How to do junior high school math moving point problems? What s the best way to do it?

In the planar Cartesian coordinate system, there is a right-angled triangle abc, the known angle acb=90 degrees, ac coincides with the y-axis, b(4,-1), ab=5, bc=4, ac=3;If the moving point P starts from C and moves in the direction of C-A-B-C with a velocity of 1 unit second, when P bisects the triangle ABC with a vertex of the triangle ABC, find the T value (three solutions) and the P coordinates. If another moving point q and p start from c at the same time, q moves along c-b-a in four-thirds of a unit second, and p and q stop moving at the same time after reaching a, when moving m seconds, the area of the triangle cpq is one-ninth of the area of the triangle abc, and all possible values of m are written.

ABCD is a right-angled trapezoid, CD AB, B=90°AB=24 cm, CD=28 cm, point P starts from point D, moves to point C at a speed of 2 cm per second, point Q starts from point B, moves to point A at a speed of 1 cm per second, two moving points start at the same time, when one of the moving points reaches the endpoint, the other moving point also stops. How many seconds have elapsed from the start of the movement for the triangle apq to be an isosceles triangle with aq as the base edge?

In the right-angled trapezoidal OABC, the coordinates of the two points of OA BC, A and B are A(13,0),B (11,12) respectively, the moving points P and Q start from the two points of O and B at the same time, the point P moves along the speed of 2 units per second along the OA to the end point A, the point Q and the velocity of 1 unit per second move along the BC direction to the point C, when the point P stops moving, the point Q also stops moving, the line segment ob, bq intersects at the point D, passes through the point D as de Qa, intersects AB at the point E, and the ray QE intersects the X axis at the point F, The time of movement of the moving points p and q is t (unit, seconds) The line segment ob and pq intersect at the point d, the point d is de oa, and the intersection ab is at the point e

1, when t is what is the value, the quadrilateral pabq is the parallelogram 2, when t=3 seconds, find the area of pqf.

3,When t is the value, pqf is an isosceles triangle? Please write out the reasoning process.

In the triangle ABC, ab=ac=2, the angle a=91 degrees, o is the midpoint of bc, and the moving point e is moving freely on the ba edge.

Point F moves freely on the AC edge.

1) Can the triangle OEF become an isosceles triangle with an angle EOF=45 degrees during the movement of points e,f? If so, indicate that the triangle OEF is the time-moving point E of an isosceles triangle'f, if not, please explain why.

2) When the angle eof is equal to 45°, let be=x, cf=y, find the analytic formula between y and x, and write the value range of x.

3) When the conditions in (2) are satisfied, if the garden with O as the center of the circle is tangent to AB, try the positional relationship between the straight line EF and the circle O, and prove.

Click on the eh, the pride of the space Western Europe ielkjg, the pride of the don't know? Turn on the Karma girl tie one, hit your hobby.

Singularity. Fun. com

: The moving point problem is generally a graph geometry problem in motion, which must be the analysis of a variety of results, and the place where it is easy to lose points is the loss of solution and the lack of situation. Follow-up:

I just don't know where to start. It's very troublesome and I don't understand it: the moving point is to turn the movement into different situations, and for one situation, you have to draw the corresponding shape.

Dynamic geometric characteristics--- the background of the problem is a special figure, and the examination problem is also a special figure, so it is necessary to grasp the relationship between the general and the special.

To determine the position of the moving point, we need to turn the moving point into static and determine whether it needs to be classified and discussed;

Let the coordinates of the moving point be unknown;

Indicates the length of the relevant line segment;

The first question is, the velocity of q relative to p is 2, the velocity of m relative to p is 4, p has walked for 1 second, that is, 4 unit lengths, and then directly calculate, the distance of q m is 2 times the distance of p, that is, 8, the time of the two is the same, so the time is the distance divided by the velocity, that is, 8 (2+4), and then m has walked 4*8 (2+4) = 16 3 unit lengths.

The second question is that the velocity of q relative to p is 2, p has traveled 4 unit lengths, if pq is 1 unit length apart, then q has traveled 3 unit lengths, and it takes 3 2=.

The known correlation quantity is fully marked on the diagram with the known correlation given by the maneuvering point, and the motion law of the moving point is found, and the focus is transformed into the distance of the movement according to the time and distance in the motion or the whole process, as long as these steps are followed.

The method of information infiltration of the first moving point problem is summarized as follows:

1. The distance between two points on the number axis can be expressed by the absolute value, that is, the absolute value of the number difference represented by the two points.

2. A moving point letter on the number axis indicates that it can be solved by addition or subtraction of rational numbers, that is, the number represented by the starting point adds or subtracts the distance of the moving point, and adds in the positive direction and subtracts in the negative direction.

3. Find the midpoint of the line segment between any two points on the number axis, and divide by 2 by the sum of the numbers represented by the two points, if the number represented by the point on the number axis is a, b, then the number represented by the midpoint of the sliding branch ridge of the line segment AB is (a+b) 2.

4. The distance between two points on the number axis is the absolute value of the coordinate difference corresponding to these two points, that is, the difference between the number on the right and the number on the left is subtracted. That is, the distance between two points on the number axis = the number represented by the right point - the number represented by the left point.

5. The number axis is the product of the combination of numbers and shapes, and the analysis of the movement of the points on the number axis of the partner should be analyzed in combination with the graph, and the path formed by the movement of the points on the number axis can be regarded as the sum and difference relationship of the upper section of the number axis.

The moving point problem in junior high school mathematics can be roughly divided into two kinds of moving points.

1。The moving point of the movement:

This type of moving point gives the direction of motion and the speed of motion, and we mainly represent the length of certain line segments according to the speed of motion Time = Distance. According to the position of the moving point, the line segment can be divided into those that have been traveled (represented by speed and time) and those that have not been traveled (the total distance to be moved by the moving point - those that have been traveled). Special attention should be paid to the fact that when the moving point moves on the polyline, it is necessary to remove some parts of the line segment that has been traveled in order to correspond to the desired line segment; The remaining ones that have not been walked are also represented by changing their end positions as the moving points move to different line segments.

When the represented line segment is different from the direction of movement of the moving point, the similarity knowledge is generally used to find out the similarity ratio with some line segments whose length can be calculated and the direction is consistent with the direction of the desired line segment.

2。Indefinite: This kind of moving point generally appears in conjunction with existential problems, that is, whether there is a point p that makes the problem satisfy some conclusions or when some conclusions exist, the position of the moving point p is found.

At this time, the solution can take the situation that the problem requirements are satisfied as a condition of use, so that p is just in the position where the requirements are met, and then combine the geometric knowledge to solve the problem.

For example, when the problem asks for the presence or absence of a point p, make a triangle 20 in area. Let's first use the algebraic expression to represent the area of the triangle, and then make it a value of 20.

In short, there are many types of questions about moving points, and it is difficult to explain them here. Pay more attention to the combination of algebraic simplification and geometric knowledge when solving problems, and you can slowly explore some of these rules.

Find the special point through the motion of the moving point, and solve it according to the special graph of the special point.

Hello, you can take a picture of the question to me.

The moving point problem is talking about whether a quadratic function finds a parallelogram.

Questions. <>

<> could you please wait a minute, reading the title for your calculations.

What are the specific questions?

Questions. Two questions, which one can be made and which one can be done.

Two complete big questions.

Okay, wait a minute.

Questions. Hurry, hurry up thank you.

Okay, I'll try.

<> hope mine is helpful to you, if you think mine is satisfactory to you, please give it a thumbs up.

Recognize the change of quantity and quantity, and develop dynamic thinking!

In general, the issue of moving points should be discussed in a categorical manner.

1.Solution: ab=80-(-40)=120

Set x seconds after the encounter, according to the title:

3x+2x=120

x=2480-3x=80-72=8

Answer: The number corresponding to c is 4

2.Set Y seconds to meet later, according to the title:

3x-2x=120

x=12080+2x=320

Answer: The number corresponding to d is 320

This kind of problem is a disguised travel problem, but it is much more difficult to go to the real geometric moving point on the number axis.

For travel issues, keep in mind.

When traveling in the opposite direction, the sum of the strokes = the original distance, and when catching up, the difference of the strokes = the original distance.

I wish you progress in your studies o( o

One by one, one by one.

Find the area of the triangle AQP, in fact the area of AQP is equal to the area of the quadrilateral ABCD minus the area of the three small triangles.

The velocity of the point is 1 unit of second, starting x seconds, so bp=x

s aqp = s rectangular abcd-s abp-s adq-s cpq

ab*bc-1/2*ab*bp-1/2*ad*dq-1/2*cp*cq

8-x-2-(4-x)/2

4-x 2BC = 4, P moving along BC, BC 1 = 4 sec.

Therefore, the value range of x is 0 x 4

2.The triangle AQP is an isosceles triangle, then there are three cases, AQ=QP, AP=PQ, AP=AQ

In the first case, aq=qp, it is not difficult to see from the graph that p and b coincide at this time, that is, x=0

In the second case, ap=pq, ap=root(ab+bp), pq=root(cp+cq), bp=x, cp=4-x, cq=1

The solution is x=13 8

In the third case, ap=aq, aq=root(ad+dq)=root17

ap = root number (ab + bp) = root number 17

x = root number 13

Turn left|Turn right.

pb = root number 35, bq = 2 root number 35, so pq = 5 root number 7 cm.

Your adoption is my motivation to move forward!

If you have any new questions, please ask me for help, it's not easy to answer, thank you for your support!

Grasp the invariant quantity relationship of the graph in the process of motion, and solve it by column equations (groups), inequalities (groups) or mathematical models such as functions. t generally refers to the time spent exercising.

The moving point is a point that moves according to a fixed trajectory, and with the change of the point, it causes the change of the line, and then causes the change of the graph.

t generally refers to time.

The key to solving the moving point problem is to build a functional model or find invariants.

Analysis: Def moves at a constant speed from the position in Figure 1 along Cb to ABC at a speed of 1 unit per second, while point P moves from A to point B at a constant speed of 1 unit per second along AB, and the right-angled edge of AC and def intersects at Q

P, E move at the same velocity, and at the same time, the Q point also moves from C to Ca at the same velocity.

During the move, |pa|=|ec|=|cq|Remain the same when |eq|=|ed|, Q is the intersection of AC and ED, after which Q will become the intersection of AC and DF, and Q will no longer move towards A, but will return to C in the direction of AC

The above is the trajectory of the movement of the point q.

Movement speed v=1, s=t

ac=8,bc=6, ∠acb=∠edf=90°，∠def=45°，ef=10

pa|=|ec|=|cq|=t

aq|=8-t

When pqe=90, apq=90

At this point, |ap|=|aq|cos∠paq=(8-t)*4/5=t

The solution yields t=32 9<5

There is that pqe is a right-angled triangle, t=32 9 def starts from the starting position and moves at a constant speed along cb to abc at a speed of 1 unit per second, and the point p starts from a and moves at a uniform speed along ab at a constant speed of 1 UNIT PER SECOND TOWARDS POINT B, AND THE RIGHT-ANGLED EDGES OF AC AND DEF INTERSECT AT POINT Q

P, E move at equal velocity, and at the same time, point Q also moves from C along Ca to A at the same speed.

During the move, |pa|=|ec|=|cq|Unchanged, when|eq|=|ed|, Q is the intersection of AC and ED, after which Q will become the intersection of AC and DF, and Q will no longer move towards A, but will return to C in the direction of AC

The above is the trajectory of the movement of the point q.

Movement speed v=1

s=tac=8,bc=6, ∠acb=∠edf=90°，∠def=45°，ef=10

pa|=|ec|=|cq|=t

aq|=8-t

When pqe=90, apq=90

At this point, |ap|=|aq|cos∠paq=(8-t)4/5=t

The solution yields t=32 9<5

The presence of pqe is a right triangle, t=32 9