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. Direct a corresponds 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:set 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 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 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 the system has infinite set 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 - 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 coordinate plane O x y z equations of intersecting planes, then there is a straight line a, which can be determined using 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, common points Same. 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 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 lines are defined using cononic or parametric equations, you need to reduce it 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 basic minor 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 mathematical sign intersections, it will occur 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

To figure out mutual arrangement 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 generally 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 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. The shortest way is 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 familiar to you from 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.

Graphic 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 point of intersection itself may be located somewhere in the thirtieth kingdom outside the notebook sheet.

Therefore, it is more expedient to search for the intersection point using the 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.

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

Complete solution and the 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 check solutions:

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.

Is our an amusing trip continues:

Distance from point to line

In front of us is a straight strip of the river 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 in the formulas by which we will find angles, it can easily turn out negative result, and it shouldn't 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 equations V general view:

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

The most close attention let's reverse it to the denominator - this is exactly scalar 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 scalar product directing vectors of straight lines:
, which means the lines are not perpendicular.

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

By using inverse function It's easy to find the corner itself. In this case, we use the oddness of the arctangent (see. Graphs and properties of elementary functions):

Answer:

In your 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 be of 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 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 .

Intersecting on the x-axis, it is necessary to solve the equation y₁=y₂, that is, k₁x+b₁=k₂x+b₂.

Transform this inequality to obtain k₁x-k₂x=b₂-b₁. Now express x: x=(b₂-b₁)/(k₁-k₂). This way you will find the point of intersection of the graphs, which is located on the OX axis. Find the intersection point on the ordinate axis. Just substitute the x value you found earlier into any of the functions.

The previous option is suitable for graphs. If the function is , use following instructions. In the same way as with linear function, find the value of x. To do this, solve a quadratic equation. In the equation 2x² + 2x - 4=0, find (the equation is given as an example). To do this, use the formula: D= b² – 4ac, where b is the value before X, and c is the numeric value.

Substituting numeric values, get an expression of the form D= 4 + 4*4= 4+16= 20. The equations depend on the value of the discriminant. Now to the value of the variable b with the “-” sign, add or subtract (in turn) the root of the resulting discriminant, and divide by double product coefficient a. This way you will find the roots of the equation, that is, the coordinates of the intersection points.

Function graphs have a peculiarity: the OX axis will intersect twice, that is, you will find two x-axis coordinates. If you receive periodic value dependence of X on Y, then know that the graph intersects the x-axis at an infinite number of points. Check if you have found the intersection points. To do this, substitute the values ​​of X into the equation f(x)=0.

Sources:

• Finding the points of intersection of lines

If you know the value of a, then you can say that you have solved the quadratic equation, because its roots will be found very easily.

You will need

• -discriminant formula for a quadratic equation;
• -knowledge of multiplication tables

Instructions

Video on the topic

Helpful advice

The discriminant of a quadratic equation can be positive, negative, or equal to 0.

Sources:

• Solution quadratic equations
• discriminant even

Tip 3: How to find the coordinates of the intersection points of the graph of a function

The graph of the function y = f (x) is the set of all points in the plane, coordinates x, which satisfy the relation y = f (x). A function graph clearly illustrates the behavior and properties of a function. To construct a graph, several values ​​of the argument x are usually selected and the corresponding values ​​of the function y=f(x) are calculated for them. To create a graph more accurately and visually, it is useful to find its points of intersection with the coordinate axes.

Instructions

When crossing the abscissa axis (X axis), the value of the function is 0, i.e. y=f(x)=0. To calculate x, you need to solve the equation f(x)=0. In the case of a function, we obtain the equation ax+b=0, and find x=-b/a.

Thus, the X axis intersects at the point (-b/a,0).

In more difficult cases, for example, in the case of a quadratic dependence of y on x, the equation f(x)=0 has two roots, therefore, the x-axis intersects twice. In the case of y depending on x, for example y=sin(x), it has an infinite number of intersection points with the X axis.

To check the correctness of finding the coordinates of the intersection points of the function graph with the X axis, it is necessary to substitute the found x values ​​f(x). The value of the expression for any of the calculated x must be equal to 0.

Instructions

First, it is necessary to discuss the choice of a coordinate system convenient for solving the problem. Typically, in problems of this kind, one of the triangle is placed on the 0X axis so that one point coincides with the origin. Therefore, you should not deviate from the generally accepted canons of the solution and do the same (see Fig. 1). The method of defining the triangle itself does not play a fundamental role, since you can always move from one of them to (as you will be able to verify later).

Let the required triangle be specified by two vectors of its sides AC and AB a(x1, y1) and b(x2, y2), respectively. Moreover, by construction, y1=0. The third side of BC corresponds to c=a-b, c(x1-x2,y1 -y2), according to this illustration. Point A is placed at the origin of coordinates, that is, it coordinates A(0, 0). It is also easy to notice that coordinates B (x2, y2), a C (x1, 0). From this we can conclude that defining a triangle with two vectors automatically coincided with defining it with three points.

Next, you should complete the required triangle to the corresponding parallelogram ABDC in size. Moreover, that at the point intersections diagonals of a parallelogram they are divided so that AQ is the median of triangle ABC, descends from A to side BC. The diagonal vector s contains this one and is, according to the parallelogram rule, geometric sum a and b. Then s = a + b, and its coordinates s(x1+x2, y1+y2)= s(x1+x2, y2). The same coordinates will also be at the point D(x1+x2, y2).

Now we can proceed to compiling the equation of a straight line containing s, the median AQ and, most importantly, the desired point intersections median H. Since the vector s itself is a guide for a given line, and the point A(0, 0) belonging to it is also known, the simplest thing is to use the equation of a plane line in canonical form: (x-x0)/m =(y-y0)/n.Here (x0, y0) coordinates arbitrary point straight line (point A(0, 0)), and (m, n) – coordinates s (vector (x1+x2, y2). And so, the desired straight line l1 will look like: x/(x1+x2)=y/ y2.

The best way to find it is at the intersection. Therefore, you should find another straight line containing so-called N. To do this, in Fig. 1 construction of another parallelogram APBC, the diagonal of which g=a+c =g(2x1-x2, -y2) contains the second median CW, lowered from C to side AB. This diagonal contains point C(x1, 0), coordinates which will play the role of (x0, y0), and the direction vector here will be g(m, n)=g(2x1-x2, -y2). Hence l2 is given by the equation: (x-x1)/(2 x1-x2)=y/(- y2).

IN old times I was interested in computer graphics, both 2D and 3D, including mathematical visualizations. What is called just for fun, as a student, I wrote a program that visualizes N-dimensional figures rotating in any dimensions, although practically I was only able to determine the points for a 4-D hypercube. But this is just a saying. My love for geometry has remained with me from then to this day, and I still love solving interesting tasks in interesting ways.
I came across one of these problems in 2010. The task itself is quite trivial: you need to find whether two 2-D segments intersect, and if they do, find the point of their intersection. A more interesting solution is one that, I think, turned out to be quite elegant, and which I want to offer to the reader. I don’t claim the originality of the algorithm (although I would like to), but I couldn’t find similar solutions on the Internet.

Task

Given two segments, each of which is defined by two points: (v11, v12), (v21, v22). It is necessary to determine whether they intersect, and if they intersect, find the point of their intersection.

Solution

First you need to determine whether the segments intersect. Necessary and sufficient condition The intersection that must be observed for both segments is the following: the end points of one of the segments must lie in different half-planes if the plane is divided by the line on which the second of the segments lies. Let's demonstrate this with a drawing.

The left figure (1) shows two segments, for both of which the condition is met, and the segments intersect. In the right (2) figure, the condition is met for segment b, but for segment a it is not met, and accordingly the segments do not intersect.
It may seem that determining which side of the line a point lies on is a non-trivial task, but fear has big eyes, and everything is not so difficult. We know that vector multiplication of two vectors gives us a third vector, the direction of which depends on whether the angle between the first and second vector is positive or negative, respectively, such an operation is anticommutative. And since all vectors lie on X-Y plane, then their vector product (which must be perpendicular to the vectors being multiplied) will only have a non-zero component Z, and accordingly, the difference between the products of vectors will only be in this component. Moreover, when changing the order of multiplication of vectors (read: the angle between the multiplied vectors), it will consist solely in changing the sign of this component.
Therefore, we can multiply the vector of the dividing segment in pairs by vectors directed from the beginning of the dividing segment to both points of the segment being checked.

If the Z components of both products have different sign, which means one of the angles is less than 0 but greater than -180, and the second is greater than 0 and less than 180, respectively, the points lie along different sides from the straight line. If the Z components of both products have same sign, therefore they lie on one side of the straight line.
If one of the components of Z is zero, then we have a borderline case when the point lies exactly on the line being tested. Let's leave it to the user to determine if they want to consider this an intersection.
Then we need to repeat the operation for another segment and line, and make sure that the location of its end points also satisfies the condition.
So, if everything is fine and both segments satisfy the condition, then the intersection exists. Let's find it, and the vector product will also help us with this.
Since in the vector product we only have a non-zero component Z, then its modulus (vector length) will be numerically equal to exactly this component. Let's see how to find the intersection point.

The length of the vector product of vectors a and b (as we found out, is numerically equal to its component Z) is equal to the product of the absolute values ​​of these vectors and the sine of the angle between them (|a| |b| sin(ab)). Accordingly, for the configuration in the figure we have the following: |AB x AC| = |AB||AC|sin(α), and |AB x AD| = |AB||AD| sin(β). |AC|sin(α) is a perpendicular from point C to segment AB, and |AD|sin(β) is a perpendicular from point D to segment AB (leg ADD"). Since angles γ and δ are vertical angles, then they are equal, which means the triangles PCC" and PDD" are similar, and, accordingly, the lengths of all their sides are proportional in equal proportions.
Having Z1 (AB x AC, which means |AB||AC|sin(α)) and Z2 (AB x AD, which means |AB||AD|sin(β)), we can calculate CC"/DD" ( which will be equal to Z1/Z2), and also knowing that CC"/DD" = CP/DP, you can easily calculate the location of point P. Personally, I do it as follows:

Px = Cx + (Dx-Cx)*|Z1|/|Z2-Z1|;
Py = Cy + (Dy-Cy)*|Z1|/|Z2-Z1|;

That's all. I think it's really very simple and elegant. In conclusion, I would like to provide the function code that implements this algorithm. The function uses a homemade vector template , which is an int-sized vector template with components of type typename. Those interested can easily adjust the function to their vector types.

1 template 2 bool are_crossing(vector const &v11, vector const &v12, vector const &v21, vector const &v22, vector *crossing) 3 ( 4 vector cut1(v12-v11), cut2(v22-v21); 5 vector prod1, prod2; 6 7 prod1 = cross(cut1 * (v21-v11)); 8 prod2 = cross(cut1 * (v22-v11)); 9 10 if(sign(prod1[Z]) == sign(prod2[Z]) || (prod1[Z] == 0) || (prod2[Z] == 0)) // We also cut off borderline cases 11 return false; 12 13 prod1 = cross(cut2 * (v11-v21)); 14 prod2 = cross(cut2 * (v12-v21)); 15 16 if(sign(prod1[Z]) == sign(prod2[Z]) || (prod1[Z] == 0) || (prod2[Z] == 0)) // We also cut off borderline cases 17 return false; 18 19 if(crossing) ( // Check whether it is necessary to determine the intersection location 20 (*crossing)[X] = v11[X] + cut1[X]*fabs(prod1[Z])/fabs(prod2[Z]- prod1[Z]); 21 (*crossing)[Y] = v11[Y] + cut1[Y]*fabs(prod1[Z])/fabs(prod2[Z]-prod1[Z]); 22 ) 23 24 return true; 25)

Lesson from the series “Geometric algorithms”

Hello dear reader!

Let's continue to get acquainted with geometric algorithms. In the last lesson, we found the equation of a straight line using the coordinates of two points. We got an equation of the form:

Today we will write a function that, using the equations of two straight lines, will find the coordinates of their intersection point (if there is one). To check the equality of real numbers, we will use the special function RealEq().

Points on the plane are described by a pair of real numbers. When using a real type, it is better to implement comparison operations using special functions.

The reason is known: on the Real type in the Pascal programming system there is no order relation, so records of the form a = b, where a and b real numbers, it is better not to use.
Today we will introduce the RealEq() function to implement the “=” (strictly equal) operation:

Function RealEq(Const a, b:Real):Boolean; (strictly equal) begin RealEq:=Abs(a-b)<=_Eps End; {RealEq}

Task. The equations of two straight lines are given: and . Find the point of their intersection.

Solution. The obvious solution is to solve the system of line equations: Let's rewrite this system a little differently:
(1)

Let us introduce the following notation: , , . Here D is the determinant of the system, and are the determinants resulting from replacing the column of coefficients for the corresponding unknown with a column of free terms. If , then system (1) is definite, that is, it has a unique solution. This solution can be found using the following formulas: , which are called Cramer formulas. Let me remind you how the second-order determinant is calculated. The determinant distinguishes two diagonals: the main and the secondary. The main diagonal consists of elements taken in the direction from the upper left corner of the determinant to the lower right corner. Side diagonal - from the upper right to the lower left. The second-order determinant is equal to the product of the elements of the main diagonal minus the product of the elements of the secondary diagonal.

The code uses the RealEq() function to check equality. Calculations on real numbers are performed with an accuracy of _Eps=1e-7.

Program geom2; Const _Eps: Real=1e-7;(calculation accuracy) var a1,b1,c1,a2,b2,c2,x,y,d,dx,dy:Real; Function RealEq(Const a, b:Real):Boolean; (strictly equal) begin RealEq:=Abs(a-b)<=_Eps End; {RealEq} Function LineToPoint(a1,b1,c1,a2,b2,c2: real; var x,y:real):Boolean; {Определение координат точки пересечения двух линий. Значение функции равно true, если точка пересечения есть, и false, если прямые параллельны. } var d:real; begin d:=a1*b2-b1*a2; if Not(RealEq(d,0)) then begin LineToPoint:=True; dx:=-c1*b2+b1*c2; dy:=-a1*c2+c1*a2; x:=dx/d; y:=dy/d; end else LineToPoint:=False End;{LineToPoint} begin {main} writeln("Введите коэффициенты уравнений: a1,b1,c1,a2,b2,c2 "); readln(a1,b1,c1,a2,b2,c2); if LineToPoint(a1,b1,c1,a2,b2,c2,x,y) then writeln(x:5:1,y:5:1) else writeln("Прямые параллельны."); end.

We have compiled a program with which you can, knowing the equations of lines, find the coordinates of their intersection points.