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In the ladder ABCD, AB//CD, DC: AB=1:2, E, F are the midpoint of the two waist BC, AD respectively, then EF: AB is equal to () A.1:4 B.1:3 C.1:2 D.3:4 In the trapezoid ABCD, AB//CD, DC: AB=1:2, E, F are the midpoint of two waist BC, AD respectively, then EF: AB is equal to () A.1:4 B.1:3 C.1:2 D.3:4
DC: AB=1:2,
Set DC=x, AB=2x,
E and F were the midpoint of BC and AD, respectively,
EF=1
2(AB+CD)=1
2(2X+x)=3
2X,
EF: AB=3
2X:2x=3:4.
Therefore, D.
Known vector AB=(6,1), BC=(x, y), CD=(-2,-3), when vector BC∥ In case of DA, the relation shall be satisfied by the number x, y. Known vector AB=(6,1), BC=(x, y), CD=(-2,-3), when vector BC∥ In the case of DA, the relation shall be satisfied by the number x, y. Known vector AB=(6,1), BC=(x, y), CD=(-2,-3), when vector BC∥ In the case of DA, the relation that the number x, y should satisfy.
∵
AB=(6,1),
BC=(x, y),
CD=(-2,-3),
∴
AD=(4+x,-2+y),
Vector
BC∥
DA,
X (-2+y)-(4+x) y=0
X+2y=0.
Given the vector AB=(4,2), the vector BC=(x, y), the vector CD=(3,5), when the vector BC//DA is a relation corresponding to x, y
Vector AB=(4,2), Vector BC=(x, y) Vector CD=(3,5)
Vector DA=AB+BC+CD=(7+x,7+y)
Vector BC//DA
X (7+y)-y (7+x)=0
7X-7y=0
Y = x
That is, the real number x, y corresponds to the relation y=x
In a given quadrilateral ABCD, vector AB=(6,1), vector BC=(x, y), vector CD=(-2,-3) (1) If the vector BC parallel vector DA is the analytic expression of y=f (x) (2) Under the condition of (1), if vector AC⊥ vector BD, find the value of x, y and the area of quadrilateral ABCD
1) According to the vector DA=(m, n), because it is a quadrilateral and has AB+BC+CD+DA=0(all pointing quantities), so AB+BC+C=-DA. According to the vector coordinate operation,(6,1)+(x, y)+(-2,-3)=(-m,-n), so we can get 4+x=-my-2=-n.
In quadrilateral ABCD, the vector AB=(6,1), the vector BC=(x, y), the vector CD=(-2,-3), the vector BC‖DA, the vector AC⊥BD, and the vector bc. Thank you for the details of the process. In quadrilateral ABCD, the vector AB=(6,1), the vector BC=(x, y), the vector CD=(-2,-3), the vector BC‖DA, the vector AC⊥BD, the vector bc coordinate. Thank you for the details of the process.
From the known conditions of the subject matter:
Vector BD=(X-2, Y-3), Vector AC=(X+6, Y+1), Vector
AD=(X+4, Y-2), vector BC=(X, Y)
Because BC‖DA,(X+4) Y=X (Y-2), the solution is X=-2Y
And because AC⊥BD,(X+6)(Y-3)+(Y+1)(X-2)=0, we get:
X=-2 Y
Get:
(X, Y)=(-6,3) or (2,-1)
Therefore, the coordinates of vector BC are (-6,3) or (2-1).
Please rate,
In quadrilateral ABCD, vector BC‖AD, vector AB=(6,1), vector BC=(x, y). Vector CD=(-2,-3)(1) Find the relation between x and y; In quadrilateral ABCD, vector BC‖AD, vector AB=(6,1), vector BC=(x, y). vector CD=(-2,-3) (1) Find the relation between x and y; (2) If vector AC is perpendicular to vector BD, find the values of x and y and the area of quadrilateral ABCD.
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