There are two points A and B on the plane, and the two points A B in the plane are symmetrical

Any position on the line segment connected by point A and point B is eligible, that is, when the point C is on the ** section AB, the sum of the distances from point C to point A point B is 7 cm. Because the line segment between two points is the shortest, the distance between the two points A and B refers to the length of the line segment AB, and the sum of the distances from point C to A and B is also 7 cm, so it can only be on the ** segment AB.

It is impossible to be less than 7 cm, assuming that there is a point that makes the sum of CA and CB less than 7, and C can no longer be on AB as mentioned above, when C is on AB extension or reverse extension line, no matter how the point is taken, when the sum of the distances is calculated, it will be greater than 7 (because the length of a line segment AB must be added).

When C and AB are not collinear, the triangle ABC can be formed, by the nature of the triangle, the sum of the two sides is greater than the third side, then the length of AC+BC must be greater than the length of AB, so the hypothetical point does not exist, so the sum of the distances between the two points from point C to A and B cannot be less than 7 cm.

2.No, point C is outside the straight line AB and can be larger than 7 cm, but not less than 7 cm.

Our teacher has already talked about it. Hope it helps.

The line segment between two points is the shortest, the distance between A and B is 7 cm, which is already the shortest, and you must be on the ** segment AB when you are looking for point C, there should be countless of them, but there will be no sum less than 7 cm. The sum of the distance from point C to point A and point B on section AB will not be greater than 7 cm, and if it is greater than 7 cm, point C can only be outside section AB.

The perpendicular bisector of line segment ab is symmetrical with respect to the perpendicular bisector of the line segment connecting them So the answer is: the perpendicular bisector of line segment ab

Summary. Point A belongs to the figure of plane A, point A belongs to plane B.

1) "Point A is in the plane, but point B is outside the plane" is expressed as: a, b, as shown in the figure: ; 2) "Line A passes through a point m outside the plane" is denoted as:

Let p(x,y) get (x-1) 2+(x+1) 2+2y 2=2(x 2+y 2)+2, so it is necessary to have x 2+y 2 max, that is, 3x+4y max finger dust. The point sought is the tangent point of a straight line with a slope of -3 4 and the Yunling circle.

Let the coordinates of p point (x,y) and p be on the circumference, so p satisfies (x-3) +y-4) =4

pa²=(x+10)²+y² pb²=(x-10)²+y²pa²+pb²=2x²+2y²+200

Put the equation of the circle x -6x+9+y -8y+16=4 x +y =6x+8y-21

pa +pb =2(6x+8y-21)+200 6x+8y4 12xy=4 3xy, and when 3x=4y, substitute the equation for a circle gives p(21 5, 28 5).

If the sum of the distances between two points a(0,3),b(4,0), p to a,b in the plane is 6, then the p trajectory equation is.

Because |ab|=5 , so the trajectory of p is an ellipse with a and b as the focus.

Let p(x,y), then [x 2+(y-3) 2]+ x-4) 2+y 2]=6 , move and square to get x 2+y 2-6y+9=36+x 2-8x+16+y 2-12* [x-4) 2+y 2], move and merge to get 12* [x-4) 2+y 2]=-8x+6y+43, squared to get 144*(x 2-8x+16+y 2)=64x 2+36y 2+1849-96xy-688x+516y, Simplify to 80x 2+96xy+108y 2-464x-516y+455=0. Agree with 0|Comments.

In a plane, it can be connected into a straight line

The diagram of the infiltration and defect of the transport can be seen that the option shouts shirt d correctly;

Therefore, the side argument is D

<> according to the diagram, the point p is ridiculous to be on the straight line ab, and it can also be outside the straight line and the tomb ab, but not on the absolute ascending line section ab

Therefore, d