The equation parabolas is quadratic function. There are several options for constructing this equation. It all depends on what parameters are presented in the problem statement.

Instructions

A parabola is a curve that resembles an arc in shape and is a graph power function. Regardless of the characteristics of a parabola, this one is even. Such a function is called even; for all values ​​of the argument from the definition, when the sign of the argument changes, the value does not change: f (-x) = f (x) Start with the simplest function: y = x^2. From its appearance we can conclude that it is both positive and negative negative values argument x. The point at which x=0, and at the same time, y =0 is considered a point.

Below are all the main options for constructing this function and its . As a first example, below we consider a function of the form: f(x)=x^2+a, where a is an integer. In order to construct a graph of this function, it is necessary to shift the graph of the function f(x) by a units. An example is the function y=x^2+3, where along the y-axis the function is shifted by two units. If a function with the opposite sign is given, for example y=x^2-3, then its graph is shifted down along the y-axis.

Another type of function that can be given a parabola is f(x)=(x +a)^2. In such cases, the graph, on the contrary, shifts along the abscissa axis (x axis) by a units. For example, we can consider the functions: y=(x +4)^2 and y=(x-4)^2. In the first case, where there is a function with a plus sign, the graph is shifted along the x-axis to the left, and in the second case - to the right. All these cases are shown in the figure.

Equation of a line passing through a given point at in this direction. Equation of a line passing through two given points. The angle between two straight lines. The condition of parallelism and perpendicularity of two straight lines. Determining the point of intersection of two lines

1. Equation of a line passing through a given point A(x 1 , y 1) in a given direction, determined by the slope k,

y - y 1 = k(x - x 1). (1)

This equation defines a pencil of lines passing through a point A(x 1 , y 1), which is called the beam center.

2. Equation of a line passing through two points: A(x 1 , y 1) and B(x 2 , y 2), written like this:

The angular coefficient of a straight line passing through two given points is determined by the formula

3. Angle between straight lines A And B is the angle by which the first straight line must be rotated A around the point of intersection of these lines counterclockwise until it coincides with the second line B. If two straight lines are given by equations with a slope

y = k 1 x + B 1 ,

Let two points be given M 1 (x 1,y 1) And M 2 (x 2,y 2). Let us write the equation of the line in the form (5), where k still unknown coefficient:

Since the point M 2 belongs to a given line, then its coordinates satisfy equation (5): . Expressing from here and substituting it into equation (5), we obtain the required equation:

If this equation can be rewritten in a form that is more convenient for memorization:

(6)

Example. Write down the equation of a straight line passing through points M 1 (1,2) and M 2 (-2,3)

Solution. . Using the property of proportion and performing the necessary transformations, we get general equation straight:

Angle between two straight lines

Consider two straight lines l 1 And l 2:

l 1: , , And

l 2: , ,

φ is the angle between them (). From Fig. 4 it is clear: .

From here , or

Using formula (7) you can determine one of the angles between straight lines. The second angle is equal to .

Example. Two straight lines are given by the equations y=2x+3 and y=-3x+2. find the angle between these lines.

Solution. From the equations it is clear that k 1 =2, and k 2 =-3. Substituting these values ​​into formula (7), we find

. Thus, the angle between these lines is equal to .

Conditions for parallelism and perpendicularity of two straight lines

If straight l 1 And l 2 are parallel, then φ=0 And tgφ=0. from formula (7) it follows that , whence k 2 =k 1. Thus, the condition for parallelism of two lines is the equality of their angular coefficients.

If straight l 1 And l 2 are perpendicular, then φ=π/2, α 2 = π/2+ α 1 . . Thus, the condition for the perpendicularity of two straight lines is that their angular coefficients are inverse in magnitude and opposite in sign.

Distance from point to line

Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Bу + C = 0 is determined as

Proof. Let point M 1 (x 1, y 1) be the base of the perpendicular dropped from point M to a given straight line. Then the distance between points M and M 1:

The coordinates x 1 and y 1 can be found by solving the system of equations:

The second equation of the system is the equation of a line passing through a given point M 0 perpendicular to a given line.

If we transform the first equation of the system to the form:

A(x – x 0) + B(y – y 0) + Ax 0 + By 0 + C = 0,

then, solving, we get:

Substituting these expressions into equation (1), we find:

The theorem has been proven.

Example. Determine the angle between the lines: y = -3x + 7; y = 2x + 1.

k 1 = -3; k 2 = 2 tanj= ; j = p/4.

Example. Show that the lines 3x – 5y + 7 = 0 and 10x + 6y – 3 = 0 are perpendicular.

We find: k 1 = 3/5, k 2 = -5/3, k 1 k 2 = -1, therefore, the lines are perpendicular.

Example. Given are the vertices of the triangle A(0; 1), B(6; 5), C(12; -1). Find the equation of the height drawn from vertex C.

We find the equation of side AB: ; 4x = 6y – 6;

2x – 3y + 3 = 0;

The required height equation has the form: Ax + By + C = 0 or y = kx + b.

k= . Then y = . Because height passes through point C, then its coordinates satisfy this equation: whence b = 17. Total: .

Answer: 3x + 2y – 34 = 0.

The distance from a point to a line is determined by the length of the perpendicular drawn from the point to the line.

If the line is parallel to the projection plane (h | | P 1), then in order to determine the distance from the point A to a straight line h it is necessary to lower a perpendicular from a point A to the horizontal h.

Let's consider more complex example, when the straight line takes general position. Let it be necessary to determine the distance from a point M to a straight line A general position.

Determination task distances between parallel lines is solved similarly to the previous one. A point is taken on one line and a perpendicular is dropped from it to another line. The length of a perpendicular is equal to the distance between parallel lines.

Second order curve is a line defined by an equation of the second degree relative to the current Cartesian coordinates. In the general case, Ax 2 + 2Bxy + Su 2 + 2Dx + 2Ey + F = 0,

where A, B, C, D, E, F are real numbers and at least one of the numbers A 2 + B 2 + C 2 ≠0.

Circle

Circle center– this is the geometric locus of points in the plane equidistant from a point in the plane C(a,b).

The circle is given by the following equation:

Where x,y are the coordinates of an arbitrary point on the circle, R is the radius of the circle.

Sign of the equation of a circle

1. The term with x, y is missing

2. The coefficients for x 2 and y 2 are equal

Ellipse

Ellipse is called the geometric locus of points in a plane, the sum of the distances of each of which from two given points of this plane is called foci (a constant value).

The canonical equation of the ellipse:

X and y belong to the ellipse.

a – semimajor axis of the ellipse

b – semi-minor axis of the ellipse

The ellipse has 2 axes of symmetry OX and OU. The axes of symmetry of an ellipse are its axes, the point of their intersection is the center of the ellipse. The axis on which the foci are located is called focal axis. The point of intersection of the ellipse with the axes is the vertex of the ellipse.

Compression (tension) ratio: ε = s/a– eccentricity (characterizes the shape of the ellipse), the smaller it is, the less the ellipse is extended along the focal axis.

If the centers of the ellipse are not at the center C(α, β)

Hyperbola

Hyperbole is called the geometric locus of points in a plane, the absolute value of the difference in distances, each of which from two given points of this plane, called foci, is a constant value different from zero.

Canonical hyperbola equation

A hyperbola has 2 axes of symmetry:

a – real semi-axis of symmetry

b – imaginary semi-axis of symmetry

Asymptotes of a hyperbola:

Parabola

Parabola is the locus of points in the plane equidistant from a given point F, called the focus, and a given line, called the directrix.

The canonical equation of a parabola:

У 2 =2рх, where р is the distance from the focus to the directrix (parabola parameter)

If the vertex of the parabola is C (α, β), then the equation of the parabola (y-β) 2 = 2р(x-α)

If the focal axis is taken as the ordinate axis, then the equation of the parabola will take the form: x 2 =2qу

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 junior classes, and today we will learn how to deal with it using the methods of analytical geometry. To master the material, you must be able to build a straight line; know what equation defines a straight line, in particular, a straight line passing through the origin of coordinates 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 the linear function turned out to be very successful and detailed. Therefore, dear teapots, warm up there first. In addition, you need to have basic knowledge about vectors, otherwise the understanding of the material will be incomplete.

On this lesson We will look at ways in which you can create an equation of a straight line on a plane. I recommend not to neglect practical examples (even if it seems very simple), since I will provide them with elementary and important facts, technical methods, which will be required in the future, including in other sections of higher mathematics.

• How to write an equation of a straight line with an angle coefficient?
• How ?
• How to find a direction vector using the general equation of a straight line?
• How to write an equation of a straight line from a point and a normal vector?

and we begin:

Equation of a straight line with slope

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

In a geometry course it is proven that the slope of the straight line is equal to tangent of the angle between positive axis directionand this line: , and the angle “unscrews” counterclockwise.

In order not to clutter the drawing, I drew angles only for two straight lines. Let's consider the “red” line and its slope. According to the above: (the “alpha” angle is indicated by a green arc). For the “blue” straight line with the angle coefficient, the equality is true (the “beta” angle is indicated by a brown arc). And if the tangent of the angle is known, then if necessary it is easy to find and the corner itself by using inverse function– arctangent. As they say, a trigonometric table or a microcalculator in your hands. Thus, the angular coefficient characterizes the degree of inclination of the straight line to the abscissa axis.

The following cases are possible:

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

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

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

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

The greater the slope coefficient in absolute value, the steeper the straight line graph goes..

For example, consider two straight lines. Here, therefore, the straight line has a steeper slope. Let me 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 .

Conversely: the smaller the slope coefficient in absolute value, the flatter the straight line.

For straight lines the inequality is true, thus the straight line is flatter. Children's slide, so as not to give yourself bruises and bumps.

Why is this necessary?

Prolong your torment Knowledge of the above facts allows you to immediately see your mistakes, in particular, errors when constructing graphs - if the drawing turns out to be “obviously something wrong.” It is advisable that you straightaway it was clear that, for example, the straight line is very steep and goes from bottom to top, and the straight line is very flat, pressed close to the axis and goes from top to bottom.

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

Designations: straight lines are designated in small Latin letters: . A popular option is to designate them using the same letter with natural subscripts. For example, the five lines we just looked at can be denoted by .

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

It's time to warm up a little:

How to write an equation of a straight line with an angle coefficient?

If a point belonging to a certain line and the angular coefficient of this line are known, then the equation of this line is expressed by the formula:

Example 1

Write an equation of a straight line with an angular coefficient if it is known that the point belongs to this straight line.

Solution: Let's compose the equation of the straight line using the formula . IN in this case:

Answer:

Examination is done simply. First, we look at the resulting equation and make sure that our slope is in place. Secondly, the coordinates of the point must satisfy this equation. Let's plug them into the equation:

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

Conclusion: The equation was found correctly.

A more tricky example to solve on your own:

Example 2

Write an equation for 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 have any difficulties, re-read the theoretical material. More precisely, more practical, I skip a lot of evidence.

It rang last call, the graduation party has died down, and outside the gates of our native school, analytical geometry itself awaits us. The jokes are over... Or maybe they are just beginning =)

We nostalgically wave our pen to the familiar and get acquainted with the general equation of a straight line. Because in analytical geometry this is exactly what is used:

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 the equation with the slope coefficient. First, let's 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 analytical geometry, the equation of a straight line will almost always be given in general form. Well, if necessary, it can be easily reduced to the “school” form with an angular coefficient (with the exception of straight lines parallel to the ordinate axis).

Let's ask ourselves what enough know to construct a straight line? Two points. But more about this childhood incident later; now sticks with arrows rule. Each straight line has a very specific slope, which is easy to “adapt” to. vector.

A vector that is parallel to a line is called the direction vector of that line. It is obvious that any straight line has infinitely many direction vectors, and all of them will be collinear (co-directional or not - it doesn’t matter).

I will denote the direction vector as follows: .

But one vector is not enough to construct a straight line; the vector is free and not tied to any point on 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 using a point and a direction vector?

If a certain point belonging to a line and the direction vector of this line are known, then the equation of this line can be compiled using the formula:

Sometimes it is called canonical equation of the line .

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

Example 3

Write an equation for a straight line using a point and a direction vector

Solution: Let's compose the equation of a straight line using the formula. In this case:

Using the properties of proportion we get rid of fractions:

And we bring the equation to general appearance:

Answer:

As a rule, there is no need to make a drawing in such examples, but for the sake of understanding:

In the drawing we see the starting point, the original direction vector (it can be plotted from any point on the plane) and the constructed straight line. By the way, in many cases it is most convenient to construct a straight line using an equation with an angular coefficient. It’s easy to transform our equation into form and easily select another point to construct a straight line.

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

Let's create an equation of a straight line using a point and a direction vector:

Resolving the proportion:

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

Those interested can test vectors in the same way or any other collinear vector.

Now let's solve the inverse problem:

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

Very simple:

If a line is given by a general equation in a rectangular coordinate system, 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 out of an infinite number, but we don’t need more. Although in some cases it is advisable to reduce the coordinates of the direction vectors:

Thus, the equation specifies a straight line that is parallel to the axis and the coordinates of the resulting direction vector are conveniently divided by –2, obtaining exactly the basis vector as the direction vector. Logical.

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

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

Firstly, using the equation of the straight line we restore its direction vector: – everything is fine, we have received the original vector (in some cases the result may be a collinear vector to the original one, and this is usually easy to notice 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 was obtained, which we are very happy about.

Conclusion: The task was completed correctly.

Example 4

Write an equation for a straight line using a point and a direction vector

This is an example for you to solve on your own. The solution and answer are at the end of the lesson. It is highly advisable to check using the algorithm just discussed. Try to always (if possible) check on a draft. It’s stupid to make mistakes where they can be 100% avoided.

In the event that one of the coordinates of the direction vector is zero, proceed very simply:

Example 5

Solution: The formula is not suitable since 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 directing vector of the straight line:
– the resulting vector is collinear to the original direction vector.

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

The correct equality is obtained

Conclusion: task completed correctly

The question arises, why bother with the formula if there is a universal version that will work in any case? There are two reasons. First, the formula is in the form of a fraction much better remembered. And secondly, the disadvantage of the universal formula is that the risk of getting confused increases significantly when substituting coordinates.

Example 6

Write an equation for a straight line using a point and a direction vector.

This is an example for you to solve on your own.

Let's return to the ubiquitous two points:

How to write an equation of a straight line using 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 type of formula and here's why: if two points are known, then the vector will be the direction vector of the given line. At the lesson Vectors for dummies we considered simplest task– how to find the coordinates of a vector from two points. According to this problem, the coordinates of the direction vector are:

Note : the points can be “swapped” and the formula can be used . Such a solution will be equivalent.

Example 7

Write an equation of a straight line using two points .

Solution: We use the formula:

Combing the denominators:

And shuffle the deck:

Right now it’s convenient to get rid of fractional numbers. In this case, you need to multiply both sides 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 line is written correctly.

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

It is worth noting that graphical verification in this case is difficult, since construct a straight line and see whether the points belong to it , not so simple.

I’ll note a couple more technical aspects of the solution. Perhaps in this problem it is more profitable to use the mirror formula and, at the same points make an equation:

Fewer fractions. If you want, you can carry out the solution to the end, the result should be the same equation.

The second point is to look at the final answer and figure out whether it can be simplified further? For example, if you get the equation , then it is advisable to reduce it by two: – the equation will define the same straight line. However, this is already a topic of conversation about relative position of lines.

Having received the answer in Example 7, just in case, I checked whether 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 an equation for a line passing through the points .

This is an example for an independent solution, which will allow you to better understand and practice calculation techniques.

Similar to the previous paragraph: if in the formula one of the denominators (the coordinate of the direction vector) becomes zero, then we rewrite it in the form . Again, notice how awkward and confused she looks. I don’t see much point in giving practical examples, since we have already actually solved this problem (see No. 5, 6).

Direct normal vector (normal vector)

What is normal? In simple words, normal is perpendicular. That is, the normal vector of a line is perpendicular to a given line. Obviously, any straight line has an infinite number of them (as well as direction vectors), and all the normal vectors of the straight line will be collinear (codirectional or not, it makes no difference).

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

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

If the coordinates of the direction vector have to be carefully “pulled out” from 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. Let us 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 construct an equation of a straight line given one point and a normal vector? I feel it in my gut, it’s possible. If the normal vector is known, then the direction of the straight line itself is clearly defined - this is a “rigid structure” with an angle of 90 degrees.

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

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

Here everything worked out without fractions and other surprises. This is our normal vector. Love him. And respect =)

Example 9

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

Solution: We use the formula:

The general equation of the line has been obtained, let’s check:

1) “Remove” the coordinates of the normal vector from the equation: – yes, indeed, the original vector was obtained from the condition (or a collinear vector should be obtained).

2) Let's check whether the point satisfies the equation:

True equality.

After we are convinced that the equation is composed correctly, we will complete the second, easier part of the task. We take out the directing vector of the straight line:

Answer:

In the drawing the situation looks like this:

For training purposes, a similar task for solving independently:

Example 10

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

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

Equation of a straight line in segments.
Equation of a 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 equal to zero and there is no way to get one on the right side).

This is, figuratively speaking, a “technical” type of equation. A common task is to represent the general equation of a line as an equation of a line in segments. How is it convenient? The equation of a line in segments allows you to quickly find the points of intersection of a line with coordinate axes, which can be very important in some problems of higher mathematics.

Let's find the point of intersection of the line with the axis. We reset the “y” to zero, and the equation takes the form . The desired point is obtained automatically: .

Same with the axis – the point at which the straight line intersects the ordinate axis.

This article reveals how to obtain the equation of a straight line passing through two given points in a rectangular coordinate system located on a plane. Let us derive the equation of a straight line passing through two given points in a rectangular coordinate system. We will clearly show and solve several examples related to the material covered.

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Before obtaining the equation of a line passing through two given points, it is necessary to pay attention to some facts. There is an axiom that says that through two divergent points on a plane it is possible to draw a straight line and only one. In other words, two given points on a plane are defined by a straight line passing through these points.

If the plane is defined by the rectangular coordinate system Oxy, then any straight line depicted in it will correspond to the equation of a straight line on the plane. There is also a connection with the directing vector of the straight line. This data is sufficient to compile the equation of a straight line passing through two given points.

Let's look at an example of solving a similar problem. It is necessary to create an equation for a straight line a passing through two divergent points M 1 (x 1, y 1) and M 2 (x 2, y 2), located in the Cartesian coordinate system.

In the canonical equation of a line on a plane, having the form x - x 1 a x = y - y 1 a y, a rectangular coordinate system O x y is specified with a line that intersects with it at a point with coordinates M 1 (x 1, y 1) with a guide vector a → = (a x , a y) .

It is necessary to draw up canonical equation straight line a, which will pass through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2).

Straight a has a direction vector M 1 M 2 → with coordinates (x 2 - x 1, y 2 - y 1), since it intersects points M 1 and M 2. We have obtained the necessary data in order to transform the canonical equation with the coordinates of the direction vector M 1 M 2 → = (x 2 - x 1, y 2 - y 1) and the coordinates of the points M 1 lying on them (x 1, y 1) and M 2 (x 2 , y 2) . We obtain an equation of the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 or x - x 2 x 2 - x 1 = y - y 2 y 2 - y 1.

Consider the figure below.

Following the calculations, we write down the parametric equations of a line on a plane that passes through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2). We obtain an equation of the form x = x 1 + (x 2 - x 1) · λ y = y 1 + (y 2 - y 1) · λ or x = x 2 + (x 2 - x 1) · λ y = y 2 + (y 2 - y 1) · λ .

Let's take a closer look at solving several examples.

Example 1

Write down the equation of a straight line passing through 2 given points with coordinates M 1 - 5, 2 3, M 2 1, - 1 6.

Solution

The canonical equation for a line intersecting at two points with coordinates x 1, y 1 and x 2, y 2 takes the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1. According to the conditions of the problem, we have that x 1 = - 5, y 1 = 2 3, x 2 = 1, y 2 = - 1 6. It is necessary to substitute the numerical values ​​into the equation x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1. From here we get that the canonical equation takes the form x - (- 5) 1 - (- 5) = y - 2 3 - 1 6 - 2 3 ⇔ x + 5 6 = y - 2 3 - 5 6.

Answer: x + 5 6 = y - 2 3 - 5 6.

If you need to solve a problem with a different type of equation, then first you can go to the canonical one, since it is easier to come from it to any other one.

Example 2

Compose the general equation of a straight line passing through points with coordinates M 1 (1, 1) and M 2 (4, 2) in the O x y coordinate system.

Solution

First, you need to write down the canonical equation of a given line that passes through given two points. We get an equation of the form x - 1 4 - 1 = y - 1 2 - 1 ⇔ x - 1 3 = y - 1 1 .

Let's bring the canonical equation to the desired form, then we get:

x - 1 3 = y - 1 1 ⇔ 1 x - 1 = 3 y - 1 ⇔ x - 3 y + 2 = 0

Answer: x - 3 y + 2 = 0 .

Examples of such tasks were discussed in school textbooks during algebra lessons. School problems differed in that the equation of a straight line with an angle coefficient was known, having the form y = k x + b. If you need to find the value of the slope k and the number b for which the equation y = k x + b defines a line in the O x y system that passes through the points M 1 (x 1, y 1) and M 2 (x 2, y 2) , where x 1 ≠ x 2. When x 1 = x 2 , then the angular coefficient takes on the value of infinity, and the straight line M 1 M 2 is defined by the general incomplete equation of the form x - x 1 = 0 .

Because the points M 1 And M 2 are on a straight line, then their coordinates satisfy the equation y 1 = k x 1 + b and y 2 = k x 2 + b. It is necessary to solve the system of equations y 1 = k x 1 + b y 2 = k x 2 + b for k and b.

To do this, we find k = y 2 - y 1 x 2 - x 1 b = y 1 - y 2 - y 1 x 2 - x 1 x 1 or k = y 2 - y 1 x 2 - x 1 b = y 2 - y 2 - y 1 x 2 - x 1 x 2 .

With these values ​​of k and b, the equation of a line passing through the given two points becomes y = y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 1 or y = y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 2.

Remember this right away great amount formulas won't work. To do this, it is necessary to increase the number of repetitions in solving problems.

Example 3

Write down the equation of a straight line with an angular coefficient passing through points with coordinates M 2 (2, 1) and y = k x + b.

Solution

To solve the problem, we use a formula with an angular coefficient of the form y = k x + b. The coefficients k and b must take such a value that given equation corresponded to a straight line passing through two points with coordinates M 1 (- 7, - 5) and M 2 (2, 1).

Points M 1 And M 2 are located on a straight line, then their coordinates must make the equation y = k x + b a true equality. From this we get that - 5 = k · (- 7) + b and 1 = k · 2 + b. Let's combine the equation into the system - 5 = k · - 7 + b 1 = k · 2 + b and solve.

Upon substitution we get that

5 = k · - 7 + b 1 = k · 2 + b ⇔ b = - 5 + 7 k 2 k + b = 1 ⇔ b = - 5 + 7 k 2 k - 5 + 7 k = 1 ⇔ ⇔ b = - 5 + 7 k k = 2 3 ⇔ b = - 5 + 7 2 3 k = 2 3 ⇔ b = - 1 3 k = 2 3

Now the values ​​k = 2 3 and b = - 1 3 are substituted into the equation y = k x + b. We find that the required equation passing through the given points will be an equation of the form y = 2 3 x - 1 3 .

This method of solution predetermines spending large quantity time. There is a way in which the task is solved in literally two steps.

Let us write the canonical equation of the line passing through M 2 (2, 1) and M 1 (- 7, - 5), having the form x - (- 7) 2 - (- 7) = y - (- 5) 1 - (- 5) ⇔ x + 7 9 = y + 5 6 .

Now let's move on to the slope equation. We get that: x + 7 9 = y + 5 6 ⇔ 6 · (x + 7) = 9 · (y + 5) ⇔ y = 2 3 x - 1 3.

Answer: y = 2 3 x - 1 3 .

If in three-dimensional space there is a rectangular coordinate system O x y z with two given non-coinciding points with coordinates M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), the straight line M passing through them 1 M 2 , it is necessary to obtain the equation of this line.

We have that canonical equations of the form x - x 1 a x = y - y 1 a y = z - z 1 a z and parametric equations of the form x = x 1 + a x · λ y = y 1 + a y · λ z = z 1 + a z · λ are able to define a line in the coordinate system O x y z, passing through points having coordinates (x 1, y 1, z 1) with a direction vector a → = (a x, a y, a z).

Straight M 1 M 2 has a direction vector of the form M 1 M 2 → = (x 2 - x 1, y 2 - y 1, z 2 - z 1), where the straight line passes through the point M 1 (x 1, y 1, z 1) and M 2 (x 2 , y 2 , z 2), hence the canonical equation can be of the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 or x - x 2 x 2 - x 1 = y - y 2 y 2 - y 1 = z - z 2 z 2 - z 1, in turn parametric x = x 1 + (x 2 - x 1) λ y = y 1 + (y 2 - y 1) λ z = z 1 + (z 2 - z 1) λ or x = x 2 + (x 2 - x 1) λ y = y 2 + (y 2 - y 1) · λ z = z 2 + (z 2 - z 1) · λ .

Consider a drawing that shows 2 given points in space and the equation of a straight line.

Example 4

Write the equation of a line defined in a rectangular coordinate system O x y z of three-dimensional space, passing through given two points with coordinates M 1 (2, - 3, 0) and M 2 (1, - 3, - 5).

Solution

It is necessary to find the canonical equation. Since we are talking about three-dimensional space, it means that when a line passes through given points, the desired canonical equation will take the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 .

By condition we have that x 1 = 2, y 1 = - 3, z 1 = 0, x 2 = 1, y 2 = - 3, z 2 = - 5. It follows that the necessary equations will be written as follows:

x - 2 1 - 2 = y - (- 3) - 3 - (- 3) = z - 0 - 5 - 0 ⇔ x - 2 - 1 = y + 3 0 = z - 5

Answer: x - 2 - 1 = y + 3 0 = z - 5.

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