Point, line, straight line, ray, segment, broken line. How to quickly draw a straight and beautiful line in Photoshop How to draw a straight line parallel to a given point through a point

Given a circle with a center ABOUT and dot A outside the circle. A) The diameter of the circle has been drawn. Using only a ruler*, lower the perpendicular from a point A for this diameter. b) through the dot A a straight line is drawn without common points with a circle. Using only the ruler lower the perpendicular from a point ABOUT to this line.

*Note. By “ruler” in construction tasks, it is always meant 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 attaching one side of the ruler to two points and drawing a line along the other side.

Hint 1

Use the ends of the diameter, not 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 heights in the triangle formed by the ends of the diameter and the point A.

Hint 3

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

Solution

A) Let Sun- given diameter (Fig. 1). To solve the problem, just remember the first two clues: 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 we get two heights of this triangle. And since the altitudes of the triangle intersect at one point, the 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 diameters, connect their ends and get a rectangle ABCD(Fig. 2, in which, for simplicity, the dot A marked on the circle), but how does this get us closer to building a perpendicular from the center of the circle?

But how: since the triangle AOB isosceles, then perpendicular (height) OK will go through the middle K sides AB. So, the task was reduced to finding the middle of this side. Surprisingly, we no longer need a circle at all, and a point D also, in general, "extra". And here is the cut CD- not superfluous, but on it we need not some specific point, but a completely arbitrary point E! If designated as L intersection point BE And AC(Fig. 3) and then extend AE before the intersection with the continuation BC at the point M, then the straight line LM is the solution to all our worries and problems!

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

So, we have learned to find the midpoint of a segment AB, which means that 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?

We will try to reduce the problem to the already solved one. This can be done, for example, like this.

First, we construct a line symmetrical to the given one with respect 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 (two blue dots can be taken on circles quite arbitrarily). At the same time we will draw through the center ABOUT another straight line perpendicular to one of the sides of the resulting rectangle in a circle in order to get two segments of equal length on this straight line.

Having two parallel lines, on one of which two ends and the middle of the segment are already marked, we take an arbitrary point T(for example, on a circle) and construct such a point S, which is straight TS will be parallel to the existing two lines. This construction is shown in Fig. 6.

Thus, we obtained a chord of a circle parallel to a given straight line, that is, we 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 prove the fact that we used above.

quadrilateral ABCE in fig. 3 - trapezoid, L is the intersection point of its diagonals, and M- the point of intersection of the extensions of its lateral sides. According to the well-known property of the trapezoid (it is also called wonderful property of the trapezoid; you can see how it is proved) direct ML passes through the midpoints of the bases of the trapezium.

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 with one 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 the spelling with two “e” has long been fixed in Russian literature). We already once talked about his mathematical achievements in the problem "In short, Sklifosovsky". In the book "Geometric constructions performed with a straight line and a fixed circle" Steiner proved a theorem according to which any construction that can be performed using a compass and straightedge can be performed without a compass if only one circle is given and its center is marked. . Steiner's proof is reduced to demonstrating the possibility of carrying out basic constructions, usually performed with a compass, in particular, to drawing parallel and perpendicular lines. Our task, as is easy to see, is a special case of this demonstration.

However, Steiner did not lead to some problems. the only way solutions. We present the second method.

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

But that's not all. Despite all 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 by a compass or straightedge is counted as one operation). Constructions that require the minimum among the known number of operations, the French mathematician Emile Lemoine (Émile Lemoine, 1840–1912) called geometric(see: Geometrography).

So, your attention is invited to the geometric solution of 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, perhaps, should also be called natural.

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

1) Let A is an arbitrary point on the line, and C is an arbitrary point on the circle. We draw a straight line AC.

2)–3) Draw the diameter OC(secondary crossing the circle at a point D) and direct AD. We mark the second points of intersection of lines AC And AD with a circle B And E, respectively.

4)–6) We spend BE, BD And CE. Direct CD And BE intersect at a point H, A BD And CE- at the point G(Fig. 9).

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

And now the promised awesome build steps:

7) We conduct GH to the intersection with the given line at the point I.
8) We spend CI to the intersection with the circle at a point J.
9) We spend b.j., which intersects with GH... Where? That's right, at the red dot, which is located on the vertical diameter of the circle (Fig. 10).

10) Draw a vertical diameter.

Instead of step 8, one could draw a straight line DI, and then in step 9 connect the second point of its intersection with the circle with a point E. The result would be the same red dot. Is it really amazing? Moreover, it is not even clear what is more surprising - that the red dot is the same for the two construction methods, or that it lies on the required perpendicular. However, geometry is not “the art of fact”, but “the art of proof”. So try to prove it.

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

You will need

- paper;
- pencil;
- ruler;
- computer with AutoCAD software.

Instruction

Start with a classic build. Determine the plane in which you will draw the line. Let this be the plane of a sheet of paper. Arrange the points depending on the conditions of the problem. They can be arbitrary, but it is possible that some coordinate system is given. Arbitrary points put 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 set aside the desired number along the x-axis and draw a straight line parallel to the y-axis through the marked point. Then plot a value equal to y1 along the corresponding axis. Draw a perpendicular from the marked point until it intersects with the first. The place of their intersection will be point A. In the same way, find point B, the coordinates of which can be denoted as (x2; y2). Connect both points with a straight line.
In AutoCAD, a straight line can be drawn in several ways. The "two-point" function is usually set by default. Find the "Home" tab in the top menu. You will see the Drawing panel in front of you. Find the button with the straight line and click on it.
A straight line from two points in this program can be constructed in two ways. Place the cursor on the desired point on the screen and click the left mouse button. Then define the second point, drag a line there and click the mouse too.
AutoCAD also allows you to set the coordinates of both points. Type in the command line below (_xline). Press Enter. Enter the coordinates of the first point and also press enter. Define the second point in the same way. It can also be specified with a mouse click by 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 a straight line and then the Angle option. The starting point can be set by mouse click or by coordinates, as in the previous method. Then set the corner size and hit enter. By default, the line will be positioned at the desired angle to the horizontal.

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

Constructing parallel lines with a compass and straightedge

Consider principle of constructing a parallel line passing through given point , using a compass and ruler.

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

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

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

Consider steps for constructing a parallel line:

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

Construction of parallel lines using a square and a ruler

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

Attach the square to the straight line $a$ with a diagonal (see the figure), and attach a ruler to its larger leg.
Move the square along the ruler until the given point $M$ is on the diagonal of the square.
Draw the desired 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 the lines $a$ and $b$ is evident from the equality of the corresponding angles, which are marked in the figure by the letters $\alpha$ and $\beta$.

Construction of a parallel line at a given 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 line $MN$ and a distance $a$ be given.

We mark an arbitrary point on the given line $MN$ and call it $B$.
Through the point $B$ we draw a line perpendicular to the line $MN$ and call it $AB$.
On the line $AB$ from the point $B$ we plot the segment $BC=a$.
With the help of a square and a straightedge, draw the line $CD$ through the point $C$, which will be parallel to the given line $AB$.

If we postpone the segment $BC=a$ on the line $AB$ from the point $B$ to the other side, then we get one more line parallel to the given one, spaced from it by the given distance $a$.

Other ways to draw parallel lines

Another way to build parallel lines is to build with a T-square. Most often, this method is used in drawing practice.

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

Content:

Parallel lines are lines whose distance 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 line parallel to the given one by eye, but there is no guarantee that the constructed line will be parallel to the given one. With the help of geometric laws and a compass, additional points can be plotted through which a real parallel line will pass.

Steps

1 Construction of perpendiculars

1 The given point does not lie on the given line - most likely, it is above or below the line. Designate this line as m 2 Draw an arc that intersects the given line at two points. To do this, set the compass needle at point A 3 Draw the first small arc opposite this point. First increase the compass solution. Set the compass needle at point B 4 Draw a second small arc that will intersect the first small arc. Do not change the compass solution. Set the compass needle at point C 5 Draw a line passing through the point of intersection of the two arcs and the given point. Designate this line as n
- Remember that a perpendicular is a line segment (in this case, a line) that intersects another line segment (a line) at an angle of 90 degrees.
6 Draw an arc that intersects a perpendicular line at two points. To do this, set the compass needle at point A 7 Draw the first small arc to the right (or left) of this point. Increase the compass solution. Set 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. Set the compass needle at point F 9 Draw a line through the point of intersection of the two arcs and the given point. The resulting line will be perpendicular to the line n Thus, the resulting line is parallel to the given line m
2 Building a rhombus
1. 1 Label this line and this point. The given point does not lie on the given line - most likely, it is above or below the line. Think of this point as the vertex of the 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 rhombus. Set the compass needle at this point and draw an arc that will intersect 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 intersects a given straight line at any point.
    - Draw an arc so that it not only intersects the given line, but also goes just above the given point.
    - For example, set the compass needle at point A 3 Find the third vertex of the rhombus. Without changing the solution of the compass, set its needle at the second vertex and draw an arc that will intersect this straight line at a new point. Do not change the compass solution.
      - Draw a short arc so that it only intersects the given line.
      - For example, set the compass needle at point B 4 Find the fourth vertex of the rhombus. Without changing the solution of the compass, set its needle at the third vertex and draw an arc that will intersect the first arc (which you drew by setting the compass needle at this point, and with which you found the second vertex).
        Draw a short arc so that it just crosses the first arc.
        
        For example, set the compass needle at point C 5 Draw a straight line through the first and fourth vertices of the rhombus. This line passes through the given point and is parallel to the given line, because these lines are opposite sides rhombus.
        For example, a line passing through points A
        3 Construction of the corresponding corners
        
        1 Label this line and this point. The given point does not lie on the given line - most likely, it is above or below the line.
        If the line and point are not already marked, do so so as not to get confused.
        
        For example, denote this line as m 2 Draw a line through the given point and any point that lies on the 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 the compass. Make the width of the compass solution 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 be 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 Build the first corner. Set the compass needle at the point of intersection of the secant line with this line. Draw an arc that intersects the secant 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, set 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 the compass. Make the width of the compass opening equal to the width of the constructed (first) angle.
        For example, the constructed angle is the angle C B D 7 Build the appropriate angle. The opening of the compass should be equal to the width of the first corner. Set the compass needle at a point that lies on the secant line above this point, and draw an arc that will intersect the second arc.
        For example, set the compass needle at point P 8 Draw a line through the given 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) . You get a line f (displaystyle f) , parallel to line m (displaystyle m) .
        
        What will you need
        
        Pen or pencil
        
        Ruler
        
        Compass