Equation of a straight line in four forms. General equation of a line

Canonical equations of a line in space are equations that define a 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 line l only if the vectors and are collinear, i.e., the condition is satisfied for them:

The above equations are the canonical equations of the straight 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 simultaneously be equal to zero. But one or two of them may turn out to be zero. In analytical geometry, for example, the following entry is allowed:

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

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

Solution. Let's find the point of intersection of this plane with the axis Oz. Since any point lying 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 this 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 directing vector of the straight line can be the normal vector given plane.

Now let’s write down the required equations of a straight line passing through a point A= (0; 0; 2) in the direction of the vector:

Equations of a 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 determine a line passing through two given points.

Example 2. Write an equation for a line in space passing through the points and .

Solution. Let us write down the required equations of the straight line in the form given above in the theoretical reference:

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

Straight as the line of intersection of planes

A straight line in space can be defined as the line of intersection of two non-parallel planes and, i.e., as a set of points satisfying 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 line in space given by general equations

Solution. To write the canonical equations of a line or, what is the same, the equations of a line passing through two given points, you need to find the coordinates of any two points on the 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 and 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) the desired line. Then assuming 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 let's write down the equations of the line passing through the points A(0; 2; 6) and B (-2; 0; 0) :

or after dividing the denominators by -2:

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 = k(x - x 0) (1)

Where k is the slope of the line.

Alternative formula:
A 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 No. 1. Write an equation for 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 straight line 2x+3y -7 = 0.
Solution . Let's represent the slope equation 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 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 No. 2. Write the equation of a line parallel to the 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. Let's find the intersection points of the desired line with the coordinate axes:
;
.
So, A(-C/2,0), B(0,-C/5). Let's substitute it into the formula for area: . We get two solutions: 2x + 5y + 10 = 0 and 2x + 5y – 10 = 0.

Example No. 3. Write an equation for a line passing through the point (-2; 5) and parallel to the 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. Having solved example 3 (A=5, B=-7) using formula (2), we find 5(x+2)-7(y-5)=0.

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

Let the line pass through the points M 1 (x 1; y 1) and M 2 (x 2; y 2). The equation of a straight line passing through 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 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 is x = x 1 .

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

Equation of a line in segments

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

Equation of a 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 non-zero vector n = (A; B).

Let's take an arbitrary point M(x; y) on the 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 equal to zero: that is

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

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

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

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

where A and B are the coordinates of the normal vector, C = -Ax o - Vu o is the free term. Equation (10.9) is the general equation of the line(see Fig. 2).

Fig.1 Fig.2

Canonical equations of the line

Where
- coordinates of the point through which the line passes, and
- 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.

Canonical equation of a circle of radius R centered at a point
:

In particular, if the center of the stake coincides with the origin of coordinates, 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, is a constant quantity
, greater than the distance between foci
.

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

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

Note that the equation of a line can be expressed parametrically, 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, the role of the parameter is played by time.

Equation of a straight line on a plane.

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

Ax + Wu + C = 0,

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

A = 0, B  0, C  0 (By + C = 0) - straight line parallel to the Ox axis

B = 0, A  0, C  0 (Ax + C = 0) – straight line 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 from a point and a normal vector.

Definition. In the Cartesian rectangular coordinate system, a vector with components (A, B) is perpendicular to the straight line given by the equation Ax + By + C = 0.

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

With A = 3 and B = -1, let’s compose the equation of the 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 required equation: 3x – y – 1 = 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 the line passing through these points is:

If any of the denominators is zero, the corresponding numerator should be set equal 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 line passing through points A(1, 2) and B(3, 4).

Applying the formula written above, we get:

Equation of a straight line using a point and slope.

If the general equation of the straight line Ax + By + C = 0 is reduced to the form:

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

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

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

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

Ax + Wu + C = 0.

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

We will look for the equation of the desired 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 the straight line has the form: Ax + Ay + C = 0, or x + y + C/A = 0.

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

Equation of a straight line in segments.

If in the general equation of the straight line Ах + Ву + С = 0 С 0, then, dividing by –С, 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 line with the Ox axis, and b– the coordinate of the point of intersection of the straight line with the Oy axis.

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

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

Normal equation of a line.

If both sides of the equation Ax + By + C = 0 are divided by the number
which is called normalizing factor, then we get

xcos + ysin - p = 0 –

normal equation of a 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. The general equation of the line 12x – 5y – 65 = 0 is given. It is required to write various types of equations for this line.

equation of this line in segments:

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

normal equation of a line:

; 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 of coordinates.

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

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

a = -4 is not suitable according to the conditions of the problem.

Total:
or x + y – 4 = 0.

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

The equation of the straight line is:
, where x 1 = y 1 = 0; x 2 = -2; y 2 = -3.

The angle between straight lines on a plane.

Definition. If two lines are given y = k 1 x + b 1, y = k 2 x + b 2, then the acute angle between these lines will be defined as

Two lines are parallel if k 1 = k 2.

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

Theorem. Direct lines Ax + Wu + 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. A straight line passing through the point M 1 (x 1, y 1) and perpendicular to the straight line y = kx + b is represented by the equation:

Distance from a point to a line.

Theorem. If the point M(x) is given 0 , y 0 ), then the distance to the straight line Ах + Ву + С =0 is defined as

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

The coordinates x 1 and y 1 can be found by solving the system of equations:

The second equation of the system is the equation of a line passing through a given point M 0 perpendicular to a 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.

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

k 1 = -3; k 2 = 2 tg =
;  = /4.

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

We find: k 1 = 3/5, k 2 = -5/3, k 1 k 2 = -1, therefore, the lines are perpendicular.

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

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

2x – 3y + 3 = 0;

The required height equation has the form: Ax + By + C = 0 or y = kx + b.

k = . Then y =
. Because 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.

Equation of a line in space.

Equation of a line in space given a point and

direction vector.

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

On the straight line we 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 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 line.

This vector equation can be represented in coordinate form:

By 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 using the formulas:

;

.

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

The numbers m, n, p are called angle coefficients straight. Because is a non-zero vector, then m, n and p cannot be equal to zero at the same time, but one or two of these numbers can be equal to zero. In this case, in the equation of the line, the corresponding numerators should be set equal 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 we can write:

Solving these equations together, we get:

This is the equation of a 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 specified by the equation:

+ D = 0, where

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

This article reveals the derivation of the equation of a straight line passing through two given points in a rectangular coordinate system located on a plane. Let us derive the equation of a straight line passing through two given points in a rectangular coordinate system. We will clearly show and solve several examples related to the material covered.

Yandex.RTB R-A-339285-1

Before obtaining the equation of a line passing through two given points, it is necessary to pay attention to some facts. There is an axiom that says that through two divergent points on a plane it is possible to draw a straight line and only one. In other words, two given points on a plane are defined by a straight line passing through these points.

If the plane is defined by the rectangular coordinate system Oxy, then any straight line depicted in it will correspond to the equation of a straight line on the plane. There is also a connection with the directing vector of the straight line. This data is sufficient to compile the equation of a straight line passing through two given points.

Let's look at an example of solving a similar problem. It is necessary to create an equation for a straight line a passing through two divergent points M 1 (x 1, y 1) and M 2 (x 2, y 2), located in the Cartesian coordinate system.

In the canonical equation of a line on a plane, having the form x - x 1 a x = y - y 1 a y, a rectangular coordinate system O x y is specified with a line that intersects with it at a point with coordinates M 1 (x 1, y 1) with a guide vector a → = (a x , a y) .

It is necessary to create a canonical equation of a straight line a, which will pass through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2).

Straight a has a direction vector M 1 M 2 → with coordinates (x 2 - x 1, y 2 - y 1), since it intersects the points M 1 and M 2. We have obtained the necessary data in order to transform the canonical equation with the coordinates of the direction vector M 1 M 2 → = (x 2 - x 1, y 2 - y 1) and the coordinates of the points M 1 lying on them (x 1, y 1) and M 2 (x 2 , y 2) . We obtain an equation of the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 or x - x 2 x 2 - x 1 = y - y 2 y 2 - y 1.

Consider the figure below.

Following the calculations, we write down the parametric equations of a line on a plane that passes through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2). We obtain an equation of the form x = x 1 + (x 2 - x 1) · λ y = y 1 + (y 2 - y 1) · λ or x = x 2 + (x 2 - x 1) · λ y = y 2 + (y 2 - y 1) · λ .

Let's take a closer look at solving several examples.

Example 1

Write down the equation of a straight line passing through 2 given points with coordinates M 1 - 5, 2 3, M 2 1, - 1 6.

Solution

The canonical equation for a line intersecting at two points with coordinates x 1, y 1 and x 2, y 2 takes the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1. According to the conditions of the problem, we have that x 1 = - 5, y 1 = 2 3, x 2 = 1, y 2 = - 1 6. It is necessary to substitute the numerical values into the equation x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1. From here we get that the canonical equation takes the form x - (- 5) 1 - (- 5) = y - 2 3 - 1 6 - 2 3 ⇔ x + 5 6 = y - 2 3 - 5 6.

Answer: x + 5 6 = y - 2 3 - 5 6.

If you need to solve a problem with a different type of equation, then first you can go to the canonical one, since it is easier to come from it to any other one.

Example 2

Compose the general equation of a straight line passing through points with coordinates M 1 (1, 1) and M 2 (4, 2) in the O x y coordinate system.

Solution

First, you need to write down the canonical equation of a given line that passes through given two points. We get an equation of the form x - 1 4 - 1 = y - 1 2 - 1 ⇔ x - 1 3 = y - 1 1 .

Let's bring the canonical equation to the desired form, then we get:

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

Answer: x - 3 y + 2 = 0 .

Examples of such tasks were discussed in school textbooks during algebra lessons. School problems differed in that the equation of a straight line with an angle coefficient was known, having the form y = k x + b. If you need to find the value of the slope k and the number b for which the equation y = k x + b defines a line in the O x y system that passes through the points M 1 (x 1, y 1) and M 2 (x 2, y 2) , where x 1 ≠ x 2. When x 1 = x 2 , then the angular coefficient takes on the value of infinity, and the straight line M 1 M 2 is defined by a general incomplete equation of the form x - x 1 = 0 .

Because the points M 1 And M 2 are on a straight line, then their coordinates satisfy the equation y 1 = k x 1 + b and y 2 = k x 2 + b. It is necessary to solve the system of equations y 1 = k x 1 + b y 2 = k x 2 + b for k and b.

To do this, we find k = y 2 - y 1 x 2 - x 1 b = y 1 - y 2 - y 1 x 2 - x 1 x 1 or k = y 2 - y 1 x 2 - x 1 b = y 2 - y 2 - y 1 x 2 - x 1 x 2 .

With these values of k and b, the equation of a line passing through the given two points becomes y = y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 1 or y = y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 2.

It is impossible to remember such a huge number of formulas at once. To do this, it is necessary to increase the number of repetitions in solving problems.

Example 3

Write down the equation of a straight line with an angular coefficient passing through points with coordinates M 2 (2, 1) and y = k x + b.

Solution

To solve the problem, we use a formula with an angular coefficient of the form y = k x + b. The coefficients k and b must take such a value that this equation corresponds to a straight line passing through two points with coordinates M 1 (- 7, - 5) and M 2 (2, 1).

Points M 1 And M 2 are located on a straight line, then their coordinates must make the equation y = k x + b a true equality. From this we get that - 5 = k · (- 7) + b and 1 = k · 2 + b. Let's combine the equation into the system - 5 = k · - 7 + b 1 = k · 2 + b and solve.

Upon substitution we get that

5 = k · - 7 + b 1 = k · 2 + b ⇔ b = - 5 + 7 k 2 k + b = 1 ⇔ b = - 5 + 7 k 2 k - 5 + 7 k = 1 ⇔ ⇔ b = - 5 + 7 k k = 2 3 ⇔ b = - 5 + 7 2 3 k = 2 3 ⇔ b = - 1 3 k = 2 3

Now the values k = 2 3 and b = - 1 3 are substituted into the equation y = k x + b. We find that the required equation passing through the given points will be an equation of the form y = 2 3 x - 1 3 .

This method of solution predetermines the waste of a lot of time. There is a way in which the task is solved in literally two steps.

Let us write the canonical equation of the line passing through M 2 (2, 1) and M 1 (- 7, - 5), having the form x - (- 7) 2 - (- 7) = y - (- 5) 1 - (- 5) ⇔ x + 7 9 = y + 5 6 .

Now let's move on to the slope equation. We get that: x + 7 9 = y + 5 6 ⇔ 6 · (x + 7) = 9 · (y + 5) ⇔ y = 2 3 x - 1 3.

Answer: y = 2 3 x - 1 3 .

If in three-dimensional space there is a rectangular coordinate system O x y z with two given non-coinciding points with coordinates M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), the straight line M passing through them 1 M 2 , it is necessary to obtain the equation of this line.

We have that canonical equations of the form x - x 1 a x = y - y 1 a y = z - z 1 a z and parametric equations of the form x = x 1 + a x · λ y = y 1 + a y · λ z = z 1 + a z · λ are able to define a line in the coordinate system O x y z, passing through points having coordinates (x 1, y 1, z 1) with a direction vector a → = (a x, a y, a z).

Straight M 1 M 2 has a direction vector of the form M 1 M 2 → = (x 2 - x 1, y 2 - y 1, z 2 - z 1), where the straight line passes through the point M 1 (x 1, y 1, z 1) and M 2 (x 2 , y 2 , z 2), hence the canonical equation can be of the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 or x - x 2 x 2 - x 1 = y - y 2 y 2 - y 1 = z - z 2 z 2 - z 1, in turn parametric x = x 1 + (x 2 - x 1) λ y = y 1 + (y 2 - y 1) λ z = z 1 + (z 2 - z 1) λ or x = x 2 + (x 2 - x 1) λ y = y 2 + (y 2 - y 1) · λ z = z 2 + (z 2 - z 1) · λ .

Consider a drawing that shows 2 given points in space and the equation of a straight line.

Example 4

Write the equation of a line defined in a rectangular coordinate system O x y z of three-dimensional space, passing through given two points with coordinates M 1 (2, - 3, 0) and M 2 (1, - 3, - 5).

Solution

It is necessary to find the canonical equation. Since we are talking about three-dimensional space, it means that when a line passes through given points, the desired canonical equation will take the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 .

By condition we have that x 1 = 2, y 1 = - 3, z 1 = 0, x 2 = 1, y 2 = - 3, z 2 = - 5. It follows that the necessary equations will be written as follows:

x - 2 1 - 2 = y - (- 3) - 3 - (- 3) = z - 0 - 5 - 0 ⇔ x - 2 - 1 = y + 3 0 = z - 5

Answer: x - 2 - 1 = y + 3 0 = z - 5.

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