Equation of a straight line on a plane.
The direction vector is straight. Normal vector

A straight line on a plane is one of the simplest geometric shapes, familiar to you since the elementary grades, and today we will learn how to deal with it using the methods of analytical geometry. To master the material, it is necessary to be able to build a straight line; know which equation defines a straight line, in particular, a straight line passing through the origin and straight lines parallel to the coordinate axes. This information can be found in the manual. Graphs and properties of elementary functions, I created it for matan, but the section about linear function turned out to be very successful and detailed. Therefore, dear teapots, first warm up there. In addition, you need to have basic knowledge of vectors otherwise the understanding of the material will be incomplete.

In this lesson, we will look at ways in which you can write the equation of a straight line in a plane. I recommend not neglecting practical examples (even if it seems very simple), as I will supply them with elementary and important facts, technical methods that will be required in the future, including in other sections of higher mathematics.

• How to write the equation of a straight line with a slope?
• How ?
• How to find the direction vector by the general equation of a straight line?
• How to write an equation of a straight line given a point and a normal vector?

and we start:

Line Equation with Slope

The well-known "school" form of the equation of a straight line is called equation of a straight line with a slope. For example, if a straight line is given by the equation, then its slope: . Consider the geometric meaning of this coefficient and how its value affects the location of the line:

In the course of geometry it is proved that the slope of the straight line is tangent of an angle between positive axis directionand given line: , and the corner is “unscrewed” counterclockwise.

In order not to clutter up the drawing, I drew angles for only two straight lines. Consider the "red" straight line and its slope. According to the above: (angle "alpha" is indicated by a green arc). For the "blue" straight line with the slope, equality is true (the angle "beta" is indicated by the brown arc). And if the tangent of the angle is known, then if necessary it is easy to find and the corner using the inverse function - arc tangent. As they say, a trigonometric table or a calculator in hand. In this way, the slope characterizes the degree of inclination of the straight line to the x-axis.

In this case, the following cases are possible:

1) If the slope is negative: , then the line, roughly speaking, goes from top to bottom. Examples are "blue" and "crimson" straight lines in the drawing.

2) If the slope is positive: , then the line goes from bottom to top. Examples are "black" and "red" straight lines in the drawing.

3) If the slope is equal to zero: , then the equation takes the form , and the corresponding line is parallel to the axis. An example is the "yellow" line.

4) For a family of straight lines parallel to the axis (there is no example in the drawing, except for the axis itself), the slope does not exist (tangent of 90 degrees not defined).

The greater the slope modulo, the steeper the line graph goes.

For example, consider two straight lines. Here , so the straight line has a steeper slope. I remind you that the module allows you to ignore the sign, we are only interested in absolute values angular coefficients.

In turn, a straight line is steeper than straight lines. .

Vice versa: the smaller the slope modulo, the straight line is flatter.

For straight lines the inequality is true, thus, the straight line is more than a canopy. Children's slide, so as not to plant bruises and bumps.

Why is this needed?

Prolong your torment Knowing the above facts allows you to immediately see your mistakes, in particular, errors when plotting graphs - if the drawing turned out “clearly something is wrong”. It is desirable that you straightaway it was clear that, for example, a straight line is very steep and goes from bottom to top, and a straight line is very flat, close to the axis and goes from top to bottom.

In geometric problems, several straight lines often appear, so it is convenient to denote them somehow.

Notation: straight lines are indicated by small Latin letters: . A popular option is the designation of the same letter with natural subscripts. For example, the five lines that we have just considered can be denoted by .

Since any straight line is uniquely determined by two points, it can be denoted by these points: etc. The notation quite obviously implies that the points belong to the line.

Time to loosen up a bit:

How to write the equation of a straight line with a slope?

If a point is known that belongs to a certain line, and the slope of this line, then the equation of this line is expressed by the formula:

Example 1

Compose the equation of a straight line with a slope if it is known that the point belongs to this straight line.

Solution: We will compose the equation of a straight line according to the formula . AT this case:

Answer:

Examination performed elementarily. First, we look at the resulting equation and make sure that our slope is in its place. Second, the coordinates of the point must satisfy the given equation. Let's plug them into the equation:

The correct equality is obtained, which means that the point satisfies the resulting equation.

Conclusion: Equation found correctly.

A trickier example for independent solution:

Example 2

Write the equation of a straight line if it is known that its angle of inclination to the positive direction of the axis is , and the point belongs to this straight line.

If you're having trouble, re-read theoretical material. More precisely, more practical, I miss many proofs.

The last bell rang, the graduation ball died down, and behind the gates of our native school, in fact, analytical geometry is waiting for us. Jokes are over... Maybe it's just getting started =)

Nostalgically we wave the handle to the familiar and get acquainted with the general equation of a straight line. Since in analytic geometry it is precisely this that is in use:

The general equation of a straight line has the form: , where are some numbers. At the same time, the coefficients simultaneously are not equal to zero, since the equation loses its meaning.

Let's dress in a suit and tie an equation with a slope. First, we move all the terms to the left side:

The term with "x" must be put in first place:

In principle, the equation already has the form , but according to the rules of mathematical etiquette, the coefficient of the first term (in this case ) must be positive. Changing signs:

Remember this technical feature! We make the first coefficient (most often ) positive!

In analytic geometry, the equation of a straight line will almost always be given in general form. Well, if necessary, it is easy to bring it to a “school” form with a slope (with the exception of straight lines parallel to the y-axis).

Let's ask ourselves what enough know to build a straight line? Two points. But about this childhood case later, now sticks with arrows rule. Each straight line has a well-defined slope, to which it is easy to "adapt" vector.

A vector that is parallel to a line is called the direction vector of that line.. Obviously, any straight line has infinitely many direction vectors, and all of them will be collinear (co-directed or not - it does not matter).

I will denote the direction vector as follows: .

But one vector is not enough to build a straight line, the vector is free and is not attached to any point of the plane. Therefore, it is additionally necessary to know some point that belongs to the line.

How to write an equation of a straight line given a point and a direction vector?

If some point belonging to the line is known, and the direction vector of this line , then the equation of this straight line can be composed by the formula:

Sometimes it is called canonical equation of the line .

What to do when one of the coordinates is zero, we will look into practical examples below. By the way, note - both at once coordinates cannot be zero, since the zero vector does not specify a specific direction.

Example 3

Write an equation of a straight line given a point and a direction vector

Solution: We will compose the equation of a straight line according to the formula. In this case:

Using the properties of proportion, we get rid of fractions:

And we bring the equation to a general form:

Answer:

Drawing in such examples, as a rule, is not necessary, but for the sake of understanding:

In the drawing, we see the starting point, the original direction vector (it can be postponed from any point on the plane) and the constructed line. By the way, in many cases, the construction of a straight line is most conveniently carried out using the slope equation. Our equation is easy to convert to the form and without any problems pick up one more point to build a straight line.

As noted at the beginning of the section, a line has infinitely many direction vectors, and they are all collinear. For example, I drew three such vectors: . Whichever direction vector we choose, the result will always be the same straight line equation.

Let's compose the equation of a straight line by a point and a directing vector:

Breaking down the proportion:

Divide both sides by -2 and get the familiar equation:

Those who wish can similarly test vectors or any other collinear vector.

Now let's solve the inverse problem:

How to find the direction vector by the general equation of a straight line?

Very simple:

If the line is given by the general equation, then the vector is the direction vector of this line.

Examples of finding direction vectors of straight lines:

The statement allows us to find only one direction vector from an infinite set, but we don’t need more. Although in some cases it is advisable to reduce the coordinates of the direction vectors:

So, the equation specifies a straight line that is parallel to the axis and the coordinates of the resulting steering vector are conveniently divided by -2, getting exactly the basis vector as the steering vector. Logically.

Similarly, the equation defines a straight line parallel to the axis, and dividing the coordinates of the vector by 5, we get the ort as the direction vector.

Now let's execute check example 3. The example went up, so I remind you that in it we made up the equation of a straight line using a point and a direction vector

Firstly, according to the equation of a straight line, we restore its directing vector: - everything is fine, we got the original vector (in some cases, it can turn out to be collinear to the original vector, and this is usually easy to see by the proportionality of the corresponding coordinates).

Secondly, the coordinates of the point must satisfy the equation . We substitute them into the equation:

The correct equality has been obtained, which we are very pleased with.

Conclusion: Job completed correctly.

Example 4

Write an equation of a straight line given a point and a direction vector

This is a do-it-yourself example. Solution and answer at the end of the lesson. It is highly desirable to make a check according to the algorithm just considered. Try to always (if possible) check on a draft. It is foolish to make mistakes where they can be 100% avoided.

In the event that one of the coordinates of the direction vector is zero, it is very simple to do:

Example 5

Solution: The formula is invalid because the denominator on the right side is zero. There is an exit! Using the properties of proportion, we rewrite the formula in the form , and the rest rolled along a deep rut:

Answer:

Examination:

1) Restore the direction vector of the straight line:
– the resulting vector is collinear to the original direction vector.

2) Substitute the coordinates of the point in the equation:

The correct equality is obtained

Conclusion: job completed correctly

The question arises, why bother with the formula if there is a universal version that will work anyway? There are two reasons. First, the fractional formula much better to remember. And secondly, the disadvantage of the universal formula is that markedly increased risk of confusion when substituting coordinates.

Example 6

Compose the equation of a straight line given a point and a direction vector.

This is a do-it-yourself example.

Let's return to the ubiquitous two points:

How to write the equation of a straight line given two points?

If two points are known, then the equation of a straight line passing through these points can be compiled using the formula:

In fact, this is a kind of formula, and here's why: if two points are known, then the vector will be the direction vector of this line. On the lesson Vectors for dummies we considered the simplest task– how to find the coordinates of a vector from two points. According to this problem, the coordinates of the direction vector:

Note : Points can be "reversed roles" and use the formula . Such a decision would be equal.

Example 7

Write the equation of a straight line from two points .

Solution: Use the formula:

We comb the denominators:

And shuffle the deck:

It is now convenient to get rid of fractional numbers. In this case, you need to multiply both parts by 6:

Open the brackets and bring the equation to mind:

Answer:

Examination is obvious - the coordinates of the initial points must satisfy the resulting equation:

1) Substitute the coordinates of the point:

True equality.

2) Substitute the coordinates of the point:

True equality.

Conclusion: the equation of the straight line is correct.

If a at least one of points does not satisfy the equation, look for an error.

It is worth noting that the graphical verification in this case is difficult, because to build a line and see if the points belong to it , not so easy.

I will note a couple of technical points of the solution. Perhaps in this problem it is more advantageous to use the mirror formula and, for the same points make an equation:

There are fewer fractions. If you want, you can complete the solution to the end, the result should be the same equation.

The second point is to look at the final answer and see if it can be further simplified? For example, if an equation is obtained, then it is advisable to reduce it by two: - the equation will set the same straight line. However, this is already a topic of conversation about mutual arrangement of straight lines.

Having received an answer in Example 7, just in case, I checked if ALL coefficients of the equation are divisible by 2, 3 or 7. Although, most often such reductions are made during the solution.

Example 8

Write the equation of a straight line passing through the points .

This is an example for an independent solution, which will just allow you to better understand and work out the calculation technique.

Similar to the previous paragraph: if in the formula one of the denominators (direction vector coordinate) vanishes, then we rewrite it as . And again, notice how awkward and confused she began to look. I don’t see much point in giving practical examples, since we have already actually solved such a problem (see Nos. 5, 6).

Straight line normal vector (normal vector)

What is normal? In simple words, the normal is the perpendicular. That is, the normal vector of a line is perpendicular to the given line. It is obvious that any straight line has an infinite number of them (as well as directing vectors), and all the normal vectors of the straight line will be collinear (codirectional or not - it does not matter).

Dealing with them will be even easier than with direction vectors:

If a straight line is given by a general equation in a rectangular coordinate system, then the vector is the normal vector of this straight line.

If the coordinates of the direction vector have to be carefully “pulled out” of the equation, then the coordinates of the normal vector can be simply “removed”.

The normal vector is always orthogonal to the direction vector of the line. We will verify the orthogonality of these vectors using dot product:

I will give examples with the same equations as for the direction vector:

Is it possible to write an equation of a straight line, knowing one point and a normal vector? It feels like it's possible. If the normal vector is known, then the direction of the straightest line is also uniquely determined - this is a “rigid structure” with an angle of 90 degrees.

How to write an equation of a straight line given a point and a normal vector?

If some point belonging to the line and the normal vector of this line are known, then the equation of this line is expressed by the formula:

Here everything went without fractions and other surprises. Such is our normal vector. Love it. And respect =)

Example 9

Compose the equation of a straight line given a point and a normal vector. Find the direction vector of the straight line.

Solution: Use the formula:

The general equation of the straight line is obtained, let's check:

1) "Remove" the coordinates of the normal vector from the equation: - yes, indeed, the original vector is obtained from the condition (or the vector should be collinear to the original vector).

2) Check if the point satisfies the equation:

True equality.

After we are convinced that the equation is correct, we will complete the second, easier part of the task. We pull out the direction vector of the straight line:

Answer:

In the drawing, the situation is as follows:

For the purposes of training, a similar task for an independent solution:

Example 10

Compose the equation of a straight line given a point and a normal vector. Find the direction vector of the straight line.

The final section of the lesson will be devoted to less common, but also important types of equations of a straight line in a plane

Equation of a straight line in segments.
Equation of a straight line in parametric form

The equation of a straight line in segments has the form , where are nonzero constants. Some types of equations cannot be represented in this form, for example, direct proportionality (since the free term is zero and there is no way to get one on the right side).

This is, figuratively speaking, a "technical" type of equation. The usual task is to represent the general equation of a straight line as an equation of a straight line in segments. Why is it convenient? The equation of a straight line in segments allows you to quickly find the points of intersection of a straight line with coordinate axes, which is very important in some problems of higher mathematics.

Find the point of intersection of the line with the axis. We reset the “y”, and the equation takes the form . The desired point is obtained automatically: .

Same with axis is the point where the line intersects the y-axis.

The task is to construct a straight line passing through it using the given coordinates of the end of the segment.

We assume that the segment is non-degenerate, i.e. has a length greater than zero (otherwise, of course, infinitely many different lines pass through it).

2D case

Let a segment be given, i.e. the coordinates of its ends , , , are known.

Required to build equation of a straight line in a plane passing through this segment, i.e. find the coefficients , , in the straight line equation:

Note that the desired triples passing through the given segment, infinitely many: you can multiply all three coefficients by an arbitrary non-zero number and get the same straight line. Therefore, our task is to find one of these triples.

It is easy to verify (by substituting these expressions and the coordinates of the points into the equation of a straight line) that the following set of coefficients is suitable:

integer case

An important advantage of this method of constructing a straight line is that if the coordinates of the ends were integer, then the resulting coefficients will also be integer. In some cases, this allows you to perform geometric operations without resorting to real numbers at all.

However, there is also a small drawback: for the same straight line, different triples of coefficients can be obtained. To avoid this, but not to get away from integer coefficients, you can apply the following trick, often called rationing. Find the greatest common divisor of the numbers , , , divide all three coefficients by it, and then normalize the sign: if or , then multiply all three coefficients by . As a result, we will come to the conclusion that identical triples of coefficients will be obtained for identical lines, which will make it easy to check lines for equality.

real-valued case

When working with real numbers, you should always be aware of errors.

The coefficients and are obtained for us on the order of the initial coordinates, the coefficient is already on the order of a square from them. This can already be quite large numbers, and, for example, when crossing straight lines, they will become even larger, which can lead to large rounding errors even at initial coordinates of order .

Therefore, when working with real numbers, it is desirable to perform the so-called normalization direct: namely, make the coefficients such that . To do this, you need to calculate the number:

and divide all three coefficients , , by it.

Thus, the order of the coefficients and will no longer depend on the order of the input coordinates, and the coefficient will be of the same order as the input coordinates. In practice, this leads to a significant improvement in the accuracy of calculations.

Finally, let's mention comparison straight lines - after all, after such a normalization, for the same straight line, only two triples of coefficients can be obtained: up to multiplication by . Accordingly, if we perform an additional normalization taking into account the sign (if or , then multiply by ), then the resulting coefficients will be unique.

And we will analyze in detail a special form of the equation of a straight line -. Let's start with the form of the equation of a straight line in segments and give an example. After that, we will focus on the construction of a straight line, which is given by the equation of a straight line in segments. In conclusion, we show how the transition from the complete general equation of a straight line to the equation of a straight line in segments is carried out.

Page navigation.

Equation of a straight line in segments - description and example.

Let Oxy be fixed on the plane.

Equation of a straight line in segments on a plane in a rectangular coordinate system Oxy has the form , where a and b are some non-zero real numbers.

The equation of a straight line in segments did not accidentally get such a name - the absolute values ​​​​of the numbers a and b are equal to the lengths of the segments that the straight line cuts off on the coordinate axes Ox and Oy, counting from the origin.

Let's explain this point. We know that the coordinates of any point on a straight line satisfy the equation of this straight line. Then it is clearly seen that the straight line given by the equation of the straight line in the segments passes through the points and , since and . And the points and are just located on the coordinate axes Ox and Oy, respectively, and are removed from the origin by a and b units. The signs of the numbers a and b indicate the direction in which the segments should be laid. The "+" sign means that the segment is plotted in the positive direction of the coordinate axis, the "-" sign means the opposite.

Let's draw a schematic drawing explaining all of the above. It shows the location of lines relative to a fixed rectangular coordinate system Oxy, depending on the values ​​of the numbers a and b in the equation of a line in segments.

Now it has become clear that the equation of a straight line in segments makes it easy to construct this straight line in a rectangular coordinate system Oxy. To build a straight line, which is given by the equation of a straight line in segments of the form , one should mark points and in a rectangular coordinate system on the plane, and then connect them with a straight line using a ruler.

Let's take an example.

Example.

Construct a straight line given by the equation of a straight line in segments of the form .

Solution.

According to the given equation of the straight line in the segments, it can be seen that the straight line passes through the points . Mark them and connect with a straight line.

Reduction of the general equation of a straight line to the equation of a straight line in segments.

When solving some problems related to a straight line in a plane, it is convenient to work with the equation of a straight line in segments. However, there are other types of equations that define a straight line on a plane. Therefore, it is necessary to carry out the transition from the given equation of a straight line to the equation of this straight line in segments.

In this paragraph, we will show how to obtain the equation of a line in segments, if the complete general equation of a line is given.

Let us know the complete general equation of a straight line in the plane . Since A, B and C are not equal to zero, then you can transfer the number C to the right side of the equality, divide both parts of the resulting equality by -C, and send the coefficients at x and y to the denominators:
.

(In the last passage we used the equality ).

So we are from the general equation of the straight line passed to the equation of a straight line in segments , where .

Example.

A line in the Oxy rectangular coordinate system is given by the equation . Write the equation of this line in segments.

Solution.

Let's move one second to the right side of the given equality: . Now we divide into both parts of the resulting equality: . It remains to transform the resulting equality to the desired form: . So we got the required equation of a straight line in segments.

Answer:

If the line defines

Task 1 #6713

Task level: Equal to the Unified State Examination

\[\begin(cases) \sqrt((x+2)^2+y^2)+\sqrt(x^2+(y-a)^2)=\sqrt(4+a^2)\\ 5y= |6-a^2| \end(cases)\]

has a unique solution.

(Task from subscribers)

Consider the second equation of the system: it defines a family of lines \(y=0,2|6-a^2|\) , parallel to the \(Ox\) axis and lying in the upper half-plane (including the \(Ox\) axis) for any value parameter \(a\) (because the modulus is always non-negative).

Consider the first equation. Let \(A(x;y)\) , \(B(-2;0)\) , \(C(0;a)\) be points. Then \(BA=\sqrt((x+2)^2+y^2)\) , \(AC=\sqrt(x^2+(y-a)^2)\) , \(BC=\sqrt( 4+a^2)\) .
Thus, the first equation of the system looks like this: \(BA+AC=BC\) . Hence, it defines the locus of points \(A\) lying on the segment \(BC\) .

In order for this system to have a unique solution, the line \(y=0,2|6-a^2|\) must intersect the segment \(BC\) at one point.

1) Let \(a<0\) , то есть точка \(C\) лежит на отрицательной части оси \(Oy\) . Единственный случай, когда прямая \(y=0,2|6-a^2|\) будет иметь с отрезком одну общую точку, – когда прямая \(y=0,2|6-a^2|\) будет проходить через точку \(B\) , то есть совпадать с осью абсцисс. Отсюда \(0,2|6-a^2|=0\) , следовательно, \(a=\pm \sqrt6\) . Так как \(a<0\) , то \(a=-\sqrt6\) .

2) Let \(a=0\) . Then the segment \(BC\) lies on the x-axis, the line \(y=0.2|6-a^2|\) lies in the upper half-plane, and they have no common points.

3) Let \(a>0\) . Then \(C\) lies on the positive direction of the y-axis.

The line \(y=0,2|6-a^2|\) intersects the y-axis at the point \(D\) . In order for the line to intersect the segment \(BC\) , it is necessary that the point \(C\) is not lower than the point \(D\) , that is, \

Let's solve this inequality. Because \(a>0\) , then we have: \[|6-a^2|\leqslant 5a\quad\Leftrightarrow\quad -5a\leqslant 6-a^2\leqslant 5a\quad\Leftrightarrow\quad 1\leqslant a\leqslant 6.\]

Answer:

\(a\in\(-\sqrt6\)\cup\)

Task 2 #3978

Task level: Equal to the Unified State Examination

Find all values ​​of the parameter \(a\) , for each of which the system \[\begin(cases) y^2-(2a+1)y+a^2+a-2=0\\ \sqrt((x-a)^2+y^2)+\sqrt((x-a)^ 2+(y-3)^2)=3 \end(cases)\] has exactly one solution.

We transform the first equation of the system. Note that \(a^2+a-2=(a+2)(a-1)\) . Note also that \(a+2+a-1=2a+1\) , therefore, by the Vieta theorem, the roots of this equation will be \(y=a+2\) and \(y=a-1\) . This means that the graph of the first equation will be two lines \(y=a+2\) and \(y=a-1\) , parallel to the x-axis.

Let's transform the second equation. Consider the points \(A(a;0)\) , \(B(a;3)\) , \(C(x;y)\) . Then \(AB=\sqrt((a-a)^2+(0-3)^2)=3\), \(AC=\sqrt((x-a)^2+y^2)\) and \(CB=\sqrt((x-a)^2+(y-3)^2)\) . Therefore, the second equation of the system can be rewritten as \(AC+CB=AB\) . Hence, it defines the set of points \(C\) that lie on the segment \(AB\) . Note that since the points \(A\) and \(B\) have the same abscissa, the segment \(AB\) is perpendicular to the abscissa axis.

Schematically, the graphs of both equations look like this:

For the system to have a unique solution, the green graph must cross the segment \(AB\) at one point. Therefore, either the line \(y=a+2\) intersects the segment, and the line \(y=a-1\) does not intersect it, or vice versa: \[\left[\begin(gathered)\begin(aligned) &\begin(cases) 0\leqslant a+2\leqslant 3\\ a-1<0 \end{cases} \\ &\begin{cases} 0\leqslant a-1\leqslant 3\\ a+2>3\end(cases) \end(aligned)\end(gathered)\right.\quad\Leftrightarrow\quad a\in [-2;1)\cup(1;4]\]

Answer:

\([-2;1)\cup(1;4]\)

Task 3 #3979

Task level: Equal to the Unified State Examination

Find the smallest value of the parameter \(a\) for which the equation \[\sqrt((x+8)^2+(x+2)^2)+\sqrt((x+14)^2+(x+3)^2)=13a\] has at least one root.

1 way.

Consider \(f(x)=\sqrt((x+8)^2+(x+2)^2)+\sqrt((x+14)^2+(x+3)^2)\).
Then the equation will take the form \(f(x)=13a\) . Then we need to find the smallest value \(a\) for which the line \(y=13a\) intersects the graph \(y=f(x)\) at least at one point. Exploring \(f(x)\) . To do this, we first find its derivative: \[\begin(aligned) &f"(x)=\dfrac(2(x+8)+2(x+2))(2\sqrt((x+8)^2+(x+2)^2 ))+ \dfrac(2(x+14)+2(x+3))(2\sqrt((x+14)^2+(x+3)^2))=\\ &=\dfrac( 2x+10)(\sqrt((x+8)^2+(x+2)^2))+ \dfrac(2x+17)(\sqrt((x+14)^2+(x+3) ^2)) \end(aligned)\] Let's find the zeros of the derivative: \[\begin(aligned) &\dfrac(2x+10)(\sqrt((x+8)^2+(x+2)^2))+ \dfrac(2x+17)(\sqrt((x +14)^2+(x+3)^2))=0 \quad\Leftrightarrow\\ &\sqrt(\dfrac((x+14)^2+(x+3)^2)((x+ 8)^2+(x+2)^2))=-\dfrac(2x+17)(2x+10) \quad\Leftrightarrow\\ &\begin(cases) \dfrac((x+14)^2 +(x+3)^2)((x+8)^2+(x+2)^2)=\left(\dfrac(2x+17)(2x+10)\right)^2 \qquad ( *)\\ \dfrac(2x+17)(2x+10)\leqslant 0\end(cases) \quad\Leftrightarrow\\ &\begin(cases) 85x^2+598x+424=0\\ x\in \left[-8,5; -5\right) \end(cases) \quad\Leftrightarrow\\ & x=-\dfrac(106)(17) \end(aligned)\]

Let's define the signs of the derivative:

Therefore, the schematic graph of the function looks like this:

Therefore, the smallest value of the parameter \(a\) is when the straight line \(y=13a\) passes through the extremum point of the function \(f(x)\) : \

2 way.

Note that in the first method there were a lot of calculations and in fact we were lucky that when solving the equation \((*)\) the terms with \(x^4\) and \(x^3\) mutually annihilated and we came to quadratic equation. But what if the numbers are not so well chosen and we do not end up with a “beautiful” equation that we can solve?
Let's look at the second way to solve such equations.

Consider three points: \(A(x;x)\) , \(B(-8; -2)\) , \(C(-14; -3)\) . Then the equation will take the form \ If we need to find the smallest value of the parameter \(a\) at which the equation has at least one solution, then we need to find the point \(A\) at which the sum of the lengths of the segments \(AB\) and \ (AC\) will be the smallest.
Where is the point \(A\) ? This point “runs” along the straight line \(y=x\) . Graphically it looks like this:

Here we will use the classical idea of ​​planimetry. Reflect the point \(B\) symmetrically with respect to the line \(y=x\) (that is, draw \(BB"\perp y=x\) , where \(BH=HB"\) ):

Then \(AB+AC=AB"+AC\) . Note that by the triangle rule, if the point \(A\) does not lie on the segment \(B"C\) , then \(AB"+AC>B" c\) . Therefore, the smallest sum of lengths \(AB"+AC\) will be achieved when \(A\in B"C\) .

Thus, we conceptually understood where the point \(A\) should be. Now it remains to find its coordinates.

1) Find the coordinates of the point \(B"\) .
To do this, first find the equation of the line \(BB"\)... Since \(BB"\perp y=x\) , then if the equation of the line \(BB"\) has the form \(y=kx+b\) , then \(k\cdot 1=-1\) (the product of the slopes of two mutually perpendicular lines is \(-1\) ) Therefore, \(y=-x+b\) .
In order to find the number \(b\) , you need to substitute the coordinates of the point \(B\) in the equation of a straight line: \[-2=-1\cdot (-8)+b\quad\Leftrightarrow\quad b=-10\] Therefore, the equation of a straight line has the form \(y=-x-10\) .
Find the coordinates of the point \(H\) - this is the intersection point of the lines \(y=x\) and \(y=-x-10\) : \[\begin(cases) y=x\\ y=-x-10\end(cases)\quad\Leftrightarrow\quad x=y=-5\quad\Rightarrow\quad H(-5;-5)\ ]\(H\) is the midpoint of the segment \(BB"\) . So, if the coordinates of the point \(B"\) are \((x_0;y_0)\) , then \[\begin(cases) -5=\dfrac(-8+x_0)2\\ -5=\dfrac(-2+y_0)2\end(cases)\quad\Leftrightarrow\quad \begin(cases) x_0 =-2\\ y_0=-8\end(cases)\] So \(B"(-2;-8)\) .

2) Find the equation of the line \(B"C\)... If the equation of this line in general looks like \(y=mx+n\) , then \[\begin(cases) -8=-2m+n\\ -3=-14m+n\end(cases)\quad\Leftrightarrow\quad \begin(cases) m=-\dfrac5(12)\\ n =-\dfrac(53)6\end(cases)\] Therefore, \(y=-\frac5(12)x-\frac(53)6\) . Now you can find the coordinates of the point \(A\) - this is the intersection point of the lines \(y=x\) and \(B"C\) : \[\begin(cases) y=x\\ y=-\frac5(12)x-\frac(53)6\end(cases) \quad\Leftrightarrow\quad x=y=-\dfrac(106)( 17)\]

3) Now you can find the value of the parameter \(a\) . \

What is good about this method? First, it is more elegant. Secondly, in the course of solving, we only encountered linear equations, which are much easier to solve.

Answer:

\(a=1\)

Task 4 #3909

Task level: More difficult than the exam

Find all values ​​of the parameter \(a\) for which the system \[\begin(cases) x^2+|x^2-2x|=y^2+|y^2-2y|\\ x+y=a\end(cases)\]

has more than two solutions.

Let's plot the first equation. To do this, consider the following cases:

1) \(x^2-2x\geqslant 0\) , \(y^2-2y\geqslant 0\) . Then the equation will take the form \ Then in this case we get the following graph:

2) \(x^2-2x\leqslant 0\) , \(y^2-2y\leqslant 0\) . Then: \ So, the graph for the first two cases will look like this:

3) \(x^2-2x\geqslant 0\) , \(y^2-2y\leqslant 0\) . Then the equation will take the form: \ Therefore, more will be added:

4) \(x^2-2x\leqslant 0\) , \(y^2-2y\geqslant 0\) . Then we have: \ The graph will be the same parabola as in item 3, only with the axes changed:

The graph \(x+y=a\) for each fixed \(a\) is the line \(y=-x+a\) , that is, the line parallel to \(y=-x\) (and also parallel to the part of the line \(y=1-x\) from item 1).
In order for the system to have more than two solutions, it is necessary that the line \(y=-x+a\) be in positions from (1) (not inclusive) to (2) (inclusive):

Indeed, when the line is in position (2), then the system will have an infinite number of solutions (namely, the part of the line \(y=1-x\) with \(x\in (-\infty;-1]\cup\]

Let some affine coordinate system OXY be given.

Theorem 2.1. Any straight l coordinate system OX is given by a linear equation of the form

BUT x+ B y+ C = O, (1)

where A, B, C R and A 2 + B 2 0. Conversely, any equation of the form (1) defines a straight line.

Equation of the form (1) - general equation of a straight line .

Let in equation (1) all the coefficients A, B and C be non-zero. Then

Ah-By=-C, and .

Let's denote -C/A=a, -C/B=b. Get

-line segment equation .

Indeed, the numbers |a| and |b| indicate the size of the segments cut off by a straight line l on the OX and OY axes, respectively.

Let the line l is given by the general equation (1) in a rectangular coordinate system and let the points M 1 (x 1, y 1) and M 2 (x 2, y 2) belong l. Then

BUT x 1 + B at 1 + C = A X 2 + B at 2 + C, that is, A( x 1 -x 2) + B( at 1 -at 2) = 0.

The last equality means that the vector \u003d (A, B) is orthogonal to the vector \u003d (x 1 -x 2, y 1 -y 2). those. The vector (A, B) is called normal vector of the line l.

Consider the vector = (-B, A). Then

A(-B)+BA=0. those. ^ .

Therefore, the vector \u003d (-B, A) is the direction vector of the spicy l.

Parametric and canonical equations of a straight line

Equation of a straight line passing through two given points

Let the straight line be given in the affine coordinate system (0, X, Y) l, its direction vector = (m,n) and the point M 0 ( x 0 ,y 0) owned l. Then for an arbitrary point M ( x,at) of this line we have

and since then .

If we designate and

The radius vectors of the points M and M 0 , respectively, then

- equation of a straight line in vector form.

Because =( X,at), =(X 0 ,at 0), then

x= x 0 + mt,

y= y 0 + nt

- parametric equation of a straight line .

Hence it follows that

- canonical equation of a straight line .

Finally, if on a straight line l two points M 1 ( X 1 ,at 1) and

M2( x 2 ,at 2), then the vector =( X 2 -X 1 ,y 2 -at 1) is guiding straight line vector l. Then

- equation of a line passing through two given points.

Mutual arrangement of two straight lines.

Let straight l 1 and l 2 are given by their general equations

l 1: A 1 X+ In 1 at+ С 1 = 0, (1)

l 2: A 2 X+ B 2 at+ C 2 = 0.

Theorem. Let straight l 1 and l 2 are given by equations (1). Then and only then:

1) lines intersect when there is no number λ such that

A 1 =λA 2 , B 1 =λB 2 ;

2) the lines coincide when there is a number λ such that

A 1 =λA 2 , B 1 =λB 2 , C 1 =λC 2 ;

3) lines are distinct and parallel when there is a number λ such that

A 1 \u003d λA 2, B 1 \u003d λB 2, C 1 λC 2.

bundle of straight lines

A bunch of straight lines is the collection of all lines in the plane passing through some point called center beam.

To specify the beam equation, it suffices to know any two straight lines l 1 and l 2 passing through the center of the beam.

Let the lines in the affine coordinate system l 1 and l 2 are given by the equations

l 1:A1 x+B1 y+ C1 = 0,

l 2:A2 x+ B2 y+ C2 = 0.

The equation:

A 1 x+B1 y+ С + λ (A 2 X+ B 2 y+ C) = 0

- the equation of a pencil of lines defined by the equations l 1 and l 2.

In what follows, by the coordinate system we mean the rectangular coordinate system .

Conditions for parallelism and perpendicularity of two lines

Let the lines l 1 and l 2. with their general equations; = (A 1 ,B 1), = (A 2 ,B 2) are the normal vectors of these lines; k 1 = tanα 1 , k 2 = tgα 2 – slope coefficients; =( m 1 ,n 1), (m 2 ,n 2) are direction vectors. Then, direct l 1 and l 2 are parallel if and only if one of the following conditions is true:

or either k 1 =k 2 or .

Let it be straight now l 1 and l 2 are perpendicular. Then, obviously, , that is, A 1 A 2 + B 1 B 2 = 0.

If straight l 1 and l 2 are given respectively by the equations

l 1: at=k 1 x+ b 1 ,

l 2: at=k 2 x+ b 2 ,

then tgα 2 = tg(90º+α) = .

Hence it follows that

Finally, if and are the direction vectors of the lines, then ^ , i.e.

m 1 m 2 + n 1 n 2 = 0

The last relation expresses the necessary and sufficient condition for two planes to be perpendicular.

Angle between two lines

At an angle φ between two lines l 1 and l 2 we will understand the smallest angle through which one line must be rotated so that it becomes parallel to another line or coincides with it, that is, 0 £ φ £

Let the lines be given by general equations. It's obvious that

cosφ=

Let it be straight now l 1 and l 2 is given by equations with slope coefficients k 1 in k 2 respectively. Then

Obviously, that is ( X-X 0) + B( at-at 0) + C( z-z 0) = 0

Let's open the brackets and denote D \u003d -A x 0 - B at 0 - C z 0 . Get

A x+ B y+ C z+ D = 0 (*)

- general plane equation or general plane equation.

Theorem 3.1 Linear Equation(*) (A 2 +B 2 +C 2 ≠ 0) is an equation of the plane and vice versa, any equation of the plane is linear.

1) D = 0, then the plane passes through the origin.

2) A \u003d 0, then the plane is parallel to the OX axis

3) A \u003d 0, B \u003d 0, then the plane is parallel to the OXY plane.

Let all the coefficients in the equation be nonzero.

- equation of a plane in segments. The numbers |a|, |b|, |c| indicate the size of the segments cut off by the plane on the coordinate axes.