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How to calculate the problem of moving point on the first axis of junior high school
The problem of moving point on the number axis is a very important and difficult problem in grade seven. When students encounter it, they just need to learn the word "dizzy". But this knowledge point has to be learned, because it is comprehensive and abstract, and it is a very common and important comprehensive problem, which requires students' comprehensive ability to use knowledge, involving "the geometric meaning of absolute value, the representation of numbers on the number axis, and the distance problem"

Basic knowledge:

How to express the distance between two points on the 1. number axis?

It can be expressed in absolute value, that is, the absolute value of the difference between the numbers represented by two points. For example, if the numbers represented by points A and B on the number axis are A and B, then AB=|a-b| or | B-A |.

2. How to use letters to represent a moving point on the number axis?

It can be solved by addition or subtraction of rational numbers, that is, from the starting point, the moving distance of the moving point is increased and decreased in the positive direction. For example, the number corresponding to point A on the number axis is-1, and point P starts from point A and moves to the right at a speed of 2 unit lengths per second. Let the moving time be t and the number represented by point p be-1+.

3. How to find the midpoint of the line segment between any two points on the axis?

The sum of the numbers represented by two points is divided by two. If the points on the number axis represent numbers A and B, then the number represented by the midpoint of the line segment AB is (a+b)/2.

Strategic approach:

To solve the moving point problem, we must first understand the meaning of the problem, understand the whole process of motion and the change of the graph, then draw the graph according to the classification discussion of the motion process, and finally find the equivalent relationship equation for different situations to solve.

For the moving point problem based on the number axis, there are often two different solutions because of the characteristics of the number axis itself. One is to find the equivalent relationship according to the analysis of the relationship between "shape", that is, to solve it by using the equation of the quantitative relationship between each line segment; The other is to find the equivalent relationship from the aspect of "number", that is, to form equations by using the internal relationship between numbers represented by points on the number axis.

Study on the law of 1 axis

Trick: use ideas from special to general.

Example 1. (At the end of Yinzhou District in the spring of 20 18) As shown in the figure, the initial position of point A is at the point of 1 on the number axis. Now, move point A as follows: 1 time, it moves 3 unit lengths to the left to point B, the second time, it moves 6 unit lengths from point B to point C, and the third time, it moves 9 unit lengths from point C to the left.

Analysis: this question examines the number axis, and the quantities with opposite meanings can be represented by positive numbers and negative numbers. Also, the coordinate change and translation law of points on the number axis (left MINUS right plus) are investigated, and the law of a column number is investigated. It is the key to solve this problem to explore the odd and even terms of this column respectively.

According to the coordinate change and translation law of a point on the number axis (left minus right plus), the numbers corresponding to the point are calculated respectively, and then the distance from the point to the origin is calculated; Then explore the odd and even terms respectively, find out the law (the difference between two adjacent numbers is 3), and write expressions to solve the problem.

Answer: At the 1 th time point, A moves to the left by 3 unit lengths to point B, then the number represented by B is1-3 =-2;

Move from point B to point C for the second time by 6 unit lengths, and the number represented by c is-2+6 = 4;

If you move 9 unit lengths from point C to point D for the third time, the number represented by D is 4-9 =-5;

Move 12 unit lengths from point D to point E for the fourth time, and the number represented by point E is-5+12 = 7;

Move 15 unit lengths from point e to point f for the fifth time, and the number represented by f is 7-15 =-8;

…;

According to the above data, when the number of movements is odd, the number represented by the points on the number axis satisfies:-1/2 (3n+ 1).

When the number of movements is even, the number represented by a point on the number axis satisfies: 1/2(3n+2),

When the number of movements is odd,-1/2 (3n+1) =-2018, n= 1345,

When the number of moves is even, 1/2(3n+2)=20 18, and n=4034/3 (irrelevant).

So the answer is: 1345.

Perception: a point on the number axis represents the number A, and after moving to the left by B units, it represents the number A-B; After B units are moved to the right, the number is A+B. When exploring the change law by using this characteristic, we should pay attention to the direction change in the process of circular reciprocating motion.

Distance problem on the number axis of type 2

Trick: Use the idea of number and shape classification and combination.

Example 2. (Huangpu District 20 17 autumn) It is known that M and N are on the number axis, the number corresponding to M is -3, the point N is on the right side of M, which is 4 unit lengths away from M, and the points P and Q are two moving points on the number axis;

(1) Write the number corresponding to point n directly;

(2) When the sum of the distances from point P to point M and point N is 5 units, what is the number corresponding to point P?

(3) If P and Q start from point M and point N, respectively, and both move to the left along the number axis, point P takes 2 unit lengths per second, starting with 5 seconds, and point Q takes 3 unit lengths per second. When the two points of P and Q are separated by 2 unit lengths, what are the corresponding points of P and Q?

The distance between two points and the number axis are checked in this analysis. The mathematical idea of "classified discussion" should be adopted when solving problems.

(1) can be solved according to the distance formula between two points;

(2) There are two situations: ① Point P is to the left of point M; ② Point P is on the right of point N; Discuss and solve;

(3) There are two situations: ① Point P is to the left of point Q; ② Point P is to the right of point Q; Discuss and solve.

Solution (1)-3+4 = 1.

So the number corresponding to point n is1;

(2)(5﹣4)÷2=0.5,

①﹣3﹣0.5=﹣3.5,

② 1+0.5= 1.5.

So the number corresponding to point P is -3.5 or 1.5.

(3)①(4+2×5﹣2)÷(3﹣2)

= 12÷ 1

= 12 (seconds),

The number corresponding to point P is -3-5× 2- 12× 2 =-37, and the number corresponding to point Q is-37+2 =-35;

②(4+2×5+2)÷(3﹣2)

= 16÷ 1

= 16 (seconds);

The number corresponding to point P is -3-5× 2- 16× 2 =-45, and the number corresponding to point Q is -45-2 =-47.

Traveling Problem on Type 3 Number Axis

Trick: Equality and Classification Thought

Example 3. (20 17 autumn moon city end point) As shown in figure 1, there are two moving points A and B on the line segment MN doing uninterrupted reciprocating uniform motion (that is, as long as the moving point coincides with one end of the line segment MN, it will immediately turn around and move to the other end of the line segment MN at the same speed until it coincides with the end point, and the moving direction and speed of the moving point will change). It is known that the speed of a is 3 m/s.

(1) It is known that MN= 100 meters, if B starts from point M first, when MB=5 meters, A starts from point M, and coincides with B for the first time _ _ _ _ seconds after A starts;

(2) Given MN= 100 m, if A and B start from point M at the same time, A and B will overlap for the first time after _ _ _ _ seconds;

(3) As shown in Figure 2, if A and B start from point M at the same time, A and B coincide with point E for the first time and point F for the second time, EF = 20m, let MN = s·M, and let the equation find S. 。

This paper analyzes the application of linear equation of one variable and the number axis. The key to solving the problem is to understand the meaning of the topic, find out the appropriate equivalence relationship, list the equations according to the conditions given by the topic, and then solve it.

(1) Let A coincide with B for the first time after departure, and solve the equation according to the equivalence relation: distance difference = speed difference × time;

(2) It can be assumed that A and B overlap for the first time after y seconds, and the equation can be listed and solved according to the equivalence relation: distance sum = speed and x time;

(3) If A and B start from point M at the same time, A and B overlap for the first time * * * and overlap for the second time * * * leaving four Mns, so we can get ME=2/(3+2)×2MN=4/5MN and MF = 2mn ﹣ 2/(3+2).

The answer (1) makes A coincide with B for the first time after x seconds.

(3 ~ 2) x = 5, and the solution is x = 5.

Answer: After 5 seconds, A and B coincide for the first time;

(2) suppose that A and B coincide for the first time after y seconds.

(3+2)x= 100×2,

The solution is x = 40.

A: After 40 seconds, A and B coincide for the first time;

(3) If A and B start from point M at the same time, A and B overlap for the first time * * * and overlap for the second time * * * leaving four Mns, so we can get ME=2/(3+2)×2MN=4/5MN and MF = 2mn ﹣ 2/(3+2).

According to the question: 4/5s-2/5s = 20,

The solution is s = 50.

A: s=50 meters.

The author regards this question as a review question for the senior high school entrance examination in grade seven, especially item (3). Students either feel dizzy and can't do it, or use the arithmetic method of primary school competition. Many students can't understand it with the arithmetic method of primary school competition, but it seems "so easy" to solve it with "letters representing moving points"

New definition problem on type 4 number axis

Unique skill: transformation, equation, classification thought

Example 4. Reading Comprehension (jurong city Autumn Senior High School Entrance Examination 20 17)

Points a, b and c are three points on the number axis. If point C is between A and B, and the distance to A is three times that from point C to B, then we call point C the singularity of {A, B}.

For example, as shown in figure 1, the number represented by point A is-3, and the number represented by point B is 1. The distance from point C representing 0 to point A is 3, and the distance from point B is 1, so point C is the singularity of {A, B}; For another example, the distance from point D to point A representing ﹣2 is 1, and the distance from point B is 3, so point D is not the singularity of {A, B}, but the singularity of {B, A}.

Knowledge application

As shown in Figure 2, M and N are two points on the number axis, the number represented by point M is -3, and the number represented by point N is 5.

The point represented by (1) number _ _ _ is the singularity of {M, N}; The point represented by the number _ _ _ _ is the singularity of {N, M};

(2) As shown in Figure 3, A and B are two points on the number axis, the number represented by point A is-50, and the number represented by point B is 30. When the existing moving point P starts from point B and stops at point A, exactly one of P, A and B is the singularity of the other two points?

This paper analyzes this problem and examines the distance and moving point of two points on the number axis, and carefully understands the new definition: the distance between the number represented by singularity and the previous point A is three times that of the latter point B, and the result can be obtained by formula.

(1) According to the definition, it is found that in the number {M, N} expressed by singularity, the distance between the front point m and the back number n is three times, and the conclusion is drawn; According to the definition, it is found that in the number {N, M} represented by the singularity, the front point n is three times the distance to the back number m, and the conclusion is drawn;

(2) The distance from point A to point B is 6. According to the definition of singularity, there are two formulas: ① Pb = 3pa; ②PA = 3PB; You can draw a conclusion.

Answer (1) 5-(-3) = 8,

8÷(3+ 1)=2,

5﹣2=3;

﹣3+2=﹣ 1.

So the point represented by the number 3 is the singularity of {M, N}; The number-1 represents the singularity of {N, M};

(2)30﹣(﹣50)=80,80÷(3+ 1)=20,

30﹣20= 10,﹣50+20=﹣30.

Therefore, when point P moves to the position of -30 or 10 on the number axis, just one of P, A and B is the singularity of the other two points.

So the answer is: 3; ﹣ 1.

Finally, summarize a few words:

The first step is to express the moving points on the number axis with letters;

Second, according to the needs of the topic, write algebraic expressions about letters;

The third step is to list the equations and solve them according to the meaning of the topic.

The essence of mathematics learning is to simplify "complex problems". When solving the fixed point problem, the first difficulty is to analyze the motion process of the fixed point, and the spatial imagination and logical analysis ability are not enough. When solving problems, especially in the exam process, my suggestion is to do more work and draw more pictures during exercise. For many different moments of movement, draw more graphics for comparison, and you can often see the movement trend of the departure point. For example, we can draw graphs of nodes at special moments, and find the laws of motion through observation and comparison. For some special positions of moving points, we need to draw graphs, such as the coincidence of two points or the arrival of a point at a special position, which is often the focus of classified discussion. Through drawing, we can master the whole process of movement, and then we can discuss it according to different situations and find the equation calculation of equal relationship. The key of this step is to express each quantity in the graph by algebraic expression, mainly the length of each line segment in the graph, and finally find out the equivalent relationship between each line segment and list the equations to be solved.