answer:(I) Since AB is parallel to l and passes through the point (0,0), the equation of the line containing AB is y=x. Let the coordinates of points A and B be (x_{1},y_{1}) and (x_{2},y_{2}), respectively. From begin{cases}x^{2}+3y^{2}=4 y=xend{cases}, we get x=±1. Thus, |AB|= sqrt {2}|x_{1}-x_{2}|=2 sqrt {2}. The altitude h of triangle ABC equals the distance from the origin to line l. Thus, h= sqrt {2}, and consequently, S_{triangle ABC}= dfrac {1}{2}|AB|⋅h=boxed{2}. (II) Let the equation of the line containing AB be y=x+m. From begin{cases}x^{2}+3y^{2}=4 y=x+mend{cases}, we obtain 4x^{2}+6mx+3m^{2}-4=0. Since A and B lie on the ellipse, the discriminant must be positive: triangle =-12m^{2}+64 > 0. Let the coordinates of A and B be (x_{1},y_{1}) and (x_{2},y_{2}), respectively. Then, x_{1}+x_{2}=- dfrac {3m}{2}, and x_{1}x_{2}= dfrac {3m^{2}-4}{4}. Consequently, |AB|= sqrt {2}|x_{1}-x_{2}|= dfrac { sqrt {32-6m^{2}}}{2}. The length of BC equals the distance from point (0,m) to line l, i.e., |BC|= dfrac {|2-m|}{ sqrt {2}}. Thus, |AC|^{2}=|AB|^{2}+|BC|^{2}=-m^{2}-2m+10=-(m+1)^{2}+11. Hence, when m=-1, AC has the maximum length (and triangle =-12+64 > 0). In this case, the equation of the line containing AB is y=x-1, leading to boxed{y=x-1} as the final answer.

answer:Given that triangles ABC and DEF have side lengths of 5, 12, 13 and 8, 15, 17 respectively, we first recognize that both sets of side lengths satisfy the Pythagorean theorem. This means both triangles are right-angled, with the hypotenuses being the longest sides, 13 and 17 respectively. For a right-angled triangle, the area can be calculated using the formula frac{1}{2} times text{base} times text{height}. In triangle ABC, the base and height can be considered as 5 and 12 (or vice versa), and in triangle DEF, they can be considered as 8 and 15 (or vice versa). Calculating the areas, we have: - Area of triangle ABC = frac{1}{2} times 5 times 12 = 30 - Area of triangle DEF = frac{1}{2} times 8 times 15 = 60 The ratio of the area of triangle ABC to the area of triangle DEF is therefore: [ frac{text{Area of } ABC}{text{Area of } DEF} = frac{30}{60} = frac{1}{2} ] Thus, the ratio of the area of triangle ABC to the area of triangle DEF is boxed{frac{1}{2}}.

answer:In an octagon, labeled with points A, B, C, D, E, F, G, H, diagonals are line segments connecting non-adjacent vertices. Any pair of vertices can either form an edge of the octagon or a diagonal. The total number of ways to choose any two vertices from the octagon (without regard to order) is given by the combination formula binom{8}{2} = frac{8 times 7}{2} = 28. Since the octagon has 8 sides, subtracting these from the 28 total pairings leaves 28 - 8 = 20 diagonals. Therefore, there are boxed{20} diagonals possible in an octagon.

answer:1. **Identify Characteristics of the Polynomial:** Since the roots are all negative integers, assume g(x) = (x + m_1)(x + m_2)(x + m_3)(x + m_4) with each m_i > 0. 2. **Determine the Value of g(1):** Given that p + q + r + s = 8091, we have: [g(1) = 1 + p + q + r + s = 8092.] Choose prime factorization of 8092 that can be decomposed into four products of (1 + m_i). The factorization of 8092 is 2^2 times 3^2 times 13 times 29. 3. **Configure the Roots:** The possible 1 + m_i values that multiply to 8092 are 4, 9, 13, 29 (from the factor components). Therefore, 4 = 1 + 3, 9 = 1 + 8, 13 = 1 + 12, 29 = 1 + 28. So, the roots m_1, m_2, m_3, m_4 are 3, 8, 12, 28. 4. **Calculate Polynomial and Determine s:** Construct g(x): [ g(x) = (x + 3)(x + 8)(x + 12)(x + 28). ] Calculating s, the constant term, [ s = 3 times 8 times 12 times 28 = 8064. ] Conclusively, [ boxed{s = 8064}. ]