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Given that 1, x1, X2, 7 are equal difference sequence, 1, Y1, Y2, 8 are equal ratio sequence, and points m (x1, Y1), n (X2, Y2), then the equation of the middle perpendicular of line Mn is______ ．
∵ 1, x1, X2, 7 into arithmetic sequence, ∵ d = 7 − 14 − 1 = 2 ∵ X1 = 1 + 2 = 3, X2 = 1 + 4 = 5 and ∵ 1, Y1, Y2, 8 into arithmetic sequence, ∵ q = 2 ∵ Y1 = 1 × 2 = 2, y2 = 1 × 4 = 4, then the midpoint of M (3,2), n (5,4), ∵ Mn is (4,3), Kmn = 4 − 25 − 3 = 1, then the equation of the middle perpendicular of M (3,2), n (5,4), ? Mn is Y -
Take two points P1 (x1, Y1) and P2 (X2, Y2) in the rectangular coordinate system, make 1, x1, X2, 7 into arithmetic sequence in turn, and 1, Y1, Y2, 8 into proportional sequence in turn. If P1 and P2 are symmetrical with respect to line L, then the equation of line L is ()
A. x+y+1=0B. x-y-1=0C. x+y-7=0D. 2x-y-5=0
∵ 1, x1, X2, 7 in turn into an arithmetic sequence, ∵ X1 = 3, X2 = 5 ∵ 1, Y1, Y2, 8 in turn into an arithmetic sequence ∵ Y1 = 2, y2 = 4, ∵ P1 (3, 2), P2 (5, 4) ∵ P1, P2 two points are symmetrical about the line L, ∵ the slope of the line between p1p2 two points is 4 − 25 − 3 = 1, ∵ the slope of the line L is - 1, and the line L passes through the point (....)
If a, x1, X2, B and a, Y1, Y2, Y3, B are all arithmetic sequences, then (y3-y1) / (x2-x1) =?
x2-x1=(b-a)/3
y3-y1=2(y3-y2)=2(b-a)/4
So (y3-y1) / (x2-x1) = 3 / 2
RELATED INFORMATIONS
- 1. If a ≠ B, and a, x1, X2, B and a, Y1, Y2, Y3, B are arithmetic sequences, then (x1-x2) / (y1-y2)=
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- 3. Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if S9 = 72, then A2 + A4 + A9 = () A. 12B. 18C. 24D. 36
- 4. Given the first n terms of {an} and Sn = an2 + BN (a is not equal to 0), find the general term formula of {an} and prove that the sequence is arithmetic sequence
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- 6. It is known that {an} is an equal ratio sequence with the first term A1, the common ratio Q, (q is not equal to 1, greater than 0), the sum of the first n terms is Sn, 5 * S2 = 4 * S4, let BN = q + Sn (1) It is known that {an} is an equal ratio sequence whose first term is A1, common ratio is Q, (q is not equal to 1, greater than 0). The sum of the first n terms is Sn, and 5 * S2 = 4 * S4. Let BN = q + Sn (1) find out whether the Q (2) sequence BN can be an equal ratio sequence. If we can find out Q, we can't give the reason,
- 7. Let the sum of the first n terms of the sequence {an} be Sn, and (3-m) Sn + 2man = m + 3 (n ∈ n *). Where m is a constant, m ≠ - 3 and m ≠ 0, please solve one step (2) If the common ratio of sequence {an} satisfies q = f (m) and B1 = A1, BN = 32 F (bn-1) (n ∈ n *, n ≥ 2) bn }For the arithmetic sequence, and BN If B 1 = a 1 = 1, q = f (m) = 2m m + 3, n ∈ N and N ≥ 2, BN = 32 f (bn-1) = 32 & # 8226; 2bn-1 bn-1 + 3, can we explain it? (# 8658); we get bnbn-1 + 3bn = 3bn-1 & # 8658; 1bn-1 bn-1 = 13.; {1bn} is the first term of 1, and 13 is the arithmetic sequence of tolerance, so BN = 3N + 2 Proving {1 / BN} as arithmetic sequence And find BN
- 8. In the sequence, A1 = 8, A4 = 2, and satisfy an + 2-2an + 1 + an = 0. It is proved that {an} is an arithmetic sequence
- 9. Given the sequence {an}, A1 = 1, a (n + 1) = an + 2 / an + 1. Prove that when n is greater than or equal to 2 and N belongs to positive integer, there is 1
- 10. 1. Given the sequence {an} A1 = - 2, an + 1 = (1 + an) / (1-an), then a2011= A.-2 B.-1/3 C.-0.5 D.0.5
- 11. Given that the average of two groups of data x1, X2,. Xn is 5, and the average of Y1, Y2,. Yn is 2, then the average of ax1 + by1, AX2 + BY2. Axn + byn And explain the reason
- 12. It is known that x1, x2. Xn, the average number is x-pull, while ax1 + B, AX2 + B. axn + B, the average number is y-pull, which is expressed by the algebraic formula containing x-pull
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- 20. Let f (x) = – 2x + 2, F 1 (x) = f (x), FN (x) = f [fn-1 (x)], n ≥ 2, n ∈ n, then the image of function y = FN (x) is always over the fixed point