 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.

Analytical geometry in space.

Equation of a line in space.

Equation of a straight line in space along a point and

guiding vector.

Take an arbitrary line and a vector (m, n, p) parallel to the given line. Vector called direction vector straight.

On the straight line, take two arbitrary points M 0 (x 0, y 0, z 0) and M (x, y, z).

M 1

Let us denote the radius vectors of these points as and , it's obvious that - =
.

Because vectors
and collinear, then the relation
= t, where t is some parameter.

Total, you can write: = + t.

Because this equation is satisfied by the coordinates of any point of the straight line, then the resulting equation - parametric equation of a straight line.

This vector equation can be represented in coordinate form:

Transforming this system and equating the values of the parameter t, we obtain canonical equations straight in space:

Definition. Direction cosines straight line are the direction cosines of the vector , which can be calculated by the formulas:

;

.

From here we get: m: n: p = cos: cos: cos.

The numbers m, n, p are called slopes straight. Because is a nonzero vector, then m, n and p cannot be zero at the same time, but one or two of these numbers can be zero. In this case, in the equation of the straight line, the corresponding numerators should be equated to zero.

Equation of a straight line in space passing

through two points.

If on a straight line in space we mark two arbitrary points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), then the coordinates of these points must satisfy the straight line equation obtained above:

In addition, for point M 1 you can write:

Solving these equations together, we get:

This is the equation of a straight line passing through two points in space.

General equations of a straight line in space.

The equation of a straight line can be considered as the equation of the line of intersection of two planes.

As discussed above, a plane in vector form can be given by the equation:

+ D = 0, where

- plane normal; - radius vector of an arbitrary point of the plane.

This article continues the topic of the equation of a straight line on a plane: consider such a form of equation as the general equation of a straight line. Let us define a theorem and give its proof; Let's figure out what an incomplete general equation of a straight line is and how to make transitions from a general equation to other types of equations of a straight line. We will consolidate the whole theory with illustrations and solving practical problems.

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Let a rectangular coordinate system O x y be given on the plane.

Theorem 1

Any equation of the first degree, having the form A x + B y + C = 0, where A, B, C are some real numbers (A and B are not equal to zero at the same time) defines a straight line in a rectangular coordinate system on a plane. In turn, any straight line in a rectangular coordinate system on a plane is determined by an equation that has the form A x + B y + C = 0 for a certain set of values A, B, C.

Proof

the indicated theorem consists of two points, we will prove each of them.

Let us prove that the equation A x + B y + C = 0 defines a straight line on the plane.

Let there exist some point М 0 (x 0, y 0), the coordinates of which correspond to the equation A x + B y + C = 0. Thus: A x 0 + B y 0 + C = 0. Subtract from the left and right sides of the equations A x + B y + C = 0 the left and right sides of the equation A x 0 + B y 0 + C = 0, we obtain a new equation of the form A (x - x 0) + B (y - y 0) = 0. It is equivalent to A x + B y + C = 0.

The resulting equation A (x - x 0) + B (y - y 0) = 0 is a necessary and sufficient condition for the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0 ). Thus, the set of points M (x, y) defines a straight line in a rectangular coordinate system perpendicular to the direction of the vector n → = (A, B). We can assume that this is not so, but then the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) would not be perpendicular, and the equality A (x - x 0 ) + B (y - y 0) = 0 would not be true.

Therefore, the equation A (x - x 0) + B (y - y 0) = 0 defines some straight line in a rectangular coordinate system on the plane, and hence the equivalent equation A x + B y + C = 0 defines the same straight line. This is how we proved the first part of the theorem.

Let us give a proof that any straight line in a rectangular coordinate system on a plane can be defined by an equation of the first degree A x + B y + C = 0.

Let us set the straight line a in a rectangular coordinate system on the plane; point M 0 (x 0, y 0) through which this line passes, as well as the normal vector of this line n → = (A, B).

Let there also be some point M (x, y) - a floating point of a straight line. In this case, the vectors n → = (A, B) and M 0 M → = (x - x 0, y - y 0) are perpendicular to each other, and their scalar product is zero:

n →, M 0 M → = A (x - x 0) + B (y - y 0) = 0

Rewrite the equation A x + B y - A x 0 - B y 0 = 0, define C: C = - A x 0 - B y 0 and in the end result we get the equation A x + B y + C = 0.

So, we have proved the second part of the theorem, and we have proved the whole theorem as a whole.

Definition 1

An equation of the form A x + B y + C = 0 - this is general equation of the line on a plane in a rectangular coordinate systemO x y.

Based on the proven theorem, we can conclude that a straight line and its general equation given on a plane in a fixed rectangular coordinate system are inextricably linked. In other words, the initial straight line corresponds to its general equation; the general equation of a straight line corresponds to a given straight line.

It also follows from the proof of the theorem that the coefficients A and B for the variables x and y are the coordinates of the normal vector of the straight line, which is given by the general equation of the straight line A x + B y + C = 0.

Consider specific example general equation of the line.

Let the equation 2 x + 3 y - 2 = 0 be given, which corresponds to a straight line in a given rectangular coordinate system. The normal vector of this line is the vector n → = (2, 3). Draw a given straight line in the drawing.

It is also possible to assert the following: the straight line that we see in the drawing is determined by the general equation 2 x + 3 y - 2 = 0, since the coordinates of all points of a given straight line correspond to this equation.

We can obtain the equation λ · A x + λ · B y + λ · C = 0 by multiplying both sides of the general equation of the line by a nonzero number λ. The resulting equation is equivalent to the original general equation, therefore, it will describe the same straight line on the plane.

Definition 2

Complete general equation of the line- such a general equation of the straight line A x + B y + C = 0, in which the numbers A, B, C are nonzero. Otherwise the equation is incomplete.

Let us examine all the variations of the incomplete general equation of the line.

When A = 0, B ≠ 0, C ≠ 0, the general equation becomes B y + C = 0. Such an incomplete general equation defines in a rectangular coordinate system O x y a straight line that is parallel to the O x axis, since for any real value of x, the variable y will take the value - C B. In other words, the general equation of the straight line A x + B y + C = 0, when A = 0, B ≠ 0, defines the locus of points (x, y), the coordinates of which are equal to the same number - C B.
If A = 0, B ≠ 0, C = 0, the general equation takes the form y = 0. This incomplete equation defines the abscissa axis O x.
When A ≠ 0, B = 0, C ≠ 0, we obtain an incomplete general equation A x + C = 0, defining a straight line parallel to the ordinate axis.
Let A ≠ 0, B = 0, C = 0, then the incomplete general equation will take the form x = 0, and this is the equation of the coordinate line O y.
Finally, for A ≠ 0, B ≠ 0, C = 0, the incomplete general equation takes the form A x + B y = 0. And this equation describes a straight line that passes through the origin. Indeed, the pair of numbers (0, 0) corresponds to the equality A x + B y = 0, since A · 0 + B · 0 = 0.

Let us graphically illustrate all the above types of the incomplete general equation of a straight line.

Example 1

It is known that a given straight line is parallel to the ordinate axis and passes through the point 2 7, - 11. It is necessary to write down the general equation of a given straight line.

Solution

A straight line parallel to the ordinate axis is given by an equation of the form A x + C = 0, in which A ≠ 0. Also, the condition specifies the coordinates of the point through which the line passes, and the coordinates of this point meet the conditions of the incomplete general equation A x + C = 0, i.e. the equality is true:

A · 2 7 + C = 0

It is possible to determine C from it by giving A some non-zero value, for example, A = 7. In this case, we get: 7 · 2 7 + C = 0 ⇔ C = - 2. We know both coefficients A and C, we substitute them into the equation A x + C = 0 and we obtain the required equation of the straight line: 7 x - 2 = 0

Answer: 7 x - 2 = 0

Example 2

The drawing shows a straight line, it is necessary to write down its equation.

Solution

The given drawing allows us to easily take the initial data for solving the problem. We see in the drawing that the given line is parallel to the O x axis and passes through the point (0, 3).

The straight line, which is parallel to the eyes of the abscissa, determines the incomplete general equation B y + C = 0. Let's find the values of B and C. The coordinates of the point (0, 3), since a given straight line passes through it, will satisfy the equation of the straight line B y + C = 0, then the equality is valid: B · 3 + C = 0. Let's set for B some value other than zero. Suppose B = 1, in this case from the equality B 3 + C = 0 we can find C: C = - 3. We use the known values of B and C, we obtain the required equation of the straight line: y - 3 = 0.

Answer: y - 3 = 0.

General equation of a straight line passing through a given point of the plane

Let the given line pass through the point М 0 (x 0, y 0), then its coordinates correspond to the general equation of the line, i.e. the equality is true: A x 0 + B y 0 + C = 0. Subtract the left and right sides of this equation from the left and right sides of the general complete equation straight. We get: A (x - x 0) + B (y - y 0) + C = 0, this equation is equivalent to the original general, passes through the point М 0 (x 0, y 0) and has a normal vector n → = (A, B).

The result that we have obtained makes it possible to write down the general equation of the straight line with the known coordinates of the normal vector of the straight line and the coordinates of a certain point of this straight line.

Example 3

Given a point М 0 (- 3, 4), through which a straight line passes, and a normal vector of this straight line n → = (1, - 2). It is necessary to write down the equation of a given straight line.

Solution

The initial conditions allow us to obtain the necessary data for drawing up the equation: A = 1, B = - 2, x 0 = - 3, y 0 = 4. Then:

A (x - x 0) + B (y - y 0) = 0 ⇔ 1 (x - (- 3)) - 2 y (y - 4) = 0 ⇔ ⇔ x - 2 y + 22 = 0

The problem could have been solved differently. General equation the line has the form A x + B y + C = 0. A given normal vector allows you to get the values of the coefficients A and B, then:

A x + B y + C = 0 ⇔ 1 x - 2 y + C = 0 ⇔ x - 2 y + C = 0

Now we find the value of C, using the point M 0 (- 3, 4) specified by the condition of the problem, through which the straight line passes. The coordinates of this point correspond to the equation x - 2 y + C = 0, i.e. - 3 - 2 4 + C = 0. Hence C = 11. The required equation of the straight line takes the form: x - 2 y + 11 = 0.

Answer: x - 2 y + 11 = 0.

Example 4

A straight line 2 3 x - y - 1 2 = 0 and a point М 0 lying on this straight line are given. Only the abscissa of this point is known, and it is equal to - 3. It is necessary to determine the ordinate of the given point.

Solution

Let's set the designation of the coordinates of the point М 0 as x 0 and y 0. The initial data indicates that x 0 = - 3. Since a point belongs to a given straight line, then its coordinates correspond to the general equation of this straight line. Then the equality will be true:

2 3 x 0 - y 0 - 1 2 = 0

Determine y 0: 2 3 (- 3) - y 0 - 1 2 = 0 ⇔ - 5 2 - y 0 = 0 ⇔ y 0 = - 5 2

Answer: - 5 2

The transition from the general equation of a straight line to other types of equations of a straight line and vice versa

As we know, there are several types of equations for the same straight line on the plane. The choice of the type of equation depends on the conditions of the problem; it is possible to choose the one that is more convenient for solving it. This is where the skill of converting an equation of one kind into an equation of another kind comes in handy.

To begin with, consider the transition from the general equation of the form A x + B y + C = 0 to the canonical equation x - x 1 a x = y - y 1 a y.

If А ≠ 0, then we transfer the term B y to right side general equation. On the left side, place A outside the parentheses. As a result, we get: A x + C A = - B y.

This equality can be written as a proportion: x + C A - B = y A.

If В ≠ 0, we leave only the term A x on the left side of the general equation, transfer the rest to the right side, we get: A x = - B y - C. We take out - B outside the brackets, then: A x = - B y + C B.

Let's rewrite equality as a proportion: x - B = y + C B A.

Of course, there is no need to memorize the resulting formulas. It is enough to know the algorithm of actions in the transition from the general equation to the canonical one.

Example 5

The general equation of the straight line is given: 3 y - 4 = 0. It is necessary to transform it into a canonical equation.

Solution

Rewrite the original equation as 3 y - 4 = 0. Next, we act according to the algorithm: the term 0 x remains on the left side; and on the right side we take out - 3 outside the brackets; we get: 0 x = - 3 y - 4 3.

Let's write the resulting equality as a proportion: x - 3 = y - 4 3 0. So, we got an equation of the canonical form.

Answer: x - 3 = y - 4 3 0.

To transform the general equation of the straight line into parametric ones, one first makes the transition to the canonical form, and then the transition from the canonical equation of the straight line to the parametric equations.

Example 6

The straight line is given by the equation 2 x - 5 y - 1 = 0. Write down the parametric equations of this straight line.

Solution

Let's make the transition from the general equation to the canonical one:

2 x - 5 y - 1 = 0 ⇔ 2 x = 5 y + 1 ⇔ 2 x = 5 y + 1 5 ⇔ x 5 = y + 1 5 2

Now we take both sides of the resulting canonical equation equal to λ, then:

x 5 = λ y + 1 5 2 = λ ⇔ x = 5 λ y = - 1 5 + 2 λ, λ ∈ R

Answer:x = 5 λ y = - 1 5 + 2 λ, λ ∈ R

The general equation can be transformed into an equation of a straight line with the slope y = k x + b, but only if B ≠ 0. For the transition on the left, we leave the term B y, the rest are transferred to the right. We get: B y = - A x - C. Divide both sides of the resulting equality by B, different from zero: y = - A B x - C B.

Example 7

The general equation of the straight line is given: 2 x + 7 y = 0. You must convert that equation to a slope equation.

Solution

Let's perform the necessary actions according to the algorithm:

2 x + 7 y = 0 ⇔ 7 y - 2 x ⇔ y = - 2 7 x

Answer: y = - 2 7 x.

From the general equation of a straight line, it is enough to simply obtain an equation in segments of the form x a + y b = 1. To make such a transition, we transfer the number C to the right-hand side of the equality, divide both sides of the resulting equality by - С and, finally, transfer the coefficients for the variables x and y to the denominators:

A x + B y + C = 0 ⇔ A x + B y = - C ⇔ ⇔ A - C x + B - C y = 1 ⇔ x - C A + y - C B = 1

Example 8

It is necessary to transform the general equation of the straight line x - 7 y + 1 2 = 0 into the equation of the straight line in segments.

Solution

Move 1 2 to the right side: x - 7 y + 1 2 = 0 ⇔ x - 7 y = - 1 2.

Divide both sides of the equality by -1/2: x - 7 y = - 1 2 ⇔ 1 - 1 2 x - 7 - 1 2 y = 1.

Answer: x - 1 2 + y 1 14 = 1.

In general, the reverse transition is also easy: from other types of equations to the general one.

The equation of a straight line in segments and an equation with a slope coefficient can be easily transformed into a general one, simply by collecting all the terms on the left side of the equality:

x a + y b ⇔ 1 a x + 1 b y - 1 = 0 ⇔ A x + B y + C = 0 y = k x + b ⇔ y - k x - b = 0 ⇔ A x + B y + C = 0

The canonical equation is transformed to the general one according to the following scheme:

x - x 1 ax = y - y 1 ay ⇔ ay (x - x 1) = ax (y - y 1) ⇔ ⇔ ayx - axy - ayx 1 + axy 1 = 0 ⇔ A x + B y + C = 0

To switch from parametric, first, the transition to the canonical is carried out, and then to the general:

x = x 1 + a x λ y = y 1 + a y λ ⇔ x - x 1 a x = y - y 1 a y ⇔ A x + B y + C = 0

Example 9

Parametric equations of the straight line x = - 1 + 2 · λ y = 4 are given. It is necessary to write down the general equation of this straight line.

Solution

Let's make the transition from parametric equations to canonical:

x = - 1 + 2 λ y = 4 ⇔ x = - 1 + 2 λ y = 4 + 0 λ ⇔ λ = x + 1 2 λ = y - 4 0 ⇔ x + 1 2 = y - 4 0

Let's move from the canonical to the general:

x + 1 2 = y - 4 0 ⇔ 0 (x + 1) = 2 (y - 4) ⇔ y - 4 = 0

Answer: y - 4 = 0

Example 10

The equation of a straight line in segments x 3 + y 1 2 = 1 is given. The transition to general view equations.

Solution:

Let's just rewrite the equation in the required form:

x 3 + y 1 2 = 1 ⇔ 1 3 x + 2 y - 1 = 0

Answer: 1 3 x + 2 y - 1 = 0.

Drawing up the general equation of a straight line

Above we said that the general equation can be written with the known coordinates of the normal vector and the coordinates of the point through which the straight line passes. Such a straight line is determined by the equation A (x - x 0) + B (y - y 0) = 0. We also analyzed the corresponding example there.

Now we will consider more complex examples, in which first it is necessary to determine the coordinates of the normal vector.

Example 11

A straight line parallel to the straight line 2 x - 3 y + 3 3 = 0 is given. Also known is the point M 0 (4, 1), through which the given line passes. It is necessary to write down the equation of a given straight line.

Solution

The initial conditions tell us that the straight lines are parallel, then, as the normal vector of the straight line, the equation of which is to be written, we take the directing vector of the straight line n → = (2, - 3): 2 x - 3 y + 3 3 = 0. Now we know all the necessary data to compose the general equation of the straight line:

A (x - x 0) + B (y - y 0) = 0 ⇔ 2 (x - 4) - 3 (y - 1) = 0 ⇔ 2 x - 3 y - 5 = 0

Answer: 2 x - 3 y - 5 = 0.

Example 12

The specified line passes through the origin perpendicular to the line x - 2 3 = y + 4 5. It is necessary to draw up a general equation for a given straight line.

Solution

The normal vector of the given line will be the direction vector of the line x - 2 3 = y + 4 5.

Then n → = (3, 5). The straight line passes through the origin, i.e. through the point O (0, 0). Let's compose the general equation of a given straight line:

A (x - x 0) + B (y - y 0) = 0 ⇔ 3 (x - 0) + 5 (y - 0) = 0 ⇔ 3 x + 5 y = 0

Answer: 3 x + 5 y = 0.

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Canonical equations of a straight line in space are equations that define a straight line passing through a given point collinear to the direction vector.

Let a point and a direction vector be given. An arbitrary point lies on a straight line l only if the vectors and are collinear, i.e., they satisfy the condition:

The above equations are the canonical equations of the straight line.

Numbers m , n and p are the projections of the direction vector onto the coordinate axes. Since the vector is nonzero, then all numbers m , n and p cannot be zero at the same time. But one or two of them may turn out to be equal to zero... In analytical geometry, for example, the following notation is allowed:

which means that the projection of the vector on the axis Oy and Oz are equal to zero. Therefore, both the vector and the straight line given by the canonical equations are perpendicular to the axes Oy and Oz, i.e., the plane yOz .

Example 1. Write the equations of a straight line in space perpendicular to the plane and passing through the point of intersection of this plane with the axis Oz .

Solution. Find the point of intersection of this plane with the axis Oz... Since any point lying on the axis Oz, has coordinates, then, setting in the given equation the plane x = y = 0, we get 4 z- 8 = 0 or z= 2. Therefore, the point of intersection of this plane with the axis Oz has coordinates (0; 0; 2). Since the sought line is perpendicular to the plane, it is parallel to its normal vector. Therefore, the directing vector of the straight line can be the normal vector a given plane.

Now we write down the sought equations of the straight line passing through the point A= (0; 0; 2) in the direction of the vector:

Equations of a straight line passing through two given points

A straight line can be specified by two points lying on it and In this case, the direction vector of the straight line can be a vector. Then the canonical equations of the straight line take the form

The above equations determine the straight line passing through two given points.

Example 2. Make the equation of a straight line in space passing through the points and.

Solution. Let us write the sought equations of the straight line in the form given above in the theoretical note:

Since, the sought line is perpendicular to the axis Oy .

Straight as a line of intersection of planes

A straight line in space can be defined as the line of intersection of two non-parallel planes and, that is, as a set of points satisfying a system of two linear equations

The equations of the system are also called general equations of a straight line in space.

Example 3. Write the canonical equations of a straight line in space given by general equations

Solution. To write the canonical equations of a straight line or, which is the same, the equations of a straight line passing through two given points, you need to find the coordinates of any two points of the straight line. They can be the points of intersection of a straight line with any two coordinate planes, for example yOz and xOz .

Point of intersection of a straight line with a plane yOz has an abscissa x= 0. Therefore, assuming in the given system of equations x= 0, we get a system with two variables:

Her decision y = 2 , z= 6 together with x= 0 defines a point A(0; 2; 6) the required straight line. Then, setting in the given system of equations y= 0, we get the system

Her decision x = -2 , z= 0 together with y= 0 defines a point B(-2; 0; 0) intersections of a straight line with a plane xOz .

Now we write down the equations of the straight line passing through the points A(0; 2; 6) and B (-2; 0; 0) :

or after dividing the denominators by -2:

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).

The straight line passing through the point K (x 0; y 0) and parallel to the straight line y = kx + a is found by the formula:

y - y 0 = k (x - x 0) (1)

Where k is the slope of the straight line.

Alternative formula:
The straight line passing through the point M 1 (x 1; y 1) and parallel to the straight line Ax + By + C = 0 is represented by the equation

A (x-x 1) + B (y-y 1) = 0. (2)

Example # 1. Make up the equation of the straight line passing through the point M 0 (-2,1) and at the same time:
a) parallel to the straight line 2x + 3y -7 = 0;
b) perpendicular to the straight line 2x + 3y -7 = 0.
Solution ... We represent the equation with the slope as y = kx + a. To do this, move all values except y to the right side: 3y = -2x + 7. Then we divide the right side by a factor of 3. We get: y = -2 / 3x + 7/3
Find the equation NK passing through the point K (-2; 1), parallel to the line y = -2 / 3 x + 7/3
Substituting x 0 = -2, k = -2 / 3, y 0 = 1 we get:
y-1 = -2 / 3 (x - (- 2))
or
y = -2 / 3 x - 1/3 or 3y + 2x +1 = 0

Example No. 2. Write the equation of a 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. Area right triangle, where a and b are his legs. Find the intersection points of the desired straight 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 No. 3. Make the equation of the straight line passing through the point (-2; 5) and parallel to the straight 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 required straight line is y - 5 = 5/7 (x - (-2)), i.e. 7 (y-5) = 5 (x + 2) or 5x-7y + 45 = 0.

Example No. 4. Solving example 3 (A = 5, B = -7) using formula (2), we find 5 (x + 2) -7 (y-5) = 0.

Example No. 5. Equate the straight line passing through the point (-2; 5) and parallel to the 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 inapplicable, since given equation cannot be resolved relative to y (this line is parallel to the ordinate).