A.Maths Straight Lines

x+y-2+k(y-2x-2)=0
x+y-2+ky-2kx-2k=0
(1-2k)x+(1+k)y-2-2k=0
(1-2k)x+(1+k)y=2+2k
[(1-2k)/(2+2k)]x+(1+k)/(2+2k)y=1
x/[(2+2k)/1-2k]+y/[(2+2k)/(1+k)]=1
(2+2k)/(1-2k)+[(2+2k)/(1+k)]=6
(2+2k)(1+k)+(2+2k)(1-2k)=6(1-2k)(1+k)
(2+2k)[(1+k+1-2k)=6(1+k-2k-2k^2)
(2+2k)(2-k)=6-6k-12k^2
4-2k+4k-2k^2=6-6k-12k^2
12k^2-2k^2+6k+2k+4-6=0
10k^2+8k-2=0
5k^2+4k-1=0
(5k+1)(k-1)=0
k=-1/5 or k=1
So, whenk=-1/5
x+y-2+(-1/5)(y-2x-2)=0
x+y-2-(1/5)y+(2/5)x+2/5=0
5x+5y-10-y+2x+2=0
7x-4y-8=0
when k=1
x+y-2+(1)(y-2x-2)=0
x+y-2+y-2x-2=0
-x+2y-4=0
x-2y+4=0

2008-06-14 11:42:15 補充：
concurrent即是共點~~~~

2008-06-14 11:57:53 補充：
Sorry, there are sth wrong, I need to correct
5k^2+4k-1=0
(5k-1)(k+1)=0
k=0.2 or -1
when k=0.2
x+y-2+(0.2)(y-2x-2)=0
5x+5y-10+y-2x-2=0
3x+6y-12=0
x+2y-4=0
when k=-1
x+y-2+(-1)(y-2x-2)=0
x+y-2-y+2x+2=0
3x=0
x=0

concurrent 即係 相交於一點


x +y -2 = 0
-2x +y -2 = 0

唔準用手計, 練習齋用眼, 睇個答案出黎

x +y -2 = 0
-2x +y -2 = 0

x 係 0 的話

前面
x
-2x 全部 0 晒
得番後面
y -2 = 0
y -2 = 0

y 明顯 = 2

(x, y) = (0, 2) 你要揾0既嗰條線, 一定要穿過 (0, 2)

slope int form:
x/a + y/b =1

0/a + 2/b =1 代 (0, 2)得

a +b = 6

solve 以上兩式, 得到 a, b