當 y=5 x=3 , y 正比於 x,當y=7 x=????
我想知點計呢？
另外兩直線A(-1,-3)B(3,5)平行C(x,y)D(1,7)，C點既X軸截距為0，求C點座標
依題係咪用斜率去計呢？係既話又係點計呢？
中學數數題
2008-01-15 9:58 pm
回答 (3)
2008-01-15 10:21 pm
✔ 最佳答案
1. 因為y正比於x , 所以設y = kx (k是常數)y = kx
當y=5, x=3:
5 = 3k
k = 5/3
所以y = (5/3)x
所以當y=7:
7 = (5/3)x
x = 21/5
2. 因為AB平衡CD, 所以AB斜率 = CD斜率
所以 (y - 7) / (0 - 1) = (5 + 3) / (3 + 1)
-(y - 7) = 2
y - 7 = -2
y = 5
所以C = (0 , 5)
2008-01-16 5:49 am
y = kx
當y=5, x=3:
5 = 3k
k = 5/3
所以y = (5/3)x
所以當y=7:
7 = (5/3)x
x = 21/5
2. 因為AB平衡CD, 所以AB斜率 = CD斜率
所以 (y - 7) / (0 - 1) = (5 + 3) / (3 + 1)
-(y - 7) = 2
y - 7 = -2
y = 5
當y=5, x=3:
5 = 3k
k = 5/3
所以y = (5/3)x
所以當y=7:
7 = (5/3)x
x = 21/5
2. 因為AB平衡CD, 所以AB斜率 = CD斜率
所以 (y - 7) / (0 - 1) = (5 + 3) / (3 + 1)
-(y - 7) = 2
y - 7 = -2
y = 5
2008-01-15 10:37 pm
1.
y 正比於 x
therefore, y = kx
5 = k * 3
k = 5/3 ....... we get the constant is 5/3
so, y = 5x/3
when y = 7
7 = 5x/3
21 = 5x
x = 21/5
2.
C 點既 x 軸截距為 0
So, x = 0
依題係咪用斜率去計
∵ AB // CD
∴ m(AB) = m(CD)
m(AB) = [ 5 - ( - 3 )] / [ 3 - ( - 1 )] = 8/4 = 2
m(CD) = ( 7 - y ) / ( 1 - x ) = 7 - y
m(AB) = m(CD)
2 = 7 - y
y = 5
hence, the coordinate of C is (0,5)
y 正比於 x
therefore, y = kx
5 = k * 3
k = 5/3 ....... we get the constant is 5/3
so, y = 5x/3
when y = 7
7 = 5x/3
21 = 5x
x = 21/5
2.
C 點既 x 軸截距為 0
So, x = 0
依題係咪用斜率去計
∵ AB // CD
∴ m(AB) = m(CD)
m(AB) = [ 5 - ( - 3 )] / [ 3 - ( - 1 )] = 8/4 = 2
m(CD) = ( 7 - y ) / ( 1 - x ) = 7 - y
m(AB) = m(CD)
2 = 7 - y
y = 5
hence, the coordinate of C is (0,5)
參考: 自己
收錄日期: 2021-04-13 19:43:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080115000051KK01335