How to quickly draw a straight and beautiful line in Photoshop. Using one ruler How to draw a line parallel to a given point through a point

Given a circle with center ABOUT and period A outside the circle. A) The diameter of the circle is drawn. Using only a ruler*, lower the perpendicular from point A to this diameter. b) Through the point A a straight line is drawn that has no common points with the circle. Using only a ruler, lower the perpendicular from point ABOUT to this straight line.

*Note. In construction tasks, a “ruler” always means not a measuring tool, but a geometric one - with its help you can only draw straight lines (through two existing points), but not measure the distance between points. In addition, a geometric ruler is considered one-sided - it cannot be used to draw a parallel line by simply applying one side of the ruler to two points and drawing a line along the other side.

Hint 1

Use the ends of the diameter rather than the center of the circle.

Hint 2

An angle with a vertex on a circle based on its diameter is a right angle. Knowing this, you can construct two altitudes in a triangle formed by the ends of the diameter and the point A.

Hint 3

Try to solve first a simpler case than the one given in paragraph b), - when a given line intersects a circle.

Solution

A) Let Sun- given diameter (Fig. 1). To solve the problem, just remember the first two tips: if you draw straight lines AB And AC, and then connect the points of their intersection with the circle with the desired vertices of the triangle ABC, then you get two heights of this triangle. And since the altitudes of the triangle intersect at one point, then the straight line CH will be the third height, that is, the desired perpendicular from A to diameter Sun.

b) The solution to this point, however, even in the case given in the third hint, does not seem simpler: yes, we can draw the diameters, connect their ends and get a rectangle ABCD(Fig. 2, in which, for simplicity, the point A marked on the circle), but how does this bring us closer to constructing a perpendicular from the center of the circle?

Here's how: since the triangle AOB isosceles, then perpendicular (height) OK will go through the middle K sides AB. This means that the task is reduced to finding the middle of this side. Surprisingly, we no longer need a circle at all, and period D also, in general, “superfluous.” And here is the segment CD- not superfluous, but on it we will need not some specific point, but a completely arbitrary point E! If we designate as L intersection point BE And A.C.(Fig. 3) and then extend A.E. until the intersection with the continuation B.C. at the point M, then straight L.M.- this is the solution to all our worries and problems!

Is it true, is very similar, What L.M. crosses AB in the middle? This is true. Try to prove it. We will postpone the proof until the end of the problem.

So, we have learned to find the midpoint of a segment AB, which means we have learned to lower the perpendicular to AB from the center of the circle. But what to do with the original problem in which the given line does not intersect the circle, as in Fig. 4?

Let's try to reduce the problem to something already solved. This can be done, for example, like this.

First, we construct a straight line symmetrical to the given one relative to the center of the circle. The construction is clear from Fig. 5, on which this straight line is horizontal under the circle, and the one constructed symmetrical to it is highlighted in red (the two blue points can be taken on the circle completely arbitrarily). At the same time we will lead you through the center ABOUT another straight line perpendicular to one of the sides of the resulting rectangle in a circle in order to obtain on this straight line two segments of equal length.

Having two parallel lines, on one of which two ends and the middle of the segment are already marked, let’s take an arbitrary point T(for example, on a circle) and construct such a point S, which is straight T.S. will be parallel to the existing two straight lines. This construction is shown in Fig. 6.

Thus, we have obtained a chord of the circle parallel to the given line, that is, we have reduced the problem to the previously solved version, because we already know how to draw a perpendicular to such a chord from the center of the circle.

It remains to provide a proof of the fact that we used above.

Quadrangle ABCE in Fig. 3 - trapezoid, L is the point of intersection of its diagonals, and M- the point of intersection of the extensions of its sides. According to the well-known property of a trapezoid (it is also called remarkable property of the trapezoid; you can see how it is proven) direct M.L. passes through the middle of the bases of the trapezoid.

Actually, once again we actually relied on the same theorem already in the last subtask, when we drew the third parallel line.

Afterword

The theory of geometric constructions using a single ruler, when an auxiliary circle with a center is given, was developed by the remarkable German geometer of the 19th century Jacob Steiner (it is more correct to pronounce his surname Steiner as “Steiner”, but in Russian literature the spelling with two “e” has long been established). We have already talked about his mathematical achievements once in the problem “In short, Sklifosovsky”. In the book “Geometric Constructions Performed with a Straight Line and a Fixed Circle,” Steiner proved the theorem according to which any construction that can be performed with a compass and ruler can be performed without a compass if only one circle is given and its center is marked. . Steiner's proof boils down to demonstrating the possibility of carrying out basic constructions usually performed using a compass - in particular, drawing parallel and perpendicular lines. Our task, as is easy to see, is a special case of this demonstration.

However, Steiner’s solution to some problems was not the only one. We will also present the second method.

Take two arbitrary points on this line A And B(Fig. 7). First we construct a perpendicular from A to the (blue) straight line B.O.- this is actually the solution to our first problem, because this straight line contains the diameter of the circle; all corresponding constructions in Fig. 7 are in blue. Then we construct a perpendicular from B to the (green) straight line A.O.- this is exactly the same solution to exactly the same problem, the constructions are made in green. Thus we got two heights of the triangle AOB. The third altitude of this triangle passes through the center O and the point of intersection of the other two heights. It is the desired perpendicular to the line AB.

But that's not all. Despite the (relative) simplicity of the second method, it is “excessively long”. This means that there is another construction method that requires fewer operations (in construction problems, each line drawn with a compass or ruler is counted as one operation). Constructions that require the minimum number of operations among the known ones were called by the French mathematician Emile Lemoine (1840–1912) geometric(see: Geometrography).

So, we bring to your attention a geometric solution to the point b). It only requires 10 steps, with the first six being “natural” and the next three being “amazing”. The very last step, drawing a perpendicular, should perhaps also be called natural.

We want to draw a red dotted perpendicular (Fig. 8), for this we need to find some point on it other than ABOUT. Go.

1) Let A is an arbitrary point on a line, and C- an arbitrary point on a circle. We carry out a direct A.C..

2)–3) We draw the diameter O.C.(secondarily intersecting the circle at the point D) and straight line AD. Mark the second points of intersection of the lines A.C. And AD with a circle - B And E, respectively.

4)–6) We carry out BE, BD And C.E.. Direct CD And BE crossed at a point H, A BD And C.E.- at the point G(Fig. 9).

By the way, could it happen that BE would be parallel CD? Yes, definitely. In case the diameter CD perpendicular A.O., then this is exactly what happens: BE And CD are parallel and the points A, O And G lie on the same straight line. But the opportunity to take the point C arbitrarily assumes our ability to choose it so that CO And A.O. were not perpendicular!

And now the promised amazing construction steps:

7) Conduct G.H. until it intersects a given line at a point I.
8) Conduct C.I. until it intersects the circle at the point J.
9) Conduct B.J., which intersects with G.H.... Where? That's right, at the red point, which is located on the vertical diameter of the circle (Fig. 10).

10) Draw the vertical diameter.

Instead of step 8, you could draw a straight line D.I., and then in step 9 connect the second point of its intersection with the circle with the point E. The result would be the same red dot. Isn't this surprising? Moreover, it is not even clear what is more surprising - the fact that the red dot turns out to be the same for the two construction methods, or the fact that it lies on the desired perpendicular. However, geometry is not the “art of fact”, but the “art of proof”. So try to prove it.

A point is an abstract object that has no measuring characteristics: no height, no length, no radius. Within the scope of the task, only its location is important

The point is indicated by a number or a capital (capital) Latin letter. Several dots - with different numbers or different letters so that they can be distinguished

point A, point B, point C

A B C

point 1, point 2, point 3

1 2 3

You can draw three dots “A” on a piece of paper and invite the child to draw a line through the two dots “A”. But how to understand through which ones? A A A

A line is a set of points. Only the length is measured. It has no width or thickness

Indicated by lowercase (small) Latin letters

line a, line b, line c

a b c

The line may be

1. closed if its beginning and end are at the same point,
2. open if its beginning and end are not connected

closed lines

open lines

You left the apartment, bought bread at the store and returned back to the apartment. What line did you get? That's right, closed. You are back to your starting point. You left the apartment, bought bread at the store, went into the entrance and started talking with your neighbor. What line did you get? Open. You haven't returned to your starting point. You left the apartment and bought bread at the store. What line did you get? Open. You haven't returned to your starting point.

1. self-intersecting
2. without self-intersections

self-intersecting lines

lines without self-intersections

1. straight
2. broken
3. crooked

straight lines

broken lines

curved lines

A straight line is a line that is not curved, has neither beginning nor end, it can be continued endlessly in both directions

Even when a small section of a straight line is visible, it is assumed that it continues indefinitely in both directions

Indicated by a lowercase (small) Latin letter. Or two capital (capital) Latin letters - points lying on a straight line

straight line a

a

straight line AB

B A

Direct may be

1. intersecting if they have a common point. Two lines can intersect only at one point.
   • perpendicular if they intersect at right angles (90°).
2. Parallel, if they do not intersect, do not have a common point.

parallel lines

intersecting lines

perpendicular lines

A ray is a part of a straight line that has a beginning but no end; it can be continued indefinitely in only one direction

The ray of light in the picture has its starting point as the sun.

Sun

A point divides a straight line into two parts - two rays A A

The beam is designated by a lowercase (small) Latin letter. Or two capital (capital) Latin letters, where the first is the point from which the ray begins, and the second is the point lying on the ray

ray a

a

beam AB

B A

The rays coincide if

1. located on the same straight line
2. start at one point
3. directed in one direction

rays AB and AC coincide

rays CB and CA coincide

C B A

A segment is a part of a line that is limited by two points, that is, it has both a beginning and an end, which means its length can be measured. The length of a segment is the distance between its starting and ending points

Through one point you can draw any number of lines, including straight lines

Through two points - an unlimited number of curves, but only one straight line

curved lines passing through two points

B A

straight line AB

B A

A piece was “cut off” from the straight line and a segment remained. From the example above you can see that its length is the shortest distance between two points. ✂ B A ✂

A segment is denoted by two capital (capital) Latin letters, where the first is the point at which the segment begins, and the second is the point at which the segment ends

segment AB

B A

Problem: where is the line, ray, segment, curve?

A broken line is a line consisting of consecutively connected segments not at an angle of 180°

A long segment was “broken” into several short ones

The links of a broken line (similar to the links of a chain) are the segments that make up the broken line. Adjacent links are links in which the end of one link is the beginning of another. Adjacent links should not lie on the same straight line.

The vertices of a broken line (similar to the tops of mountains) are the point from which the broken line begins, the points at which the segments that form the broken line are connected, and the point at which the broken line ends.

A broken line is designated by listing all its vertices.

broken line ABCDE

vertex of polyline A, vertex of polyline B, vertex of polyline C, vertex of polyline D, vertex of polyline E

broken link AB, broken link BC, broken link CD, broken link DE

link AB and link BC are adjacent

link BC and link CD are adjacent

link CD and link DE are adjacent

A B C D E 64 62 127 52

The length of a broken line is the sum of the lengths of its links: ABCDE = AB + BC + CD + DE = 64 + 62 + 127 + 52 = 305

Task: which broken line is longer, A which has more vertices? The first line has all the links of the same length, namely 13 cm. The second line has all links of the same length, namely 49 cm. The third line has all links of the same length, namely 41 cm.

A polygon is a closed polygonal line

The sides of the polygon (the expressions will help you remember: “go in all four directions”, “run towards the house”, “which side of the table will you sit on?”) are the links of a broken line. Adjacent sides of a polygon are adjacent links of a broken line.

The vertices of a polygon are the vertices of a broken line. Adjacent vertices are the endpoints of one side of the polygon.

A polygon is denoted by listing all its vertices.

closed polyline without self-intersection, ABCDEF

polygon ABCDEF

polygon vertex A, polygon vertex B, polygon vertex C, polygon vertex D, polygon vertex E, polygon vertex F

vertex A and vertex B are adjacent

vertex B and vertex C are adjacent

vertex C and vertex D are adjacent

vertex D and vertex E are adjacent

vertex E and vertex F are adjacent

vertex F and vertex A are adjacent

polygon side AB, polygon side BC, polygon side CD, polygon side DE, polygon side EF

side AB and side BC are adjacent

side BC and side CD are adjacent

CD side and DE side are adjacent

side DE and side EF are adjacent

side EF and side FA are adjacent

A B C D E F 120 60 58 122 98 141

The perimeter of a polygon is the length of the broken line: P = AB + BC + CD + DE + EF + FA = 120 + 60 + 58 + 122 + 98 + 141 = 599

A polygon with three vertices is called a triangle, with four - a quadrilateral, with five - a pentagon, etc.

The methods for constructing parallel lines using various tools are based on the signs of parallel lines.

Constructing parallel lines using a compass and ruler

Let's consider the principle of constructing a parallel line passing through a given point, using a compass and ruler.

Let a line be given and some point A that does not belong to the given line.

It is necessary to construct a line passing through a given point $A$ parallel to the given line.

In practice, it is often necessary to construct two or more parallel lines without a given line and point. In this case, it is necessary to draw a straight line arbitrarily and mark any point that will not lie on this straight line.

Let's consider stages of constructing a parallel line:

In practice, they also use the method of constructing parallel lines using a drawing square and a ruler.

Constructing parallel lines using a square and ruler

For constructing a line that will pass through point M parallel to a given line a, necessary:

1. Apply the square to the straight line $a$ diagonally (see figure), and attach a ruler to its larger leg.
2. Move the square along the ruler until the given point $M$ is on the diagonal of the square.
3. Draw the required straight line $b$ through the point $M$.

We have obtained a line passing through a given point $M$, parallel to a given line $a$:

$a \parallel b$, i.e. $M \in b$.

The parallelism of straight lines $a$ and $b$ is evident from the equality of the corresponding angles, which are marked in the figure with the letters $\alpha$ and $\beta$.

Construction of a parallel line spaced at a specified distance from a given line

If it is necessary to construct a straight line parallel to a given straight line and spaced from it at a given distance, you can use a ruler and a square.

Let a straight line $MN$ and a distance $a$ be given.

1. Let us mark an arbitrary point on the given line $MN$ and call it $B$.
2. Through the point $B$ we draw a line perpendicular to the line $MN$ and call it $AB$.
3. On the straight line $AB$ from the point $B$ we plot the segment $BC=a$.
4. Using a square and a ruler, we draw a straight line $CD$ through the point $C$, which will be parallel to the given straight line $AB$.

If we plot the segment $BC=a$ on the straight line $AB$ from point $B$ in the other direction, we obtain another parallel line to the given one, spaced from it at a given distance $a$.

Other ways to construct parallel lines

Another way to construct parallel lines is to construct using a crossbar. Most often this method is used in drawing practice.

When performing carpentry work for marking and constructing parallel lines, a special drawing tool is used - a clapper - two wooden planks that are fastened with a hinge.

Construction of straight lines is the basis of technical drawing. Nowadays this is increasingly done with the help of graphic editors, which provide the designer with great opportunities. However, some principles of construction remain the same as in classical drawing - using a pencil and ruler.

You will need

• - paper;
• - pencil;
• - ruler;
• - computer with AutoCAD program.

Instructions

• Start with the classic construction. Determine the plane in which you will build the line. Let this be the plane of a sheet of paper. Depending on the conditions of the problem, arrange the points. They can be arbitrary, but it is possible that some kind of coordinate system is specified. Place arbitrary points where you like best. Label them A and B. Use a ruler to connect them. According to the axiom, it is always possible to draw a straight line through two points, and only one.
• Draw a coordinate system. Let you be given the coordinates of point A (x1; y1). To find them, you need to plot the required number along the x-axis and draw a straight line parallel to the y-axis through the marked point. Then plot the value equal to y1 along the corresponding axis. From the marked point, draw a perpendicular until it intersects with the first one. The place of their intersection will be point A. In the same way, find point B, the coordinates of which can be designated as (x2; y2). Connect both points with a straight line.
• In AutoCAD, a straight line can be constructed in several ways. The two-point function is usually installed by default. Find the “Home” tab in the top menu. You will see the Draw panel in front of you. Find the button with the image of a straight line and click on it.
• A straight line from two points can be constructed in two ways in this program. Place the cursor at the desired point on the screen and click the left mouse button. Then determine the second point, draw a line there and click the mouse too.
• AutoCAD also allows you to specify the coordinates of both points. Type (_xline) in the command line below. Press Enter. Enter the coordinates of the first point and also press enter. Determine the second point in the same way. It can also be specified by clicking the mouse, placing the cursor at the desired point on the screen.
• In AutoCAD, you can build a straight line not only by two points, but also by the angle of inclination. From the Draw context menu, select Line and then the Angle option. The starting point can be set by clicking the mouse or using coordinates, as in the previous method. Then set the angle size and press enter. By default, the straight line will be located at the desired angle to the horizontal.

Content:

Parallel lines are lines the distance between which does not change and which never intersect. In some problems, you are given a line and a point through which you need to draw a line parallel to the given one. Of course, you can take a ruler and draw a straight line parallel to the given one by eye, but there is no guarantee that the constructed straight line will be parallel to the given one. Using geometric laws and a compass, you can plot additional points through which a real parallel line will pass.

Steps

1 Construction of perpendiculars

1. 1 This point does not lie on this line - most likely, it is located above or below the line. Designate this line as m 2 Draw an arc that intersects this line at two points. To do this, install the compass needle at point A 3 Draw the first small arc opposite this point. First increase the compass solution. Place the compass needle at point B 4 Draw a second minor arc that will intersect the first minor arc. Do not change the compass solution. Place the compass needle at point C 5 Draw a line passing through the intersection point of the two arcs and the given point. Label this line as n
   • Remember that a perpendicular is a segment (in this case a straight line) that intersects another segment (a straight line) at an angle of 90 degrees.
2. 6 Draw an arc that intersects a perpendicular line at two points. To do this, install the compass needle at point A 7 Draw the first small arc to the right (or left) of this point. Increase the compass solution. Place the compass needle at point E 8 Draw a second small arc to the right (or left) of this point. Do not change the compass solution. Install the compass needle at point F 9 Draw a line through the intersection point of the two arcs and the given point. The resulting straight line will be perpendicular to the straight line n. Thus, the resulting straight line is parallel to the given straight line m

   2 Construction of a rhombus

   1. 1 Label this line and this point. This point does not lie on this line; most likely, it is located above or below the line. Consider this point as the vertex of a rhombus. Since opposite sides of a rhombus are parallel, by constructing a rhombus you will get a parallel line.
      • Find the second vertex of the diamond. Place the compass needle at a given point and draw an arc that intersects the given line at one point. Do not change the compass solution.
        • The width of the compass opening is not important - the main thing is to draw an arc that will intersect a given straight line at any point.
        • Draw an arc so that it not only intersects this line, but also goes just above this point.
        • For example, set the compass needle at point A 3 Find the third vertex of the diamond. Without changing the angle of the compass, install its needle at the second vertex and draw an arc that intersects this line at a new point. Do not change the compass solution.
          • Draw a short arc so that it only intersects this line.
          • For example, set the compass needle at point B 4 Find the fourth vertex of the diamond. Without changing the angle of the compass, install its needle at the third vertex and draw an arc that intersects the first arc (which you drew by installing the compass needle at this point, and with the help of which you found the second vertex).
            • Draw a short arc so that it just intersects the first arc.
            • For example, set the compass needle at point C 5 Draw a line through the first and fourth vertices of the rhombus. This line passes through a given point and is parallel to a given line, because these lines are opposite sides of a rhombus.
              • For example, a straight line passing through points A

                3 Construction of corresponding angles

                1. 1 Label this line and this point. This point does not lie on this line; most likely, it is located above or below the line.
                   • If the straight line and the point are not yet marked, do so to avoid confusion.
                   • For example, denote this line as m 2 Draw a line through a given point and any point that lies on a given line. Using such a secant line, you can construct the corresponding angles, and then draw a parallel line.
                     • Draw a long secant line so that it goes beyond the given point.
                     • For example, through point A 3 Take a compass. Make the width of the compass opening less than half the length of the resulting segment.
                       • The exact width of the compass opening does not matter - the main thing is that it is less than half the length of the resulting segment.
                       • For example, make the width of the compass opening less than half the length of the segment A B 4 Construct the first corner. Place the compass needle at the point of intersection of the secant line with the given line. Draw an arc that intersects the secant line and the given line. Do not change the compass solution.
                         • For example, set the compass needle at point B 5 Draw a second arc. Without changing the solution of the compass, install its needle at this point. Draw an arc that intersects the secant line above the given point and goes just below the given point.
                           • For example, set the compass needle at point A 6 Take a compass. Make the width of the compass opening equal to the width of the constructed (first) angle.
                             • For example, the constructed angle is angle C B D 7 Construct the corresponding angle. The opening of the compass should be equal to the width of the first corner. Place the compass needle at a point that lies on the secant line above this point, and draw an arc that intersects the second arc.
                               • For example, set the compass needle at point P 8 Draw a line through this point and the intersection point of the two arcs. This line is parallel to the given line and passes through the given point.
                                 • For example, draw a line through point A (displaystyle A) and point Q (displaystyle Q). The result is a straight line f (displaystyle f) parallel to a straight line m (displaystyle m).

                What you will need

                1. Pen or pencil
                2. Ruler
                3. Compass