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標題:

a.maths

發問:

Given two straight lines L1:3x+y-4=0 and L2:x+3y+4=0,let F be the family of straight lines passing through the point of intersection of L1 and L2. (a)Write down an equation of F.Hence find the equation of the line L in F which passes through the origin. 更新: (b)The point of intersection of L1 and L2 is A.C is a point in the second quadrant lying on the line L which is found in (a).Point B and D are the feet of perpendicular from C to L1 and L2 respectively. 更新 2: (i)Show that tan ∠CBA=tan∠CAD.Hence find the value of tan ∠CAB. (ii)Find the coordinates of A. (iii)If the area of quarilateral ABCD is 80, find (1)the length of BC (2)the coordinates of C.

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(a)equation of F: 3x + y - 4 + k(x + 3y + 4) = 0 ie (3 + k)x + (3k + 1)y + 4(k - 1) = 0 for a equation in F passing thru origin, x = 0 and y = 0 ie k = 1 the equation required is x + y = 0 (b)(i) (tan ∠CBA = tan 90degree = undefined, assumed as tan ∠CAB) slope of L = -1 slope of L1 = -3 slope of L2 = -1/3 tan ∠CAB = |(-1+1/3)/(1+1/3)| = 1/2 tan ∠CAD = |(-1+3)/(1+3)| = 1/2 = tan∠CAB (b)(ii) {3x + y - 4 = 0 {x + 3y + 4 = 0 ie 8y = -16, y = -2, x = 2 A = (2, -2) (b)(iii)(1) As tan ∠CAB=tan∠CAD, thus, triangle CAB=triangle CAD are congurent. tan ∠CAB = BC/AB = 2 Also, 0.5(BC)(AB) = 40 Thus AB = 2*root10, BC = 4*root10 (b)(iii)(2) By pyth. Theorem AC^2 = AB^2 + BC^2 AC = 10*root2 Let C be (-a, a)as C is on L we have 200 = (2+a)^2 + (-2 -a)^2 (2+a)^2 = 100, a = 8 Thus C = (-8, 8)

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