Maths problems [mixed 3]

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(34) x-intercept of L1 = -b/a x-intercept of L2 = -d/c Therefore -b/a = -d/c, so ad = bc (35) a2 - ab - 2b2 = 0 (a + b)(a - 2b) = 0 a = -b or a = 2b (rejected since ab < 0) (a - b)/(a - 2b) = (-b - b)/(-b- 2b) = 2/3 (36) P is point of int. between y = x + 1 and perp. bisector of AB. Perp. bisector of AB is x - 2y + 5 = 0 So ans is (3, 4) (38) Let ∠OAB = ∠OBA = θ, then ∠TAB = ∠TBA = 90 - θ ∠ATB = 2θ and ∠AOB = 180 - 2θ. Therefore , I must be true, II may not be true and III must be true. (40) (√3 - 1)√x = 2 √x = 2/(√3 - 1) = √3 + 1 x = (√3 + 1)2 = 4 + 2√3 (42) Let AB= x, then MF = x/2 and AF x√2 So tan ∠MAF = 1/(2√2) ∠MAF = 19.5 (44) Join BT, then △BAT and △TAC are similar for the reason of AAA. So, BA/TA = BT/TC = AT/AC 9/AT = AT/25 AT = 15 cm (47) △CDE and △BAE are similar for the reason of AAA. So, CD/AB = CE/BE = DE/AE In △DAE, ∠EDA is 90 since AB is a diameter. So, DE/AE = cos θ, i.e. CD/AB = cos θ (22) x-intercept of L is -1 So the equation is: (y - 0)/(x + 1) = 1 y = x + 1 x - y + 1 = 0 (27) (I) Slope of L1 = 1/2 and slope of L2 = 2, thus they are NOT perp. (II) is true. (III) Solving for x - 2y = 1 and 2x - y = 2, we have x = 1 and y = 0. So it is true. 2009-07-02 22:37:28 補充： 2/(√3 - 1) = 2(√3 + 1)/[(√3 + 1)(√3 - 1)] = 2(√3 + 1)/(3 - 1) = 2(√3 + 1)/2 = √3 + 1

thx so much, but 我唔係太明點解 2/(√3 - 1) = √3 + 1 2009-07-05 19:15:49 補充： 明啦! thx~

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