How to find the point of intersection of two lines. The relative position of lines in space

IN two-dimensional space two lines intersect only at one point, defined by coordinates (x,y). Since both lines pass through their point of intersection, the coordinates (x,y) must satisfy both equations that describe these lines. With some additional skills, you can find the intersection points of parabolas and other quadratic curves.

Steps

The point of intersection of two lines

Write the equation of each line, isolating the variable “y” on the left side of the equation. Other terms of the equation should be placed on right side equations Perhaps the equation given to you will contain the variable f(x) or g(x) instead of “y”; in this case, isolate such a variable. To isolate a variable, perform the appropriate math on both sides of the equation.

• If the equations of the lines are not given to you, based on the information you know.
• Example. Given straight lines described by equations and y − 12 = − 2 x (\displaystyle y-12=-2x). To isolate the “y” in the second equation, add the number 12 to both sides of the equation:

1. You are looking for the point of intersection of both lines, that is, a point whose coordinates (x, y) satisfy both equations. Since the variable “y” is on the left side of each equation, the expressions located on the right side of each equation can be equated. Write down a new equation.

   • Example. Because y = x + 3 (\displaystyle y=x+3) And y = 12 − 2 x (\displaystyle y=12-2x), then we can write the following equality: .

2. Find the value of the variable "x". The new equation contains only one variable, "x". To find "x", isolate that variable on the left side of the equation by performing the appropriate math on both sides of the equation. You should get an equation of the form x = __ (if you can't do this, see this section).

   • Example. x + 3 = 12 − 2 x (\displaystyle x+3=12-2x)
   • Add 2 x (\displaystyle 2x) to each side of the equation:
   • 3 x + 3 = 12 (\displaystyle 3x+3=12)
   • Subtract 3 from each side of the equation:
   • 3 x = 9 (\displaystyle 3x=9)
   • Divide each side of the equation by 3:
   • x = 3 (\displaystyle x=3).

3. Use the found value of the variable "x" to calculate the value of the variable "y". To do this, substitute the found value of “x” into the equation (any) of the straight line.

   • Example. x = 3 (\displaystyle x=3) And y = x + 3 (\displaystyle y=x+3)
   • y = 3 + 3 (\displaystyle y=3+3)
   • y = 6 (\displaystyle y=6)

4. Check the answer. To do this, substitute the value of “x” into the other equation of the line and find the value of “y”. If you receive different meaning"y", check the correctness of your calculations.

   • Example: x = 3 (\displaystyle x=3) And y = 12 − 2 x (\displaystyle y=12-2x)
   • y = 12 − 2 (3) (\displaystyle y=12-2(3))
   • y = 12 − 6 (\displaystyle y=12-6)
   • y = 6 (\displaystyle y=6)
   • You got the same value for y, so there are no errors in your calculations.

5. Write down the coordinates (x,y). Having calculated the values ​​of “x” and “y”, you have found the coordinates of the point of intersection of two lines. Write down the coordinates of the intersection point in (x,y) form.

   • Example. x = 3 (\displaystyle x=3) And y = 6 (\displaystyle y=6)
   • Thus, two straight lines intersect at a point with coordinates (3,6).

6. Calculations in special cases. In some cases, the value of the variable "x" cannot be found. But that doesn't mean you made a mistake. Special case occurs when one of the following conditions is met:

   • If two lines are parallel, they do not intersect. In this case, the variable “x” will simply be reduced, and your equation will turn into a meaningless equality (for example, 0 = 1 (\displaystyle 0=1)). In this case, write down in your answer that the lines do not intersect or there is no solution.
   • If both equations describe one straight line, then there will be an infinite number of intersection points. In this case, the variable “x” will simply be reduced, and your equation will turn into a strict equality (for example, 3 = 3 (\displaystyle 3=3)). In this case, write down in your answer that the two lines coincide.

   Problems with quadratic functions

   1. Definition of a quadratic function. In a quadratic function, one or more variables have a second degree (but not higher), for example, x 2 (\displaystyle x^(2)) or y 2 (\displaystyle y^(2)). The graphs of quadratic functions are curves that may not intersect or may intersect at one or two points. In this section, we will tell you how to find the intersection point or points of quadratic curves.

   2. Rewrite each equation by isolating the variable “y” on the left side of the equation. The other terms of the equation should be placed on the right side of the equation.

      • Example. Find the point(s) of intersection of the graphs x 2 + 2 x − y = − 1 (\displaystyle x^(2)+2x-y=-1) And
      • Isolate the variable "y" on the left side of the equation:
      • And y = x + 7 (\displaystyle y=x+7) .
      • In this example, you are given one quadratic function and one linear function. Remember that if you are given two quadratic functions, the calculations are similar to the steps outlined below.

   3. Equate the expressions on the right side of each equation. Since the variable “y” is on the left side of each equation, the expressions located on the right side of each equation can be equated.

      • Example. y = x 2 + 2 x + 1 (\displaystyle y=x^(2)+2x+1) And y = x + 7 (\displaystyle y=x+7)

   4. Transfer all terms of the resulting equation to its left side, and write 0 on the right side. To do this, do some basic math. This will allow you to solve the resulting equation.

      • Example. x 2 + 2 x + 1 = x + 7 (\displaystyle x^(2)+2x+1=x+7)
      • Subtract "x" from both sides of the equation:
      • x 2 + x + 1 = 7 (\displaystyle x^(2)+x+1=7)
      • Subtract 7 from both sides of the equation:

   5. Decide quadratic equation. By moving all the terms of the equation to its left side, you get a quadratic equation. It can be solved in three ways: using a special formula, and.

      • Example. x 2 + x − 6 = 0 (\displaystyle x^(2)+x-6=0)
      • When you factor an equation, you get two binomials, which, when multiplied, give you the original equation. In our example, the first term x 2 (\displaystyle x^(2)) can be decomposed to x * x. Write this down: (x)(x) = 0
      • In our example, the free term -6 can be factorized into the following factors: − 6 ∗ 1 (\displaystyle -6*1), − 3 ∗ 2 (\displaystyle -3*2), − 2 ∗ 3 (\displaystyle -2*3), − 1 ∗ 6 (\displaystyle -1*6).
      • In our example, the second term is x (or 1x). Add each pair of factors of the dummy term (in our example -6) until you get 1. In our example, the appropriate pair of factors of the dummy term are the numbers -2 and 3 ( − 2 ∗ 3 = − 6 (\displaystyle -2*3=-6)), because − 2 + 3 = 1 (\displaystyle -2+3=1).
      • Fill in the blanks with the found pair of numbers: .

   6. Don't forget about the second point of intersection of the two graphs. If you solve the problem quickly and not very carefully, you may forget about the second intersection point. Here's how to find the x coordinates of two intersection points:

      • Example (factorization). If in Eq. (x − 2) (x + 3) = 0 (\displaystyle (x-2)(x+3)=0) one of the expressions in brackets will be equal to 0, then the entire equation will be equal to 0. Therefore, we can write it like this: x − 2 = 0 (\displaystyle x-2=0) → x = 2 (\displaystyle x=2) And x + 3 = 0 (\displaystyle x+3=0) → x = − 3 (\displaystyle x=-3) (that is, you found two roots of the equation).
      • Example (using a formula or completing a perfect square). When using one of these methods, the solution will appear square root. For example, the equation from our example will take the form x = (− 1 + 25) / 2 (\displaystyle x=(-1+(\sqrt (25)))/2). Remember that when taking the square root you will get two solutions. In our case: 25 = 5 ∗ 5 (\displaystyle (\sqrt (25))=5*5), And 25 = (− 5) ∗ (− 5) (\displaystyle (\sqrt (25))=(-5)*(-5)). So write two equations and find two values ​​of x.

   7. The graphs intersect at one point or do not intersect at all. Such situations occur if the following conditions are met:

      • If the graphs intersect at one point, then the quadratic equation is decomposed into identical factors, for example, (x-1) (x-1) = 0, and the square root of 0 appears in the formula ( 0 (\displaystyle (\sqrt (0)))). In this case, the equation has only one solution.
      • If the graphs do not intersect at all, then the equation cannot be factorized, and the square root of negative number(For example, − 2 (\displaystyle (\sqrt (-2)))). In this case, write in your answer that there is no solution.

Perpendicular line

This task is probably one of the most popular and in demand in school textbooks. The tasks based on this topic are varied. This is the definition of the point of intersection of two lines, this is also the definition of the equation of a line passing through a point on the original line at any angle.

We will cover this topic using in our calculations the data obtained using

It was there that the transformation of the general equation of a straight line into an equation with an angular coefficient and vice versa was considered, and the determination of the remaining parameters of the straight line according to given conditions.

What do we lack in order to solve the problems that this page is dedicated to?

1. Formulas for calculating one of the angles between two intersecting lines.

If we have two lines that are given by the equations:

then one of the angles is calculated like this:

2. Equation of a straight line with a slope passing through a given point

From formula 1, we can see two borderline states

a) when then and therefore these two given lines are parallel (or coincide)

b) when , then , and therefore these lines are perpendicular, that is, intersect at right angles.

What could be the initial data for solving such problems, other than the given straight line?

A point on a straight line and the angle at which the second straight line intersects it

Second equation of the line

What problems can a bot solve?

1. Two lines are given (explicitly or indirectly, for example, by two points). Calculate the point of intersection and the angles at which they intersect.

2. Given one straight line, a point on a straight line and one angle. Determine the equation of a straight line that intersects a given line at a specified angle

Examples

Two lines are given by equations. Find the point of intersection of these lines and the angles at which they intersect

line_p A=11;B=-5;C=6,k=3/7;b=-5

We get the following result

Equation of the first line

y = 2.2 x + (1.2)

Equation of the second line

y = 0.4285714285714 x + (-5)

Angle of intersection of two straight lines (in degrees)

-42.357454705937

The point of intersection of two lines

x = -3.5

y = -6.5

Don't forget that the parameters of two lines are separated by a comma, and the parameters of each line are separated by a semicolon.

A straight line passes through two points (1:-4) and (5:2). Find the equation of the line that passes through the point (-2:-8) and intersects the original line at an angle of 30 degrees.

We know one straight line because we know the two points through which it passes.

It remains to determine the equation of the second line. We know one point, but instead of the second, the angle at which the first line intersects the second is indicated.

It seems that everything is known, but the main thing here is not to make mistakes. It's about about the angle (30 degrees) not between the x-axis and the line, but between the first and second line.

This is why we post like this. Let's determine the parameters of the first line and find out at what angle it intersects the x-axis.

line xa=1;xb=5;ya=-4;yb=2

General equation Ax+By+C = 0

Coefficient A = -6

Factor B = 4

Factor C = 22

Coefficient a= 3.6666666666667

Coefficient b = -5.5

Coefficient k = 1.5

Angle of inclination to the axis (in degrees) f = 56.309932474019

Coefficient p = 3.0508510792386

Coefficient q = 2.5535900500422

Distance between points=7.211102550928

We see that the first line intersects the axis at an angle 56.309932474019 degrees.

The source data does not say exactly how the second line intersects the first. You can, after all, construct two lines that satisfy the conditions, the first one rotated 30 degrees CLOCKWISE, and the second 30 degrees COUNTERclockwise.

Let's count them

If the second line is rotated 30 degrees COUNTERclockwise, then the second line will have the degree of intersection with the x-axis 30+56.309932474019 = 86 .309932474019 degrees

line_p xa=-2;ya=-8;f=86.309932474019

Parameters of a straight line according to specified parameters

General equation Ax+By+C = 0

Coefficient A = 23.011106998916

Coefficient B = -1.4840558255286

Coefficient C = 34.149767393603

Equation of a straight line in segments x/a+y/b = 1

Coefficient a= -1.4840558255286

Coefficient b = 23.011106998916

Equation of a straight line with angular coefficient y = kx + b

Coefficient k = 15.505553499458

Angle of inclination to the axis (in degrees) f = 86.309932474019

Normal equation of the line x*cos(q)+y*sin(q)-p = 0

Coefficient p = -1.4809790664999

Coefficient q = 3.0771888256405

Distance between points=23.058912962428

Distance from a point to a straight line li =

that is, our second line equation is y= 15.505553499458x+ 23.011106998916

In order to solve a geometric problem using the coordinate method, an intersection point is needed, the coordinates of which are used in the solution. A situation arises when you need to look for the coordinates of the intersection of two lines on a plane or determine the coordinates of the same lines in space. This article considers cases of finding the coordinates of points where given lines intersect.

Yandex.RTB R-A-339285-1

It is necessary to define the points of intersection of two lines.

The section on the relative position of lines on a plane shows that they can coincide, be parallel, intersect at one common point, or intersect. Two lines in space are called intersecting if they have one common point.

The definition of the point of intersection of lines sounds like this:

Definition 1

The point at which two lines intersect is called their point of intersection. In other words, the point of intersecting lines is the point of intersection.

Let's look at the figure below.

Before finding the coordinates of the point of intersection of two lines, it is necessary to consider the example below.

If the plane has a coordinate system O x y, then two straight lines a and b are specified. Line a corresponds to a general equation of the form A 1 x + B 1 y + C 1 = 0, for line b - A 2 x + B 2 y + C 2 = 0. Then M 0 (x 0 , y 0) is a certain point of the plane; it is necessary to determine whether the point M 0 will be the point of intersection of these lines.

To solve the problem, it is necessary to adhere to the definition. Then the lines must intersect at a point whose coordinates are the solution to the given equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0. This means that the coordinates of the intersection point are substituted into all given equations. If, upon substitution, they give the correct identity, then M 0 (x 0 , y 0) is considered their point of intersection.

Example 1

Given two intersecting lines 5 x - 2 y - 16 = 0 and 2 x - 5 y - 19 = 0. Will the point M 0 with coordinates (2, - 3) be an intersection point.

Solution

For the intersection of lines to be valid, it is necessary that the coordinates of the point M 0 satisfy the equations of the lines. This can be checked by substituting them. We get that

5 2 - 2 (- 3) - 16 = 0 ⇔ 0 = 0 2 2 - 5 (- 3) - 19 = 0 ⇔ 0 = 0

Both equalities are true, which means M 0 (2, - 3) is the intersection point of the given lines.

Let's depict this decision on the coordinate line of the figure below.

Answer: the given point with coordinates (2, - 3) will be the intersection point of the given lines.

Example 2

Will the lines 5 x + 3 y - 1 = 0 and 7 x - 2 y + 11 = 0 intersect at point M 0 (2, - 3)?

Solution

To solve the problem, you need to substitute the coordinates of the point into all equations. We get that

5 2 + 3 (- 3) - 1 = 0 ⇔ 0 = 0 7 2 - 2 (- 3) + 11 = 0 ⇔ 31 = 0

The second equality is not true, it means that the given point does not belong to the line 7 x - 2 y + 11 = 0. From this we have that point M 0 is not the point of intersection of lines.

The drawing clearly shows that M 0 is not the point of intersection of lines. They have a common point with coordinates (- 1, 2).

Answer: the point with coordinates (2, - 3) is not the intersection point of the given lines.

We proceed to finding the coordinates of the points of intersection of two lines using the given equations on the plane.

Two intersecting lines a and b are specified by equations of the form A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0, located at O ​​x y. When designating the intersection point M 0, we find that we should continue searching for coordinates using the equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0.

From the definition it is obvious that M 0 is the common point of intersection of lines. In this case, its coordinates must satisfy the equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0. In other words, this is the solution to the resulting system A 1 x + B 1 y + C 1 = 0 A 2 x + B 2 y + C 2 = 0.

This means that to find the coordinates of the intersection point, it is necessary to add all the equations to the system and solve it.

Example 3

Given two straight lines x - 9 y + 14 = 0 and 5 x - 2 y - 16 = 0 on the plane. it is necessary to find their intersection.

Solution

Data on the conditions of the equation must be collected into the system, after which we obtain x - 9 y + 14 = 0 5 x - 2 y - 16 = 0. To solve it, solve the first equation for x and substitute the expression into the second:

x - 9 y + 14 = 0 5 x - 2 y - 16 = 0 ⇔ x = 9 y - 14 5 x - 2 y - 16 = 0 ⇔ ⇔ x = 9 y - 14 5 9 y - 14 - 2 y - 16 = 0 ⇔ x = 9 y - 14 43 y - 86 = 0 ⇔ ⇔ x = 9 y - 14 y = 2 ⇔ x = 9 2 - 14 y = 2 ⇔ x = 4 y = 2

The resulting numbers are the coordinates that needed to be found.

Answer: M 0 (4, 2) is the intersection point of the lines x - 9 y + 14 = 0 and 5 x - 2 y - 16 = 0.

Finding coordinates comes down to solving a system of linear equations. If by condition a different type of equation is given, then it should be reduced to normal form.

Example 4

Determine the coordinates of the points of intersection of the lines x - 5 = y - 4 - 3 and x = 4 + 9 · λ y = 2 + λ, λ ∈ R.

Solution

First you need to bring the equations to general appearance. Then we get that x = 4 + 9 · λ y = 2 + λ , λ ∈ R is transformed as follows:

x = 4 + 9 · λ y = 2 + λ ⇔ λ = x - 4 9 λ = y - 2 1 ⇔ x - 4 9 = y - 2 1 ⇔ ⇔ 1 · (x - 4) = 9 · (y - 2) ⇔ x - 9 y + 14 = 0

Then we take the equation of the canonical form x - 5 = y - 4 - 3 and transform it. We get that

x - 5 = y - 4 - 3 ⇔ - 3 x = - 5 y - 4 ⇔ 3 x - 5 y + 20 = 0

From here we have that the coordinates are the point of intersection

x - 9 y + 14 = 0 3 x - 5 y + 20 = 0 ⇔ x - 9 y = - 14 3 x - 5 y = - 20

Let's use Cramer's method to find coordinates:

∆ = 1 - 9 3 - 5 = 1 · (- 5) - (- 9) · 3 = 22 ∆ x = - 14 - 9 - 20 - 5 = - 14 · (- 5) - (- 9) · ( - 20) = - 110 ⇒ x = ∆ x ∆ = - 110 22 = - 5 ∆ y = 1 - 14 3 - 20 = 1 · (- 20) - (- 14) · 3 = 22 ⇒ y = ∆ y ∆ = 22 22 = 1

Answer: M 0 (- 5 , 1) .

There is also a way to find the coordinates of the intersection point of lines located on a plane. It is applicable when one of the lines is given by parametric equations of the form x = x 1 + a x · λ y = y 1 + a y · λ , λ ∈ R . Then instead of the value x we ​​substitute x = x 1 + a x λ and y = y 1 + a y λ, where we get λ = λ 0, corresponding to the intersection point having coordinates x 1 + a x λ 0, y 1 + a y λ 0 .

Example 5

Determine the coordinates of the point of intersection of the line x = 4 + 9 · λ y = 2 + λ, λ ∈ R and x - 5 = y - 4 - 3.

Solution

It is necessary to perform a substitution in x - 5 = y - 4 - 3 with the expression x = 4 + 9 · λ, y = 2 + λ, then we get:

4 + 9 λ - 5 = 2 + λ - 4 - 3

When solving, we find that λ = - 1. It follows that there is an intersection point between the lines x = 4 + 9 · λ y = 2 + λ, λ ∈ R and x - 5 = y - 4 - 3. To calculate the coordinates, you need to substitute the expression λ = - 1 into the parametric equation. Then we get that x = 4 + 9 · (- 1) y = 2 + (- 1) ⇔ x = - 5 y = 1.

Answer: M 0 (- 5 , 1) .

To fully understand the topic, you need to know some nuances.

First you need to understand the location of the lines. When they intersect, we will find the coordinates; in other cases, there will be no solution. To avoid this check, you can create a system of the form A 1 x + B 1 y + C 1 = 0 A 2 x + B 2 + C 2 = 0 If there is a solution, we conclude that the lines intersect. If there is no solution, then they are parallel. When a system has an infinite number of solutions, then they are said to coincide.

Example 6

Given lines x 3 + y - 4 = 1 and y = 4 3 x - 4. Determine whether they have a common point.

Solution

Simplifying the given equations, we obtain 1 3 x - 1 4 y - 1 = 0 and 4 3 x - y - 4 = 0.

The equations should be collected into a system for subsequent solution:

1 3 x - 1 4 y - 1 = 0 1 3 x - y - 4 = 0 ⇔ 1 3 x - 1 4 y = 1 4 3 x - y = 4

From this we can see that the equations are expressed through each other, then we get an infinite number of solutions. Then the equations x 3 + y - 4 = 1 and y = 4 3 x - 4 define the same line. Therefore there are no points of intersection.

Answer: the given equations define the same straight line.

Example 7

Find the coordinates of the point of intersecting lines 2 x + (2 - 3) y + 7 = 0 and 2 3 + 2 x - 7 y - 1 = 0.

Solution

According to the condition, this is possible, the lines will not intersect. It is necessary to create a system of equations and solve. To solve, it is necessary to use the Gaussian method, since with its help it is possible to check the equation for compatibility. We get a system of the form:

2 x + (2 - 3) y + 7 = 0 2 (3 + 2) x - 7 y - 1 = 0 ⇔ 2 x + (2 - 3) y = - 7 2 (3 + 2) x - 7 y = 1 ⇔ ⇔ 2 x + 2 - 3 y = - 7 2 (3 + 2) x - 7 y + (2 x + (2 - 3) y) · (- (3 + 2)) = 1 + - 7 · (- (3 + 2)) ⇔ ⇔ 2 x + (2 - 3) y = - 7 0 = 22 - 7 2

We received an incorrect equality, which means the system has no solutions. We conclude that the lines are parallel. There are no intersection points.

Second solution.

First you need to determine the presence of intersection of lines.

n 1 → = (2, 2 - 3) is the normal vector of the line 2 x + (2 - 3) y + 7 = 0, then the vector n 2 → = (2 (3 + 2) , - 7 is the normal vector for the line 2 3 + 2 x - 7 y - 1 = 0 .

It is necessary to check the collinearity of the vectors n 1 → = (2, 2 - 3) and n 2 → = (2 (3 + 2) , - 7). We obtain an equality of the form 2 2 (3 + 2) = 2 - 3 - 7. It is correct because 2 2 3 + 2 - 2 - 3 - 7 = 7 + 2 - 3 (3 + 2) 7 (3 + 2) = 7 - 7 7 (3 + 2) = 0. It follows that the vectors are collinear. This means that the lines are parallel and have no points of intersection.

Answer: there are no points of intersection, the lines are parallel.

Example 8

Find the coordinates of the intersection of the given lines 2 x - 1 = 0 and y = 5 4 x - 2.

Solution

To solve, we compose a system of equations. We get

2 x - 1 = 0 5 4 x - y - 2 = 0 ⇔ 2 x = 1 5 4 x - y = 2

Let's find the determinant of the main matrix. For this, 2 0 5 4 - 1 = 2 · (- 1) - 0 · 5 4 = - 2. Since it is not equal to zero, the system has 1 solution. It follows that the lines intersect. Let's solve a system for finding the coordinates of intersection points:

2 x = 1 5 4 x - y = 2 ⇔ x = 1 2 4 5 x - y = 2 ⇔ x = 1 2 5 4 1 2 - y = 2 ⇔ x = 1 2 y = - 11 8

We found that the intersection point of the given lines has coordinates M 0 (1 2, - 11 8).

Answer: M 0 (1 2 , - 11 8) .

Finding the coordinates of the point of intersection of two lines in space

In the same way, the points of intersection of straight lines in space are found.

When straight lines a and b are given in the coordinate plane O x y z by equations of intersecting planes, then there is a straight line a, which can be determined using the given system A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 1 = 0 and straight line b - A 3 x + B 3 y + C 3 z + D 3 = 0 A 4 x + B 4 y + C 4 z + D 4 = 0.

When point M 0 is the point of intersection of lines, then its coordinates must be solutions of both equations. We get linear equations in the system:

A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0 A 3 x + B 3 y + C 3 z + D 3 = 0 A 4 x + B 4 y + C 4 z + D 4 = 0

Let's look at similar tasks using examples.

Example 9

Find the coordinates of the intersection point of the given lines x - 1 = 0 y + 2 z + 3 = 0 and 3 x + 2 y + 3 = 0 4 x - 2 z - 4 = 0

Solution

We compose the system x - 1 = 0 y + 2 z + 3 = 0 3 x + 2 y + 3 = 0 4 x - 2 z - 4 = 0 and solve it. To find the coordinates, you need to solve through the matrix. Then we obtain the main matrix of the form A = 1 0 0 0 1 2 3 2 0 4 0 - 2 and the extended matrix T = 1 0 0 1 0 1 2 - 3 4 0 - 2 4 . We determine the Gaussian rank of the matrix.

We get that

1 = 1 ≠ 0 , 1 0 0 1 = 1 ≠ 0 , 1 0 0 0 1 2 3 2 0 = - 4 ≠ 0 , 1 0 0 1 0 1 2 - 3 3 2 0 - 3 4 0 - 2 4 = 0

It follows that the rank of the extended matrix has the value 3. Then the system of equations x - 1 = 0 y + 2 z + 3 = 0 3 x + 2 y + 3 = 0 4 x - 27 - 4 = 0 results in only one solution.

The basis minor has the determinant 1 0 0 0 1 2 3 2 0 = - 4 ≠ 0 , then the last equation does not apply. We obtain that x - 1 = 0 y + 2 z + 3 = 0 3 x + 2 y + 3 = 0 4 x - 2 z - 4 = 0 ⇔ x = 1 y + 2 z = - 3 3 x + 2 y - 3. Solution of the system x = 1 y + 2 z = - 3 3 x + 2 y = - 3 ⇔ x = 1 y + 2 z = - 3 3 1 + 2 y = - 3 ⇔ x = 1 y + 2 z = - 3 y = - 3 ⇔ ⇔ x = 1 - 3 + 2 z = - 3 y = - 3 ⇔ x = 1 z = 0 y = - 3 .

This means that the intersection point x - 1 = 0 y + 2 z + 3 = 0 and 3 x + 2 y + 3 = 0 4 x - 2 z - 4 = 0 has coordinates (1, - 3, 0).

Answer: (1 , - 3 , 0) .

System of the form A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0 A 3 x + B 3 y + C 3 z + D 3 = 0 A 4 x + B 4 y + C 4 z + D 4 = 0 has only one solution. This means that lines a and b intersect.

In other cases, the equation has no solution, that is, no common points either. That is, it is impossible to find a point with coordinates, since it does not exist.

Therefore, a system of the form A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0 A 3 x + B 3 y + C 3 z + D 3 = 0 A 4 x + B 4 y + C 4 z + D 4 = 0 is solved by the Gaussian method. If it is incompatible, the lines are not intersecting. If there is an infinite number of solutions, then they coincide.

You can solve by calculating the basic and extended rank of the matrix, and then apply the Kronecker-Capelli theorem. We get one, many or complete absence decisions.

Example 10

The equations of the lines x + 2 y - 3 z - 4 = 0 2 x - y + 5 = 0 and x - 3 z = 0 3 x - 2 y + 2 z - 1 = 0 are given. Find the intersection point.

Solution

First, let's create a system of equations. We get that x + 2 y - 3 z - 4 = 0 2 x - y + 5 = 0 x - 3 z = 0 3 x - 2 y + 2 z - 1 = 0. We solve it using the Gaussian method:

1 2 - 3 4 2 - 1 0 - 5 1 0 - 3 0 3 - 2 2 1 ~ 1 2 - 3 4 0 - 5 6 - 13 0 - 2 0 - 4 0 - 8 11 - 11 ~ ~ 1 2 - 3 4 0 - 5 6 - 13 0 0 - 12 5 6 5 0 0 7 5 - 159 5 ~ 1 2 - 3 4 0 - 5 6 - 13 0 0 - 12 5 6 5 0 0 0 311 10

Obviously, the system has no solutions, which means the lines do not intersect. There is no intersection point.

Answer: there is no intersection point.

If the lines are given using cononical or parametric equations, you need to reduce them to the form of equations of intersecting planes, and then find the coordinates.

Example 11

Given two lines x = - 3 - λ y = - 3 λ z = - 2 + 3 λ, λ ∈ R and x 2 = y - 3 0 = z 5 in O x y z. Find the intersection point.

Solution

We define straight lines by equations of two intersecting planes. We get that

x = - 3 - λ y = - 3 λ z = - 2 + 3 λ ⇔ λ = x + 3 - 1 λ = y - 3 λ = z + 2 3 ⇔ x + 3 - 1 = y - 3 = z + 2 3 ⇔ ⇔ x + 3 - 1 = y - 3 x + 3 - 1 = z + 2 3 ⇔ 3 x - y + 9 = 0 3 x + z + 11 = 0 x 2 = y - 3 0 = z 5 ⇔ y - 3 = 0 x 2 = z 5 ⇔ y - 3 = 0 5 x - 2 z = 0

We find the coordinates 3 x - y + 9 = 0 3 x + z + 11 = 0 y - 3 = 0 5 x - 2 z = 0, for this we calculate the ranks of the matrix. The rank of the matrix is ​​3, and the basis minor is 3 - 1 0 3 0 1 0 1 0 = - 3 ≠ 0, which means that the last equation must be excluded from the system. We get that

3 x - y + 9 = 0 3 x + z + 11 = 0 y - 3 = 0 5 x - 2 z = 0 ⇔ 3 x - y + 9 = 0 3 x + z + 11 = 0 y - 3 = 0

Let's solve the system using Cramer's method. We get that x = - 2 y = 3 z = - 5. From here we get that the intersection of the given lines gives a point with coordinates (- 2, 3, - 5).

Answer: (- 2 , 3 , - 5) .

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