ce maths 00年LQ 16題b)ii)
2008-03-30 11:07 am
有無人可以話我知16題b)ii)點做!我見solution 答案係3x+4y-75=0,但我唔明白,因為唔係計左QO斜率,再利用90度(相乘=-1)關係搵到RQ斜率,再用點斜式,（y-8)/(x-6)=-3/4就得架啦咩?
回答 (1)
2008-03-30 4:13 pm
✔ 最佳答案
或者我重頭做一次,00 LQ 16(b)(ii)
OC的長度 = sqrt (6^2 8^2) = 10
CQ = CP (圓的半徑)
= OC sin30
= 10 sin30
= 5
設Q = (a, b)
OC : CQ = 10 : 5
(10a 5*0)/(10 5) = 6
10a = 90
a = 9
(10b 5*0)/(10 5)=8
10b = 120
b = 12
所以Q的坐標是(9, 12)
OC的斜率= (8-0)/(6-0) = 4/3
RQ的斜率= -1/4/3 (因為OC垂直RQ)
= -3/4
利用點斜式, RQ的方程 :
(y -12)/(x - 9) = -3/4
-4y 48 = 3x - 27
3x 4y - 75 = 0
收錄日期: 2021-04-24 23:25:07
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