This video lesson will be useful for those who want to independently study the topic “Distance from a point to a line. Distance between parallel lines." During the lesson you will learn how to calculate the distance from a point to a line. Then the teacher will give the definition of the distance between parallel lines.

In this lesson we will get acquainted with the concept "distance" generally. We also specify this concept in the case of calculating distances between two points, a point and a line, parallel lines

Let's look at Figure 1. It shows 2 points A and B. The distance between two points A and B is a segment having ends at given points, that is, segment AB

Rice. 1. AB - distance between points

It is noteworthy that distance cannot be considered a curve or broken line connecting two points. Distance- this is the shortest path from one point to another. It is the segment AB that is the smallest of all possible lines connecting points A and B

Consider Figure 2, which shows the straight line A, and point A, which does not belong to this line. Distance from point A to a straight line will be the length of the perpendicular AN.

Rice. 2. AN - distance between a point and a line

It is important to note that AN is shortest distance, since in the triangle AMN this segment is a leg, and an arbitrary other segment connecting point A and the line A(V in this case- this is AM) will be the hypotenuse. As you know, the leg is always less than the hypotenuse

Distance designation:

Let's consider parallel lines a and b shown in Figure 3

Rice. 3. Parallel lines a and b

Let's fix two points on a straight line a and drop perpendiculars from them onto a line parallel to it b. Let us prove that if ,

Let us draw segment AM for convenience of proof. Let us consider the resulting triangles ABM and ANM. Since , and , then . Likewise, . These right triangles () have a common side AM. It is the hypotenuse in both triangles. Angles AMN and AMB are internal cross angles with parallel straight lines AB and NM and secant AM. By known property, .

From all of the above it follows that . From the equality of triangles it follows that AN = BM

So, we have proven that in Figure 3 the segments AN and BM are equal. It means that distance between parallel lines is the length of their common perpendicular, and the choice of perpendicular can be arbitrary. Thus,

The converse is also true: a set of points that are at the same distance from a certain line form a line parallel to the given one.

Let's consolidate our knowledge and solve several problems

Example 1: Problem 272 from the textbook “Geometry 7-9”. Author - Atanasyan L.S.

In an equilateral triangle ABC, the bisector AD is drawn. The distance from point D to straight line AC is 6 cm. Find the distance from point A to straight line BC

Rice. 4. Drawing for example 1

Solution:

An equilateral triangle is a triangle with three equal sides (and therefore three equal angles, that is, 60 0 each). An equilateral triangle is a special case of an isosceles triangle, therefore all the properties inherent in an isosceles triangle also apply to an equilateral triangle. Therefore, AD is not only a bisector, but also a height, therefore AD ⊥BC

Since the distance from point D to line AC is the length of the perpendicular drawn from point D to line AC, then DH is this distance. Consider the triangle AND. In it, the angle H = 90 0, since DH is perpendicular to AC (by definition of the distance from a point to a straight line). In addition, in this triangle the leg DH lies opposite the angle, so AD = (cm) (By property)

The distance from point A to straight line BC is the length of the perpendicular dropped onto straight line BC. According to the proven AD ⊥BC, it means .

Answer: 12 cm.

Example 2: Problem 277 from the textbook “Geometry 7-9”. Author - Atanasyan L.S.

The distance between parallel lines a and b is 3 cm, and the distance between parallel lines a and c is 5 cm. Find the distance between parallel lines b and c

Solution:

Rice. 5. Drawing for example 2 (first case)

Since , then = 5 - 3 = 2 (cm).

However, this answer is incomplete. There is another option for locating straight lines on a plane:

Rice. 6. Drawing for example 2 (second case)

In this case .

1. Single collection of digital educational resources ().
2. Math tutor ().

1. No. 280, 283. Atanasyan L. S., Butuzov V. F., Kadomtsev S. B., Poznyak E. G., Yudina I. I. edited by Tikhonov A. N. Geometry grades 7-9. M.: Enlightenment. 2010
2. Sum of hypotenuse SE and leg SC right triangle SKE is equal to 31 cm, and their difference is 3 cm. Find the distance from vertex C to straight line KE
3. Based on AB isosceles triangle ABC is taken at point M, equidistant from the sides. Prove that CM is the height of triangle ABC
4. Prove that all points of the plane located on one side of a given line and equidistant from it lie on a line parallel to the given one

Along with a point and a plane. This is an infinite figure that can connect any two points in space. A straight line always belongs to some plane. Based on the location of the two straight lines, one should use different methods finding the distance between them.

There are three options for the location of two lines in space relative to each other: they are parallel, intersect or. The second option is possible only if they are in the same plane; it does not exclude belonging to two parallel planes. The third situation suggests that the lines lie in different parallel planes.

To find the distance between two parallel lines, you need to determine the length of the perpendicular segment connecting them at any two points. Since straight lines have two identical coordinates, which follows from the definition of their parallelism, the equations of straight lines in two-dimensional coordinate space can be written as follows:
L1: a x + b y + c = 0;
L2: a x + b y + d = 0.
Then you can find the length of the segment using the formula:
s = |c - d|/√(a² + b²), and it is easy to notice that when C = D, i.e. If the lines coincide, the distance will be zero.

It is clear that the distance between intersecting lines in two-dimensional coordinates does not make sense. But when they are located in different planes, it can be found as the length of a segment lying in a plane perpendicular to both of them. The ends of this segment will be points that are projections of any two points of lines onto this plane. In other words, its length is equal to the distance between parallel planes containing these lines. Thus, if the planes are given by general equations:
α: A1 x + B1 y + C1 z + E = 0,
β: A2 x + B2 y + C2 z + F = 0,
the distance between straight lines can be used by the formula:
s = |E – F|/√(|A1 A2| + B1 B2 + C1 C2).

note

Straight lines in general and crossed lines in particular are of interest not only to mathematicians. Their properties are useful in many other areas: in construction and architecture, in medicine and in nature itself.

Tip 2: How to find the distance between two parallel lines

Determining the distance between two objects located in one or more planes is one of the most common problems in geometry. Using generally accepted methods, you can find the distance between two parallel lines.

Instructions

Parallel lines are lines lying in the same plane that either do not intersect or coincide. To find the distance between parallel lines, select an arbitrary point on one of them, and then drop a perpendicular to the second line. Now all that remains is to measure the length of the resulting segment. The length of the perpendicular connecting two parallel lines will be the distance between them.

Pay attention to the order in which the perpendicular is drawn from one parallel line to another, since the accuracy of the calculated distance depends on this. To do this, use a right-angle triangle drawing tool. Select a point on one of the lines, attach to it one of the sides of the triangle adjacent to right angle(leg), and align the second side with the other straight line. Using a sharp pencil, draw a line along the first leg so that it reaches the opposite straight line.

In this article, we will examine the issue of finding the distance between two parallel lines, in particular, using the coordinate method. Analysis of typical examples will help consolidate the acquired theoretical knowledge.

Yandex.RTB R-A-339285-1 Definition 1

Distance between two parallel lines is the distance from some arbitrary point of one of the parallel lines to the other line.

Here is an illustration for clarity:

The drawing shows two parallel lines a And b. Point M 1 belongs to line a, from it a perpendicular is dropped onto the line b. The resulting segment M 1 H 1 is the distance between two parallel lines a And b.

The specified definition of the distance between two parallel lines is valid both on the plane and for lines in three-dimensional space. Besides, this definition is interconnected with the following theorem.

Theorem

When two lines are parallel, all points on one of them are equidistant from the other line.

Proof

Let us be given two parallel lines a And b. Let's set it on a straight line A points M 1 and M 2, drop perpendiculars from them to the straight line b, designating their bases as H 1 and H 2, respectively. M 1 H 1 is the distance between two parallel lines by definition, and we need to prove that | M 1 N 1 | = | M 2 N 2 | .

Let there also be some secant that intersects two given parallel lines. The condition of parallelism of lines, discussed in the corresponding article, gives us the right to assert that in this case, the internal crosswise angles formed when the secant of the given lines intersect are equal: ∠ M 2 M 1 H 2 = ∠ H 1 H 2 M 1 . The straight line M 2 H 2 is perpendicular to the straight line b by construction, and, of course, perpendicular to the straight line a. The resulting triangles M 1 H 1 H 2 and M 2 M 1 H 2 are rectangular and equal to each other in hypotenuse and acute angle: M 1 H 2 – common hypotenuse, ∠ M 2 M 1 H 2 = ∠ H 1 H 2 M 1 . Based on the equality of triangles, we can talk about the equality of their sides, i.e.: | M 1 N 1 | = | M 2 N 2 | . The theorem has been proven.

Note that the distance between two parallel lines is the smallest of the distances from the points of one line to the points of the other.

Finding the distance between parallel lines

We have already found out that, in fact, in order to find the distance between two parallel lines, it is necessary to determine the length of the perpendicular dropped from a certain point of one line to another. There are several ways to do this. In some problems it is convenient to use the Pythagorean theorem; others involve the use of signs of equality or similarity of triangles, etc. In cases where lines are specified in a rectangular coordinate system, it is possible to calculate the distance between two parallel lines using the coordinate method. Let's take a closer look at it.

Let's set the conditions. Suppose we have a fixed rectangular coordinate system in which two parallel lines a and b are given. It is necessary to determine the distance between given straight lines.

The solution to the problem will be based on determining the distance between parallel lines: to find the distance between two given parallel lines it is necessary:

Find the coordinates of a certain point M 1 belonging to one of the given lines;

Calculate the distance from point M 1 to a given line to which this point does not belong.

Based on the skills of working with the equations of a straight line on a plane or in space, it is easy to determine the coordinates of point M 1. When finding the distance from point M 1 to a straight line, the material in the article on finding the distance from a point to a straight line will be useful.

Let's go back to the example. Let straight line a be described by the general equation A x + B y + C 1 = 0, and straight line b by the equation A x + B y + C 2 = 0. Then the distance between two given parallel lines can be calculated using the formula:

M 1 H 1 = C 2 - C 1 A 2 + B 2

Let's derive this formula.

We use some point M 1 (x 1, y 1) belonging to the line a. In this case, the coordinates of point M 1 will satisfy the equation A x 1 + B y 1 + C 1 = 0. Thus, the equality is valid: A x 1 + B y 1 + C 1 = 0; from it we get: A x 1 + B y 1 = - C 1 .

When C 2< 0 , нормальное уравнение прямой b будет иметь вид:

A A 2 + B 2 x + B A 2 + B 2 y + C 2 A 2 + B 2 = 0

For C 2 ≥ 0, the normal equation of line b will look like this:

A A 2 + B 2 x + B A 2 + B 2 y - C 2 A 2 + B 2 = 0

And then for cases when C 2< 0 , применима формула: M 1 H 1 = A A 2 + B 2 x 1 + B A 2 + B 2 y 1 + C 2 A 2 + B 2 .

And for C 2 ≥ 0, the required distance is determined by the formula M 1 H 1 = - A A 2 + B 2 x 1 - B A 2 + B 2 y 1 - C 2 A 2 + B 2 = = A A 2 + B 2 x 1 + B A 2 + B 2 y 1 + C 2 A 2 + B 2

Thus, for any value of the number C 2, the length of the segment | M 1 N 1 | (from point M 1 to line b) is calculated by the formula: M 1 H 1 = A A 2 + B 2 x 1 + B A 2 + B 2 y 1 + C 2 A 2 + B 2

Above we received: A x 1 + B y 1 = - C 1, then we can transform the formula: M 1 H 1 = - C 1 A 2 + B 2 + C 2 A 2 + B 2 = C 2 - C 1 A 2 + B 2 . This is how we, in fact, obtained the formula specified in the coordinate method algorithm.

Let's look at the theory using examples.

Example 1

Given two parallel lines y = 2 3 x - 1 and x = 4 + 3 · λ y = - 5 + 2 · λ . It is necessary to determine the distance between them.

Solution

The original parametric equations make it possible to specify the coordinates of the point through which the straight line described by the parametric equations passes. Thus, we obtain point M 1 (4, - 5). The required distance is the distance between point M 1 (4, - 5) to the straight line y = 2 3 x - 1, let's calculate it.

Let us transform the given equation of a straight line with the slope y = 2 3 x - 1 into a normal equation of a straight line. To this end, we first make the transition to the general equation of the straight line:

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

Let's calculate the normalizing factor: 1 2 2 + (- 3) 2 = 1 13. Let's multiply both sides of the last equation by it and, finally, we will be able to write the normal equation of the line: 1 13 · 2 x - 3 y - 3 = 1 13 · 0 ⇔ 2 13 x - 3 13 y - 3 13 = 0.

For x = 4 and y = - 5, we calculate the required distance as the modulus of the value of the extreme equality:

2 13 · 4 - 3 13 · - 5 - 3 13 = 20 13

Answer: 20 13 .

Example 2

In a fixed rectangular coordinate system O x y, two parallel lines are given, defined by the equations x - 3 = 0 and x + 5 0 = y - 1 1. It is necessary to find the distance between given parallel lines.

Solution

The conditions of the problem determine one general equation, specified by one of the original straight lines: x-3=0. Let's transform the original canonical equation into a general one: x + 5 0 = y - 1 1 ⇔ x + 5 = 0. For variable x, the coefficients in both equations are equal (also equal for y – zero), and therefore we can apply the formula to find the distance between parallel lines:

M 1 H 1 = C 2 - C 1 A 2 + B 2 = 5 - (- 3) 1 2 + 0 2 = 8

Answer: 8 .

Finally, consider the problem of finding the distance between two parallel lines in three-dimensional space.

Example 3

In the rectangular coordinate system O x y z, two parallel lines are given, described by the canonical equations of a line in space: x - 3 1 = y - 1 = z + 2 4 and x + 5 1 = y - 1 - 1 = z - 2 4. It is necessary to find the distance between these lines.

Solution

From the equation x - 3 1 = y - 1 = z + 2 4, the coordinates of the point through which the line described by this equation passes are easily determined: M 1 (3, 0, - 2). Let's calculate the distance | M 1 N 1 | from point M 1 to straight line x + 5 1 = y - 1 - 1 = z - 2 4.

The straight line x + 5 1 = y - 1 - 1 = z - 2 4 passes through the point M 2 (- 5, 1, 2). Let us write the direction vector of the straight line x + 5 1 = y - 1 - 1 = z - 2 4 as b → with coordinates (1 , - 1 , 4) . Let's determine the coordinates of the vector M 2 M →:

M 2 M 1 → = 3 - (- 5 , 0 - 1 , - 2 - 2) ⇔ M 2 M 1 → = 8 , - 1 , - 4

Let's calculate the vector product of vectors:

b → × M 2 M 1 → = i → j → k → 1 - 1 4 8 - 1 - 4 = 8 · i → + 36 · j → + 7 · k → ⇒ b → × M 2 M 1 → = ( 8, 36, 7)

Let's apply the formula for calculating the distance from a point to a line in space:

M 1 H 1 = b → × M 2 M 1 → b → = 8 2 + 36 2 + 7 2 1 2 + (- 1) 2 + 4 2 = 1409 3 2

Answer: 1409 3 2 .

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

This video lesson will be useful for those who want to independently study the topic “Distance from a point to a line. Distance between parallel lines." During the lesson you will learn how to calculate the distance from a point to a line. Then the teacher will give the definition of the distance between parallel lines.

In this lesson we will get acquainted with the concept "distance" generally. We also specify this concept in the case of calculating distances between two points, a point and a line, parallel lines

Let's look at Figure 1. It shows 2 points A and B. The distance between two points A and B is a segment having ends at given points, that is, segment AB

Rice. 1. AB - distance between points

It is noteworthy that distance cannot be considered a curve or broken line connecting two points. Distance- this is the shortest path from one point to another. It is the segment AB that is the smallest of all possible lines connecting points A and B

Consider Figure 2, which shows the straight line A, and point A, which does not belong to this line. Distance from point A to a straight line will be the length of the perpendicular AN.

Rice. 2. AN - distance between a point and a line

It is important to note that AN is the shortest distance, since in the triangle AMN this segment is a leg, and an arbitrary other segment connecting point A and the line A(in this case it is AM) will be the hypotenuse. As you know, the leg is always less than the hypotenuse

Distance designation:

Let's consider parallel lines a and b shown in Figure 3

Rice. 3. Parallel lines a and b

Let's fix two points on a straight line a and drop perpendiculars from them onto a line parallel to it b. Let us prove that if ,

Let us draw segment AM for convenience of proof. Let us consider the resulting triangles ABM and ANM. Since , and , then . Likewise, . These right triangles () have a common side AM. It is the hypotenuse in both triangles. Angles AMN and AMB are internal cross angles with parallel straight lines AB and NM and secant AM. According to the well-known property, .

From all of the above it follows that . From the equality of triangles it follows that AN = BM

So, we have proven that in Figure 3 the segments AN and BM are equal. It means that distance between parallel lines is the length of their common perpendicular, and the choice of perpendicular can be arbitrary. Thus,

The converse is also true: a set of points that are at the same distance from a certain line form a line parallel to the given one.

Let's consolidate our knowledge and solve several problems

Example 1: Problem 272 from the textbook “Geometry 7-9”. Author - Atanasyan L.S.

In an equilateral triangle ABC, the bisector AD is drawn. The distance from point D to straight line AC is 6 cm. Find the distance from point A to straight line BC

Rice. 4. Drawing for example 1

Solution:

An equilateral triangle is a triangle with three equal sides (and therefore three equal angles, that is, 60 0 each). An equilateral triangle is a special case of an isosceles triangle, therefore all the properties inherent in an isosceles triangle also apply to an equilateral triangle. Therefore, AD is not only a bisector, but also a height, therefore AD ⊥BC

Since the distance from point D to line AC is the length of the perpendicular drawn from point D to line AC, then DH is this distance. Consider the triangle AND. In it, the angle H = 90 0, since DH is perpendicular to AC (by definition of the distance from a point to a straight line). In addition, in this triangle the leg DH lies opposite the angle, so AD = (cm) (By property)

The distance from point A to straight line BC is the length of the perpendicular dropped onto straight line BC. According to the proven AD ⊥BC, it means .

Answer: 12 cm.

Example 2: Problem 277 from the textbook “Geometry 7-9”. Author - Atanasyan L.S.

The distance between parallel lines a and b is 3 cm, and the distance between parallel lines a and c is 5 cm. Find the distance between parallel lines b and c

Solution:

Rice. 5. Drawing for example 2 (first case)

Since , then = 5 - 3 = 2 (cm).

However, this answer is incomplete. There is another option for locating straight lines on a plane:

Rice. 6. Drawing for example 2 (second case)

In this case .

1. Unified collection of digital educational resources ().
2. Math tutor ().

1. No. 280, 283. Atanasyan L. S., Butuzov V. F., Kadomtsev S. B., Poznyak E. G., Yudina I. I. edited by Tikhonov A. N. Geometry grades 7-9. M.: Enlightenment. 2010
2. The sum of the hypotenuse CE and leg CK of the right triangle SKE is 31 cm, and their difference is 3 cm. Find the distance from vertex C to straight line KE
3. Based on AB of the isosceles triangle ABC, point M is taken, equidistant from the lateral sides. Prove that CM is the height of triangle ABC
4. Prove that all points of the plane located on one side of a given line and equidistant from it lie on a line parallel to the given one

A parallelogram is a quadrilateral whose opposite sides are parallel, that is, they lie on parallel lines (Fig. 1).

Theorem 1. On the properties of the sides and angles of a parallelogram. In a parallelogram, opposite sides are equal, opposite angles are equal, and the sum of the angles adjacent to one side of the parallelogram is 180°.

Proof. In this parallelogram ABCD we draw a diagonal AC and get two triangles ABC and ADC (Fig. 2).

These triangles are equal, since ∠ 1 = ∠ 4, ∠ 2 = ∠ 3 (crosswise angles for parallel lines), and side AC is common. From the equality Δ ABC = Δ ADC it follows that AB = CD, BC = AD, ∠ B = ∠ D. The sum of angles adjacent to one side, for example angles A and D, is equal to 180° as one-sided for parallel lines. The theorem has been proven.

Comment. The equality of opposite sides of a parallelogram means that the segments of parallels cut off by parallel ones are equal.

Corollary 1. If two lines are parallel, then all points on one line are at the same distance from the other line.

Proof. Indeed, let a || b (Fig. 3).

Let us draw perpendiculars BA and CD to straight line a from some two points B and C of line b. Since AB || CD, then figure ABCD is a parallelogram, and therefore AB = CD.

The distance between two parallel lines is the distance from an arbitrary point on one of the lines to the other line.

According to what has been proven, it is equal to the length of the perpendicular drawn from some point of one of the parallel lines to the other line.

Example 1. The perimeter of the parallelogram is 122 cm. One of its sides is 25 cm larger than the other. Find the sides of the parallelogram.

Solution. By Theorem 1, opposite sides of a parallelogram are equal. Let's denote one side of the parallelogram by x and the other by y. Then, by condition $$\left\(\begin(matrix) 2x + 2y = 122 \\x - y = 25 \end(matrix)\right.$$ Solving this system, we obtain x = 43, y = 18. Thus Thus, the sides of the parallelogram are 18, 43, 18 and 43 cm.

Example 2.

Solution. Let Figure 4 meet the conditions of the problem.

Let us denote AB by x, and BC by y. According to the condition, the perimeter of the parallelogram is 10 cm, i.e. 2(x + y) = 10, or x + y = 5. The perimeter of triangle ABD is 8 cm. And since AB + AD = x + y = 5 then BD = 8 - 5 = 3. So BD = 3 cm.

Example 3. Find the angles of the parallelogram, knowing that one of them is 50° greater than the other.

Solution. Let Figure 5 meet the conditions of the problem.

Let us denote the degree measure of angle A by x. Then the degree measure of angle D is x + 50°.

Angles BAD and ADC are one-sided interior angles with parallel lines AB and DC and secant AD. Then the sum of these named angles will be 180°, i.e.
x + x + 50° = 180°, or x = 65°. Thus, ∠ A = ∠ C = 65°, a ∠ B = ∠ D = 115°.

Example 4. The sides of the parallelogram are 4.5 dm and 1.2 dm. From the top acute angle a bisector is drawn. What parts does it divide into? big side parallelogram?

Solution. Let Figure 6 meet the conditions of the problem.

AE is the bisector of an acute angle of a parallelogram. Therefore, ∠ 1 = ∠ 2.