Find the point of intersection of the lines given by the canonical equations. P.6.3. How to find the point of intersection of two lines

When solving some geometric problems using the coordinate method, you have to find the coordinates of the point of intersection of lines. Most often you have to look for the coordinates of the point of intersection of two lines on a plane, but sometimes there is a need to determine the coordinates of the point of intersection of two lines in space. In this article we will deal with finding the coordinates of the point at which two lines intersect.

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The point of intersection of two lines is a definition.

Let's first define the point of intersection of two lines.

In the section on the relative position of lines on a plane, it is shown that two lines on a plane can either coincide (and they have infinitely many common points), or be parallel (and two lines have no common points), or intersect, having one common point. Options relative position there are more than two lines in space - they can coincide (have infinitely many common points), they can be parallel (that is, lie in the same plane and do not intersect), they can be intersecting (not lie in the same plane), and they can also have one common point, that is, to intersect. So, two lines both on the plane and in space are called intersecting if they have one common point.

From the definition of intersecting lines it follows determining the point of intersection of lines: The point at which two lines intersect is called the point of intersection of these lines. In other words, the only common point of two intersecting lines is the point of intersection of these lines.

For clarity, we present a graphical illustration of the point of intersection of two straight lines on a plane and in space.

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Finding the coordinates of the point of intersection of two lines on a plane.

Before finding the coordinates of the intersection point of two straight lines on a plane using their known equations, consider an auxiliary problem.

Oxy a And b. We will assume that straight a corresponds to a general equation of the straight line of the form , and the straight line b– type . Let be some point on the plane, and we need to find out whether the point is M 0 the point of intersection of given lines.

Let's solve the problem.

If M0 a And b, then by definition it also belongs to the line a and straight b, that is, its coordinates must satisfy both the equation and the equation. Therefore, we need to substitute the coordinates of the point M 0 into the equations of given lines and see if this results in two correct equalities. If the coordinates of the point M 0 satisfy both equations and , then is the point of intersection of the lines a And b, otherwise M 0 .

Is the point M 0 with coordinates (2, -3) point of intersection of lines 5x-2y-16=0 And 2x-5y-19=0?

If M 0 is indeed the point of intersection of given lines, then its coordinates satisfy the equations of lines. Let's check this by substituting the coordinates of the point M 0 into the given equations:

We got two true equalities, therefore, M 0 (2, -3)- point of intersection of lines 5x-2y-16=0 And 2x-5y-19=0.

For clarity, we present a drawing that shows straight lines and the coordinates of their intersection points are visible.

yes, period M 0 (2, -3) is the point of intersection of the lines 5x-2y-16=0 And 2x-5y-19=0.

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

Let's substitute the coordinates of the point M 0 into the equations of straight lines, this action will check whether the point belongs to M 0 both straight lines at the same time:

Since the second equation, when substituting the coordinates of the point into it M 0 did not turn into a true equality, then point M 0 does not belong to the line 7x-2y+11=0. From this fact we can conclude that the point M 0 is not the point of intersection of the given lines.

The drawing also clearly shows that the point M 0 is not the point of intersection of lines 5x+3y-1=0 And 7x-2y+11=0. Obviously, the given lines intersect at a point with coordinates (-1, 2) .

M 0 (2, -3) is not the point of intersection of lines 5x+3y-1=0 And 7x-2y+11=0.

Now we can move on to the task of finding the coordinates of the point of intersection of two lines using the given equations of lines on a plane.

Let a rectangular cartesian system coordinates Oxy and given two intersecting lines a And b equations and respectively. Let us denote the point of intersection of the given lines as M 0 and solve the following problem: find the coordinates of the point of intersection of two lines a And b according to the known equations of these lines and .

Dot M0 belongs to each of the intersecting lines a And b a-priory. Then the coordinates of the point of intersection of the lines a And b satisfy both the equation and the equation . Therefore, the coordinates of the point of intersection of two lines a And b are the solution to a system of equations (see the article solving systems of linear algebraic equations).

Thus, to find the coordinates of the intersection point of two lines defined on a plane general equations, you need to solve a system made up of equations of given lines.

Let's look at the example solution.

Find the intersection point of two lines defined in a rectangular coordinate system on a plane by the equations x-9y+14=0 And 5x-2y-16=0.

We are given two general equations of lines, let's make a system out of them: . Solutions to the resulting system of equations are easily found by solving its first equation with respect to the variable x and substitute this expression into the second equation:

The found solution to the system of equations gives us the desired coordinates of the point of intersection of two lines.

M 0 (4, 2)– point of intersection of lines x-9y+14=0 And 5x-2y-16=0.

So, finding the coordinates of the point of intersection of two straight lines, defined by general equations on the plane, comes down to solving a system of two linear equations with two unknown variables. But what if lines on a plane are given not by general equations, but by equations of a different type (see types of equations of a line on a plane)? In these cases, you can first reduce the equations of lines to general appearance, and after that find the coordinates of the intersection point.

Before finding the coordinates of the intersection point of the given lines, we reduce their equations to a general form. The transition from the parametric equations of a line to the general equation of this line looks like this:

Now let's carry out the necessary actions with the canonical equation of the straight line:

Thus, the desired coordinates of the point of intersection of the lines are a solution to a system of equations of the form . We use Cramer's method to solve it:

M 0 (-5, 1)

There is another way to find the coordinates of the point of intersection of two lines on a plane. It is convenient to use when one of the lines is given by parametric equations of the form, and the other by a line equation of a different type. In this case, in another equation instead of variables x And y you can substitute the expressions and , from where you can get the value that corresponds to the intersection point of the given lines. In this case, the point of intersection of the lines has coordinates.

Let's find the coordinates of the point of intersection of the lines from the previous example using this method.

Determine the coordinates of the point of intersection of the lines and .

Let's substitute the straight line expression into the equation:

Having solved the resulting equation, we get . This value corresponds to the common point of the lines and . We calculate the coordinates of the intersection point by substituting a straight line into the parametric equations:
.

M 0 (-5, 1).

To complete the picture, one more point should be discussed.

Before finding the coordinates of the point of intersection of two lines on a plane, it is useful to make sure that the given lines actually intersect. If it turns out that the original lines coincide or are parallel, then there can be no question of finding the coordinates of the point of intersection of such lines.

You can, of course, do without such a check, but immediately create a system of equations of the form and solve it. If a system of equations has a unique solution, then it gives the coordinates of the point at which the original lines intersect. If the system of equations does not have solutions, then we can conclude that the original lines are parallel (since there is no such pair of real numbers x And y, which would simultaneously satisfy both equations of the given lines). From the presence of an infinite number of solutions to a system of equations, it follows that the original straight lines have infinitely many common points, that is, they coincide.

Let's look at examples that fit these situations.

Find out whether the lines and intersect, and if they intersect, then find the coordinates of the intersection point.

The given equations of lines correspond to the equations and . Let's solve the system made up of these equations.

It is obvious that the equations of the system are linearly expressed through each other (the second equation of the system is obtained from the first by multiplying both its parts by 4 ), therefore, the system of equations has an infinite number of solutions. Thus, the equations define the same line, and we cannot talk about finding the coordinates of the point of intersection of these lines.

equations and are defined in a rectangular coordinate system Oxy the same straight line, so we cannot talk about finding the coordinates of the intersection point.

Find the coordinates of the point of intersection of the lines and , if possible.

The condition of the problem allows that the lines may not intersect. Let's create a system from these equations. Let us apply the Gauss method to solve it, since it allows us to establish the compatibility or incompatibility of a system of equations, and if it is compatible, find a solution:

The last equation of the system after the direct passage of the Gauss method turned into an incorrect equality, therefore, the system of equations has no solutions. From this we can conclude that the original lines are parallel, and we cannot talk about finding the coordinates of the point of intersection of these lines.

Second solution.

Let's find out whether the given lines intersect.

A normal vector is a line, and a vector is a normal vector of a line. Let us check that the condition for collinearity of vectors and : the equality is true, since , therefore, the normal vectors of the given straight lines are collinear. Then these lines are parallel or coincident. Thus, we cannot find the coordinates of the intersection point of the original lines.

it is impossible to find the coordinates of the intersection point of the given lines, since these lines are parallel.

Find the coordinates of the point of intersection of the lines 2x-1=0 and , if they intersect.

Let's compose a system of equations that are general equations of given lines: . The determinant of the main matrix of this system of equations is nonzero, therefore the system of equations has a unique solution, which indicates the intersection of the given lines.

To find the coordinates of the point of intersection of the lines, we need to solve the system:

The resulting solution gives us the coordinates of the point of intersection of the lines, that is, the point of intersection of the lines 2x-1=0 And .

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

The coordinates of the point of intersection of two lines in three-dimensional space are found similarly.

Let the intersecting lines a And b specified in a rectangular coordinate system Oxyz equations of two intersecting planes, that is, a straight line a is determined by a system of the form , and the straight line b- . Let M 0– point of intersection of lines a And b. Then point M 0 by definition also belongs to the line a and straight b, therefore, its coordinates satisfy the equations of both lines. Thus, the coordinates of the point of intersection of the lines a And b represent a solution to a system of linear equations of the form . Here we will need information from the section on solving systems of linear equations in which the number of equations does not coincide with the number of unknown variables.

Let's look at the solutions to the examples.

Find the coordinates of the point of intersection of two lines defined in space by the equations and .

Let's compose a system of equations from the equations of the given lines: . The solution of this system will give us the desired coordinates of the point of intersection of lines in space. Let's find the solution to the written system of equations.

The main matrix of the system has the form , and the extended one - .

Let's determine the rank of the matrix A and matrix rank T. We use the method of bordering minors, but we will not describe in detail the calculation of determinants (if necessary, refer to the article Calculation of the determinant of a matrix):

Thus, the rank of the main matrix equal to rank extended matrix and is equal to three.

Consequently, the system of equations has a unique solution.

We will take the determinant as the basis minor, therefore the last equation should be excluded from the system of equations, since it does not participate in the formation of the basis minor. So,

The solution to the resulting system is easy to find:

Thus, the point of intersection of the lines has coordinates (1, -3, 0) .

(1, -3, 0) .

It should be noted that the system of equations has a unique solution if and only if the straight lines a And b intersect. If straight A And b parallel or crossing, then latest system has no solution equations, since in this case the lines do not have common points. If straight a And b coincide, then they have an infinite number of common points, therefore, the indicated system of equations has an infinite number of solutions. However, in these cases we cannot talk about finding the coordinates of the point of intersection of the lines, since the lines are not intersecting.

Thus, if we do not know in advance whether the given lines intersect a And b or not, then it is reasonable to create a system of equations of the form and solve it by the Gauss method. If we get a unique solution, then it will correspond to the coordinates of the point of intersection of the lines a And b. If the system turns out to be inconsistent, then the direct a And b do not intersect. If the system has an infinite number of solutions, then the straight lines a And b match up.

You can do without using the Gaussian method. Alternatively, you can calculate the ranks of the main and extended matrices of this system, and based on the data obtained and the Kronecker-Capelli theorem, conclude either the existence of a single solution, or the existence of many solutions, or the absence of solutions. It's a matter of taste.

If the lines intersect, then determine the coordinates of the intersection point.

Let's create a system from the given equations: . Let's solve it using the Gaussian method in matrix form:

It became clear that the system of equations has no solutions, therefore, the given lines do not intersect, and there can be no question of finding the coordinates of the point of intersection of these lines.

we cannot find the coordinates of the intersection point of the given lines, since these lines do not intersect.

When intersecting lines are given by canonical equations of a line in space or parametric equations of a line in space, then one should first obtain their equations in the form of two intersecting planes, and only after that find the coordinates of the intersection point.

Two intersecting lines are defined in a rectangular coordinate system Oxyz equations and . Find the coordinates of the point of intersection of these lines.

Let us define the initial straight lines by the equations of two intersecting planes:

To find the coordinates of the point of intersection of the lines, it remains to solve the system of equations. The rank of the main matrix of this system is equal to the rank of the extended matrix and is equal to three (we recommend checking this fact). Let us take as the basis minor; therefore, we can eliminate the last equation from the system. Having solved the resulting system using any method (for example, Cramer’s method), we obtain the solution. Thus, the point of intersection of the lines has coordinates (-2, 3, -5) .

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The point of intersection of two lines is a definition.

Let's first define the point of intersection of two lines.

Thus, in order to find the coordinates of the point of intersection of two straight lines defined on a plane by general equations, you need to solve a system composed of equations of given straight lines.

Let's look at the example solution.

Example.

Find the intersection point of two lines defined in a rectangular coordinate system on a plane by the equations x-9y+14=0 and 5x-2y-16=0.

Solution.

The found solution to the system of equations gives us the desired coordinates of the point of intersection of two lines.

Answer:

M 0 (4, 2) x-9y+14=0 and 5x-2y-16=0 .

So, finding the coordinates of the point of intersection of two straight lines, defined by general equations on a plane, comes down to solving a system of two linear equations with two unknown variables. But what if lines on a plane are given not by general equations, but by equations of a different type (see types of equations of a line on a plane)? In these cases, you can first reduce the equations of lines to a general form, and only after that find the coordinates of the intersection point.

Example.

And .

Solution.

Before finding the coordinates of the intersection point of the given lines, we reduce their equations to a general form. Transition from parametric straight line equations to the general equation of this line is as follows:

Now let's carry out the necessary actions with the canonical equation of the straight line:

Thus, the desired coordinates of the point of intersection of the lines are the solution to a system of equations of the form . To solve it we use:

Answer:

M 0 (-5, 1)

There is another way to find the coordinates of the point of intersection of two lines on a plane. It is convenient to use when one of the lines is given by parametric equations of the form , and the other is an equation of a straight line of a different type. In this case, in another equation, instead of the variables x and y, you can substitute the expressions And , from where it will be possible to obtain the value that corresponds to the intersection point of the given lines. In this case, the point of intersection of the lines has coordinates.

Let's find the coordinates of the point of intersection of the lines from the previous example using this method.

Example.

Determine the coordinates of the point of intersection of the lines And .

Solution.

Let's substitute the straight line expression into the equation:

Having solved the resulting equation, we get . This value corresponds to the common point of the lines And . We calculate the coordinates of the intersection point by substituting a straight line into the parametric equations:
.

Answer:

M 0 (-5, 1) .

To complete the picture, one more point should be discussed.

You can, of course, do without such a check and immediately create a system of equations of the form and solve it. If a system of equations has a unique solution, then it gives the coordinates of the point at which the original lines intersect. If the system of equations does not have solutions, then we can conclude that the original lines are parallel (since there is no pair of real numbers x and y that would simultaneously satisfy both equations of the given lines). From the presence of an infinite number of solutions to a system of equations, it follows that the original straight lines have infinitely many common points, that is, they coincide.

Let's look at examples that fit these situations.

Example.

Find out whether the lines and intersect, and if they intersect, then find the coordinates of the intersection point.

Solution.

The given equations of lines correspond to the equations And . Let's solve the system made up of these equations .

It is obvious that the equations of the system are linearly expressed through each other (the second equation of the system is obtained from the first by multiplying both its parts by 4), therefore, the system of equations has an infinite number of solutions. Thus, the equations define the same line, and we cannot talk about finding the coordinates of the point of intersection of these lines.

Answer:

The equations and define the same straight line in the rectangular coordinate system Oxy, so we cannot talk about finding the coordinates of the intersection point.

Example.

Find the coordinates of the point of intersection of the lines And , if possible.

Solution.

The condition of the problem allows that the lines may not intersect. Let's create a system from these equations. Let us apply to solve it, since it allows us to establish the compatibility or incompatibility of a system of equations, and if it is compatible, find a solution:

Second solution.

Let's find out whether the given lines intersect.

- normal line vector , and the vector is a normal line vector . Let's check the execution And : equality is true, since, therefore, the normal vectors of the given lines are collinear. Then these lines are parallel or coincident. Thus, we cannot find the coordinates of the intersection point of the original lines.

Answer:

It is impossible to find the coordinates of the intersection point of the given lines, since these lines are parallel.

Example.

Find the coordinates of the intersection point of the lines 2x-1=0 and , if they intersect.

Solution.

Let's compose a system of equations that are general equations of given straight lines: . The determinant of the main matrix of this system of equations is nonzero , therefore the system of equations has a unique solution, which indicates the intersection of the given lines.

To find the coordinates of the point of intersection of the lines, we need to solve the system:

The resulting solution gives us the coordinates of the point of intersection of the lines, that is, 2x-1=0 and .

Answer:

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

The coordinates of the point of intersection of two lines in three-dimensional space are found similarly.

Let's look at the solutions to the examples.

Example.

Find the coordinates of the intersection point of two lines given in space by the equations And .

Solution.

Let's compose a system of equations from the equations of given lines: . The solution of this system will give us the desired coordinates of the point of intersection of lines in space. Let's find the solution to the written system of equations.

The main matrix of the system has the form , and extended - .

Let's define A and the rank of the matrix T. We use

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 it is better not to use records of the form a = b, where a and b are real numbers.
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 linear 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.

If the lines intersect at a 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 of a system of two linear equations with two unknowns- these are two intersecting (most often) lines on a plane.

It is convenient to split the task into several stages. Analysis of the condition suggests that it is necessary:
1) Make an equation of one straight line.
2) Write an equation for the second line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.

Example 13.

Find the point of intersection of lines

Solution: It is advisable to search for the intersection point using the analytical method. Let's solve the system:

Answer:

P.6.4. Distance from point to line

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

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

Distance from point to a straight line expressed by the formula

Example 14.

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:

P.6.5. Angle between straight lines.

Example 15.

Find the angle between the lines.

1. Check whether the lines are perpendicular:

Let's calculate the scalar product of the direction vectors of the lines:
, which means the lines are not perpendicular.
2. Find the angle between straight lines using the formula:

Thus:

Answer:

Second order curves. Circle

Let a rectangular coordinate system 0xy be specified on the plane.

Second order curve is a line on a plane defined by an equation of the second degree relative to the current coordinates of the point M(x, y, z). In general, this equation looks like:

where coefficients A, B, C, D, E, L are any real numbers, and at least one of the numbers A, B, C is non-zero.

1.Circle is the set of points on the plane, the distance from which to a fixed point M 0 (x 0, y 0) is constant and equal to R. Point M 0 is called the center of the circle, and the number R is its radius

– equation of a circle with center at point M 0 (x 0, y 0) and radius R.

If the center of the circle coincides with the origin of coordinates, then we have:

– canonical equation of a circle.

Ellipse.

Ellipse is a set of points on a plane, for each of which the sum of the distances to two given points is a constant value (and this value is greater than the distances between these points). These points are called ellipse foci.

is the canonical equation of the ellipse.

The relationship is called eccentricity ellipse and is denoted by: , . Since then< 1.

Consequently, as the ratio decreases, it tends to 1, i.e. b differs little from a and the shape of the ellipse becomes closer to the shape of a circle. In the limiting case when , we get a circle whose equation is

x 2 + y 2 = a 2.

Hyperbola

Hyperbole is a set of points on a plane, for each of which the absolute value of the difference in distances to two given points, called tricks, is a constant quantity (provided that this quantity is less than the distance between the focuses and is not equal to 0).

Let F 1, F 2 be the foci, the distance between them will be denoted by 2c, the parameter of the parabola).

– canonical equation of a parabola.

Note that the equation for negative p also defines a parabola, which will be located to the left of the 0y axis. The equation describes a parabola, symmetrical about the 0y axis, lying above the 0x axis for p > 0 and lying below the 0x axis for p< 0.

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) .

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