1.Find the point of intersection between L1: x+3y- 4=0 and L2: 2x+ 6y+3=0
2. Show that the straight lines L1: 2x-y-4=0 and L2: 6y+3x-4=0 are perpendicular with different y-intercepts.
Math:Equation of straight line
2009-06-28 11:59 pm
回答 (2)
2009-06-29 1:06 am
✔ 最佳答案
1. Slpoe of L1=-(1/3)
Slpoe of L2=-(2/6)=-1/3
∴L1//L2
∴The point of intersection=0
2.
L2: 6y+3x-4=0
i.e. 3x+6y-4=0
y-intercepts of L1=-(-4/-1)=-4
y-intercepts of L2=-(-4/6)=2/3
Slpoe of L1=-(2/-1)=2
Slpoe of L2=-(3/6)=-1/2
(Slpoe of L1) (Slpoe of L2)=(2)(-1/2)
=-1
2009-06-29 12:13 am
先給你一點提示：
1. 只要solve 了equation 找出 x=?, y=? 就是 point of intersection
2. Slope of Equation = -A/B,
你計出 L1 及 L2 的 slope, 若他們 perpendicular, slope相乘便是 -1
y-intercept 的問題，代 x=0 入 L1, 計到的 y值便是 L1的 y-intercept
最後希望你計數多用腦筋
1. 只要solve 了equation 找出 x=?, y=? 就是 point of intersection
2. Slope of Equation = -A/B,
你計出 L1 及 L2 的 slope, 若他們 perpendicular, slope相乘便是 -1
y-intercept 的問題，代 x=0 入 L1, 計到的 y值便是 L1的 y-intercept
最後希望你計數多用腦筋
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