Find the equations of two straight lines which are concurrent with the lines
2x- y - 8=0 and 4x + y - 3=0 and make equal intercepts with the axes.
F4 A.Maths
2008-06-30 4:52 am
回答 (2)
2008-06-30 5:35 am
✔ 最佳答案
吾人可設該直線方程為x/a+y/a=1ax+ay-a^2=0
又由2x - y - 8 = 0 和 4x + y - 3 = 0
得y=2x-8, 6x=11
x=11/6,y=22/6-48/6=-26/6
代入ax+ay-a^2=0
(-15/6)a-a^2=0
a=-15/6
故所求為
-x-y-15/6=0
6x+6y+15=0
2008-06-29 21:36:13 補充：
即2x+2y+5=0
2008-07-02 9:59 pm
By family of st. line
2x- y - 8 + k(4x + y - 3)=0........(i)
(2 +4k)x+y(k-1)-(8+3k)=0
[(2 +4k)/(8+3k)]x+y[(k-1)/8+3k)]=1
They have same intercepts , such that
(8+3k)/(2 +4k)=(8+3k)/(k-1)
k=-1
sub k=-1 into (i)
2x+2y+5=0
or they have zero intercept
i.e. (x,y)=(0,0)
we have k= - 8/3
26x+11y+48=0
2x- y - 8 + k(4x + y - 3)=0........(i)
(2 +4k)x+y(k-1)-(8+3k)=0
[(2 +4k)/(8+3k)]x+y[(k-1)/8+3k)]=1
They have same intercepts , such that
(8+3k)/(2 +4k)=(8+3k)/(k-1)
k=-1
sub k=-1 into (i)
2x+2y+5=0
or they have zero intercept
i.e. (x,y)=(0,0)
we have k= - 8/3
26x+11y+48=0
收錄日期: 2021-04-25 16:59:15
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