A variable line passing through the point (5,0) intersects the lines 3x-4y=0 and 3x+4y=0
at H and K respectively. Find the equation of the locus of the mid-point of HK.
a.maths locus
2007-06-04 5:19 am
回答 (2)
2007-06-04 6:43 am
✔ 最佳答案
As follows:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazylocus1.jpg
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazylocus2.jpg
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazylocus3.jpg
2007-06-15 09:54:01 補充：
最後一部份, 由 y = 45m/(16^2 - 9) 開始已是找 equation 的 steps.不過尾三和二步有點亂子, 其實應該係:= 3/4 x [x(x-5)]^(0.5)因為分母中的 x - 5 個分母忘了乘上分子.至於下面位仁兄的答案, 可以試畫圖來驗證是不可能為直線的.
參考: My Maths knowledge
2007-06-14 12:07 am
Consider the lines 3x - 4y = 0 and 3x + 4y = 0
Let
L[1] : 3x - 4y = 0
L[2]: 3x + 4y = 0
L[2] - L[1]
3x - 3x + 4y - ( - 4y ) = 0
8y = 0
y = 0
L[2] + L[1]
3x + 3x + 4y - 4y = 0
6x = 0
x = 0
so (0,0) and (5,0) are on the mid-pt of H and K
Applying two-pt form:
y - 0 = [ ( 0 - 0 ) / ( 5 - 0 ) ] (x - 0)
y = ( 0 / 5 ) x
y = 0
So the equation is y=0
2007-06-13 16:08:18 補充：
文遠兄的答案似乎不是 "locus of the mid-point of HK"
2007-06-15 15:38:01 補充：
我知自己錯咩了，
就是計了L[1] 和L[2]的 mid point......
Let
L[1] : 3x - 4y = 0
L[2]: 3x + 4y = 0
L[2] - L[1]
3x - 3x + 4y - ( - 4y ) = 0
8y = 0
y = 0
L[2] + L[1]
3x + 3x + 4y - 4y = 0
6x = 0
x = 0
so (0,0) and (5,0) are on the mid-pt of H and K
Applying two-pt form:
y - 0 = [ ( 0 - 0 ) / ( 5 - 0 ) ] (x - 0)
y = ( 0 / 5 ) x
y = 0
So the equation is y=0
2007-06-13 16:08:18 補充：
文遠兄的答案似乎不是 "locus of the mid-point of HK"
2007-06-15 15:38:01 補充：
我知自己錯咩了，
就是計了L[1] 和L[2]的 mid point......
收錄日期: 2021-04-12 19:53:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070603000051KK04909