Find an equation for two points. Various equations of a straight line

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

Ah + Wu + C = 0,

and the constants A, B are not equal to zero at the same time. This first order equation is called the general equation of a straight line. Depending on the values of the constants A, B and C, the following special cases are possible:

C \u003d 0, A ≠ 0, B ≠ 0 - the line passes through the origin

A \u003d 0, B ≠ 0, C ≠ 0 (By + C \u003d 0) - the line is parallel to the Ox axis

B \u003d 0, A ≠ 0, C ≠ 0 ( Ax + C \u003d 0) - the line is parallel to the Oy axis

B \u003d C \u003d 0, A ≠ 0 - the straight line coincides with the Oy axis

A \u003d C \u003d 0, B ≠ 0 - the straight line coincides with the Ox axis

The equation of a straight line can be presented 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) is perpendicular to the line given by the equation Ax + By + C = 0.

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

Decision. At A = 3 and B = -1, we compose the equation of a straight line: 3x - y + C = 0. To find the coefficient C, we substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore, C = -1 . Total: the desired equation: 3x - y - 1 \u003d 0.

Equation of a 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 a straight line 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 straight line equation written above is simplified:

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

Fraction = k is called slope factor straight.

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

Decision. Applying the above formula, we get:

Equation of a straight line from a point and a slope

If the total Ax + Wu + C = 0 lead to the form:

and designate , then the resulting equation is called equation of a straight line with a slopek.

Equation of a straight line with a point and direction vector

By analogy with the paragraph considering the equation of a straight line through the normal vector, you can enter the assignment of a straight line through a point and a directing vector of a straight line.

Definition. Each non-zero vector (α 1, α 2), the components of which satisfy the condition A α 1 + B α 2 = 0 is called the directing vector of the 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).

Decision. We will look for the equation of the desired straight line in the form: Ax + By + C = 0. In accordance with the definition, the coefficients must satisfy the conditions:

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

Then the equation of a straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0. for x = 1, y = 2 we get C / A = -3, i.e. desired equation:

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

The geometric meaning of the coefficients is that the coefficient a is the coordinate of the point of intersection of the line with the x-axis, and b- the coordinate of the point of intersection of the straight line with the Oy axis.

Example. Given the general equation of the line x - y + 1 = 0. Find the equation of this line in the segments.

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

Normal equation of a straight line

If both sides of the equation Ax + Vy + C = 0 are multiplied by the 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 * С< 0. р – длина перпендикуляра, опущенного из начала координат на прямую, а φ - угол, образованный этим перпендикуляром с положительным направлением оси Ох.

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

the equation of this straight line in segments:

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

; cos φ = 12/13; sin φ= -5/13; p=5.

It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines parallel to the axes or passing through the origin.

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

Decision. The straight line equation has the form: , ab /2 = 8; ab=16; a=4, a=-4. a = -4< 0 не подходит по условию задачи. Итого: или х + у – 4 = 0.

Example. Write the equation of a straight line passing through the point A (-2, -3) and the origin.

Decision. The equation of a straight line has the form: , where x 1 \u003d y 1 \u003d 0; x 2 \u003d -2; y 2 \u003d -3.

Angle between 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 + Vy + C \u003d 0 and A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients A 1 \u003d λA, B 1 \u003d λB are proportional. If also С 1 = λС, 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.

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

Definition. The line passing through the point M 1 (x 1, y 1) and perpendicular to the line y \u003d kx + b is represented by the equation:

Distance from point to line

Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Vy + C \u003d 0 is defined as

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

(1)

The x 1 and y 1 coordinates 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 has been proven.

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

k 1 \u003d -3; k2 = 2; tgφ = ; φ= π /4.

Example. Show that the lines 3x - 5y + 7 = 0 and 10x + 6y - 3 = 0 are perpendicular.

Decision. We find: k 1 \u003d 3/5, k 2 \u003d -5/3, k 1 * k 2 \u003d -1, therefore, the 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.

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

2x – 3y + 3 = 0;

The desired height equation is: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: whence b = 17. Total: .

Answer: 3x + 2y - 34 = 0.

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.
Decision . Let's represent the slope equation as y = kx + a . To do this, we will transfer all values except y to the 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.
Decision . 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 .
Decision. 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.
Decision. 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 a first-degree equation (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 With 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).

Decision. 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).

Decision. 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).

Decision. 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 various types of 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 on a plane.

As is known, any point on the plane is determined by two coordinates in some coordinate system. Coordinate systems can be different depending on the choice of basis and origin.

Definition. Line equation is the relation y = f(x) between the coordinates of the points that make up this line.

Note that the line equation can be expressed in a parametric way, that is, each coordinate of each point is expressed through 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 line in the plane can be given by a first order equation

Ah + Wu + C = 0,

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

Depending on the values of the constants A, B and C, the following special cases are possible:

C \u003d 0, A  0, B  0 - the line passes through the origin

A \u003d 0, B  0, C  0 (By + C \u003d 0) - the line is parallel to the Ox axis

B \u003d 0, A  0, C  0 ( Ax + C \u003d 0) - the line is parallel to the Oy axis

B \u003d C \u003d 0, A  0 - the straight line coincides with the Oy axis

A \u003d C \u003d 0, B  0 - the straight line coincides with the Ox axis

The equation of a straight line can be presented 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) is perpendicular to the line given by the equation Ax + By + C = 0.

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

Let us 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 \u003d 0, therefore C \u003d -1.

Total: the desired equation: 3x - y - 1 \u003d 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 a straight line passing through these points:

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

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

if x 1  x 2 and x \u003d x 1, if x 1 \u003d x 2.

Fraction
=k is called slope factor straight.

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

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

and designate
, then the resulting equation is called equation of a straight line with a slopek.

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

Definition. Every non-zero vector ( 1 ,  2), the components of which satisfy the condition A 1 + B 2 = 0 is called the directing vector of the line

Ah + Wu + C = 0.

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

We will look for the equation of the desired straight line in the form: Ax + By + C = 0. In accordance with the definition, the coefficients must satisfy the conditions:

1A + (-1)B = 0, i.e. A = B.

Then the equation of a straight line has the form: Ax + Ay + C = 0, or x + y + C/A = 0.

at x = 1, y = 2 we get С/A = -3, i.e. desired equation:

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

Example. Given the general equation of the line x - y + 1 = 0. Find the equation of this line in the segments.

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

Normal equation of a straight line.

If both sides of the equation Ax + Wy + C = 0 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 must 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. Given the general equation of the line 12x - 5y - 65 = 0. It is required to write various types of equations for this line.

the equation of this straight line in segments:

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

normal equation of a straight 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. Write the equation of a straight line if the area of the triangle formed by these segments is 8 cm 2.

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

a = -4 does not fit the condition of the problem.

Total:
or x + y - 4 = 0.

Example. Write the equation of a straight line passing through the point A (-2, -3) and the origin.

The equation of a straight line has the form:
, where x 1 \u003d y 1 \u003d 0; x 2 \u003d -2; y 2 \u003d -3.

Angle between 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. Straight lines Ax + Vy + 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 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.

Equation of a line passing through a given point

perpendicular to this line.

Definition. The line passing through the point M 1 (x 1, y 1) and perpendicular to the line y \u003d kx + b is represented by the equation:

The distance from a point to a line.

Theorem. If a point M(x 0 , y 0 ), then the distance to the line Ax + Vy + C = 0 is defined as

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

The x 1 and y 1 coordinates 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 has been proven.

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

k 1 \u003d -3; k 2 = 2tg =
;  = /4.

Example. Show that the lines 3x - 5y + 7 = 0 and 10x + 6y - 3 = 0 are perpendicular.

We find: k 1 \u003d 3/5, k 2 \u003d -5/3, k 1 k 2 \u003d -1, therefore, the 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 desired height equation is: Ax + By + C = 0 or y = kx + b.

k = . Then y =
. Because the height passes through point C, then its coordinates satisfy this equation:
whence b = 17. Total:
.

Answer: 3x + 2y - 34 = 0.

Analytical geometry in space.

Line equation in space.

The equation of a straight line in space by a point and

direction vector.

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

Let's take two arbitrary points M 0 (x 0 , y 0 , z 0) and M(x, y, z) on the straight line.

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

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

In total, we can write: = + t.

Because this equation is satisfied by the coordinates of any point on the line, then the resulting equation is 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 the canonical equations of a straight line in space:

Definition. Direction cosines direct 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 slope factors straight. Because is a non-zero 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 a straight line, the corresponding numerators should be equated to zero.

Equation of a straight line in space passing

through two points.

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

In addition, for point M 1 we 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 a 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.

The canonical equations of a straight line in space are equations that define a straight line passing through a given point collinearly to a direction vector.

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

The above equations are the canonical equations of the line.

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

which means that the projections of the vector on the axes 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. planes yOz .

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

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

Now we write the desired 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 defined by two points lying on it and In this case, the directing vector of the straight line can be the vector . Then the canonical equations of the line take the form

The above equations define a straight line passing through two given points.

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

Decision. We write the desired equations of the straight line in the form given above in the theoretical reference:

Since , then the desired line is perpendicular to the axis Oy .

Straight as a line of intersection of planes

A straight line in space can be defined as a line of intersection of two non-parallel planes and, i.e., as a set of points that satisfy a system of two linear equations

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

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

Decision. To write the canonical equations of a straight line or, what is the same, the equation of a straight line passing through two given points, you need to find the coordinates of any two points on 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 line with a plane yOz has an abscissa x= 0 . Therefore, assuming in this 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) of the desired line. Assuming then 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) intersection of a line with a plane xOz .

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

or after dividing the denominators by -2: