In the rectangular coordinate system xoy, point B and point a (- 1,1) are symmetric about the origin o, P is a moving point, and the product of slope of straight line AP and BP is equal to - 1 / 3 Drawing and finding that APB and MPN are complementary sinAPB=sinMPN PA/PM=PN/PB.a (x0+1)/(3-x0)=(3-x0)/(x0-1).b Why can a write B

In the rectangular coordinate system xoy, point B and point a (- 1,1) are symmetric about the origin o, P is a moving point, and the product of slope of straight line AP and BP is equal to - 1 / 3 Drawing and finding that APB and MPN are complementary sinAPB=sinMPN PA/PM=PN/PB.a (x0+1)/(3-x0)=(3-x0)/(x0-1).b Why can a write B

The projection is proportional to the x-axis

As shown in the figure, in the plane rectangular coordinate system, O is the coordinate origin. The image of the quadratic function y = - x2 + BX + 3 passes through point a (- 1,0), and the vertex is B (1) Find the analytic expression of the quadratic function and write the coordinates of vertex B; (2) If the coordinate of point C is (4,0), AE ⊥ BC, the perpendicular foot is point E, point D is on the straight line AE, de = 1, find the coordinates of point D

(1) ∵ the image of the quadratic function y = - x2 + BX + 3 passes through point a (- 1,0),
∴0=-1-b+3，
The solution is: B = 2,
The analytic expression of the quadratic function is y = - x2 + 2x + 3,
Then the coordinates of vertex B of the quadratic function image are (1,4);
(2) Passing through point B as BF ⊥ X axis, perpendicular foot as point F,
In RT △ BCF, BF = 4, CF = 3, BC = 5,
∴sin∠BCF=4
5，
In RT △ ace, sin ∠ ace = AE
AC，
And ∵ AC = 5, AE can be obtained
5＝4
5，
∴AE=4，
Point D is the axis of DH ⊥ x, and the foot perpendicular is point H,
It is easy to prove △ ADH ∽ ace,
∴AH
AE=DH
CE=AD
AC，
CE = 3, AE = 4,
If the coordinates of point D are (x, y), then ah = x + 1, DH = y,
① If point D is on the extension line of AE, then ad = 5,
X + 1
4＝y
3＝5
5，
∴x=3，y=3，
So the coordinates of point D are (3, 3);
② If point D is on line AE, then ad = 3
X + 1
4＝y
3＝3
5，
∴x＝7
5，y＝9
So the coordinates of point D are (7
5，9
5）．
To sum up, the coordinates of point D are (3,3) or (7)
5，9
5）．

In the plane rectangular coordinate system, the parallelogram ABCD is symmetric about the origin, and the points a (- 2, - 4), B (3, - 5) are known Then the coordinates of point C are_________ , s parallelogram ABCD=________

Coordinates of point C (2,4) s = 45

It is known that the diagonal lines AC and BD of the parallelogram ABCD intersect at the coordinate origin o, It is known that the diagonal lines AC and BD of the parallelogram ABCD intersect at the coordinate origin o, the angle between AC and X axis is 30 degrees, DC is parallel to the X axis, AC = 8, BD = 6~

To make double heights, use the opposite side of the 30 ° angle of the right triangle = half of the hypotenuse

As shown in the figure, it is known that in the parallelogram ABCD, AC and BD intersect at the point O, and AC = 6, the plane rectangular coordinate system is established with o as the coordinate origin, and DC is parallel to the X axis And the coordinate of point C is (2 roots, 2,1), ∠ DCB = 45 ° to find the face of triangle DOA, and C coordinate (2 roots, 2,1)

OC=3
∠OBC+∠OCB=90=∠COB
∠OCB=22.5
（tan22.5）²=(1-cos45)/(1+cos45)
OB=OC*tan22.5
Area of triangle DOA = area of triangle cob = OC * ob / 2

In the plane rectangular coordinate system, the symmetry center of quadrilateral ABCD is at the origin o, and the coordinates of points a and B are a (3,1) and B (- 1,2), respectively!

C（-3,-1）,D(1,-2)

In the rectangular coordinate system, O is known as the origin. In the rectangular ABCD, the coordinates of point a, point B and point C are a (- 3.1) B (- 3.3). C (2.3) In the rectangular coordinate system, O is known as the origin. In the rectangular ABCD, the coordinates of point a, point B, point C and point D are a (- 1.1) B (- 1.3). C (4.3) d (4.1) The area of △ OBD is 3 / 2 of the area of the rectangle ABCD after a few seconds Look down here

The area of rectangle ABCD = (4 - (- 1)) * (3-1) = 10
Then s △ OBD = 15
[for the sake of simplicity and for illustrative purposes,
This solution only explains the train of thought]
When t = 1, s △ OBD = (1 + 3) * 5 / 2-1 * 5 / 2 [trapezoid minus right triangle] = 15 / 2
Obviously, t must be greater than 1
In this case, s △ OBD = (t-1) * 3 / 2 + (1 + 3) * 5 / 2 - (5 + (t-1)) * 1 / 2 [let the vertical axis of a and D on the X axis be e, f respectively, and the area of triangle = s △ OBE + s trapezoidal bdfe-s △ ODF] = 15
The solution is t = 7.5

In the plane rectangular coordinate system, given the points a (3, - 4), B (4, - 3), C (5,0), O as the coordinate origin, then the area of quadrilateral ABCD is? Please select: A.9 B.10 C.11 D.12

Select C. after the composition is completed, make a vertical line to the X axis through two points a and B, and then calculate in sections

In the plane rectangular coordinate system, if the three vertices of a parallelogram ABCD are a (1,1), B (- 3, - 1), C (3,2), find the coordinates of vertex D

Let D coordinate (x, y) vector AB = (- 3-1, - 1-1) = (- 4, - 2) vector CD = (x-3, Y-2), so - 4 = x-3, x = - 1 - 2 = Y-2, y = 0
d(-1,0)

In the plane rectangular coordinate system, the coordinates of point 1D and the coordinates of center of gravity of 2 parallelogram ABCD can be calculated by the vertices a (2,4), B (1.2), C (5,3) of the parallelogram ABCD

Coordinates of the center of gravity (2,7,6)