Find the sum of coordinates of the point of intersection of lines online. The simplest problems with a straight line on a plane

If two lines are not parallel, then they will inevitably intersect at one point. Discover coordinates points intersection of 2 lines is allowed both graphically and arithmetic, depending on what data the task provides.

You will need

• – two straight lines in the drawing;
• – equations of 2 straight lines.

Instructions

1. If the lines are already drawn on the graph, find the solution graphically. To do this, continue both or one of the lines so that they intersect. After this, mark the intersection point and lower a perpendicular from it to the x-axis (as usual, oh).

2. Using the scale marks marked on the axis, find the x value for that point. If it is in the positive direction of the axis (to the right of the zero mark), then its value will be correct; otherwise, it will be negative.

3. Correctly also find the ordinate of the intersection point. If the projection of a point is located above the zero mark, it is correct; if below, it is negative. Write down the coordinates of the point in the form (x, y) - this is the solution to the problem.

4. If the lines are given in the form of the formulas y=khx+b, you can also solve the problem graphically: draw the lines on a coordinate grid and find the solution using the method described above.

5. Try to discover the solution to the problem using these formulas. To do this, create a system from these equations and solve it. If the equations are given in the form y=khx+b, simply equate both sides with x and discover x. Then plug the value of x into one of the equations and find y.

6. You can find a solution using Cramer's method. In this case, reduce the equations to the form A1x+B1y+C1=0 and A2x+B2y+C2=0. According to Cramer's formula, x=-(C1B2-C2B1)/(A1B2-A2B1), and y=-(A1C2-A2C1)/(A1B2-A2B1). Please note that if the denominator is zero, then the lines are parallel or coincide and, accordingly, do not intersect.

7. If you are given lines in space in canonical form, before you start searching for a solution, check whether the lines are parallel. To do this, evaluate the exponents before t if they are proportional, say, x=-1+3t, y=7+2t, z=2+t and x=-1+6t, y=-1+4t, z=-5 +2t, then the lines are parallel. In addition, lines can intersect, in which case the system will not have a solution.

8. If you find out that lines intersect, find the point of their intersection. First, equate variables from different lines, conditionally replacing t with u for the first line and with v for the 2nd line. Say, if you are given lines x=t-1, y=2t+1, z=t+2 and x=t+1, y=t+1, z=2t+8 you will get expressions like u-1=v +1, 2u+1=v+1, u+2=2v+8.

9. Express u from one equation, substitute it into another and find v (in this problem u=-2,v=-4). Now, in order to find the intersection point, substitute the obtained values ​​instead of t (it doesn’t matter in the first or second equation) and get the coordinates of the point x=-3, y=-3, z=0.

To consider 2 intersecting direct It is enough to consider them in a plane, since two intersecting lines lie in the same plane. Knowing the equations of these direct, it is possible to detect the coordinate of their point intersections .

You will need

• equations of lines

Instructions

1. In Cartesian coordinates, the general equation of a line looks like this: Ax+By+C = 0. Let two lines intersect. The equation of the first line is Ax+By+C = 0, the 2nd line is Dx+Ey+F = 0. All indicators (A, B, C, D, E, F) must be specified. In order to detect a point intersections these direct it is necessary to solve the system of these 2 linear equations.

2. To solve, it is convenient to multiply the first equation by E, and the second by B. As a result, the equations will look like: AEx+BEy+CE = 0, DBx+EBy+FB = 0. After subtracting the second equation from the first, you get: (AE- DB)x = FB-CE. Hence, x = (FB-CE)/(AE-DB). By analogy, the first equation of the initial system can be multiplied by D, the second by A, after which the second can be subtracted from the first. As a result, y = (CD-FA)/(AE-DB). The resulting x and y values ​​will be the coordinates of the point intersections direct .

3. Equations direct can also be written through the angular index k, equal to the tangent of the angle of inclination of the straight line. In this case, the equation of the line has the form y = kx+b. Let now the equation of the first line be y = k1*x+b1, and the equation of the 2nd line be y = k2*x+b2.

4. If we equate the right sides of these 2 equations, we get: k1*x+b1 = k2*x+b2. From there it is easy to obtain that x = (b1-b2)/(k2-k1). After substituting this x value into any of the equations, you get: y = (k2*b1-k1*b2)/(k2-k1). The x and y values ​​will specify the coordinates of the point intersections direct.If two lines are parallel or coincident, then they do not have universal points or have an immensely large number of universal points, respectively. In these cases k1 = k2, denominators for the coordinates of the points intersections will vanish, therefore, the system will not have a classical solution. The system can have only one classical solution, which is unconditional, because two divergent and non-parallel lines can have only one point intersections .

Video on the topic

1. To find the coordinates of the intersection point of the graphs of functions, you need to equate both functions to each other, move all terms containing $ x $ to the left side, and the rest to the right side, and find the roots of the resulting equation.
2. The second method is to create a system of equations and solve it by substituting one function into another
3. The third method involves graphically constructing functions and visually determining the intersection point.

The case of two linear functions

Consider two linear functions $ f(x) = k_1 x+m_1 $ and $ g(x) = k_2 x + m_2 $. These functions are called direct. It is quite easy to construct them; you need to take any two values ​​$ x_1 $ and $ x_2 $ and find $ f(x_1) $ and $ (x_2) $. Then repeat the same with the function $ g(x) $. Next, visually find the coordinate of the intersection point of the function graphs.

You should know that linear functions have only one intersection point and only when $ k_1 \neq k_2 $. Otherwise, in the case of $ k_1=k_2 $ the functions are parallel to each other, since $ k $ is the slope coefficient. If $ k_1 \neq k_2 $ but $ m_1=m_2 $, then the intersection point will be $ M(0;m) $. It is advisable to remember this rule to quickly solve problems.

The case of two nonlinear functions

Oh-oh-oh-oh-oh... well, it’s tough, as if he was reading out a sentence to himself =) However, relaxation will help later, especially since today I bought the appropriate accessories. Therefore, let's proceed to the first section, I hope that by the end of the article I will maintain a cheerful mood.

The relative position of two straight lines

This is the case when the audience sings along in chorus. Two straight lines can:

1) match;

2) be parallel: ;

3) or intersect at a single point: .

Help for dummies : please remember the math sign intersections, it will occur very often. The notation means that the line intersects with the line at point .

How to determine the relative position of two lines?

Let's start with the first case:

Two lines coincide if and only if their corresponding coefficients are proportional, that is, there is a number “lambda” such that the equalities are satisfied

Let's consider the straight lines and create three equations from the corresponding coefficients: . From each equation it follows that, therefore, these lines coincide.

Indeed, if all the coefficients of the equation multiply by –1 (change signs), and all coefficients of the equation cut by 2, you get the same equation: .

The second case, when the lines are parallel:

Two lines are parallel if and only if their coefficients of the variables are proportional: , But.

As an example, consider two straight lines. We check the proportionality of the corresponding coefficients for the variables:

However, it is quite obvious that.

And the third case, when the lines intersect:

Two lines intersect if and only if their coefficients of the variables are NOT proportional, that is, there is NO such value of “lambda” that the equalities are satisfied

So, for straight lines we will create a system:

From the first equation it follows that , and from the second equation: , which means the system is inconsistent (no solutions). Thus, the coefficients of the variables are not proportional.

Conclusion: lines intersect

In practical problems, you can use the solution scheme just discussed. By the way, it is very reminiscent of the algorithm for checking vectors for collinearity, which we looked at in class The concept of linear (in)dependence of vectors. Basis of vectors . But there is a more civilized packaging:

Example 1

Find out the relative position of the lines:

Solution based on the study of directing vectors of straight lines:

a) From the equations we find the direction vectors of the lines: .

, which means that the vectors are not collinear and the lines intersect.

Just in case, I’ll put a stone with signs at the crossroads:

The rest jump over the stone and follow further, straight to Kashchei the Immortal =)

b) Find the direction vectors of the lines:

The lines have the same direction vector, which means they are either parallel or coincident. There is no need to count the determinant here.

It is obvious that the coefficients of the unknowns are proportional, and .

Let's find out whether the equality is true:

Thus,

c) Find the direction vectors of the lines:

Let's calculate the determinant made up of the coordinates of these vectors:
, therefore, the direction vectors are collinear. The lines are either parallel or coincident.

The proportionality coefficient “lambda” is easy to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: .

Now let's find out whether the equality is true. Both free terms are zero, so:

The resulting value satisfies this equation (any number in general satisfies it).

Thus, the lines coincide.

Answer:

Very soon you will learn (or even have already learned) to solve the problem discussed verbally literally in a matter of seconds. In this regard, I don’t see any point in offering anything for an independent solution; it’s better to lay another important brick in the geometric foundation:

How to construct a line parallel to a given one?

For ignorance of this simplest task, the Nightingale the Robber severely punishes.

Example 2

The straight line is given by the equation. Write an equation for a parallel line that passes through the point.

Solution: Let's denote the unknown line by the letter . What does the condition say about her? The straight line passes through the point. And if the lines are parallel, then it is obvious that the direction vector of the straight line “tse” is also suitable for constructing the straight line “de”.

We take the direction vector out of the equation:

Answer:

The example geometry looks simple:

Analytical testing consists of the following steps:

1) We check that the lines have the same direction vector (if the equation of the line is not simplified properly, then the vectors will be collinear).

2) Check whether the point satisfies the resulting equation.

In most cases, analytical testing can be easily performed orally. Look at the two equations, and many of you will quickly determine the parallelism of the lines without any drawing.

Examples for independent solutions today will be creative. Because you will still have to compete with Baba Yaga, and she, you know, is a lover of all sorts of riddles.

Example 3

Write an equation for a line passing through a point parallel to the line if

There is a rational and not so rational way to solve it. The shortest way is at the end of the lesson.

We worked a little with parallel lines and will return to them later. The case of coinciding lines is of little interest, so let’s consider a problem that is very familiar to you from the school curriculum:

How to find the point of intersection of two lines?

If straight intersect at 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.

Example 4

Find the point of intersection of lines

Solution: There are two ways to solve - graphical and analytical.

The graphical method is to simply draw the given lines and find out the intersection point directly from the drawing:

Here is our point: . To check, you should substitute its coordinates into each equation of the line, they should fit both there and there. In other words, the coordinates of a point are a solution to the system. Essentially, we looked at a graphical solution systems of linear equations with two equations, two unknowns.

The graphical method is, of course, not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide this way, the point is that it will take time to create a correct and ACCURATE drawing. In addition, some straight lines are not so easy to construct, and the point of intersection itself may be located somewhere in the thirtieth kingdom outside the notebook sheet.

Therefore, it is more expedient to search for the intersection point using an analytical method. Let's solve the system:

To solve the system, the method of term-by-term addition of equations was used. To develop relevant skills, take a lesson How to solve a system of equations?

Answer:

The check is trivial - the coordinates of the intersection point must satisfy each equation of the system.

Example 5

Find the point of intersection of the lines if they intersect.

This is an example for you to solve on your own. It is convenient to split the task into several stages. Analysis of the condition suggests that it is necessary:
1) Write down the equation of the straight line.
2) Write down the equation of the straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.

The development of an action algorithm is typical for many geometric problems, and I will repeatedly focus on this.

Full solution and answer at the end of the lesson:

Not even a pair of shoes were worn out before we got to the second section of the lesson:

Perpendicular lines. Distance from a point to a line.
Angle between straight lines

Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to this one, and now the hut on chicken legs will turn 90 degrees:

How to construct a line perpendicular to a given one?

Example 6

The straight line is given by the equation. Write an equation perpendicular to the line passing through the point.

Solution: By condition it is known that . It would be nice to find the directing vector of the line. Since the lines are perpendicular, the trick is simple:

From the equation we “remove” the normal vector: , which will be the directing vector of the straight line.

Let's compose the equation of a straight line using a point and a direction vector:

Answer:

Let's expand the geometric sketch:

Hmmm... Orange sky, orange sea, orange camel.

Analytical verification of the solution:

1) We take out the direction vectors from the equations and with the help scalar product of vectors we come to the conclusion that the lines are indeed perpendicular: .

By the way, you can use normal vectors, it's even easier.

2) Check whether the point satisfies the resulting equation .

The test, again, is easy to perform orally.

Example 7

Find the point of intersection of perpendicular lines if the equation is known and period.

This is an example for you to solve on your own. There are several actions in the problem, so it is convenient to formulate the solution point by point.

Our exciting journey continues:

Distance from point to line

In front of us is a straight strip of the river 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 line expressed by the formula

Example 8

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:

Let's make the drawing:

The found distance from the point to the line is exactly the length of the red segment. If you draw up a drawing on checkered paper on a scale of 1 unit. = 1 cm (2 cells), then the distance can be measured with an ordinary ruler.

Let's consider another task based on the same drawing:

The task is to find the coordinates of a point that is symmetrical to the point relative to the straight line . I suggest performing the steps yourself, but I will outline the solution algorithm with intermediate results:

1) Find a line that is perpendicular to the line.

2) Find the point of intersection of the lines: .

Both actions are discussed in detail in this lesson.

3) The point is the midpoint of the segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the midpoint of a segment we find .

It would be a good idea to check that the distance is also 2.2 units.

Difficulties may arise in calculations here, but a microcalculator is a great help in the tower, allowing you to calculate ordinary fractions. I have advised you many times and will recommend you again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for you to decide on your own. I’ll give you a little hint: there are infinitely many ways to solve this. Debriefing at the end of the lesson, but it’s better to try to guess for yourself, I think your ingenuity was well developed.

Angle between two straight lines

Every corner is a jamb:


In geometry, the angle between two straight lines is taken to be the SMALLER angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered the angle between intersecting lines. And his “green” neighbor or oppositely oriented"raspberry" corner.

If the lines are perpendicular, then any of the 4 angles can be taken as the angle between them.

How are the angles different? Orientation. Firstly, the direction in which the angle is “scrolled” is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example if .

Why did I tell you this? It seems that we can get by with the usual concept of an angle. The fact is that the formulas by which we will find angles can easily result in a negative result, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing, for a negative angle, be sure to indicate its orientation with an arrow (clockwise).

How to find the angle between two straight lines? There are two working formulas:

Example 10

Find the angle between lines

Solution And Method one

Let's consider two straight lines defined by equations in general form:

If straight not perpendicular, That oriented The angle between them can be calculated using the formula:

Let us pay close attention to the denominator - this is exactly scalar product directing vectors of straight lines:

If , then the denominator of the formula becomes zero, and the vectors will be orthogonal and the lines will be perpendicular. That is why a reservation was made about the non-perpendicularity of straight lines in the formulation.

Based on the above, it is convenient to formalize the solution in two steps:

1) 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:

Using the inverse function, it is easy to find the angle itself. In this case, we use the oddness of the arctangent (see. Graphs and properties of elementary functions ):

Answer:

In your answer, we indicate the exact value, as well as an approximate value (preferably in both degrees and radians), calculated using a calculator.

Well, minus, minus, no big deal. Here is a geometric illustration:

It is not surprising that the angle turned out to be of a negative orientation, because in the problem statement the first number is a straight line and the “unscrewing” of the angle began precisely with it.

If you really want to get a positive angle, you need to swap the lines, that is, take the coefficients from the second equation , and take the coefficients from the first equation. In short, you need to start with a direct .

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.