急!!! F.4 Maths

2010-09-24 1:18 am
1.Consider three straight lines L1:3x-y-1=0,L2:x-y+1=0 and L3:ax+by+c=0 where a,b and c are constants.If L3 passes through the point of intersection of L1 and L2,find the one possible equation of L3.

2.Find an equation of the straight line L which is perpendicular to L1:y=4x+6.

3.Consider the two straight lines L1:x+2y+3=0 and L2:ax+by-2=0.If L1//L2,describe how to find the relationship between a and b and suggest a pair of possible values of a and b.

4.A straight line L with positive slope cuts the x-axis and the y-axis at points A and B respectively such that AB=5 units and the coordinates of A and B are integers.Find all possible equations of L.

回答 (1)

2010-09-24 1:26 pm
✔ 最佳答案
1. First of all, find the coordinates of the intersection point of L1 and L2.x = 1, y = 2Then, let the slope of L3 be m.Using point-slope form,y – 2 = m (x – 1)mx – y – m + 2 = 0 You can find out one possible equation of L3 by putting different values of m.For example, if you choose m = 1, then you have x – y + 1 = 0 if you choose m = –1, then you have x + y – 3 = 0. 2.Slope of L1 = 4, Since L is perpendicular to L1, slope of L = –1/4The equation of L : y = (–1/4)x + k, where k is a real constant. 3.If L1 // L2, 1/2 = a/bb = 2aTherefore, the relationship between a and b is b = 2a.A pair of possible values of a and b is a = 1 and b = 2. 4.Let the coordinates of A and B be (a, 0) and (0, b) respectively.Ö(a – 0)2+(0 – b)2 = 5a2 + b2 = 25Since a and b are integers, the solution sets should be a = 3, b = – 4a = 3, b = 4a = –3, b = – 4a = –3, b = 4a = 4, b = – 3a = 4, b = 3a = –4, b = – 3a = –4, b = 3 As the slope of AB = –b/a is positive, the only possible sets of solutions are the ones with different signs, i.e. a = 3, b = – 4a = –3, b = 4a = 4, b = – 3a = –4, b = 3 Therefore, all possible equations of L are:(Find by using intercept form)(For a = 3, b = – 4) x/3 + y/(– 4) = 1 è 4x – 3y – 12 = 0(For a = –3, b = 4) x/(– 3) + y/4 = 1 è 4x – 3y + 12 = 0(For a = 4, b = –3) x/(4) + y/(– 3) = 1 è 3x – 4y – 12 = 0(For a = –4, b = 3) x/(– 4) + y/3 = 1 è 3x – 4y + 12 = 0


2010-09-24 05:27:32 補充:
In question 4, a2 + b2 = 25 should be a^2 + b^2 = 25
If you have any questions, please feel free to ask.
參考: My Maths knowledge


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