What is the point of intersection of two lines called? The intersection of two lines. Angle and point of intersection

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.

In two-dimensional space, two lines intersect only at one point, defined by the coordinates (x,y). Since both lines pass through their point of intersection, the coordinates (x,y) must satisfy both equations that describe these lines. With some additional skills, you can find the intersection points of parabolas and other quadratic curves.

Steps

Point of intersection of two lines

Write the equation of each line, isolating the variable “y” on the left side of the equation. The other terms of the equation should be placed on the right side of the equation. Perhaps the equation given to you will contain the variable f(x) or g(x) instead of “y”; in this case, isolate such a variable. To isolate a variable, perform the appropriate math on both sides of the equation.

• If the equations of the lines are not given to you, based on the information you know.
• Example. Given straight lines described by equations and y − 12 = − 2 x (\displaystyle y-12=-2x). To isolate the “y” in the second equation, add the number 12 to both sides of the equation:

1. You are looking for the point of intersection of both lines, that is, a point whose coordinates (x, y) satisfy both equations. Since the variable “y” is on the left side of each equation, the expressions located on the right side of each equation can be equated. Write a new equation.

   • Example. Because y = x + 3 (\displaystyle y=x+3) And y = 12 − 2 x (\displaystyle y=12-2x), then we can write the following equality: .

2. Find the value of the variable "x". The new equation contains only one variable, "x". To find "x", isolate that variable on the left side of the equation by performing the appropriate math on both sides of the equation. You should get an equation of the form x = __ (if you can't do this, see this section).

   • Example. x + 3 = 12 − 2 x (\displaystyle x+3=12-2x)
   • Add 2 x (\displaystyle 2x) to each side of the equation:
   • 3 x + 3 = 12 (\displaystyle 3x+3=12)
   • Subtract 3 from each side of the equation:
   • 3 x = 9 (\displaystyle 3x=9)
   • Divide each side of the equation by 3:
   • x = 3 (\displaystyle x=3).

3. Use the found value of the variable "x" to calculate the value of the variable "y". To do this, substitute the found value of “x” into the equation (any) of the straight line.

   • Example. x = 3 (\displaystyle x=3) And y = x + 3 (\displaystyle y=x+3)
   • y = 3 + 3 (\displaystyle y=3+3)
   • y = 6 (\displaystyle y=6)

4. Check the answer. To do this, substitute the value of “x” into the other equation of the line and find the value of “y”. If you get different y values, check that your calculations are correct.

   • Example: x = 3 (\displaystyle x=3) And y = 12 − 2 x (\displaystyle y=12-2x)
   • y = 12 − 2 (3) (\displaystyle y=12-2(3))
   • y = 12 − 6 (\displaystyle y=12-6)
   • y = 6 (\displaystyle y=6)
   • You got the same value for y, so there are no errors in your calculations.

5. Write down the coordinates (x,y). Having calculated the values ​​of “x” and “y”, you have found the coordinates of the point of intersection of two lines. Write down the coordinates of the intersection point in (x,y) form.

   • Example. x = 3 (\displaystyle x=3) And y = 6 (\displaystyle y=6)
   • Thus, two straight lines intersect at a point with coordinates (3,6).

6. Calculations in special cases. In some cases, the value of the variable "x" cannot be found. But that doesn't mean you made a mistake. A special case occurs when one of the following conditions is met:

   • If two lines are parallel, they do not intersect. In this case, the variable “x” will simply be reduced, and your equation will turn into a meaningless equality (for example, 0 = 1 (\displaystyle 0=1)). In this case, write down in your answer that the lines do not intersect or there is no solution.
   • If both equations describe one straight line, then there will be an infinite number of intersection points. In this case, the variable “x” will simply be reduced, and your equation will turn into a strict equality (for example, 3 = 3 (\displaystyle 3=3)). In this case, write down in your answer that the two lines coincide.

   Problems with quadratic functions

   1. Definition of a quadratic function. In a quadratic function, one or more variables have a second degree (but not higher), for example, x 2 (\displaystyle x^(2)) or y 2 (\displaystyle y^(2)). The graphs of quadratic functions are curves that may not intersect or may intersect at one or two points. In this section, we will tell you how to find the intersection point or points of quadratic curves.

   2. Rewrite each equation by isolating the variable “y” on the left side of the equation. The other terms of the equation should be placed on the right side of the equation.

      • Example. Find the point(s) of intersection of the graphs x 2 + 2 x − y = − 1 (\displaystyle x^(2)+2x-y=-1) And
      • Isolate the variable "y" on the left side of the equation:
      • And y = x + 7 (\displaystyle y=x+7) .
      • In this example, you are given one quadratic function and one linear function. Remember that if you are given two quadratic functions, the calculations are similar to the steps outlined below.

   3. Equate the expressions on the right side of each equation. Since the variable “y” is on the left side of each equation, the expressions located on the right side of each equation can be equated.

      • Example. y = x 2 + 2 x + 1 (\displaystyle y=x^(2)+2x+1) And y = x + 7 (\displaystyle y=x+7)

   4. Transfer all terms of the resulting equation to its left side, and write 0 on the right side. To do this, do some basic math. This will allow you to solve the resulting equation.

      • Example. x 2 + 2 x + 1 = x + 7 (\displaystyle x^(2)+2x+1=x+7)
      • Subtract "x" from both sides of the equation:
      • x 2 + x + 1 = 7 (\displaystyle x^(2)+x+1=7)
      • Subtract 7 from both sides of the equation:

   5. Solve the quadratic equation. By moving all the terms of the equation to its left side, you get a quadratic equation. It can be solved in three ways: using a special formula, and.

      • Example. x 2 + x − 6 = 0 (\displaystyle x^(2)+x-6=0)
      • When you factor an equation, you get two binomials, which, when multiplied, give you the original equation. In our example, the first term x 2 (\displaystyle x^(2)) can be decomposed to x * x. Write this down: (x)(x) = 0
      • In our example, the free term -6 can be factorized into the following factors: − 6 ∗ 1 (\displaystyle -6*1), − 3 ∗ 2 (\displaystyle -3*2), − 2 ∗ 3 (\displaystyle -2*3), − 1 ∗ 6 (\displaystyle -1*6).
      • In our example, the second term is x (or 1x). Add each pair of factors of the dummy term (in our example -6) until you get 1. In our example, the appropriate pair of factors of the dummy term are the numbers -2 and 3 ( − 2 ∗ 3 = − 6 (\displaystyle -2*3=-6)), because − 2 + 3 = 1 (\displaystyle -2+3=1).
      • Fill in the blanks with the found pair of numbers: .

   6. Don't forget about the second point of intersection of the two graphs. If you solve the problem quickly and not very carefully, you may forget about the second intersection point. Here's how to find the x coordinates of two intersection points:

      • Example (factorization). If in Eq. (x − 2) (x + 3) = 0 (\displaystyle (x-2)(x+3)=0) one of the expressions in brackets will be equal to 0, then the entire equation will be equal to 0. Therefore, we can write it like this: x − 2 = 0 (\displaystyle x-2=0) → x = 2 (\displaystyle x=2) And x + 3 = 0 (\displaystyle x+3=0) → x = − 3 (\displaystyle x=-3) (that is, you found two roots of the equation).
      • Example (using a formula or completing a perfect square). When using one of these methods, a square root will appear in the solution process. For example, the equation from our example will take the form x = (− 1 + 25) / 2 (\displaystyle x=(-1+(\sqrt (25)))/2). Remember that when taking a square root you will get two solutions. In our case: 25 = 5 ∗ 5 (\displaystyle (\sqrt (25))=5*5), And 25 = (− 5) ∗ (− 5) (\displaystyle (\sqrt (25))=(-5)*(-5)). So write down two equations and find two values ​​of x.

   7. The graphs intersect at one point or do not intersect at all. Such situations occur if the following conditions are met:

      • If the graphs intersect at one point, then the quadratic equation is decomposed into identical factors, for example, (x-1) (x-1) = 0, and the square root of 0 appears in the formula ( 0 (\displaystyle (\sqrt (0)))). In this case, the equation has only one solution.
      • If the graphs do not intersect at all, then the equation is not factorized, and the square root of a negative number appears in the formula (for example, − 2 (\displaystyle (\sqrt (-2)))). In this case, write in your answer that there is no solution.

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

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

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

Oxy L 1 and L 2:

Let's build an extended matrix:

If B" 2 =0 and WITH" 2 =0, then the system of linear equations has many solutions. Therefore straight L 1 and L 2 match. If B" 2 =0 and WITH" 2 ≠0, then the system is inconsistent and, therefore, the lines are parallel and do not have 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:

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:

(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.

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.

Page navigation.

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.

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 substituting 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.

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.

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 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 points in common, 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:

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.

- 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 required 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