Mutual arrangement of lines in space. Problems with a straight line in space. Coordinates of the point of intersection of two lines - examples of finding
Perpendicular line
This task is probably one of the most popular and in demand in school textbooks. The tasks based on this theme are manifold. This is the definition of the point of intersection of two lines, this is the definition of the equation of a straight 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 a slope and vice versa, and the determination of the remaining parameters of a straight line according to given conditions were considered.
What do we lack in order to solve the problems that this page is devoted to?
1. Formulas for calculating one of the angles between two intersecting lines.
If we have two straight lines which 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 border 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, they intersect at a right angle.
What can be the initial data for solving such problems, except for a given straight line?
A point on a line and the angle at which the second line intersects it
The second equation of the line
What tasks can a bot solve?
1. Two straight lines are given (explicitly or implicitly, 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 one at a specified angle
Examples
Two straight 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 lines (in degrees)
-42.357454705937
Point of intersection of two lines
x=-3.5
y=-6.5
Do not forget that the parameters of the two lines are separated by a comma, and the parameters of each line by a semicolon.
The line passes through two points (1:-4) and (5:2) . Find the equation of a straight line that passes through the point (-2:-8) and intersects the original line at an angle of 30 degrees.
One straight line is known to us, since two points through which it passes are known.
It remains to determine the equation of the second straight line. One point is known to us, and instead of the second, the angle at which the first line intersects the second is indicated.
Everything seems to be known, but the main thing here is not to be mistaken. We are talking about the angle (30 degrees) not between the x-axis and the line, but between the first and second lines.
For this 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
Coefficient 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 crosses the axis at an angle 56.309932474019 degrees.
The source data does not say exactly how the second line intersects the first. After all, it is possible to draw two lines that satisfy the conditions, the first rotated 30 degrees clockwise, and the second 30 degrees counterclockwise.
Let's count them
If the second line is rotated 30 degrees COUNTER-CLOCKWISE, then the second line will have a degree of intersection with the x-axis 30+56.309932474019 = 86 .309932474019 degrees
line_p xa=-2;ya=-8;f=86.309932474019
Straight line parameters according to the given parameters
General equation Ax+By+C = 0
Coefficient A = 23.011106998916
Factor 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 point to line li =
that is, our second line equation is y= 15.505553499458x+ 23.011106998916
- To find the coordinates of the intersection point of the graphs of functions, you need to equate both functions to each other, move all terms containing $ x $ to the left side, and the rest to the right side and find the roots of the resulting equation.
- The second way is to compose a system of equations and solve it by substituting one function into another
- The third method involves the graphical construction of functions and the visual definition of the intersection point.
Case of two linear functions
Consider two linear functions $ f(x) = k_1 x+m_1 $ and $ g(x) = k_2 x + m_2 $. These functions are called direct. Building them is easy enough, you just need to take any two values $x_1$ and $x_2$ and find $f(x_1)$ and $(x_2)$. Then repeat the same with the $ g(x) $ function. Next, visually find the coordinate of the intersection point of the function graphs.
You should know that linear functions have only one intersection point and only when $ k_1 \neq k_2 $. Otherwise, in the case of $ k_1=k_2 $, the functions are parallel to each other, since $ k $ is the slope factor. If $ k_1 \neq k_2 $, but $ m_1=m_2 $, then the intersection point will be $ M(0;m) $. It is desirable to remember this rule for accelerated problem solving.
| Example 1 |
| Let $ f(x) = 2x-5 $ and $ g(x)=x+3 $ be given. Find the coordinates of the intersection point of function graphs. |
| Decision |
|
How to do it? Since there are two linear functions, the first thing we look at is the coefficient of the slope of both functions $ k_1 = 2 $ and $ k_2 = 1 $. Note that $ k_1 \neq k_2 $, so there is one intersection point. Let's find it using the equation $ f(x)=g(x) $: $$ 2x-5 = x+3 $$ We move the terms from $ x $ to the left side, and the rest to the right: $$ 2x - x = 3+5 $$ We got $ x=8 $ the abscissa of the intersection point of the graphs, and now let's find the ordinate. To do this, we substitute $ x = 8 $ into any of the equations either in $ f(x) $ or in $ g(x) $: $$ f(8) = 2\cdot 8 - 5 = 16 - 5 = 11 $$ So, $ M (8;11) $ - is the intersection point of the graphs of two linear functions. If you cannot solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the progress of the calculation and gather information. This will help you get a credit from the teacher in a timely manner! |
| Answer |
| $$ M (8;11) $$ |
Case of two non-linear functions
| Example 3 |
| Find the coordinates of the intersection point of function graphs: $ f(x)=x^2-2x+1 $ and $ g(x)=x^2+1 $ |
| Decision |
|
What about two non-linear functions? The algorithm is simple: we equate the equations to each other and find the roots: $$ x^2-2x+1=x^2+1 $$ We spread the terms with $ x $ and without it on different sides of the equation: $$ x^2-2x-x^2=1-1 $$ The abscissa of the desired point was found, but it is not enough. The ordinate $ y $ is still missing. Substitute $ x = 0 $ into any of the two equations of the problem statement. For example: $$ f(0)=0^2-2\cdot 0 + 1 = 1 $$ $ M (0;1) $ - intersection point of function graphs |
| Answer |
| $$ M (0;1) $$ |
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 it is required to look for the coordinates of the intersection of two lines on the plane or to determine the coordinates of the same lines in space. This article considers cases of finding the coordinates of points where the 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 where two lines intersect is called their point of intersection. In other words, the point of intersecting lines is the point of intersection.
Consider the figure below.
Before finding the coordinates of the point of intersection of two lines, it is necessary to consider the example below.
If there is a coordinate system O x y on the plane, then two straight lines a and b are given. The line a corresponds to the general equation of the form A 1 x + B 1 y + C 1 = 0, for the line b - A 2 x + B 2 y + C 2 = 0. Then M 0 (x 0 , y 0) is some 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 of 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 they give the correct identity when substituting, then M 0 (x 0 , y 0) is considered their intersection point.
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 the point of intersection.
Decision
For the intersection of lines to be real, it is necessary that the coordinates of the point M 0 satisfy the equations of lines. This is verified 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.
We depict this solution on the coordinate line of the figure below.
Answer: the given point with coordinates (2, - 3) will be the point of intersection of the given lines.
Example 2
Will the lines 5 x + 3 y - 1 = 0 and 7 x - 2 y + 11 = 0 intersect at the point M 0 (2 , - 3) ?
Decision
To solve the problem, it is necessary to substitute the coordinates of the point in 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, which means that the given point does not belong to the line 7 x - 2 y + 11 = 0 . Hence we have that the point M 0 is not a point of intersection of lines.
The drawing clearly shows that M 0 is not the point of intersection of the lines. They have a common point with coordinates (- 1 , 2) .
Answer: the point with coordinates (2, - 3) is not the point of intersection of the given lines.
We turn 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 given by equations of the form A 1 x + B 1 y + C 1 \u003d 0 and A 2 x + B 2 y + C 2 \u003d 0 located in O x y. When designating the intersection point M 0, we get that we should continue the search for coordinates according to the equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0.
It is obvious from the definition that M 0 is a common point of intersection of the 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 of the resulting system A 1 x + B 1 y + C 1 = 0 A 2 x + B 2 y + C 2 = 0 .
This means that in order 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 lines x - 9 y + 14 = 0 and 5 x - 2 y - 16 = 0 on the plane. you need to find their intersection.
Decision
Data on the condition of the equation must be collected into a system, after which we get x - 9 y + 14 \u003d 0 5 x - 2 y - 16 \u003d 0. To solve it, the first equation is resolved for x, 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 .
The search for coordinates is reduced to solving a system of linear equations. If, according to the condition, another form of the equation is given, then it should be reduced to the 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 .
Decision
To begin with, it is necessary to bring the equations to a general form. Then we get that x = 4 + 9 λ y = 2 + λ , λ ∈ R is transformed in this way:
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
Hence 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 apply Cramer's method to find the 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 another way to find the coordinates of the point of intersection of lines located on the 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 x = x 1 + a x λ and y = y 1 + a y λ are substituted for x, 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 .
Decision
It is necessary to perform a substitution in x - 5 \u003d y - 4 - 3 by the expression x \u003d 4 + 9 λ, y \u003d 2 + λ, then we get:
4 + 9 λ - 5 = 2 + λ - 4 - 3
When solving, we obtain that λ = - 1 . This implies 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, it is necessary 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 of the 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, we can compose 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 be the same.
Example 6
Given lines x 3 + y - 4 = 1 and y = 4 3 x - 4 . Determine if they have a common point.
Decision
Simplifying the given equations, we get 1 3 x - 1 4 y - 1 = 0 and 4 3 x - y - 4 = 0 .
It is necessary to collect the equations in 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
This shows 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 straight line. Therefore, there are no intersection points.
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 .
Decision
By condition, it is possible that the lines will not intersect. Write a system of equations and solve. For the solution, it is necessary to use the Gauss 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 got the wrong equality, so the system has no solutions. We conclude that the lines are parallel. There are no intersection points.
The second solution.
First you need to determine the presence of the 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 get 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 intersection points.
Answer: there are no intersection points, the lines are parallel.
Example 8
Find the intersection coordinates of the given lines 2 x - 1 = 0 and y = 5 4 x - 2 .
Decision
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
Find the determinant of the main matrix. For this, 2 0 5 4 - 1 = 2 · (- 1) - 0 · 5 4 = - 2 . Since it is non-zero, the system has 1 solution. It follows that the lines intersect. Let's solve the system for finding the coordinates of the 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 got that the point of intersection of the given lines has the 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 the lines of space are found.
When lines a and b are given in the coordinate plane O x y z by the equations of intersecting planes, then there is a 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 \u003d 0 and the straight line b - A 3 x + B 3 y + C 3 z + D 3 \u003d 0 A 4 x + B 4 y + C 4 z + D 4 \u003d 0.
When the point M 0 is the point of intersection of the 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 consider such tasks with examples.
Example 9
Find the coordinates of the point of intersection 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
Decision
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, it is necessary to solve through the matrix. Then we get 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 rank of the matrix according to Gauss.
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 augmented matrix is 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 fit. We get 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 . System solution 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 .
So we have 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. So lines a and b intersect.
In other cases, the equation has no solution, that is, there are 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 Gauss method. With its incompatibility, the lines are not intersecting. If there are an infinite number of solutions, then they coincide.
You can make a decision by calculating the main and extended rank of the matrix, and then apply the Kronecker-Capelli theorem. We get one, many or complete absence of solutions.
Example 10
Equations of 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 point of intersection.
Decision
First, let's set up 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 Gauss 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 that the lines do not intersect. There is no intersection point.
Answer: no intersection point.
If the lines are given using cononic or parametric equations, it is necessary to bring 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 point of intersection.
Decision
We set straight lines by the 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 basic 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 by 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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With the help of this online calculator you can find the point of intersection of lines on the plane. A detailed solution with explanations is given. To find the coordinates of the point of intersection of the lines, specify the type of the equation of the lines ("canonical", "parametric" or "general"), enter the coefficients of the equations of the lines into the cells and click the "Solve" button. See the theoretical part and numerical examples below.
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Point of intersection of lines in the plane - theory, examples and solutions
1. Point of intersection of straight lines given in general form.
Oxy L 1 and L 2:
Let's build an augmented matrix:
 |
If a B" 2=0 and WITH" 2 =0, then the system of linear equations has many solutions. Hence the direct L 1 and L 2 match. If a B" 2=0 and WITH" 2 ≠0, then the system is inconsistent and, therefore, the lines are parallel and do not have a 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. Get the point of intersection of the lines L 1 and L 2: M(x, y).
2. Point of intersection of lines given in canonical form.
Let a Cartesian rectangular coordinate system be given Oxy and let lines be given in this coordinate system L 1 and L 2:
Let's open the brackets and make the transformations:
By 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 the canonical form is described above.
4. Intersection point of lines defined in different views.
Let a Cartesian rectangular coordinate system be given Oxy and let lines be given in this coordinate system L 1 and L 2:
Let's find t:
| A 1 x 2 +A 1 mt+B 1 y 2 +B 1 pt+C 1 =0, |
We solve the system of linear equations with respect to x, y. To do this, we use the Gauss 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 it is necessary to solve the system of linear equations (20) and (21). We represent the equations in matrix form.
Point of intersection of lines
Let us be given two straight lines given by their coefficients and . It is required to find their intersection point, or find out that the lines are parallel.
Decision
If two lines are not parallel, then they intersect. To find the intersection point, it is enough to compose a system of two equations of lines and solve it:
Using Cramer's formula, we immediately find a solution to the system, which will be the desired intersection point:
If the denominator is zero, i.e.
then the system of solutions has no (direct are parallel and do not coincide) or has infinitely many (direct match). If it is necessary to distinguish between these two cases, it is necessary to check that the coefficients of the lines are proportional with the same coefficient of proportionality as the coefficients and , for which it is enough to calculate two determinants, if they are both equal to zero, then the lines coincide:
Implementation
struct pt (double x, y;); struct line (double a, b, c;); constdouble EPS=1e-9; double det (double a, double b, double c, double d)(return a * d - b * c;) bool intersect (line m, line n, pt & res)(double zn = det (m.a, m.b, n.a , n.b);if(abs(zn)< EPS)returnfalse; res.x=- det (m.c, m.b, n.c, n.b)/ zn; res.y=- det (m.a, m.c, n.a, n.c)/ zn;returntrue;} bool parallel (line m, line n){returnabs(det (m.a, m.b, n.a, n.b))< EPS;} bool equivalent (line m, line n){returnabs(det (m.a, m.b, n.a, n.b))< EPS &&abs(det (m.a, m.c, n.a, n.c))< EPS &&abs(det (m.b, m.c, n.b, n.c))< EPS;}
Lesson from the series " Geometric Algorithms»
Hello dear reader.
Tip 1: How to find the coordinates of the point of intersection of two lines
Let's write three more new functions.
The LinesCross() function will determine if intersect whether two segment. In it, the relative position of the segments is determined using vector products. To calculate vector products, let's write a function - VektorMulti().
The RealLess() function will be used to implement the comparison operation “<” (строго меньше) для вещественных чисел.
Task1. Two segments are given by their coordinates. Write a program that determines Do these segments intersect? without finding the intersection point.
Decision
. The second is given by dots.
Consider a segment and points and .
The point lies to the left of the line, for which the vector product 
> 0, since the vectors are positively oriented.
The point is located to the right of the line, for it the vector product< 0, так как векторы отрицательно ориентированы.
In order for the points and , to lie on opposite sides of the line , it is sufficient that the condition< 0 (векторные произведения имели противоположные знаки).
Similar reasoning can be carried out for the segment and points and .
So if 
, then the segments intersect.
To check this condition, the LinesCross() function is used, and to calculate vector products, the VektorMulti() function is used.
ax, ay are the coordinates of the first vector,
bx, by are the coordinates of the second vector.
Program geometry4; (Do 2 segments intersect?) Const _Eps: Real=1e-4; (calculation precision) var x1,y1,x2,y2,x3,y3,x4,y4: real; var v1,v2,v3,v4: real;function RealLess(Const a, b: Real): Boolean; (Strictly less than) begin RealLess:= b-a> _Eps end; (RealLess)function VektorMulti(ax,ay,bx,by:real): real; (ax,ay - a coordinates bx,by - b coordinates) begin vektormulti:= ax*by-bx*ay; end;Function LinesCross(x1,y1,x2,y2,x3,y3,x4,y4:real): boolean; (Do the segments intersect?) begin v1:=vektormulti(x4-x3,y4-y3,x1-x3,y1-y3); v2:=vectormulti(x4-x3,y4-y3,x2-x3,y2-y3); v3:=vectormulti(x2-x1,y2-y1,x3-x1,y3-y1); v4:=vectormulti(x2-x1,y2-y1,x4-x1,y4-y1); if RealLess(v1*v2.0) and RealLess(v3*v4.0) (v1v2<0 и v3v4<0, отрезки пересекаются} then LinesCross:= true else LinesCross:= false end; {LinesCross}begin {main} writeln(‘Введите координаты отрезков: x1,y1,x2,y2,x3,y3,x4,y4’); readln(x1,y1,x2,y2,x3,y3,x4,y4); if LinesCross(x1,y1,x2,y2,x3,y3,x4,y4) then writeln (‘Да’) else writeln (‘Нет’) end.
Program execution results:
Enter the coordinates of the segments: -1 1 2 2.52 2 1 -1 3
Yes.
We have written a program that determines whether the segments given by their coordinates intersect.
In the next lesson, we will write an algorithm that can be used to determine whether a point lies inside a triangle.
Dear reader.
You have already read several lessons from the Geometric Algorithms series. Is everything available written? I will be very grateful if you leave a review about these lessons. Perhaps something else needs to be improved.
Sincerely, Vera Gospodarets.
Let two segments be given. The first one is given by dots P 1 (x 1 ;y 1) and P 2 (x 2 ;y 2). The second is given by dots P 3 (x 3 ;y 3) and P 4 (x 4 ;y 4).
The relative position of the segments can be checked using vector products: 
Consider the segment P 3 P 4 and points P1 and P2.
Dot P1 lies to the left of the line P 3 P 4, for it the vector product v1 > 0, since the vectors are positively oriented.
Dot P2 located to the right of the line, for it the vector product v2< 0
, since the vectors are negatively oriented.
To point P1 and P2 lie on opposite sides of a straight line P 3 P 4, it is sufficient that the condition v 1 v 2< 0 (vector products had opposite signs).
Similar reasoning can be carried out for the segment P 1 P 2 and points P3 and P4.
So if v 1 v 2< 0 and v 3 v 4< 0 , then the segments intersect.
The cross product of two vectors is calculated by the formula:
where:
ax, ay are the coordinates of the first vector,
bx, by are the coordinates of the second vector.
The equation of a straight line passing through two different points given by their coordinates.
Let two non-coinciding points be given on a straight line: P1 with coordinates ( x1;y1) and P2 with coordinates (x 2 ; y 2).
Line intersection
Accordingly, the vector with the origin at the point P1 and end at a point P2 has coordinates (x 2 -x 1, y 2 -y 1). If a P(x, y) is an arbitrary point on the line, then the coordinates of the vector P 1 P equal (x - x 1, y - y 1).
With the help of the cross product, the condition of collinarity of vectors P 1 P and P 1 P 2 can be written like this:
|P 1 P,P 1 P 2 |=0, i.e. (x-x 1)(y 2 -y 1)-(y-y 1)(x 2 -x 1)=0
or
(y 2 -y 1)x + (x 1 -x 2)y + x 1 (y 1 -y 2) + y 1 (x 2 -x 1) = 0
The last equation is rewritten as follows:
ax + by + c = 0, (1)
where
a \u003d (y 2 -y 1),
b \u003d (x 1 -x 2),
c \u003d x 1 (y 1 -y 2) + y 1 (x 2 -x 1)
So, the straight line can be given by an equation of the form (1).
How to find the point of intersection of lines?
The obvious solution is to solve the system of equations of lines:
ax 1 +by 1 =-c 1
ax 2 +by 2 =-c 2 (2)
Enter designations:
Here D is the determinant of the system, and D x ,D y are the determinants obtained by replacing the column of coefficients for the corresponding unknown with a column of free terms. If a D ≠ 0, then system (2) is definite, that is, it has a unique solution. This solution can be found using the following formulas: x 1 \u003d D x / D, y 1 \u003d D y / D, which are called Cramer's formulas. A little reminder of how the second order determinant is calculated. The determinant distinguishes between two diagonals: the main and 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.