Three vertices A (0,2), B (-1,-2), C (3,1) of a quadrilateral ABCD are known, and BC =2 AD, the coordinates of vertex D are () A.(2,7 2) B.(2,−1 2) C.(3,2) D.(1,3) Three vertices A (0,2), B (-1,-2), C (3,1) of the quadrilateral ABCD are known, and BC =2 AD, the coordinates of vertex D are () A.(2,7 2) B.(2,−1 2) C.(3,2) D.(1,3)

Three vertices A (0,2), B (-1,-2), C (3,1) of a quadrilateral ABCD are known, and BC =2 AD, the coordinates of vertex D are () A.(2,7 2) B.(2,−1 2) C.(3,2) D.(1,3) Three vertices A (0,2), B (-1,-2), C (3,1) of the quadrilateral ABCD are known, and BC =2 AD, the coordinates of vertex D are () A.(2,7 2) B.(2,−1 2) C.(3,2) D.(1,3)

Let the coordinates of vertex D be (x, y)
∵

BC =(4,3),

AD =(x, y−2),
And

BC =2

AD,
∴
2X=4
2Y−4=3⇒
X =2
Y =7
2
Guxuan A

Given A (2,1), B (3,2), D (-1,4)(1) find the vector AB× vector AD,(2) if the quadrilateral ABCD is a parallelogram, find the vertex C... Given A (2,1), B (3,2), D (-1,4)(1), find vector AB×vector AD,(2) if quadrilateral ABCD is parallelogram, find vertex C and diagonal intersection E

AB*AD=(1,1)*(-3,3)=-3*1+3*1=0(2) It is easy to know that AB is parallel and equal to DC, C (x, y), DC=(x+1, y-4) x+1=y-4 and (x+1)^2+(y-4)^2=1+1=2, so we can get x=0, y=5.C (0,5), Y=-2X+5, Y=-(1/2)*X+7/2, and X=1, Y=3, E (1,3).

In quadrilateral ABCD, the vector AC = vector AB + AD, find what quadrilateral ABCD is As above

A parallelogram, because ac=ab+bc and ac=ab+ad, is parallel and because it's a vector, you can rule it out. It's not a trapezoid. You can draw a graph. It's important to draw.

In the quadrilateral ABCD, AB⊥AD, BC⊥CD, AB=BC,∠ADC=120°, the vertex of the 30° angle of a sufficiently large triangular ruler MNB coincides with the vertex B of the quadrilateral. When the 30° angle (∠MBN) of the triangular ruler rotates around the point B, the two sides of the triangular ruler intersect each other respectively, and the straight line where DC is located is E, F. (1) When ∠MBN rotates around point B to AE=CF (as shown in Figure 1), please directly write the quantitative relationship among AE, CF and EF. (2) When ∠MBN rotates around point B to AE=CF (as shown in Figure 2), is the conclusion in (1) still true? If so, please provide proof; if not, what is the quantitative relationship between AE, CF and EF? Please write down your conjecture and give your reasons. (3) When ∠MBN rotates around point B to AE=CF (as shown in question 3 and 4), please directly write the quantitative relationship between AE, CF and EF of outgoing line segments.

(1) AE+CF=EF;(2) The reason is: extend EA to G, make AG=FC, GA=FC,∠GAB=∠FCB=90°, AB=CB, GAB FCB (SAS), GBA=∠FBC, GB=FB, AG=CF, FBC+∠FBA=60°, GBA+∠FBA=60°, i.e.∠GBF=60° EB...

Given that the four vertices of quadrilateral ABCD are all on circle O, and AD is parallel to BC, the radius of circle O is 6, BC=10, AD=8, the area of quadrilateral ABCD is obtained

It is obvious that ABCD is an isosceles trapezoid, connecting AO and BO, so we can find the distance from O to AD and BC. The distance from center O to AD h=√(6^2-4^2)=2√5 The distance from center O to BC g=√(6^2-5^2)=√11 If ABCD is in half circle, Sabcd=(8+10)(2√5-√11)/2=9(2√5-√11) If center is in ABCD, Sabcd=(8+...

Given vectors a=(1,2), b=(x,1), u=a+2b, v=2a-b, and u is parallel to v, find the value of x? And determine if uv is in the same direction or in the opposite direction

U=(2x+1,4), v=(2-x,3)
U is parallel to v, then:3(2x+1)-4(2-x)=0
Get: x =1/2
In this case, u=(2,4), v=(3/2,3)
Easy: u=4v/3
Therefore, the value of x is 1/2, u, v in the same direction.