有條family of straight line唔識計

(4k-1)x + (k+1)y + (k-4) = 0
(a) Distance of (0,0) from the line
= | k - 4 | / √[(4k-1)^2 + (k+1)^2] = 1
| k - 4 | = √[(4k-1)^2 + (k+1)^2]
k^2 - 8k + 16 = 16k^2 - 8k + 1 + k^2 + 2k + 1
16k^2 + 2k - 14 = 0
8k^2 + k - 7 = 0
(8k - 7)(k + 1) = 0
k = 7/8 or k = -1
The two equations are :
(4*7/8 - 1)x + (7/8 + 1)y + (7/8 - 4) = 0
(28 - 8)x + (7 + 8)y + (7 - 32) = 0
20x + 15y - 25 = 0
4x + 3y - 5 = 0 ... (A)
and
(-4 - 1)x + (-1 + 1)y + (-1 - 4) = 0
x + 1 = 0 ... (B)
(b) Slope of (A) = -4/3
tanθ = (x + 4/3)/[1 - x(-4/3)]
= (1 + 4/3x)/(1/x + 4/3)
Angle of inclination for (A) is tanx = -4/3 is an obtuse angle
The acute angle of inclination is 180 - x
Line (B) is a vertical line and perpendicular to the x-axis
Acute angle θ = 90 - (180 - x) = x - 90
tanθ = tan(x - 90)
= -tan(90 - x)
= -cotx
= -1/tanx
= 3/4

2009-12-20 01:10:02 補充：
Ignore these 3 lines :

Slope of (A) = -4/3

tanθ = (x + 4/3)/[1 - x(-4/3)]

= (1 + 4/3x)/(1/x + 4/3)