直角坐標幾何問題~緊急
回答 (3)
2012-05-05 3:00 pm
Let AP : PB = r : s.
Since P is on x - axis, so the y - coordinate of P is 0.
So 0 = [14r + (-2)s]/(r + s) (Using formula of point of division).
14r - 2s = 0(r + s) = 0
so 14r = 2s
r : s= r/s = 2/14 = 1/7 = 1 : 7.
Since P is on x - axis, so the y - coordinate of P is 0.
So 0 = [14r + (-2)s]/(r + s) (Using formula of point of division).
14r - 2s = 0(r + s) = 0
so 14r = 2s
r : s= r/s = 2/14 = 1/7 = 1 : 7.
2012-05-05 8:25 am
較簡單的方法：
設 P = (a, 0)，則 APB 的斜率：
(0 + 2)/(a - 2) = (0 - 14)/(a - 6)
a = 2.5
P (2.5, 0)
以 y 坐標計算：
AP : PB = (0 + 2) : (14 - 0) = 1 : 7 ...... (答案)
或 以 x 坐標計算：
AP : PB = (2.5 - 2) : (6 - 2.5) = 1 : 7 ...... (答案)
設 P = (a, 0)，則 APB 的斜率：
(0 + 2)/(a - 2) = (0 - 14)/(a - 6)
a = 2.5
P (2.5, 0)
以 y 坐標計算：
AP : PB = (0 + 2) : (14 - 0) = 1 : 7 ...... (答案)
或 以 x 坐標計算：
AP : PB = (2.5 - 2) : (6 - 2.5) = 1 : 7 ...... (答案)
2012-05-05 7:57 am
先求直線方程:
[y-(-2)]/(x-2)=[14-(-2)]/(6-2)
y+2 / x-2=16/4
y+2=4x-8
y=4x-10
求P點座標(x,0)
0=4x-10
4x=10
x=2.5
設AP:PB=s:r
(6s+2r)/(s+r)=2.5
6s+2r=2.5s+2.5r
3.5s=0.5r
s:r=0.5/3.5
=1/7
Ans(AP:PB)=1/7
[y-(-2)]/(x-2)=[14-(-2)]/(6-2)
y+2 / x-2=16/4
y+2=4x-8
y=4x-10
求P點座標(x,0)
0=4x-10
4x=10
x=2.5
設AP:PB=s:r
(6s+2r)/(s+r)=2.5
6s+2r=2.5s+2.5r
3.5s=0.5r
s:r=0.5/3.5
=1/7
Ans(AP:PB)=1/7
收錄日期: 2021-04-13 18:40:09
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