Formula for calculating the distance from a point to a line on a plane

If the equation of the line Ax + By + C = 0 is given, then the distance from the point M(M x , M y) to the line can be found using the following formula

Examples of problems for calculating the distance from a point to a line on a plane

Example 1.

Find the distance between the line 3x + 4y - 6 = 0 and the point M(-1, 3).

Solution. Let's substitute the coefficients of the line and the coordinates of the point into the formula

Answer: the distance from the point to the line is 0.6.

equation of a plane passing through points perpendicular to a vectorGeneral equation of a plane

A nonzero vector perpendicular to a given plane is called normal vector (or, in short, normal ) for this plane.

Let the following be given in coordinate space (in a rectangular coordinate system):

a) point ;

b) non-zero vector (Fig. 4.8, a).

You need to create an equation for a plane passing through a point perpendicular to the vector End of proof.

Let's now consider Various types equations of a straight line on a plane.

1) General equation of the planeP .

From the derivation of the equation it follows that at the same time A, B And C are not equal to 0 (explain why).

The point belongs to the plane P only if its coordinates satisfy the equation of the plane. Depending on the odds A, B, C And D plane P occupies one position or another:

- the plane passes through the origin of the coordinate system, - the plane does not pass through the origin of the coordinate system,

- plane parallel to the axis X,

X,

- plane parallel to the axis Y,

- the plane is not parallel to the axis Y,

- plane parallel to the axis Z,

- the plane is not parallel to the axis Z.

Prove these statements yourself.

Equation (6) is easily derived from equation (5). Indeed, let the point lie on the plane P. Then its coordinates satisfy the equation. Subtracting equation (7) from equation (5) and grouping the terms, we obtain equation (6). Let us now consider two vectors with coordinates respectively. From formula (6) it follows that their scalar product is equal to zero. Therefore, the vector is perpendicular to the vector. The beginning and end of the last vector are located, respectively, at points that belong to the plane P. Therefore, the vector is perpendicular to the plane P. Distance from point to plane P, whose general equation determined by the formula The proof of this formula is completely similar to the proof of the formula for the distance between a point and a line (see Fig. 2).
Rice. 2. To derive the formula for the distance between a plane and a straight line.

Indeed, the distance d between a straight line and a plane is equal

where is a point lying on the plane. From here, as in lecture No. 11, the above formula is obtained. Two planes are parallel if their normal vectors are parallel. From here we obtain the condition for parallelism of two planes - coefficients of general equations of planes. Two planes are perpendicular if their normal vectors are perpendicular, hence we obtain the condition for the perpendicularity of two planes if their general equations are known

Corner f between two planes equal to angle between their normal vectors (see Fig. 3) and can therefore be calculated using the formula
Determining the angle between planes.

(11)

Distance from a point to a plane and methods for finding it

Distance from point to plane– the length of the perpendicular dropped from a point onto this plane. There are at least two ways to find the distance from a point to a plane: geometric And algebraic.

With the geometric method You must first understand how the perpendicular from a point to a plane is located: maybe it lies in some convenient plane, is a height in some convenient (or not so convenient) triangle, or maybe this perpendicular is generally a height in some pyramid.

After this first and most complex stage, the problem breaks down into several specific planimetric problems (perhaps in different planes).

With the algebraic method in order to find the distance from a point to a plane, you need to enter a coordinate system, find the coordinates of the point and the equation of the plane, and then apply the formula for the distance from a point to the plane.

This article talks about the topic « distance from a point to a line », Discusses the definition of the distance from a point to a line with illustrated examples using the coordinate method. Each theory block at the end has shown examples of solving similar problems.

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The distance from a point to a line is found by determining the distance from point to point. Let's take a closer look.

Let there be a line a and a point M 1 that does not belong to the given line. Through it we draw a straight line b, located perpendicular to the straight line a. Let's take the point of intersection of the lines as H 1. We obtain that M 1 H 1 is a perpendicular that was lowered from point M 1 to straight line a.

Definition 1

Distance from point M 1 to straight line a is called the distance between points M 1 and H 1.

There are definitions that include the length of the perpendicular.

Definition 2

Distance from point to line is the length of the perpendicular drawn from a given point to a given line.

The definitions are equivalent. Consider the figure below.

It is known that the distance from a point to a line is the smallest of all possible. Let's look at this with an example.

If we take a point Q lying on a straight line a, which does not coincide with the point M 1, then we obtain that the segment M 1 Q is called an inclined segment, lowered from M 1 to a straight line a. It is necessary to indicate that the perpendicular from point M 1 is less than any other inclined line drawn from the point to the straight line.

To prove this, consider the triangle M 1 Q 1 H 1, where M 1 Q 1 is the hypotenuse. It is known that its length is always greater than the length of any of the legs. This means we have that M 1 H 1< M 1 Q . Рассмотрим рисунок, приведенный ниже.

The initial data for finding from a point to a line allows you to use several solution methods: through the Pythagorean theorem, determination of sine, cosine, tangent of an angle and others. Most tasks of this type are solved at school during geometry lessons.

When, when finding the distance from a point to a line, it is possible to introduce a rectangular coordinate system, then the coordinate method is used. In this paragraph, we will consider the main two methods of finding the required distance from a given point.

The first method involves searching for the distance as a perpendicular drawn from M 1 to straight line a. The second method uses the normal equation of straight line a to find the required distance.

If there is a point on the plane with coordinates M 1 (x 1 , y 1), located in a rectangular coordinate system, straight line a, and you need to find the distance M 1 H 1, you can make the calculation in two ways. Let's look at them.

First way

If there are coordinates of point H 1 equal to x 2, y 2, then the distance from the point to the line is calculated using the coordinates from the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2.

Now let's move on to finding the coordinates of point H 1.

It is known that a straight line in O x y corresponds to the equation of a straight line on the plane. Let's take the method of defining a straight line a by writing a general equation of a straight line or an equation with an angular coefficient. We compose the equation of a straight line that passes through point M 1 perpendicular to a given straight line a. Let's denote the straight line by the letter b. H 1 is the point of intersection of lines a and b, which means to determine the coordinates you need to use the article in which we're talking about about the coordinates of the points of intersection of two lines.

It can be seen that the algorithm for finding the distance from a given point M 1 (x 1, y 1) to straight line a is carried out according to the points:

Definition 3

• finding the general equation of a straight line a, having the form A 1 x + B 1 y + C 1 = 0, or an equation with an angular coefficient, having the form y = k 1 x + b 1;
• obtaining a general equation of line b, having the form A 2 x + B 2 y + C 2 = 0 or an equation with an angular coefficient y = k 2 x + b 2, if line b intersects point M 1 and is perpendicular to a given line a;
• determination of the coordinates x 2, y 2 of the point H 1, which is the intersection point of a and b, for this purpose the system of linear equations is solved A 1 x + B 1 y + C 1 = 0 A 2 x + B 2 y + C 2 = 0 or y = k 1 x + b 1 y = k 2 x + b 2 ;
• calculating the required distance from a point to a line using the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2.

Second way

The theorem can help answer the question of finding the distance from a given point to a given straight line on a plane.

Theorem

The rectangular coordinate system has O x y has a point M 1 (x 1, y 1), from which a straight line is drawn to the plane, given by the normal equation of the plane, having the form cos α x + cos β y - p = 0, equal to The absolute value obtained on the left side of the normal equation of the line, calculated at x = x 1, y = y 1, means that M 1 H 1 = cos α · x 1 + cos β · y 1 - p.

Proof

Line a corresponds to the normal equation of the plane, having the form cos α x + cos β y - p = 0, then n → = (cos α, cos β) is considered the normal vector of line a at a distance from the origin to line a with p units . It is necessary to display all the data in the figure, add a point with coordinates M 1 (x 1, y 1), where the radius vector of the point M 1 - O M 1 → = (x 1, y 1). It is necessary to draw a straight line from a point to a straight line, which we denote as M 1 H 1 . It is necessary to show the projections M 2 and H 2 of the points M 1 and H 2 onto a straight line passing through the point O with a direction vector of the form n → = (cos α, cos β), and denote the numerical projection of the vector as O M 1 → = (x 1, y 1) to the direction n → = (cos α , cos β) as n p n → O M 1 → .

The variations depend on the location of the M1 point itself. Let's look at the figure below.

We fix the results using the formula M 1 H 1 = n p n → O M → 1 - p. Then we bring the equality to this form M 1 H 1 = cos α · x 1 + cos β · y 1 - p in order to obtain n p n → O M → 1 = cos α · x 1 + cos β · y 1 .

The scalar product of vectors results in a transformed formula of the form n → , O M → 1 = n → · n p n → O M 1 → = 1 · n p n → O M 1 → = n p n → O M 1 → , which is the product in coordinate form of the form n → , O M 1 → = cos α · x 1 + cos β · y 1 . This means that we get that n p n → O M 1 → = cos α · x 1 + cos β · y 1 . It follows that M 1 H 1 = n p n → O M 1 → - p = cos α · x 1 + cos β · y 1 - p. The theorem has been proven.

We find that to find the distance from point M 1 (x 1 , y 1) to straight line a on the plane, you need to perform several actions:

Definition 4

• obtaining the normal equation of the straight line a cos α · x + cos β · y - p = 0, provided that it is not in the task;
• calculation of the expression cos α · x 1 + cos β · y 1 - p, where the resulting value takes M 1 H 1.

Let's apply these methods to solve problems with finding the distance from a point to a plane.

Example 1

Find the distance from the point with coordinates M 1 (- 1, 2) to the straight line 4 x - 3 y + 35 = 0.

Solution

Let's use the first method to solve.

To do this, it is necessary to find the general equation of the line b, which passes through a given point M 1 (- 1, 2), perpendicular to the line 4 x - 3 y + 35 = 0. From the condition it is clear that line b is perpendicular to line a, then its direction vector has coordinates equal to (4, - 3). Thus, we have the opportunity to write down the canonical equation of line b on the plane, since there are coordinates of the point M 1, which belongs to line b. Let's determine the coordinates of the directing vector of the straight line b. We get that x - (- 1) 4 = y - 2 - 3 ⇔ x + 1 4 = y - 2 - 3. The resulting canonical equation must be converted to a general one. Then we get that

x + 1 4 = y - 2 - 3 ⇔ - 3 · (x + 1) = 4 · (y - 2) ⇔ 3 x + 4 y - 5 = 0

Let us find the coordinates of the points of intersection of the lines, which we will take as the designation H 1. The transformations look like this:

4 x - 3 y + 35 = 0 3 x + 4 y - 5 = 0 ⇔ x = 3 4 y - 35 4 3 x + 4 y - 5 = 0 ⇔ x = 3 4 y - 35 4 3 3 4 y - 35 4 + 4 y - 5 = 0 ⇔ ⇔ x = 3 4 y - 35 4 y = 5 ⇔ x = 3 4 5 - 35 4 y = 5 ⇔ x = - 5 y = 5

From what was written above, we have that the coordinates of point H 1 are equal to (- 5; 5).

It is necessary to calculate the distance from point M 1 to straight line a. We have that the coordinates of the points M 1 (- 1, 2) and H 1 (- 5, 5), then we substitute them into the formula to find the distance and get that

M 1 H 1 = (- 5 - (- 1) 2 + (5 - 2) 2 = 25 = 5

Second solution.

In order to solve in another way, it is necessary to obtain the normal equation of the line. We calculate the value of the normalizing factor and multiply both sides of the equation 4 x - 3 y + 35 = 0. From here we get that the normalizing factor is equal to - 1 4 2 + (- 3) 2 = - 1 5, and the normal equation will be of the form - 1 5 4 x - 3 y + 35 = - 1 5 0 ⇔ - 4 5 x + 3 5 y - 7 = 0 .

According to the calculation algorithm, it is necessary to obtain the normal equation of the line and calculate it with the values ​​x = - 1, y = 2. Then we get that

4 5 · - 1 + 3 5 · 2 - 7 = - 5

From this we obtain that the distance from point M 1 (- 1, 2) to the given straight line 4 x - 3 y + 35 = 0 has the value - 5 = 5.

Answer: 5 .

It can be seen that in this method it is important to use the normal equation of the line, since this method is the shortest. But the first method is convenient because it is consistent and logical, although it has more calculation points.

Example 2

On the plane there is a rectangular coordinate system O x y with point M 1 (8, 0) and straight line y = 1 2 x + 1. Find the distance from a given point to a straight line.

Solution

The first method involves reducing a given equation with an angular coefficient to a general equation. To simplify, you can do it differently.

If the product of the angular coefficients of perpendicular lines has a value of - 1, then the angular coefficient of a line perpendicular to a given one y = 1 2 x + 1 has a value of 2. Now we get the equation of a line passing through a point with coordinates M 1 (8, 0). We have that y - 0 = - 2 · (x - 8) ⇔ y = - 2 x + 16 .

We proceed to finding the coordinates of point H 1, that is, the intersection points y = - 2 x + 16 and y = 1 2 x + 1. We compose a system of equations and get:

y = 1 2 x + 1 y = - 2 x + 16 ⇔ y = 1 2 x + 1 1 2 x + 1 = - 2 x + 16 ⇔ y = 1 2 x + 1 x = 6 ⇔ ⇔ y = 1 2 · 6 + 1 x = 6 = y = 4 x = 6 ⇒ H 1 (6, 4)

It follows that the distance from the point with coordinates M 1 (8, 0) to the straight line y = 1 2 x + 1 is equal to the distance from the start point and end point with coordinates M 1 (8, 0) and H 1 (6, 4) . Let's calculate and find that M 1 H 1 = 6 - 8 2 + (4 - 0) 2 20 = 2 5.

The solution in the second way is to move from an equation with a coefficient to its normal form. That is, we get y = 1 2 x + 1 ⇔ 1 2 x - y + 1 = 0, then the value of the normalizing factor will be - 1 1 2 2 + (- 1) 2 = - 2 5. It follows that the normal equation of the line takes the form - 2 5 1 2 x - y + 1 = - 2 5 0 ⇔ - 1 5 x + 2 5 y - 2 5 = 0. Let's carry out the calculation from the point M 1 8, 0 to a line of the form - 1 5 x + 2 5 y - 2 5 = 0. We get:

M 1 H 1 = - 1 5 8 + 2 5 0 - 2 5 = - 10 5 = 2 5

Answer: 2 5 .

Example 3

It is necessary to calculate the distance from the point with coordinates M 1 (- 2, 4) to the lines 2 x - 3 = 0 and y + 1 = 0.

Solution

We obtain the equation of the normal form of the straight line 2 x - 3 = 0:

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

Then we proceed to calculating the distance from the point M 1 - 2, 4 to the straight line x - 3 2 = 0. We get:

M 1 H 1 = - 2 - 3 2 = 3 1 2

The equation of the straight line y + 1 = 0 has a normalizing factor with a value equal to -1. This means that the equation will take the form - y - 1 = 0. We proceed to calculating the distance from the point M 1 (- 2, 4) to the straight line - y - 1 = 0. We find that it is equal to - 4 - 1 = 5.

Answer: 3 1 2 and 5.

Let's take a closer look at finding the distance from a given point on the plane to the coordinate axes O x and O y.

In a rectangular coordinate system, the O axis y has an equation of a straight line, which is incomplete and has the form x = 0, and O x - y = 0. The equations are normal for the coordinate axes, then it is necessary to find the distance from the point with coordinates M 1 x 1, y 1 to the lines. This is done based on the formulas M 1 H 1 = x 1 and M 1 H 1 = y 1. Let's look at the figure below.

Example 4

Find the distance from the point M 1 (6, - 7) to the coordinate lines located in the O x y plane.

Solution

Since the equation y = 0 relates to the line O x, we can find the distance from M 1 s given coordinates, to this straight line using the formula. We get that 6 = 6.

Since the equation x = 0 refers to the straight line O y, you can find the distance from M 1 to this straight line using the formula. Then we get that - 7 = 7.

Answer: the distance from M 1 to O x has a value of 6, and from M 1 to O y has a value of 7.

When in three-dimensional space we have a point with coordinates M 1 (x 1, y 1, z 1), it is necessary to find the distance from point A to straight line a.

Let's consider two methods that allow you to calculate the distance from a point to a straight line a located in space. The first case considers the distance from point M 1 to a line, where a point on the line is called H 1 and is the base of a perpendicular drawn from point M 1 to line a. The second case suggests that the points of this plane must be sought as the height of the parallelogram.

First way

From the definition we have that the distance from point M 1 located on straight line a is the length of the perpendicular M 1 H 1, then we obtain that with the found coordinates of point H 1, then we find the distance between M 1 (x 1, y 1, z 1 ) and H 1 (x 1 , y 1 , z 1) , based on the formula M 1 H 1 = x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2.

We find that the whole solution goes towards finding the coordinates of the base of the perpendicular drawn from M 1 to the straight line a. This is done as follows: H 1 is the point where straight line a intersects with the plane that passes through the given point.

This means that the algorithm for determining the distance from point M 1 (x 1, y 1, z 1) to line a in space implies several points:

Definition 5

• drawing up the equation of the plane χ as an equation of the plane passing through a given point located perpendicular to the line;
• determination of the coordinates (x 2, y 2, z 2) belonging to the point H 1, which is the intersection point of straight line a and plane χ;
• calculating the distance from a point to a line using the formula M 1 H 1 = x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2.

Second way

From the condition we have a straight line a, then we can determine the direction vector a → = a x, a y, a z with coordinates x 3, y 3, z 3 and a certain point M 3 belonging to straight a. If you have the coordinates of the points M 1 (x 1, y 1) and M 3 x 3, y 3, z 3, you can calculate M 3 M 1 →:

M 3 M 1 → = (x 1 - x 3, y 1 - y 3, z 1 - z 3)

We should set aside the vectors a → = a x , a y , a z and M 3 M 1 → = x 1 - x 3 , y 1 - y 3 , z 1 - z 3 from point M 3 , connect them and get a parallelogram figure. M 1 H 1 is the height of the parallelogram.

Let's look at the figure below.

We have that the height M 1 H 1 is the required distance, then it is necessary to find it using the formula. That is, we are looking for M 1 H 1.

Let us denote the area of ​​the parallelogram by the letter S, found by the formula using the vector a → = (a x, a y, a z) and M 3 M 1 → = x 1 - x 3. y 1 - y 3, z 1 - z 3. The area formula is S = a → × M 3 M 1 → . Also, the area of ​​the figure is equal to the product of the lengths of its sides and the height, we get that S = a → · M 1 H 1 with a → = a x 2 + a y 2 + a z 2, which is the length of the vector a → = (a x, a y, a z), being equal side parallelogram. This means that M 1 H 1 is the distance from the point to the line. It is found using the formula M 1 H 1 = a → × M 3 M 1 → a → .

To find the distance from a point with coordinates M 1 (x 1, y 1, z 1) to a straight line a in space, you need to perform several steps of the algorithm:

Definition 6

• determination of the direction vector of the straight line a - a → = (a x, a y, a z);
• calculating the length of the direction vector a → = a x 2 + a y 2 + a z 2 ;
• obtaining coordinates x 3 , y 3 , z 3 belonging to point M 3 located on straight line a;
• calculating the coordinates of the vector M 3 M 1 → ;
• finding vector product vectors a → (a x , a y , a z) and M 3 M 1 → = x 1 - x 3 , y 1 - y 3 , z 1 - z 3 as a → × M 3 M 1 → = i → j → k → a x a y a z x 1 - x 3 y 1 - y 3 z 1 - z 3 to obtain the length using the formula a → × M 3 M 1 → ;
• calculating the distance from a point to a line M 1 H 1 = a → × M 3 M 1 → a → .

Solving problems of finding the distance from a given point to a given line in space

Find the distance from the point with coordinates M 1 2, - 4, - 1 to the line x + 1 2 = y - 1 = z + 5 5.

Solution

The first method begins with writing the equation of the plane χ passing through M 1 and perpendicular to a given point. We get an expression like:

2 (x - 2) - 1 (y - (- 4)) + 5 (z - (- 1)) = 0 ⇔ 2 x - y + 5 z - 3 = 0

It is necessary to find the coordinates of the point H 1, which is the point of intersection with the χ plane to the line specified by the condition. You should move from the canonical view to the intersecting one. Then we obtain a system of equations of the form:

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

It is necessary to calculate the system x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 2 x - y + 5 z - 3 = 0 ⇔ x + 2 y = - 1 5 x - 2 z = 5 2 x - y + 5 z = 3 by Cramer’s method, then we get that:

∆ = 1 2 0 5 0 - 2 2 - 1 5 = - 60 ∆ x = - 1 2 0 5 0 - 2 3 - 1 5 = - 60 ⇔ x = ∆ x ∆ = - 60 - 60 = 1 ∆ y = 1 - 1 0 5 5 2 2 3 5 = 60 ⇒ y = ∆ y ∆ = 60 - 60 = - 1 ∆ z = 1 2 - 1 5 0 5 2 - 1 3 = 0 ⇒ z = ∆ z ∆ = 0 - 60 = 0

From here we have that H 1 (1, - 1, 0).

M 1 H 1 = 1 - 2 2 + - 1 - - 4 2 + 0 - - 1 2 = 11

The second method must begin by searching for coordinates in the canonical equation. To do this, you need to pay attention to the denominators of the fraction. Then a → = 2, - 1, 5 is the direction vector of the line x + 1 2 = y - 1 = z + 5 5. It is necessary to calculate the length using the formula a → = 2 2 + (- 1) 2 + 5 2 = 30.

It is clear that the straight line x + 1 2 = y - 1 = z + 5 5 intersects the point M 3 (- 1 , 0 , - 5), hence we have that the vector with the origin M 3 (- 1 , 0 , - 5) and its end at the point M 1 2, - 4, - 1 is M 3 M 1 → = 3, - 4, 4. Find the vector product a → = (2, - 1, 5) and M 3 M 1 → = (3, - 4, 4).

We get an expression of the form a → × M 3 M 1 → = i → j → k → 2 - 1 5 3 - 4 4 = - 4 i → + 15 j → - 8 k → + 20 i → - 8 · j → = 16 · i → + 7 · j → - 5 · k →

we find that the length of the vector product is equal to a → × M 3 M 1 → = 16 2 + 7 2 + - 5 2 = 330.

We have all the data to use the formula for calculating the distance from a point for a straight line, so let’s apply it and get:

M 1 H 1 = a → × M 3 M 1 → a → = 330 30 = 11

Answer: 11 .

If you notice an error in the text, please highlight it and press Ctrl+Enter

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 mathematical intersection sign, it will appear 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 positions 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 familiar to you from 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 systems of two linear equations in 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.

Graphic 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 the 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.

Complete solution and the 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

We have a straight strip of river in front of us 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 count common 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

Consider two straight lines given by the equations in general view:

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 scalar product directing vectors of straight 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 .