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.

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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 us depict this solution 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, the first equation is resolved for x, and the expression is substituted 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 a general form. 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. 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 obtain 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 main and extended ranks of the matrix, and then apply the Kronecker-Capelli theorem. We get one, many or no solutions at all.

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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Oh-oh-oh-oh-oh... well, it’s tough, as if he was reading out a sentence to himself =) However, relaxation will help later, especially since today I bought the appropriate accessories. Therefore, let's proceed to the first section, I hope that by the end of the article I will maintain a cheerful mood.

The relative position of two straight lines

This is the case when the audience sings along in chorus. Two straight lines can:

1) match;

2) be parallel: ;

3) or intersect at a single point: .

Help for dummies : Please remember the mathematical intersection sign, it will appear very often. The notation means that the line intersects with the line at point .

How to determine the relative position of two lines?

Let's start with the first case:

Two lines coincide if and only if their corresponding coefficients are proportional, that is, there is a number “lambda” such that the equalities are satisfied

Let's consider the straight lines and create three equations from the corresponding coefficients: . From each equation it follows that, therefore, these lines coincide.

Indeed, if all the coefficients of the equation multiply by –1 (change signs), and all coefficients of the equation cut by 2, you get the same equation: .

The second case, when the lines are parallel:

Two lines are parallel if and only if their coefficients of the variables are proportional: , But.

As an example, consider two straight lines. We check the proportionality of the corresponding coefficients for the variables:

However, it is quite obvious that.

And the third case, when the lines intersect:

Two lines intersect if and only if their coefficients of the variables are NOT proportional, that is, there is NO such value of “lambda” that the equalities are satisfied

So, for straight lines we will create a system:

From the first equation it follows that , and from the second equation: , which means the system is inconsistent(no solutions). Thus, the coefficients of the variables are not proportional.

Conclusion: lines intersect

In practical problems, you can use the solution scheme just discussed. By the way, it is very reminiscent of the algorithm for checking vectors for collinearity, which we looked at in class The concept of linear (in)dependence of vectors. Basis of vectors. But there is a more civilized packaging:

Example 1

Find out relative position direct:

Solution based on the study of directing vectors of straight lines:

a) From the equations we find the direction vectors of the lines: .

, which means that the vectors are not collinear and the lines intersect.

Just in case, I’ll put a stone with signs at the crossroads:

The rest jump over the stone and follow further, straight to Kashchei the Immortal =)

b) Find the direction vectors of the lines:

The lines have the same direction vector, which means they are either parallel or coincident. There is no need to count the determinant here.

It is obvious that the coefficients of the unknowns are proportional, and .

Let's find out whether the equality is true:

Thus,

c) Find the direction vectors of the lines:

Let's calculate the determinant made up of the coordinates of these vectors:
, therefore, the direction vectors are collinear. The lines are either parallel or coincident.

The proportionality coefficient “lambda” is easy to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: .

Now let's find out whether the equality is true. Both free terms are zero, so:

The resulting value satisfies this equation (any number in general satisfies it).

Thus, the lines coincide.

Answer:

Very soon you will learn (or even have already learned) to solve the problem discussed verbally literally in a matter of seconds. In this regard, I see no point in offering anything for independent decision, it’s better to lay another important brick in the geometric foundation:

How to construct a line parallel to a given one?

For ignorance of this simplest task, the Nightingale the Robber severely punishes.

Example 2

The straight line is given by the equation. Write an equation for a parallel line that passes through the point.

Solution: Let's denote the unknown line by the letter . What does the condition say about her? The straight line passes through the point. And if the lines are parallel, then it is obvious that the direction vector of the straight line “tse” is also suitable for constructing the straight line “de”.

We take the direction vector out of the equation:

Answer:

The example geometry looks simple:

Analytical testing consists of the following steps:

1) We check that the lines have the same direction vector (if the equation of the line is not simplified properly, then the vectors will be collinear).

2) Check whether the point satisfies the resulting equation.

In most cases, analytical testing can be easily performed orally. Look at the two equations, and many of you will quickly determine the parallelism of the lines without any drawing.

Examples for independent solutions today will be creative. Because you will still have to compete with Baba Yaga, and she, you know, is a lover of all sorts of riddles.

Example 3

Write an equation for a line passing through a point parallel to the line if

There is a rational and not so rational way to solve it. Most shortcut- at the end of the lesson.

We worked a little with parallel lines and will return to them later. The case of coinciding lines is of little interest, so let’s consider a problem that is very familiar to you from the school curriculum:

How to find the point of intersection of two lines?

If straight intersect at point , then its coordinates are the solution systems of linear equations

How to find the point of intersection of lines? Solve the system.

Here you go geometric meaning systems of two linear equations in two unknowns- these are two intersecting (most often) lines on a plane.

Example 4

Find the point of intersection of lines

Solution: There are two ways to solve - graphical and analytical.

The graphical method is to simply draw the given lines and find out the intersection point directly from the drawing:

Here is our point: . To check, you should substitute its coordinates into each equation of the line; they should fit both there and there. In other words, the coordinates of a point are a solution to the system. Essentially, we looked at a graphical solution systems of linear equations with two equations, two unknowns.

The graphical method is, of course, not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide this way, the point is that it will take time to create a correct and ACCURATE drawing. In addition, some straight lines are not so easy to construct, and the intersection point itself may be located somewhere in the thirtieth kingdom outside the notebook sheet.

Therefore, it is more expedient to look for the intersection point analytical method. Let's solve the system:

To solve the system, the method of term-by-term addition of equations was used. To develop relevant skills, take a lesson How to solve a system of equations?

Answer:

The check is trivial - the coordinates of the intersection point must satisfy each equation of the system.

Example 5

Find the point of intersection of the lines if they intersect.

This is an example for you to solve on your own. It is convenient to split the task into several stages. Analysis of the condition suggests that it is necessary:
1) Write down the equation of the straight line.
2) Write down the equation of the straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.

The development of an action algorithm is typical for many geometric problems, and I will repeatedly focus on this.

Full solution and answer at the end of the lesson:

Not even a pair of shoes were worn out before we got to the second section of the lesson:

Perpendicular lines. Distance from a point to a line.
Angle between straight lines

Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to this one, and now the hut on chicken legs will turn 90 degrees:

How to construct a line perpendicular to a given one?

Example 6

The straight line is given by the equation. Write an equation perpendicular to the line passing through the point.

Solution: By condition it is known that . It would be nice to find the directing vector of the line. Since the lines are perpendicular, the trick is simple:

From the equation we “remove” the normal vector: , which will be the directing vector of the straight line.

Let's compose the equation of a straight line using a point and a direction vector:

Answer:

Let's expand the geometric sketch:

Hmmm... Orange sky, orange sea, orange camel.

Analytical verification of the solution:

1) We take out the direction vectors from the equations and with the help scalar product of vectors we come to the conclusion that the lines are indeed perpendicular: .

By the way, you can use normal vectors, it’s even easier.

2) Check whether the point satisfies the resulting equation .

The test, again, is easy to perform orally.

Example 7

Find the point of intersection of perpendicular lines if the equation is known and period.

This is an example for you to solve on your own. There are several actions in the problem, so it is convenient to formulate the solution point by point.

Our exciting journey continues:

Distance from point to line

We have a straight strip of river in front of us and our task is to get to it by the shortest route. There are no obstacles, and the most optimal route will be to move along the perpendicular. That is, the distance from a point to a line is the length of the perpendicular segment.

Distance in geometry is traditionally denoted by the Greek letter “rho”, for example: – the distance from the point “em” to the straight line “de”.

Distance from point to line expressed by the formula

Example 8

Find the distance from a point to a line

Solution: all you need to do is carefully substitute the numbers into the formula and carry out the calculations:

Answer:

Let's make the drawing:

The found distance from the point to the line is exactly the length of the red segment. If you draw up a drawing on checkered paper on a scale of 1 unit. = 1 cm (2 cells), then the distance can be measured with an ordinary ruler.

Let's consider another task based on the same drawing:

The task is to find the coordinates of a point that is symmetrical to the point relative to the straight line . I suggest performing the steps yourself, but I will outline the solution algorithm with intermediate results:

1) Find a line that is perpendicular to the line.

2) Find the point of intersection of the lines: .

Both actions are discussed in detail in this lesson.

3) The point is the midpoint of the segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the midpoint of a segment we find .

It would be a good idea to check that the distance is also 2.2 units.

Difficulties may arise in calculations here, but a microcalculator is a great help in the tower, allowing you to calculate common fractions. I have advised you many times and will recommend you again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for you to decide on your own. I’ll give you a little hint: there are infinitely many ways to solve this. Debriefing at the end of the lesson, but it’s better to try to guess for yourself, I think your ingenuity was well developed.

Angle between two straight lines

Every corner is a jamb:


In geometry, the angle between two straight lines is taken to be the SMALLER angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered the angle between intersecting lines. And his “green” neighbor or oppositely oriented"raspberry" corner.

If the lines are perpendicular, then any of the 4 angles can be taken as the angle between them.

How are the angles different? Orientation. Firstly, the direction in which the angle is “scrolled” is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example if .

Why did I tell you this? It seems that we can get by with the usual concept of an angle. The fact is that the formulas by which we will find angles can easily result in a negative result, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing, for a negative angle, be sure to indicate its orientation with an arrow (clockwise).

How to find the angle between two straight lines? There are two working formulas:

Example 10

Find the angle between lines

Solution And Method one

Consider two straight lines given by the equations in general view:

If straight not perpendicular, That oriented The angle between them can be calculated using the formula:

Let us pay close attention to the denominator - this is exactly dot product directing vectors of straight lines:

If , then the denominator of the formula becomes zero, and the vectors will be orthogonal and the lines will be perpendicular. That is why a reservation was made about the non-perpendicularity of straight lines in the formulation.

Based on the above, it is convenient to formalize the solution in two steps:

1) Let's calculate the scalar product of the direction vectors of the lines:
, which means the lines are not perpendicular.

2) Find the angle between straight lines using the formula:

Using the inverse function it is easy to find the angle itself. In this case, we use the oddness of the arctangent (see. Graphs and properties of elementary functions):

Answer:

In the answer we indicate the exact value, as well as an approximate value (preferably in both degrees and radians), calculated using a calculator.

Well, minus, minus, no big deal. Here is a geometric illustration:

It is not surprising that the angle turned out to have a negative orientation, because in the problem statement the first number is a straight line and the “unscrewing” of the angle began precisely with it.

If you really want to get a positive angle, you need to swap the lines, that is, take the coefficients from the second equation , and take the coefficients from the first equation. In short, you need to start with a direct .

Let two lines be given and you need to find their point of intersection. Since this point belongs to each of the two given lines, its coordinates must satisfy both the equation of the first line and the equation of the second line.

Thus, in order to find the coordinates of the point of intersection of two lines, one must solve the system of equations

Example 1. Find the point of intersection of lines and

Solution. We will find the coordinates of the desired intersection point by solving the system of equations

The intersection point M has coordinates

Let's show how to construct a straight line using its equation. To construct a straight line, it is enough to know its two points. To construct each of these points, we specify an arbitrary value for one of its coordinates, and then from the equation we find the corresponding value for the other coordinate.

If in the general equation of a straight line both coefficients at the current coordinates are not equal to zero, then to construct this straight line it is best to find the points of its intersection with the coordinate axes.

Example 2. Construct a straight line.

Solution. We find the point of intersection of this line with the abscissa axis. To do this, we solve their equations together:

and we get . Thus, the point M (3; 0) of intersection of this line with the abscissa axis has been found (Fig. 40).

Then solving together the equation of this line and the equation of the ordinate axis

we find the point of intersection of the line with the ordinate axis. Finally, we construct a straight line from its two points M and

If two lines are not parallel, then they will inevitably intersect at one point. Discover coordinates points intersection of 2 lines is allowed both graphically and arithmetic, depending on what data the task provides.

You will need

• – two straight lines in the drawing;
• – equations of 2 straight lines.

Instructions

1. If the lines are already drawn on the graph, find the solution graphically. To do this, continue both or one of the lines so that they intersect. After this, mark the intersection point and lower a perpendicular from it to the x-axis (as usual, oh).

2. Using the scale marks marked on the axis, find the x value for that point. If it is in the positive direction of the axis (to the right of the zero mark), then its value will be correct; otherwise, it will be negative.

3. Correctly also find the ordinate of the intersection point. If the projection of a point is located above the zero mark, it is correct; if below, it is negative. Write down the coordinates of the point in the form (x, y) - this is the solution to the problem.

4. If the lines are given in the form of the formulas y=khx+b, you can also solve the problem graphically: draw the lines on a coordinate grid and find the solution using the method described above.

5. Try to discover the solution to the problem using these formulas. To do this, create a system from these equations and solve it. If the equations are given in the form y=khx+b, simply equate both sides with x and discover x. Then plug the value of x into one of the equations and find y.

6. You can find a solution using Cramer's method. In this case, reduce the equations to the form A1x+B1y+C1=0 and A2x+B2y+C2=0. According to Cramer's formula, x=-(C1B2-C2B1)/(A1B2-A2B1), and y=-(A1C2-A2C1)/(A1B2-A2B1). Please note that if the denominator is zero, then the lines are parallel or coincide and, accordingly, do not intersect.

7. If you are given lines in space in canonical form, before you start looking for a solution, check whether the lines are parallel. To do this, evaluate the exponents before t if they are proportional, say, x=-1+3t, y=7+2t, z=2+t and x=-1+6t, y=-1+4t, z=-5 +2t, then the lines are parallel. In addition, lines can intersect, in which case the system will not have a solution.

8. If you find out that lines intersect, find the point of their intersection. First, equate variables from different lines, conditionally replacing t with u for the first line and with v for the 2nd line. Say, if you are given lines x=t-1, y=2t+1, z=t+2 and x=t+1, y=t+1, z=2t+8 you will get expressions like u-1=v +1, 2u+1=v+1, u+2=2v+8.

9. Express u from one equation, substitute it into another and find v (in this problem u=-2,v=-4). Now, in order to find the intersection point, substitute the obtained values ​​instead of t (it makes no difference, into the first or second equation) and get the coordinates of the point x=-3, y=-3, z=0.

To consider 2 intersecting direct It is enough to consider them in a plane, since two intersecting lines lie in the same plane. Knowing the equations of these direct, it is possible to detect the coordinate of their point intersections .

You will need

• equations of lines

Instructions

1. In Cartesian coordinates, the general equation of a line looks like this: Ax+By+C = 0. Let two lines intersect. The equation of the first line is Ax+By+C = 0, the 2nd line is Dx+Ey+F = 0. All indicators (A, B, C, D, E, F) must be specified. In order to detect a point intersections these direct it is necessary to solve the system of these 2 linear equations.

2. To solve, it is convenient to multiply the first equation by E, and the second by B. As a result, the equations will look like: AEx+BEy+CE = 0, DBx+EBy+FB = 0. After subtracting the second equation from the first, you get: (AE- DB)x = FB-CE. Hence, x = (FB-CE)/(AE-DB). By analogy, the first equation initial system You can multiply by D, the second by A, then again subtract the second from the first. As a result, y = (CD-FA)/(AE-DB). The resulting x and y values ​​will be the coordinates of the point intersections direct .

3. Equations direct can also be written through the angular index k, equal to the tangent of the angle of inclination of the straight line. In this case, the equation of the line has the form y = kx+b. Let now the equation of the first line be y = k1*x+b1, and the equation of the 2nd line be y = k2*x+b2.

4. If we equate the right sides of these 2 equations, we get: k1*x+b1 = k2*x+b2. From there it is easy to obtain that x = (b1-b2)/(k2-k1). After substituting this value x into any of the equations, it turns out: y = (k2*b1-k1*b2)/(k2-k1). The x and y values ​​will specify the coordinates of the point intersections direct.If two lines are parallel or coincident, then they do not have universal points or have an immensely large number of universal points, respectively. In these cases k1 = k2, the denominators for the coordinates of the points intersections will vanish, therefore, the system will not have a classical solution. The system can have only one classical solution, which is unconditional, because two divergent and non-parallel lines can have only one point intersections .

Video on the topic

With this online calculator you can find the point of intersection of lines on a plane. Given detailed solution with explanations. To find the coordinates of the point of intersection of lines, set the type of equation of lines ("canonical", "parametric" or "general"), enter the coefficients of the equations of lines in the cells and click on the "Solve" button. See the theoretical part and numerical examples below.

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Data entry instructions. Numbers are entered as integers (examples: 487, 5, -7623, etc.), decimals (ex. 67., 102.54, etc.) or fractions. The fraction must be entered in the form a/b, where a and b (b>0) are integers or decimals. Examples 45/5, 6.6/76.4, -7/6.7, etc.

The point of intersection of lines on a plane - theory, examples and solutions

1. The point of intersection of lines given in general form.

Oxy L 1 and L 2:

Let's build an extended matrix:

If B" 2 =0 and WITH" 2 =0, then the system of linear equations has many solutions. Therefore straight L 1 and L 2 match. If B" 2 =0 and WITH" 2 ≠0, then the system is inconsistent and, therefore, the lines are parallel and do not have common point. If B" 2 ≠0, then the system of linear equations has a unique solution. From the second equation we find y: y=WITH" 2 /B" 2 and substituting the resulting value into the first equation we find x: x=−WITH 1 −B 1 y. We got the point of intersection of the lines L 1 and L 2: M(x, y).

2. The point of intersection of lines given in canonical form.

Let a Cartesian rectangular coordinate system be given Oxy and let straight lines be given in this coordinate system L 1 and L 2:

Let's open the brackets and make the transformations:

Using a similar method, we obtain the general equation of the straight line (7):

From equations (12) it follows:

How to find the intersection point of lines given in canonical form is described above.

4. The point of intersection of lines specified in different views.

Let a Cartesian rectangular coordinate system be given Oxy and let straight lines be given in this coordinate system L 1 and L 2:

We'll find t:

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

Let us solve the system of linear equations with respect to x, y. To do this, we will use the Gaussian method. We get:

Example 2. Find the point of intersection of lines L 1 and L 2:

L 1: 2x+3y+4=0,

(20)

(21)

To find the point of intersection of lines L 1 and L 2 you need to solve the system of linear equations (20) and (21). Let us present the equations in matrix form.