已知 L1: 2x - y + 3 = 0 及 L2 : x + 3y - 3 = 0。
(a) 求L1及L2的斜率。
(b) 若L1及L2所夾的銳角為θ, 求θ。(答案準確至最接近的度)
A . Maths
2008-07-07 4:55 am
回答 (3)
2008-07-07 8:21 pm
✔ 最佳答案
Convert the equation into the form of y=mx+c, where m is the slope of the equation.L1: y=2x+3, L2: y=-(1/3)x-1
∴Slope of L1=2, slope of L2=-1/3
By the equation tanθ=|(m1-m2)/(1+m1m2)|
We got the result that tanθ=|(7/3)/(1/3)|
tanθ=7
θ=82o, corr to nearest integer.
2008-07-07 8:32 pm
2008-07-07 5:28 am
L1
2x -y + 3 = 0
dy/dx = 2
L2
x + 3y -3 = 0
dy/dx = -1/3
θ= tan(-1) 2 + tan(-1)1/3
=81.86989765 deg
2x -y + 3 = 0
dy/dx = 2
L2
x + 3y -3 = 0
dy/dx = -1/3
θ= tan(-1) 2 + tan(-1)1/3
=81.86989765 deg
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