a-maths~~~~~~

(c)(ii) 中的L是甚麼？

2009-10-07 19:13:46 補充：
(2k+3)x + (k-2)y - (5k+4) = 0
(3x - 2y - 4) + k(2x + y - 5) = 0

(a) P is the intersection of 3x-2y=4 and 2x+y=5.
Therefore P=(2, 1).

(b) 2x + y - 5 = 0 is the required straight.
In general, for L1+k*L2=0, L2 is NOT in the family.

(c)(i) Slope = - 2/5, passing through P

Thus the required equation is
(y-1) / (x-2) = -2/5
2(x-2) = -5(y-1)
2x + 5y - 9 = 0

(ii) What is L???


2009-10-07 22:30:49 補充：
ok...my mistake, sorry.

2009-10-07 22:38:13 補充：
Slope of the required st. line = - (2k+3) / (k-2)
Slope of L = -2

Thus
| [(2k+3)/(k-2) - 2] / [1 + 2(2k+3)/(k-2)] | = tan45 = 1
| [(2k+3) - 2(k-2)] / [(k-2) + 2(2k+3)] | = 1
| 7 / [5k+4] | = 1
5k+4=7 or 5k+4=-7
k=3/5 or k=-11/5

2009-10-07 22:38:19 補充：
Therefore the required equations are
5(3x - 2y - 4) + 3(2x + y - 5) = 0 and 5(3x - 2y - 4) - 11(2x + y - 5) = 0
3x - y - 5 = 0 and x + 3y - 5 = 0

你看看(b)