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In the plane rectangular coordinate system, as shown in Figure 1, translate the line AB to the line CD, and connect AC and BD. (1) directly write out the equal line segments and parallel lines in the figure
AC‖BD,CD‖AB,AC＝BD,AB＝DC
Given the line AB = 8, there is a point P on the plane. If AP = 5, what is Pb, P is on the line ab
Given the line AB = 8, there is a point P on the plane
(1) If AP = 5 and Pb is equal to, P is on line ab
(2) When p is on the line AB and PA = Pb, the position of P is determined and the size of PA + Pb and ab is compared
(3) If PA + Pb = 7, do P exist? Why?
Question 1: 3 or 13, you assume that P is on the line AB, and then find Pb
The second question: according to the condition, PA = Pb = 4, you can know: PA + Pb = AB = 8
Question 3: it doesn't exist because the line segment between two points is the shortest, that is, when p is on AB, PA + Pb is the smallest = 8
It is known that line segment AB and plane a are oblique to B, and its projective length on the plane is half the length of line segment ab
Find the angle between line AB and plane a
Let the projection of AB on the plane be Bo
Then Ao ⊥ Bo triangle ABO is a right triangle
And Bo = 1 / 2Ab
Then the angle Bao = 30 and 186;
Angle ABO = 60 & # 186;
The angle between line AB and plane a = angle ABO = 60 & # 186;
There are two points a and B on the plane, and the distance between them is 5cm. We should find a point C on the plane, so that the sum of the distances from a and B is 5cm. Question: where is point C? Can the sum of the distances from C to a and B be less than 5cm? Why?
Mathematical geometry. Three questions,
If the sum of distances is 5cm, then the point C is on the straight line AB, not less than 5cm. If C is not on AB, then ABC is the vertex of the triangle, and the sum of two sides must be greater than the third side
Given the position of isosceles △ OAB in the plane rectangular coordinate system, as shown in the figure, the coordinate of point a is (- 3 radical 3,3), and the coordinate of point B is (- 6,0)
(1) If the isosceles triangle oba is folded along the x-axis to △ oba ', and the point a' just falls on the hyperbola y = K / x, the value of K is obtained
(2) If the isosceles triangle oa'b is folded again along a 'to △ oa'b', does the point B 'fall on the hyperbola
It's mainly about how to do the second question. I'll do the first one
Then I'll give you some ideas
According to the conditions given in the title, it is easy to get the angle boa '= 30 ° (the sine value is easy to see = 1 / 2)
So along the OA 'fold back angle b'ob = 60 ° ob' = ob = 6
So we can see that the coordinates of B 'are (- 3, - 3, root 3)
The first question has found K
If you are not on the hyperbola, you will know that it should be not
In the rectangular coordinate system, there is a point a whose coordinate is (3, - 3). Find a point B on the coordinate axis so that the triangle AOB is an isosceles triangle. How many ways can we find such a point B
8 (6,0) (0,-6)(3,0)(0,-3)(-3*2^0.5,0)(0,3*2^0.5)(3*2^0.5,0)(0,-3*2^0.5）
In the rectangular coordinate system, given the point a (- 2,3), determine the point B on the coordinate axis, so that the triangle AOB is isosceles triangle
A total of 8 points, I have found 8 points, but the coordinates of the two points on the vertical bisector will not be calculated
Ao midpoint coordinates (- 1,1.5)
The slope of the vertical bisector is inversely proportional to the slope of Ao, and then it is brought into the midpoint coordinate
Vertical bisector equation y = (2 / 3) x + (13 / 6)
So B (0,13 / 16) or B (- 13 / 4,0)
As shown in the figure, we know that P is a point outside the circle o, PA.PB Tangent circle O to a, B, OP and ab respectively, intersect with point m, C is a point on arc AB, try to explain that angle OPC = angle OCM holds
Connect to OA
Because PA and Pb are tangent lines of ⊙ o
So OA ⊥ PA, ab ⊥ op
So we can prove that △ OAM ∽ OPA
So OA / op = OM / OA
From OA = OC
OC/OP＝OM/OC
And ∠ cop = ∠ MOC
So △ POC ∽ com
Therefore, OPC = OCM
As shown in the figure, PA and Pb cut ⊙ o to points a and B respectively, and point E is a point on ⊙ o, and ∠ AEB = 60 °, then the degree of ∠ P is ()
A. 120°B. 90°C. 60°D. 75°
In the quadrilateral paob, because PA and Pb cut ⊙ o at points a and B respectively, ∠ OAP = ∠ OBP = 90 °, AOB = 2 ∠ e = 120 ° and ∠ P = 60 °, C is selected
It is known that PA, Pb and DC cut ⊙ o at points a, B and e respectively as shown in the figure. (1) if ∠ P = 40 °, calculate ∠ cod; (2) if PA = 10cm, calculate the perimeter of △ PCD
(1) To connect OA, OEO, ob, PA, Pb, cut \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2) ∵ pa The circumferences of PCD are: PC + PD + CD = PC + CE + PD + DB = PC + Ca + PD + DB = PA + Pb = 2PA = 2 × 10 = 20cm
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