If you know that the line segment AB is 6cm, and the line segment AB is 6cm, and the line segment AC

1) Non-existent. Imagine that when C is outside AB, then you can draw a triangle ABC, and the relationship between the sides of the triangle shows that the sum of the two sides is greater than the third side, and the sum of the distances is greater than 6, which cannot be 5; When c is on AB, it cannot be less than 6 either

2) Exists. Such a point c should be on the ** segment AB, including two endpoints A and B. (Note that it is not on a straight line ab, not on the extension line of the line segment), there are countless such cs.

1 Impossible, because it takes 6cm even on top of AB

2 On ab there are an infinite number ...

1.There are two of them, and the abc is connected in turn to form an isosceles triangle. 2.There is, there is one, is the midpoint of AB.

1) It does not exist, because "the straight line between two points is the shortest"! 、

2) There are countless points C, and point C is any point other than A and B on the AB line segment!

1) 2 pieces, the radius is 4cm, the diameter is 8cm, which is greater than the length of the line segment AB, and can pass through two points A and B up and down the diameter.

2) 1, the radius is 3cm, then the diameter is 6cm, which is equal to the length of the line segment AB, and can pass through two points A and B on the diameter.

3) 0, the radius is 2cm, the diameter is 4cm, less than the length of the line AB, can not pass through A, B two points.

The line segment AB=6cm, AC=2cm, if A and B are on the side of C's trembling and leaky chain, the length of BC is 6-2=4cm;

If A and B are on both sides of C, then BC is 6+2=8cm; The length of the BC is 8cm or 4cm

So the answer is 4 or 8

Draw 2 circles with a radius of 4 cm.

Draw 1 circle with a radius of 3 cm.

Draw a circle with a radius of 2 cm and a 0

(1) Two (2) One (3) Zero.

It should be, you see.

There are countless of them, and there are no two points.

(1) When the sum of the distances from point C to points A and B is equal to 6cm, the position of point C should be in **? Why?

C is on AB, because only in this way can the sum of the distances from C to A and B be equal to 6cm (2) When the sum of the distances from point C to A and B is equal to 7cm, the position of point C should be in **?

c on the extension line of ab or ba.

(1) C is any point between AB (including points A and B) because: AC+CB=AB=6cm

2) C is a point outside AB.

Let c be on the left side of a, then: ca+cb=ca+ca+ab=2ca+ab=7ca=(7-ab) 2=

Let c be to the right of b, then ac+bc=ab+bc+bc=ab+2bc=7bc=(7-ab) 2=

When the sum of the distances from point C to points A and B is equal to 6cm, the position of point C should be between points A and B, and can coincide with Point A B.

Because the straight-line distance between two points is the shortest.

When the sum of the distances from points C to A and B is equal to 7cm, it is an elliptical trajectory

The Wolf Hunter team will answer for you.

1.Whether there is a point C, such that the sum of the distances from the two points to AB is equal to 5cm, does not exist. The line segment between two points is the shortest, and ab=6 is the shortest.

2.Is there a point C, so that the sum of the distances from the two points to AB is equal to 6cm, and the position of point C should be at **?

The sum of the distances from any point on the line AB to a and B points is equal to 6, and there are an infinite number of such points C.

1.There is no such thing as a c because the shortest distance between c to a and b is equal to 6

2.There is , c at any point in the straight line between ab.

(1) The third points of line segment AB are 2, and together with the 2 endpoints of line segment AB, there are 4 points, which can form (4*3) 2=6 line segments. There are 3 line segments of length 2, 2 line segments of length 4, and 1 line segment of length 6, and the sum of the length of these line segments is 2*3+4*2+6=20.

2) For the convenience of narration, P is recorded as the equinox in the line segment AB.

Take A as the starting point, and mark the position of the M point with the length of the line segment MA.

The third equinox is ;

The quartile points are;

The sixth equinox is ;

Excluding the number of repeats, there are 1,,2,3,4,5 points, including a and b, a total of 9, divided into line segments 9 * 8 2 = 36.

There are 4 of length, 6 of length 1, 4 of length, 5 of length 2, 2 of length, 5 of length 3, 2 of length, 3 of length, 3 of length 4, 2 of length, 2 of length 5, and 1 of 6.

The total length of these segments.

When c is in the middle of a and b:

ac=ab-bc=16-6=10㎝

c is on the outside of a, b:

ac=ab+bc=16+6=22㎝