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In the plane rectangular coordinate system x0y, the coordinates of the parallelogram A and B are (2,3), (5,1), respectively (1) If point C is on the positive half axis of the X axis and D is on the positive half axis of the Y axis, write the coordinates of the vertices C and D (2) If (1) condition is changed to point C and D on the coordinate axis, write the coordinates of C and D Ask God to answer, correctly add points
If the slope of lab calculated by C (3,0) d (0,2) is - 2 / 3, the slope of LCD should also be - 2 / 3. Let C (C, 0) d (0, d) get klcd = (- D) / C = - 2 / 3, and ab length is 13 ^ 1 / 2, then CD is also 13 ^ 1 / 2, that is C (3,0) d (0,2)
2：（1）C（3,0） D（0,2） （2）C（-3,0） D（0,-2）
In the plane rectangular coordinate system, we know the coordinates of three vertices of the parallelogram ABCD: a (0,0), B (3, √ 3), C (4,0). Find the equation of the line where the edge CD is fast (2) It is proved that the parallelogram ABCD is a rectangle and its area is calculated
The slope of the line AB is k = √ 3 / 3, so the equation of the line CD is: y-0 = √ 3 / 3 (x-4), that is: √ 3x-3y-4 √ 3 = 0 is the height be on AC side of triangle ABC, then be = √ 3, CE = 4-3 = 1, TGC = ∠ 3 / 1 = √ 3 ∠ C = 60 degrees and TGA = k = √ 3 / 3 ∠ a = 30 degrees ∠ B = 90 degrees ◇ ABCD is a rectangle
In the plane rectangular coordinate system, the coordinates of three vertices of the parallelogram ABCD are known: a (0,0), B (3, √ 3), C (4,0) Find the equation of the straight line where the edges BC and CD are located (the result is written in general form) (2) It is proved that the parallelogram ABCD is a rectangle and its area is calculated Uh
BC：
K=(√3-0)/(3-4)=-√3
y=-√3(x-4)
∴√3x+y-4√3=0
∵ the figure is a parallelogram
∴AB//CD
∴Kab=Kcd
ν CD slope k = √ 3 / 3
The CD linear equation is: y = √ 3 / 3 (x-4)
Because AB is parallel and equal to CD, and KBC * KCD = - 1
So it's a rectangle
S=4*√3=4√3
In the plane rectangular coordinate system, O is the origin of the coordinate, and the point P (m, n) is on the image of the inverse scale function y = k X. if OP = 3, M + n = 4, then the value of K is?
(1) According to the meaning of the question, K-2 = K / k = 1, k = 3. (2)
It is known that in the same plane rectangular coordinate system, the images of the inverse scale function y = m + 2 / X and y = - M / X completely coincide, then M=—— (1) If the abscissa and ordinate of point P are reciprocal of each other, is point P necessarily on the above function image? Why? (2) If the abscissa and ordinate of point q are opposite to each other, is point Q not necessarily on the above image? Why?
(1) let the coordinates of point p be (x, 1 / x), and bring P into the inverse ratio function. No matter what value x is, it is always true
As shown in the figure, in the plane rectangular coordinate system xoy, a (2, 1). If point P is on the coordinate axis, so that the area of △ Pao is 3, find the coordinates of point P
When the point P is on the x-axis, the area of △ Pao = 1
2OP×1=3，
OP = 6,
Therefore, the coordinates of point P are (6, 0) or (- 6, 0),
When the point P is on the y-axis, the area of △ Pao = 1
2OP×2=3，
OP = 3,
Therefore, the coordinates of point P are (0, 3) or (0, - 3),
To sum up, the coordinates of point P are (6,0) or (- 6,0) or (0,3) or (0, - 3)
As shown in the figure, in the plane rectangular coordinate system, △ ABC is a right triangle, ∠ ACB = 90 ° and the coordinates of points a and C are respectively a (- 3,0), C（1,0）,tan∠BAC= 3/4 (1) Find a point D on the x-axis and connect BD so that △ ADB is similar to △ ABC (excluding congruence), and calculate the coordinates of point D; (2) Find the sine value of angle BDC under the condition of (1) (no need to do the first question)
Because, Tan ∠ BAC = 3 / 4, according to the principle that the corresponding angles of similar triangles are equal and the corresponding sides are proportional
So the sine of the angle BDC is 4 / 5
As shown in the figure, in the plane rectangular coordinate system, the triangle ABC, BC = AC, angle ACB = 90 degrees, point C and point B are on the X axis and Y axis respectively 1. If the coordinates of point B are (0,4) and the coordinates of point a are (- 2, - 2), find the coordinates of point C; 2. If angle ABO = angle CBO, AC intersects Y-axis at point D, passing through point a as AE, perpendicular to Y-axis at point E. if AE = 3, find the length of BD? 3. When point a is in the fourth quadrant, AF is made through point a perpendicular to y axis and to point F. the quantitative relationship among segments AF, OB and OC is obtained and the reasons are explained
1. Point C is on the vertical bisector of line AB, and the intersection of the vertical bisector and X axis is point C;
The slope of the horizontal line (- 1,2) is equal to - 1,2,
Therefore, the vertical bisector of line AB is Y-1 = - 1 (x + 1) / 3; the intersection point with X axis is (2,1), i.e. C (2,1);
2. It is proved that the triangles ACh and BCD are congruent when the extended segments AE and BC intersect at the H point;
AB intersection x-axis and point G
Ah is parallel to the x-axis and be is the angular bisector of ∠ ABH, so Ag = HC can be obtained (because AB = Hb, BG = BC);
Δ BDG ≌ △ BDC, △ ADG is an isosceles right triangle, and Ag = DG = DC is obtained
So HC = DC, right angle, AC = BC, we get congruence, BD = ah,
AE=3,AH=6,
3. Relationship: OB = 0C + AF;
It is proved that △ BOC ≌ AHC is △ BOC ≌ AHC
BC = AC (known), right angle, ∠ OBC = ∠ OCA = ∠ HAC (corner or HL)
AH=OC+AF.
As shown in the figure, in the plane rectangular coordinate system, the image of primary function y=kx+5 passes through point A (1, 4), and point B is the image of primary function y=kx+5 and the positive scale function y = 2 The intersection of 3x images (1) Find the coordinates of point B (2) Find the area of △ AOB
(1) Substituting a (1,4) into y = KX + 5, 4 = K + 5,
The solution is: k = - 1,
Then the analytic formula of the first order function is y = - x + 5,
y＝−x+5
y＝2
3x ，
The solution
x＝3
y＝2 ，
So the coordinates of point B are (3,2);
(2) When y = 0, - x + 5 = 0,
The solution is: x = 5,
Then E (0, 5),
S△AOB=S△BOE-S△AOE=1
2×5×3-1
2×5×1=5．
In the plane rectangular coordinate system, a (0,1) B (2,0) C (4,3 (1) is known to calculate the area of triangle ABC. The rest is shown in the picture (2) Let point p be in the coordinate axis, and the area of triangle ABP and triangle ABC are equal, and find the coordinates of point P, there is no picture,
If the height of P on the x-axis is equal to 2, then the bottom is AP, then p (0,5) or (0, - 3); when p is on the Y axis, the height is Ao = 1, then BP = 8, so p (10,0) or (- 6,0)
RELATED INFORMATIONS
- 1. Given the point a (- 4,0) B (2,0), if the point is on the image of the linear function y = - 2 / 1X + 2 and the triangle ABC is a right triangle, several conditions of point C are satisfied
- 2. As shown in the figure, in the plane rectangular coordinate system xoy, the vertex e coordinates of rectangular oefg are (4,0), and the vertex g coordinates are (0,2). Rotate the rectangular oefg anticlockwise around point O, so that point F falls on the point n of the Y axis, and the rectangle omnp, OM and GF intersect at point a (1) Judge whether △ OGA and △ NPO are similar and explain the reasons; (2) The analytic formula of inverse proportional function of point a is obtained; (3) If the image of the inverse scaling function obtained in (2) intersects with EF at point B, please explore whether the line AB is perpendicular to OM and explain the reason
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