Coordinates of the intersection point of function graphs. Intersection of two lines

In order to solve a geometric problem using the coordinate method, an intersection point is needed, the coordinates of which are used in the solution. A situation arises when it is required to look for the coordinates of the intersection of two lines on the plane or to determine the coordinates of the same lines in space. This article considers cases of finding the coordinates of points where the given lines intersect.

Yandex.RTB R-A-339285-1

It is necessary to define the points of intersection of two lines.

The section on the relative position of lines on a plane shows that they can coincide, be parallel, intersect at one common point, or intersect. Two lines in space are called intersecting if they have one common point.

The definition of the point of intersection of lines sounds like this:

Definition 1

The point where two lines intersect is called their point of intersection. In other words, the point of intersecting lines is the point of intersection.

Consider the figure below.

Before finding the coordinates of the point of intersection of two lines, it is necessary to consider the example below.

If there is a coordinate system O x y on the plane, then two straight lines a and b are given. The line a corresponds to the general equation of the form A 1 x + B 1 y + C 1 = 0, for the line b - A 2 x + B 2 y + C 2 = 0. Then M 0 (x 0 , y 0) is some point of the plane, it is necessary to determine whether the point M 0 will be the point of intersection of these lines.

To solve the problem, it is necessary to adhere to the definition. Then the lines must intersect at a point whose coordinates are the solution of the given equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0 . This means that the coordinates of the intersection point are substituted into all given equations. If they give the correct identity when substituting, then M 0 (x 0 , y 0) is considered their intersection point.

Example 1

Given two intersecting lines 5 x - 2 y - 16 = 0 and 2 x - 5 y - 19 = 0 . Will the point M 0 with coordinates (2, - 3) be the point of intersection.

Solution

For the intersection of lines to be real, it is necessary that the coordinates of the point M 0 satisfy the equations of lines. This is verified by substituting them. We get that

5 2 - 2 (- 3) - 16 = 0 ⇔ 0 = 0 2 2 - 5 (- 3) - 19 = 0 ⇔ 0 = 0

Both equalities are true, which means M 0 (2, - 3) is the intersection point of the given lines.

We depict this solution on the coordinate line of the figure below.

Answer: the given point with coordinates (2, - 3) will be the point of intersection of the given lines.

Example 2

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

Solution

To solve the problem, it is necessary to substitute the coordinates of the point in all equations. We get that

5 2 + 3 (- 3) - 1 = 0 ⇔ 0 = 0 7 2 - 2 (- 3) + 11 = 0 ⇔ 31 = 0

The second equality is not true, which means that the given point does not belong to the line 7 x - 2 y + 11 = 0 . Hence we have that the point M 0 is not a point of intersection of lines.

The drawing clearly shows that M 0 is not the point of intersection of the lines. They have a common point with coordinates (- 1 , 2) .

Answer: the point with coordinates (2, - 3) is not the point of intersection of the given lines.

We turn to finding the coordinates of the points of intersection of two lines using the given equations on the plane.

Two intersecting lines a and b are given by equations of the form A 1 x + B 1 y + C 1 \u003d 0 and A 2 x + B 2 y + C 2 \u003d 0 located in O x y. When designating the intersection point M 0, we get that we should continue the search for coordinates according to the equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0.

It is obvious from the definition that M 0 is a common point of intersection of the lines. In this case, its coordinates must satisfy the equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0 . In other words, this is the solution of the resulting system A 1 x + B 1 y + C 1 = 0 A 2 x + B 2 y + C 2 = 0 .

This means that in order to find the coordinates of the intersection point, it is necessary to add all the equations to the system and solve it.

Example 3

Given two lines x - 9 y + 14 = 0 and 5 x - 2 y - 16 = 0 on the plane. you need to find their intersection.

Solution

Data on the condition of the equation must be collected into a system, after which we get x - 9 y + 14 \u003d 0 5 x - 2 y - 16 \u003d 0. To solve it, the first equation is resolved for x, the expression is substituted into the second:

x - 9 y + 14 = 0 5 x - 2 y - 16 = 0 ⇔ x = 9 y - 14 5 x - 2 y - 16 = 0 ⇔ ⇔ x = 9 y - 14 5 9 y - 14 - 2 y - 16 = 0 ⇔ x = 9 y - 14 43 y - 86 = 0 ⇔ ⇔ x = 9 y - 14 y = 2 ⇔ x = 9 2 - 14 y = 2 ⇔ x = 4 y = 2

The resulting numbers are the coordinates that needed to be found.

Answer: M 0 (4 , 2) is the intersection point of the lines x - 9 y + 14 = 0 and 5 x - 2 y - 16 = 0 .

The search for coordinates is reduced to solving a system of linear equations. If, according to the condition, another form of the equation is given, then it should be reduced to the normal form.

Example 4

Determine the coordinates of the points of intersection of the lines x - 5 = y - 4 - 3 and x = 4 + 9 · λ y = 2 + λ , λ ∈ R .

Solution

To begin with, it is necessary to bring the equations to a general form. Then we get that x = 4 + 9 λ y = 2 + λ , λ ∈ R is transformed in this way:

x = 4 + 9 λ y = 2 + λ ⇔ λ = x - 4 9 λ = y - 2 1 ⇔ x - 4 9 = y - 2 1 ⇔ ⇔ 1 (x - 4) = 9 (y - 2) ⇔ x - 9 y + 14 = 0

Then we take the equation of the canonical form x - 5 = y - 4 - 3 and transform. We get that

x - 5 = y - 4 - 3 ⇔ - 3 x = - 5 y - 4 ⇔ 3 x - 5 y + 20 = 0

Hence we have that the coordinates are the point of intersection

x - 9 y + 14 = 0 3 x - 5 y + 20 = 0 ⇔ x - 9 y = - 14 3 x - 5 y = - 20

Let's apply Cramer's method to find the coordinates:

∆ = 1 - 9 3 - 5 = 1 (- 5) - (- 9) 3 = 22 ∆ x = - 14 - 9 - 20 - 5 = - 14 (- 5) - (- 9) ( - 20) = - 110 ⇒ x = ∆ x ∆ = - 110 22 = - 5 ∆ y = 1 - 14 3 - 20 = 1 (- 20) - (- 14) 3 = 22 ⇒ y = ∆ y ∆ = 22 22 = 1

Answer: M 0 (- 5 , 1) .

There is another way to find the coordinates of the point of intersection of lines located on the plane. It is applicable when one of the lines is given by parametric equations of the form x = x 1 + a x · λ y = y 1 + a y · λ , λ ∈ R . Then x = x 1 + a x λ and y = y 1 + a y λ are substituted for x, where we get λ = λ 0 corresponding to the intersection point having coordinates x 1 + a x λ 0, y 1 + a y λ 0 .

Example 5

Determine the coordinates of the point of intersection of the line x = 4 + 9 · λ y = 2 + λ , λ ∈ R and x - 5 = y - 4 - 3 .

Solution

It is necessary to perform a substitution in x - 5 \u003d y - 4 - 3 by the expression x \u003d 4 + 9 λ, y \u003d 2 + λ, then we get:

4 + 9 λ - 5 = 2 + λ - 4 - 3

When solving, we obtain that λ = - 1 . This implies that there is an intersection point between the lines x = 4 + 9 λ y = 2 + λ , λ ∈ R and x - 5 = y - 4 - 3 . To calculate the coordinates, it is necessary to substitute the expression λ = - 1 into the parametric equation. Then we get that x = 4 + 9 (- 1) y = 2 + (- 1) ⇔ x = - 5 y = 1 .

Answer: M 0 (- 5 , 1) .

To fully understand the topic, you need to know some of the nuances.

First you need to understand the location of the lines. When they intersect, we will find the coordinates, in other cases there will be no solution. To avoid this check, we can compose a system of the form A 1 x + B 1 y + C 1 = 0 A 2 x + B 2 + C 2 = 0 If there is a solution, we conclude that the lines intersect. If there is no solution, then they are parallel. When a system has an infinite number of solutions, then they are said to be the same.

Example 6

Given lines x 3 + y - 4 = 1 and y = 4 3 x - 4 . Determine if they have a common point.

Solution

Simplifying the given equations, we get 1 3 x - 1 4 y - 1 = 0 and 4 3 x - y - 4 = 0 .

It is necessary to collect the equations in a system for subsequent solution:

1 3 x - 1 4 y - 1 = 0 1 3 x - y - 4 = 0 ⇔ 1 3 x - 1 4 y = 1 4 3 x - y = 4

This shows that the equations are expressed through each other, then we get an infinite number of solutions. Then the equations x 3 + y - 4 = 1 and y = 4 3 x - 4 define the same straight line. Therefore, there are no intersection points.

Answer: the given equations define the same straight line.

Example 7

Find the coordinates of the point of intersecting lines 2 x + (2 - 3) y + 7 = 0 and 2 3 + 2 x - 7 y - 1 = 0 .

Solution

By condition, it is possible that the lines will not intersect. Write a system of equations and solve. For the solution, it is necessary to use the Gauss method, since with its help it is possible to check the equation for compatibility. We get a system of the form:

2 x + (2 - 3) y + 7 = 0 2 (3 + 2) x - 7 y - 1 = 0 ⇔ 2 x + (2 - 3) y = - 7 2 (3 + 2) x - 7 y = 1 ⇔ ⇔ 2 x + 2 - 3 y = - 7 2 (3 + 2) x - 7 y + (2 x + (2 - 3) y) (- (3 + 2)) = 1 + - 7 (- (3 + 2)) ⇔ ⇔ 2 x + (2 - 3) y = - 7 0 = 22 - 7 2

We got the wrong equality, so the system has no solutions. We conclude that the lines are parallel. There are no intersection points.

The second solution.

First you need to determine the presence of the intersection of lines.

n 1 → = (2 , 2 - 3) is the normal vector of the line 2 x + (2 - 3) y + 7 = 0 , then the vector n 2 → = (2 (3 + 2) , - 7 is the normal vector for the line 2 3 + 2 x - 7 y - 1 = 0 .

It is necessary to check the collinearity of the vectors n 1 → = (2, 2 - 3) and n 2 → = (2 (3 + 2) , - 7) . We get an equality of the form 2 2 (3 + 2) = 2 - 3 - 7 . It is correct because 2 2 3 + 2 - 2 - 3 - 7 = 7 + 2 - 3 (3 + 2) 7 (3 + 2) = 7 - 7 7 (3 + 2) = 0 . It follows that the vectors are collinear. This means that the lines are parallel and have no intersection points.

Answer: there are no intersection points, the lines are parallel.

Example 8

Find the intersection coordinates of the given lines 2 x - 1 = 0 and y = 5 4 x - 2 .

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

Find the determinant of the main matrix. For this, 2 0 5 4 - 1 = 2 · (- 1) - 0 · 5 4 = - 2 . Since it is non-zero, the system has 1 solution. It follows that the lines intersect. Let's solve the system for finding the coordinates of the intersection points:

2 x = 1 5 4 x - y = 2 ⇔ x = 1 2 4 5 x - y = 2 ⇔ x = 1 2 5 4 1 2 - y = 2 ⇔ x = 1 2 y = - 11 8

We got that the point of intersection of the given lines has the coordinates M 0 (1 2 , - 11 8) .

Answer: M 0 (1 2 , - 11 8) .

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

In the same way, the points of intersection of the lines of space are found.

When the lines a and b in the coordinate plane O x y z are given by the equations of intersecting planes, then there is a line a, which can be determined using the given system A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 1 \u003d 0 and the straight line b - A 3 x + B 3 y + C 3 z + D 3 \u003d 0 A 4 x + B 4 y + C 4 z + D 4 \u003d 0.

When the point M 0 is the point of intersection of the lines, then its coordinates must be solutions of both equations. We obtain linear equations in the system:

A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0 A 3 x + B 3 y + C 3 z + D 3 = 0 A 4 x + B 4 y + C 4 z + D 4 = 0

Let's consider such tasks with examples.

Example 9

Find the coordinates of the point of intersection of the given lines x - 1 = 0 y + 2 z + 3 = 0 and 3 x + 2 y + 3 = 0 4 x - 2 z - 4 = 0

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, it is necessary to solve through the matrix. Then we get the main matrix of the form   A = 1 0 0 0 1 2 3 2 0 4 0 - 2 and the extended matrix T = 1 0 0 1 0 1 2 - 3 4 0 - 2 4 . We determine the rank of the matrix according to Gauss.

We get that

1 = 1 ≠ 0 , 1 0 0 1 = 1 ≠ 0 , 1 0 0 0 1 2 3 2 0 = - 4 ≠ 0 , 1 0 0 1 0 1 2 - 3 3 2 0 - 3 4 0 - 2 4 = 0

It follows that the rank of the augmented matrix is ​​3 . Then the system of equations x - 1 = 0 y + 2 z + 3 = 0 3 x + 2 y + 3 = 0 4 x - 27 - 4 = 0 results in only one solution.

The basis minor has the determinant 1 0 0 0 1 2 3 2 0 = - 4 ≠ 0 , then the last equation does not fit. We get that x - 1 = 0 y + 2 z + 3 = 0 3 x + 2 y + 3 = 0 4 x - 2 z - 4 = 0 ⇔ x = 1 y + 2 z = - 3 3 x + 2 y - 3 . System solution x = 1 y + 2 z = - 3 3 x + 2 y = - 3 ⇔ x = 1 y + 2 z = - 3 3 1 + 2 y = - 3 ⇔ x = 1 y + 2 z = - 3 y = - 3 ⇔ ⇔ x = 1 - 3 + 2 z = - 3 y = - 3 ⇔ x = 1 z = 0 y = - 3 .

So we have that the intersection point x - 1 = 0 y + 2 z + 3 = 0 and 3 x + 2 y + 3 = 0 4 x - 2 z - 4 = 0 has coordinates (1 , - 3 , 0) .

Answer: (1 , - 3 , 0) .

System of the form A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0 A 3 x + B 3 y + C 3 z + D 3 = 0 A 4 x + B 4 y + C 4 z + D 4 = 0 has only one solution. So lines a and b intersect.

In other cases, the equation has no solution, that is, there are no common points either. That is, it is impossible to find a point with coordinates, since it does not exist.

Therefore, a system of the form A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0 A 3 x + B 3 y + C 3 z + D 3 = 0 A 4 x + B 4 y + C 4 z + D 4 = 0 is solved by the Gauss method. With its incompatibility, the lines are not intersecting. If there are an infinite number of solutions, then they coincide.

You can make a decision by calculating the main and extended rank of the matrix, and then apply the Kronecker-Capelli theorem. We get one, many or complete absence of solutions.

Example 10

Equations of lines x + 2 y - 3 z - 4 = 0 2 x - y + 5 = 0 and x - 3 z = 0 3 x - 2 y + 2 z - 1 = 0 are given. Find the point of intersection.

Solution

First, let's set up a system of equations. We get that x + 2 y - 3 z - 4 = 0 2 x - y + 5 = 0 x - 3 z = 0 3 x - 2 y + 2 z - 1 = 0 . We solve it using the Gauss method:

1 2 - 3 4 2 - 1 0 - 5 1 0 - 3 0 3 - 2 2 1 ~ 1 2 - 3 4 0 - 5 6 - 13 0 - 2 0 - 4 0 - 8 11 - 11 ~ ~ 1 2 - 3 4 0 - 5 6 - 13 0 0 - 12 5 6 5 0 0 7 5 - 159 5 ~ 1 2 - 3 4 0 - 5 6 - 13 0 0 - 12 5 6 5 0 0 0 311 10

Obviously, the system has no solutions, which means that the lines do not intersect. There is no intersection point.

Answer: no intersection point.

If the lines are given using cononic or parametric equations, it is necessary to bring them to the form of equations of intersecting planes, and then find the coordinates.

Example 11

Given two lines x = - 3 - λ y = - 3 · λ z = - 2 + 3 · λ , λ ∈ R and x 2 = y - 3 0 = z 5 in O x y z . Find the point of intersection.

Solution

We set straight lines by the equations of two intersecting planes. We get that

x = - 3 - λ y = - 3 λ z = - 2 + 3 λ ⇔ λ = x + 3 - 1 λ = y - 3 λ = z + 2 3 ⇔ x + 3 - 1 = y - 3 = z + 2 3 ⇔ ⇔ x + 3 - 1 = y - 3 x + 3 - 1 = z + 2 3 ⇔ 3 x - y + 9 = 0 3 x + z + 11 = 0 x 2 = y - 3 0 = z 5 ⇔ y - 3 = 0 x 2 = z 5 ⇔ y - 3 = 0 5 x - 2 z = 0

We find the coordinates 3 x - y + 9 = 0 3 x + z + 11 = 0 y - 3 = 0 5 x - 2 z = 0 , for this we calculate the ranks of the matrix. The rank of the matrix is ​​​​3, and the basic minor is 3 - 1 0 3 0 1 0 1 0 = - 3 ≠ 0, which means that the last equation must be excluded from the system. We get that

3 x - y + 9 = 0 3 x + z + 11 = 0 y - 3 = 0 5 x - 2 z = 0 ⇔ 3 x - y + 9 = 0 3 x + z + 11 = 0 y - 3 = 0

Let's solve the system by Cramer's method. We get that x = - 2 y = 3 z = - 5 . From here we get that the intersection of the given lines gives a point with coordinates (- 2 , 3 , - 5) .

Answer: (- 2 , 3 , - 5) .

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

In two-dimensional space, two lines intersect only at one point, given 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 advanced skills, you can find the intersection points of parabolas and other quadratic curves.

Steps

Point of intersection of two lines

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

• If the equations of the lines are not given to you, on the basis of information known to you.
• Example. Given straight lines described by the 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 intersection point of both lines, that is, the point whose (x, y) coordinates satisfy both equations. Since the variable "y" is on the left side of each equation, the expressions on the right side of each equation can be equated. Write down 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 this variable on the left side of the equation by doing the appropriate math on both sides of the equation. You should end up with an equation like x = __ (if you can't do that, see this section).

   • Example. x + 3 = 12 − 2 x (\displaystyle x+3=12-2x)
   • Add 2x (\displaystyle 2x) to each side of the equation:
   • 3x + 3 = 12 (\displaystyle 3x+3=12)
   • Subtract 3 from each side of the equation:
   • 3x=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 "x" in the equation (any) 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" in another equation of a straight 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 "y" value, so there are no errors in your calculations.

5. Write down the coordinates (x, y). By calculating the values ​​\u200b\u200bof "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 the form (x, y).

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

6. Computations 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)). Graphs of quadratic functions are curves that may not intersect or intersect at one or two points. In this section, we will tell you how to find the point or points of intersection of quadratic curves.

   2. Rewrite each equation by isolating the variable "y" on the left side of the equation. 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 the same as the steps 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 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 the terms of the resulting equation to its left side, and write 0 on the right side. To do this, perform basic mathematical operations. 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 transferring 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 factoring the equation, you get two binomials, which, when multiplied, give the original equation. In our example, the first member x 2 (\displaystyle x^(2)) can be decomposed into x*x. Make the following entry: (x)(x) = 0
      • In our example, the intercept -6 can be factored as follows: − 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 intercept factors (in our example -6) until you get 1. In our example, the correct pair of intercept factors are -2 and 3 ( − 2 ∗ 3 = − 6 (\displaystyle -2*3=-6)), because − 2 + 3 = 1 (\displaystyle -2+3=1).
      • Fill in the gaps 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 can forget about the second intersection point. Here's how to find the "x" coordinates of two intersection points:

      • Example (factoring). If in the equation (x − 2) (x + 3) = 0 (\displaystyle (x-2)(x+3)=0) one of the expressions in brackets will be equal to 0, then the whole 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 (use formula or complete 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 the 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 x values.

   7. Graphs intersect at one point or do not intersect at all. Such situations occur when the following conditions are met:

      • If the graphs intersect at one point, then the quadratic equation is decomposed into equal 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 does not factorize, and the square root of a negative number appears in the formula (for example, − 2 (\displaystyle (\sqrt(-2)))). In this case, write in the answer that there is no solution.

Lesson from the series "Geometric Algorithms"

Hello dear reader!

We continue to get acquainted with geometric algorithms. In the last lesson, we found the equation of a straight line in the coordinates of two points. We have 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 any). 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 the real type, it is better to arrange the comparison operations with special functions.

The reason is known: there is no order relation on the Real type in the Pascal programming system, 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. Equations of two straight lines are given: and . Find their point of intersection.

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

We introduce the notation: , , . Here D is the determinant of the system, and are the determinants obtained by 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 by the following formulas: , , which are called Cramer's formulas. Let me remind you how the second order determinant is calculated. The determinant distinguishes between two diagonals: the main and secondary. The main diagonal consists of elements taken in the direction from the upper left corner of the determinant to the lower right corner. Side diagonal - from the upper right to the lower left. The second order determinant is equal to the product of the elements of the main diagonal minus the product of the elements of the secondary diagonal.

The code uses the RealEq() function to check for equality. Calculations over real numbers are made with accuracy up to _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 the lines, find the coordinates of their intersection point.

When solving some geometric problems using the coordinate method, it is necessary to find the coordinates of the point of intersection of lines. Most often, one has to look for the coordinates of the point of intersection of two lines on the plane, but sometimes it becomes necessary 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 lines defined on the plane by general equations, it is necessary to solve a system composed of equations of given lines.

Let's consider an example solution.

Example.

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

Solution.

We are given two general equations of lines, we will compose a system from them: . The solutions of the resulting system of equations are easily found if its first equation is solved with respect to the variable x and this expression is substituted into the second equation:

The found solution of 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 lines, defined by general equations on the plane, is reduced to solving a system of two linear equations with two unknown variables. But what if the straight lines on the plane are given not by general equations, but by equations of a different type (see the types of the equation of a straight line on the plane)? In these cases, you can first bring 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 point of intersection of the given lines, we bring their equations to a general form. Transition from parametric equations to a straight line to the general equation of this straight line is as follows:

Now we will carry out the necessary actions with the canonical equation of the line:

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

Answer:

M 0 (-5, 1)

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

Example.

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

Solution.

Substitute in the equation of the direct expression:

Solving 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 the 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 in the plane, it is useful to make sure that the given lines really 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 intersection point of such lines.

You can, of course, do without such a check, and immediately compose a system of equations of the form and solve it. If the 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 has no solutions, then we can conclude that the original lines are parallel (since there is no such pair of real numbers x and y that would simultaneously satisfy both equations of given lines). From the presence of an infinite set of solutions to the system of equations, it follows that the original lines have infinitely many points in common, that is, they coincide.

Let's look at examples that fit these situations.

Example.

Find out if 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 composed 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 and define the same line, and we cannot talk about finding the coordinates of the point of intersection of these lines.

Answer:

The equations and determine 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 admits that the lines may not intersect. Let's compose a system of these equations. Applicable for its solution, since it allows you to establish the compatibility or inconsistency of the system of equations, and if it is compatible, find a solution:

The last equation of the system after the direct course 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.

The second solution.

Let's find out if the given lines intersect.

- normal line vector , and the vector is a normal vector of the line . 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 coincide. Thus, we cannot find the coordinates of the point of intersection of the original lines.

Answer:

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

Example.

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

Solution.

We compose a system of equations that are general equations of given lines: . The determinant of the main matrix of this system of equations is different from zero , so 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 consider examples.

Example.

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

Solution.

We 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 us find the solution of the written system of equations.

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

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