If there are two points A 2,2,B4,1,Ca,0 in the plane

Solution: Point C is on the x-axis.

1) Make point a, point a, symmetry point a, with respect to the x-axis'（2，-2）。ca=ca'.

When a', b, and c can form a triangle when they are not on the same straight line. From the trilateral relationship of the triangle, it can be seen that ca'+cb>ba'

When a', b, c three points in the same straight line, ca'+cb=ba'

a'When , b, and c are on the same line, ca+cb is the smallest.

i.e.: vector a'c is in the same direction as the vector cb. (a-2,2)=in(4-a,1)a=10 3

2) When a, b, and c points are not on the same straight line, a triangle can be formed. From the trilateral relationship of the triangle, it can be seen that when ca-cb is on the same line, ca-cb = aba, b, and c are on the same line, ca-cb is the largest.

That is, the vector Ca is in the same direction as the vector Cb. (2-a,2)=in(4-a,1)a=6.

1) Let the point where A is symmetrical with respect to the x-axis be A1, then the point where A1B intersects with the X-axis is point C (the shortest straight line between two points) A=10 3

2) The point at which ray AB extends to intersect the x-axis is point C, A=6

Here's how it works, but I don't know if it's right because I'm more annoyed with calculations.

ab = 6 - (-2) = 8, 6, (xa-xb) plus (ya-yb) squared and finally squared.

Pythagorean theorem ab = 2 times the root macro sail number 10

2, the distance from A to the origin is 6, the distance from B to the origin is 2, and the two points are on both sides of the origin, so the distance between A and B is 8

i.e. |ab|=8,1,8,1,If a(x,y) b(x',y')

ab is equal to the distance between two points a and b.

i.e.: ab = root number under < (x-x')^2+(y-y')^2>

8,1,8,0, what is ab equal to two points on a given plane: a(6,0) b(-2,0)?

Knowing how much ab is equal to two points a(6,0) b(-2,0) on a plane?

3 collinear tease. So bring the coordinates of the three points of the stool to y=kx+b, so that we can get three linear equations.

a=k+b①

a^2=2k+b②

a^3=3k+b③

：k=a^2+a ④

：k=a^3-a^2 ⑤

：1=(a+1)/(a^2-a)

The solution yields a=1 2

a>0.The state of Zaozhou is a=1 + 2

Let p coordinates: (3+2cos, 4+2sin).

ap)^2+(bp)^2=(4+2cosα)^2+2(4+2sinα)^2+(2+2cosα)^2=60+24cosα+32sinα

60+8 5 (3 5cosa+4 5sina)=60+8 5sin ( + where arctan3 4

When sin( +=-1 takes the minimum value, that is, sina = -4 5, cos = -3 5

So p coordinates: (9 5, 12 5).

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

pa²=(x+1)²+y² pb²=(x-1)²+y²

pa²+pb²=2x²+2y²+2

Put the equation of the circle x -6x + 9 + y -8y + 16 = 4 x + y + 1 = 6x + 8y - 20

pa²+pb²=4(3x+4y-10)

3x+4y 2 12xy=4 3xy and when 3x=4y.

Substituting the equation of the circle gives us p(9 5, 12 5).

On the circle, assume that the p-point coordinates are: (x,y).

p is on the circumference and (x-3) +y-4) = 4

and pa = (x+1) +y pb =(x-1) +y pa +pb =2x +2y +2

i.e. x -6x+9+y -8y+16=4 x +y +1=6x+8y-20

pa²+pb²=4(3x+4y-10)

3x+4y≥2√12xy

That is, 3x+4y 4 3xy

When 3x=4y.

Substituting the equation of the circle gives us p(9 5, 12 5).

Why and when 3x=4y.

Substituting the equation of the circle gives us p(9 5, 12 5).

If the points a, b, c planes, points a, b, c planes, and abc three points are not collinear, then the plane and plane coincide

Let p coordinates: (3+2cos, 4+2sin)ap) 2+(bp) 2=(4+2cos) 2+2(4+2sin) 2+(2+2cos) 2=60+24cos +32sin

60+8 5 (3 5cosa+4 5sina)=60+8 5sin ( + where arctan3 4

When sin( +1 takes the minimum value, i.e., sina = -4 5, cos = -3 5 so p coordinates: (9 round 5, 12 5).

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

pa²=(x+1)²+y²

PB = (x-1) socks + y

pa²+pb²=2x²+2y²+2

Put the equation of the circle x -6x + 9 + y -8y + 16 = 4 x + y + 1 = 6x + 8y - 20

pa²+pb²=4(3x+4y-10)

3x+4y 2 12xy=4 3xy and when 3x=4y.

Substituting the equation of the circle gives us p(9 5, 12 5).

Answer: Circumference (x-3) + y-4 pants banquet pei = 4 circle center m(3,4), radius r=2

Let p(x,y).

x-3）²+y-4）²=4

ap²+bp²=(x+1)²+y²+(x-1)²+y²=2(x²+y²)+2

Xiang Chain. The geometric meaning of (x +y) is the distance from p to o.

The minimum value is |mp|-r=5-2=3

The minimum value of ap+bp is 2*9+2=20

In this case, p is the intersection point of the line segment om and the circle.

The equation for om is Hu Wei y=(4 3)x

with x + y = 9

The coordinates of p are found to be (9 5, 12 5).

Vector ab = (1, 1), vector ad = (-3, 3) ab·ad = -3 + 3 = 0

ab ad set point c coordinates as (a, b).

Vector ba = vector cd

then (-1,-1)=(a+1,b-4).

a=-2,b=3

c coordinates are (-2,3).

The linear od equation y=-4x, let the coordinates of the m point be (t,-4t) then ma· mb=2t -8t+8=2(t-2) when t=2 ma· mb takes the minimum value of 0, and the coordinates of om are (2,-8).

(1) The straight line ab: y=x-1, the straight line ad: y=-x+3k1 and k2 are opposite to each other, so the straight line ab and the straight line ad are perpendicular to each other (2) b and a comparison, the abscissa +1, the ordinate +1

Comparing c with d, the abscissa is +1 and the ordinate is +1, then c is (0,5).

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).