Perpendicular line
This task is probably one of the most popular and in demand in school textbooks. The tasks based on this topic are varied. This is the definition of the point of intersection of two lines, this is also the definition of the equation of a line passing through a point on the original line at any angle.
We will cover this topic using in our calculations the data obtained using
It was there that the transformation of the general equation of a straight line into an equation with an angular coefficient and vice versa was considered, and the determination of the remaining parameters of the straight line according to given conditions.
What do we lack in order to solve the problems to which this page is devoted?
1. Formulas for calculating one of the angles between two intersecting lines.
If we have two lines that are given by the equations:
then one of the angles is calculated like this:
2. Equation of a straight line with a slope passing through a given point
From formula 1, we can see two borderline states
a) when then and therefore these two given lines are parallel (or coincide)
b) when , then , and therefore these lines are perpendicular, that is, intersect at right angles.
What could be the initial data for solving such problems, other than the given straight line?
A point on a straight line and the angle at which the second straight line intersects it
Second equation of the line
What problems can a bot solve?
1. Two lines are given (explicitly or indirectly, for example, by two points). Calculate the point of intersection and the angles at which they intersect.
2. Given one straight line, a point on a straight line and one angle. Determine the equation of a straight line that intersects a given line at a specified angle
Examples
Two lines are given by equations. Find the point of intersection of these lines and the angles at which they intersect
line_p A=11;B=-5;C=6,k=3/7;b=-5
We get the following result
Equation of the first line
y = 2.2 x + (1.2)
Equation of the second line
y = 0.4285714285714 x + (-5)
Angle of intersection of two straight lines (in degrees)
-42.357454705937
Point of intersection of two lines
x = -3.5
y = -6.5
Don't forget that the parameters of two lines are separated by a comma, and the parameters of each line are separated by a semicolon.
A straight line passes through two points (1:-4) and (5:2). Find the equation of the line that passes through the point (-2:-8) and intersects the original line at an angle of 30 degrees.
We know one straight line because we know the two points through which it passes.
It remains to determine the equation of the second line. One point is known to us, and instead of the second, the angle at which the first line intersects the second is indicated.
It seems that everything is known, but the main thing here is not to make mistakes. We are talking about the angle (30 degrees) not between the x-axis and the line, but between the first and second line.
This is why we post like this. Let's determine the parameters of the first line and find out at what angle it intersects the x-axis.
line xa=1;xb=5;ya=-4;yb=2
General equation Ax+By+C = 0
Coefficient A = -6
Factor B = 4
Factor C = 22
Coefficient a= 3.6666666666667
Coefficient b = -5.5
Coefficient k = 1.5
Angle of inclination to the axis (in degrees) f = 56.309932474019
Coefficient p = 3.0508510792386
Coefficient q = 2.5535900500422
Distance between points=7.211102550928
We see that the first line intersects the axis at an angle 56.309932474019 degrees.
The source data does not say exactly how the second line intersects the first. You can, after all, construct two lines that satisfy the conditions, the first one rotated 30 degrees CLOCKWISE, and the second 30 degrees COUNTERclockwise.
Let's count them
If the second line is rotated 30 degrees COUNTERclockwise, then the second line will have the degree of intersection with the x-axis 30+56.309932474019 = 86 .309932474019 degrees
line_p xa=-2;ya=-8;f=86.309932474019
Parameters of a straight line according to specified parameters
General equation Ax+By+C = 0
Coefficient A = 23.011106998916
Coefficient B = -1.4840558255286
Coefficient C = 34.149767393603
Equation of a straight line in segments x/a+y/b = 1
Coefficient a= -1.4840558255286
Coefficient b = 23.011106998916
Equation of a straight line with angular coefficient y = kx + b
Coefficient k = 15.505553499458
Angle of inclination to the axis (in degrees) f = 86.309932474019
Normal equation of the line x*cos(q)+y*sin(q)-p = 0
Coefficient p = -1.4809790664999
Coefficient q = 3.0771888256405
Distance between points=23.058912962428
Distance from a point to a straight line li =
that is, our second line equation is y= 15.505553499458x+ 23.011106998916
- 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 method is to create a system of equations and solve it by substituting one function into another
- The third method involves graphically constructing functions and visually determining the intersection point.
The 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. It is quite easy to construct them; you 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 function $ g(x) $. 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 coefficient. If $ k_1 \neq k_2 $ but $ m_1=m_2 $, then the intersection point will be $ M(0;m) $. It is advisable to remember this rule to quickly solve problems.
| Example 1 |
| Let $ f(x) = 2x-5 $ and $ g(x)=x+3 $ be given. Find the coordinates of the intersection point of the function graphs. |
| Solution |
|
How to do this? Since two linear functions are presented, the first thing we look at is the slope coefficient of both functions $ k_1 = 2 $ and $ k_2 = 1 $. We 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 with $ x $ to the left side, and the rest to the right: $$ 2x - x = 3+5 $$ We have obtained $ x=8 $ the abscissa of the intersection point of the graphs, and now let’s find the ordinate. To do this, let’s 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 point of intersection of the graphs of two linear functions. If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner! |
| Answer |
| $$ M (8;11) $$ |
The case of two nonlinear functions
| Example 3 |
| Find the coordinates of the intersection point of the function graphs: $ f(x)=x^2-2x+1 $ and $ g(x)=x^2+1 $ |
| Solution |
|
What about two nonlinear functions? The algorithm is simple: we equate the equations to each other and find the roots: $$ x^2-2x+1=x^2+1 $$ We distribute terms with and without $ x $ on different sides of the equation: $$ x^2-2x-x^2=1-1 $$ The abscissa of the desired point has been found, but it is not enough. The ordinate $y$ is still missing. We substitute $ x = 0 $ into any of the two equations of the problem condition. 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 you need to look for the coordinates of the intersection of two lines on a plane or determine the coordinates of the same lines in space. This article considers cases of finding the coordinates of points where given lines intersect.
Yandex.RTB R-A-339285-1
It is necessary to define the points of intersection of two lines.
The section on the relative position of lines on a plane shows that they can coincide, be parallel, intersect at one common point, or intersect. Two lines in space are called intersecting if they have one common point.
The definition of the point of intersection of lines sounds like this:
Definition 1
The point at which two lines intersect is called their point of intersection. In other words, the point of intersecting lines is the point of intersection.
Let's look at the figure below.
Before finding the coordinates of the point of intersection of two lines, it is necessary to consider the example below.
If the plane has a coordinate system O x y, then two straight lines a and b are specified. Line a corresponds to a general equation of the form A 1 x + B 1 y + C 1 = 0, for line b - A 2 x + B 2 y + C 2 = 0. Then M 0 (x 0 , y 0) is a certain point of the plane; it is necessary to determine whether the point M 0 will be the point of intersection of these lines.
To solve the problem, it is necessary to adhere to the definition. Then the lines must intersect at a point whose coordinates are the solution to the given equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0. This means that the coordinates of the intersection point are substituted into all given equations. If, upon substitution, they give the correct identity, then M 0 (x 0 , y 0) is considered their point of intersection.
Example 1
Given two intersecting lines 5 x - 2 y - 16 = 0 and 2 x - 5 y - 19 = 0. Will the point M 0 with coordinates (2, - 3) be an intersection point.
Solution
For the intersection of lines to be valid, it is necessary that the coordinates of the point M 0 satisfy the equations of the lines. This can be checked by substituting them. We get that
5 2 - 2 (- 3) - 16 = 0 ⇔ 0 = 0 2 2 - 5 (- 3) - 19 = 0 ⇔ 0 = 0
Both equalities are true, which means M 0 (2, - 3) is the intersection point of the given lines.
Let 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, 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 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 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 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) .
If you notice an error in the text, please highlight it and press Ctrl+Enter
Using this online calculator you can find the point of intersection of lines on a plane. A detailed solution with explanations is given. 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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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 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. 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.
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
