![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
(1) In the plane rectangular coordinate system, move point a (- 3,4) to the right 5 units to point A1, and then rotate point A1 clockwise 90 ° around the origin of the coordinate to point A2. Directly write out the coordinates of points A1 and A2; (2) in the plane rectangular coordinate system, move point B (a, b) in the second quadrant to the right m units to point B1 in the first quadrant, and then rotate point B1 clockwise 90 ° around the origin of the coordinate to point B 2. Directly write out the coordinates of points B1 and B2; (3) in the plane rectangular coordinate system, translate point P (C, d) n units along the horizontal direction to point P1, and then rotate point P1 90 ° clockwise around the origin of the coordinate to point P2, directly write out the coordinates of point P2
(1) As shown in the figure, ∵ move point a (- 3, 4) to the right 5 units to point A1, the coordinates of ∵ A1 are (2, 4), ∵ rotate point A1 clockwise 90 ° around the origin of coordinates to point A2, ≌ oma1 ≌ om1a2, ≌ A2 coordinates (4, - 2). (2) according to the rule in (1), the coordinates of B1 are (a + m, b)
Try Cramer's rule to solve the following linear equations {X1 + X3 = 1 2x1 + 2x2 + 3x3 = 3 x2 + X3 = - 1}
D＝1,Dx＝4 Dy＝2.Dz＝-3 ∴x＝4,y＝2,z＝-3.
As shown in the triangle ABC in the figure, AE bisection ∠ BAC intersects BC at e, D is the midpoint of BC, DF is parallel to AE intersects AC at F, ab = 1, AC = 2, find the length of CF
Do BH / / AE through B and cross the extension line of CA to H point
Because, ∠ BAE = ∠ ABH (internal stagger angle), ∠ EAC = ∠ H (appositive angle)
And because AE is an angular bisector,
So, BAE = EAC
Therefore, ∠ ABH = ∠ h,
So, ah = AB = 1
So HC = HA + AC = AB + AC = 1 + 2 = 3
Because DF / / AE, D is the midpoint of BC,
So, f is the midpoint of HC
So, ch = 2 / 3, CF = 1
The sum of binomial coefficients of the last three terms of the expansion of (x ^ 2 + 1 / x) ^ n is 22, and the constant term is obtained
Hello! I have calculated CN (n-2) + CN (n-1) + CNN = 22. I don't know what to do next. Please help me. I want to know CN (n-2) =? CN (n-1) =? CNN =?
C(n,n-2)+C(n,n-1)+C(n,n)
=C(n,2)+C(n,1)+1
=(n-1)n/2+n+1
=22
Easy to get
N=6
The constant term is C (6,2) = 15
In the plane rectangular coordinate system, the origin of O coordinate is known. Points a (3.0), B (0.4) take point a as the rotation center, and rotate the triangle ABO clockwise,
The ACD of triangle is obtained, and the rotation angle is α, and ∠ ABO is β
(1) When the point d just falls on the edge of AB after rotation, calculate the coordinates of point D; (II) when the rotation satisfies BC / / X axis, calculate the quantitative relationship between α and β; (III) when the rotation satisfies ∠ AOD = β, calculate the analytical formula of straight line CD
Turn △ ABO clockwise
The point after B rotation is C, and the point after o rotation is d?
Let AB = - 3,4, Ao = (3,0). So | ab | = 5, | Ao | = 3, let the clockwise rotation angle be α, so AC = 5 (COS α, - sin α), ad = 3 (COS α, - sin α)
(1) : if D falls on the edge of AB, that is, ad is parallel to AB, - cos α = - 3sin α / 4, then Tan α = 4 / 3, sin α = 4 / 5 (or - 4 / 5), cos α = 3 / 5 (or - 3 / 5), so ad = 3 (3 / 5, - 4 / 5) (or 3 (- 3 / 5,4 / 5)), then the coordinates of point D can be obtained (omitted)
(2) If BC is parallel to the x-axis, then BC = (5cos α + 3, - 5sin α - 4), if BC is parallel to the x-axis, - 5sin α - 4 = 0, sin α = - 4 / 5, and cos β = 4 / 5, then α = - (PI / 2 - β) + 2K PI (or 3pi / 2 - β + 2K PI), where k is an integer
(3) Od = (3cos α - 3, - 3sin α). So - Tan β = - 3sin α / (3cos α - 3), so sin α and cos α can be obtained, and the coordinates of point D can be obtained. Finally, the analytical expression of CD is obtained
The following linear equations are solved by Cramer's Law: x1-2x2 + 3x3-4x4 = 4,0 + x2-x3 + X4 = - 3, X1 + 3x3 + 0 + X4 = 1,0-7x2 + 3x3 + X4 = - 3
x1-2x2+3x3-4x4=4,0+x2-x3+x4=-3,x1+3x3+0+x4=1,0-7x2+3x3+x4=-3D=1 -2 3 -40 1 -1 11 3 0 10 -7 3 1=16D1=4 -2 3 -4-3 1 -1 11 3 0 1-3 -7 3 1=-128D2=1 4 3 -40 -3 -1 11 1 0 10 -3 3 1=48D3=1 -2 4 -40 1 -3 11 3...
Ex2. P is a point on the plane of triangle ABC. If the vector PA &; Pb = Pb &; PC = PC &; PA, then p is the center of triangle ABC
It's the heart:
From PA &; Pb = Pb &; PC, Pb &; (pa-pc) = 0, i.e. Pb &; CA = 0, i.e. Pb is perpendicular to AC,
For the same reason Pb &; PC = PC &; PA, PC is perpendicular to AB, that is, P is the intersection of the two heights of the triangle, so p is perpendicular!
In the (x ^ 2-1 / x) ^ n expansion, the sum of all binomial coefficients is 512, and the constant term is obtained
The sum of all binomial coefficients in the expansion is 512
So, 2 ^ n = 512
So, n = 9
Constant term = C (6,9) = 84
(1 / 2) if three points o (0,0), a (- 1,1), B (- 1,0) in the plane rectangular coordinate system are known, and the triangle AOB is rotated 135 degrees clockwise around the point O, then
(1 / 2) given the three points o (0,0), a (- 1,1), B (- 1,0) in the plane rectangular coordinate system, if the triangle AOB is rotated 135 degrees clockwise around the point O, the pairing of the points a and B is correct
It is known that the square root of the equation (x-3) (X-2) - P about x = 0 (1). This paper proves that no matter what the value of P is, the equation always has two unequal real roots
∠AOB=45
The complement is 135
It is equal to AO and coincides with the positive x axis after turning 135 clockwise
A0=√2
Then a'o = A0 = √ 2
A(√2,0)
AOB is an isosceles right triangle
So △ a'ob 'is an isosceles right triangle
Distance from b'to a'o = √ 2 / 2
B(√2/2,√2/2)
How to prove Cramer's law of linear equations?
RELATED INFORMATIONS
- 1. [help] there is a point P in the triangle ABC, which satisfies "vector PA + vector Pb + vector PC = 0". Can you explain that P is the center of gravity of the triangle ABC?
- 2. The midpoint o (0,0) and a (2,0) B of the plane rectangular coordinate system are the midpoint of the line OA. Rotate OA clockwise about point O for 30 degrees. Mark the corresponding point of point B as C, and find the location of point C
- 3. As shown in the figure, in △ ABC, ∠ B = 90 °, M is the point on AB, so that am = BC, n is the point on BC, so that CN = BM, connecting an and cm intersects at P, try to find the degree of ∠ APM
- 4. Given that the negative integer solution of inequality x-m > 0 is - 1, - 2, - 3, - 4, find the value range of M (and draw the number axis) It's OK not to draw, but it's not easy to write
- 5. If the value of the algebraic formula 2x + 33 is nonnegative, then the value range of X is () A. x≥32B. x≥-32C. x＞32D. x＞-32
- 6. Inequality system: if the solution set of 2x − a ＜ 0x + 3 ＞ 2x − 1 is all negative numbers, then the value range of a is () A. a=0B. a＜8C. a＜6D. a＜4
- 7. Given that the solution of inequality 3 + K (x + 2) - 4x ＞ K (x + 3) about X is negative, then the value range of K is negative
- 8. Given that the maximum value of X satisfying the inequality | x2-4x + P | + | x-3 | ≤ 5 is 3, then the value of real number P is () A. - 2b. 8C. - 2 or 8D. Not sure
- 9. The solution set of quadratic inequality KX ^ 2 + KX + 4 ≥ 0 with one variable is r, and the value range of real number k is obtained
- 10. It is known that the solution set of inequality KX ^ 2 + 2kx-1 > 0 about X is an empty set, and the value range of real number k is obtained
- 11. In △ ABC, extend the midline BD on the AC side to F, so that DF = BD, extend the midline CE on the AB side to g, so that eg = CE
- 12. Let an and BN satisfy A1 = B1 = 6, A2 = B2 = 4, A3 = B3 = 3, and a (n + 1) - an (n belongs to positive integer) is an arithmetic sequence Let an and BN satisfy A1 = B1 = 6, A2 = B2 = 4, A3 = B3 = 3, and a (n + 1) - an (n is a positive integer) is an arithmetic sequence, Sn is the sum of the first several terms of the sequence {BN}, and Sn = 2n-bn + 101) is the general term formula of an and BN (2) Does K belong to a positive integer, so that AK BK belongs to (0,1 / 2)? If so, find K; if not, explain the reason
- 13. In the RT triangle ABC, the angle BAC = 90, ad ⊥ BC at point D, e is the midpoint of AD, the extension line connecting ed and intersecting AB at point F, AB / AC = DF / AF is proved E is the midpoint of AC
- 14. It is known that [an] and {BN} are arithmetic sequences, and CN = 3an + 2. It is proved that {CN} is also arithmetic sequences (2) The second question CN = an + BN (AB is not equal to 0): the sequence {CN} is an arithmetic sequence
- 15. As shown in the figure, BD and CE are the two heights of triangle ABC, and m and N are the midpoint of BC and de respectively
- 16. In the arithmetic sequence {an}, if A1 = 3, A100 = 36, then A37 =?
- 17. As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ AB is at e, DF ⊥ AC is at F, and be = CF
- 18. The N-1 power {BN} of an and Sn = 2-1 / 2 is the arithmetic sequence A1 = B1, A2 * (b2-b1) = A1 to find the general term of BN? Let CN = BN / an to find the general term of CN
- 19. In the RT triangle ABC, the angle c = 90, BC = a, AC = B, a + B = 16, then the function analysis of the area s of the RT triangle with respect to the variable length a When s = 32, a=______
- 20. It is known that the equal ratio sequence satisfies A1 = 2, A2 + a3 = 12 and A4 > 0.1. Find the general term formula of sequence an. 2. If BN = anlog2an, the first n term of sequence BN is SN The main thing is to ask the second question!