A math-straight lines 3條(15分)

1.
The equation of the line passing through A is:
x + y - 5 + k(2x - 3y) = 0
(1 + 2k)x + (1 - 3k)y - 5 = 0

Distance from the origin (0, 0) to the line:
[(1 + 2k)(0) + (1 - 3k)(0) - 5]/√[(1 + 2k)2 + (1 - 3k)2] = 2
-5/√[1 + 4k + 4k2 + 1 - 6k + 9k2] = 2
-5/√[2 - 2k + 13k2] = 2
2√[2 - 2k + 13k2] = -5
4(2 - 2k + 13k2) = 25
52k2 - 8k + 8 = 25
52k2 - 8k - 17 = 0
(26k - 17)(2k + 1) = 0
k = 17/26 or k = -1/2

When k = 17/26:
[1 + 2(17/26)]x + [1 - 3(17/26)]y - 5 = 0
(60/26)x - (25/26)y - 5 = 0
(12/26)x - (5/26)y - 1 = 0
12x - 5y - 26 = 0

When k = -1/2:
[1 + 2(-1/2)]x + [1 - 3(-1/2)]y - 5 = 0
(5/2)y - 5 = 0
y - 2 = 0

Ans: 12x - 5y - 26 = 0 ooroy - 2 = 0

2.
(a)
The equation of the line: y - 3 + k(x - y + 1) = 0
kx + (1 - k)y + (k - 3) = 0
When y = 0, x = -(k - 3)/k

x-intercept = -(k - 3)/k = 5
-k + 3 = 5k
6k = 3
k = 1/2

L1: (1/2)x + [1 - (1/2)]y + [(1/2) - 3] = 0
L1: (1/2)x + (1/2)y + (-5/2) = 0
L1: x + y - 5 = 0

(b)
The equation of the line: y - 3 + k(x - y + 1) = 0
kx + (1 - k)y + (k - 3) = 0

When the line // x-axis:
k = 0

L2: (0)x + [1 - (0)]y + [(0) - 3] = 0
L2: y - 3 = 0

(c)
Slope of L1 = -(1/1) = -1
Angle between L1 and x-axis = tan-1|-1|= 45o

Since L2 // x-axis, angle between L1 and L2 = 45o

3.
(a)
Slope of L = 2

tan45o = (m - 2)/(1 + 2m) ooro tan45o = (2 - m)/(1 + 2m)
1 = (m - 2)/(1 + 2m) ooro 1 = (2 - m)/(1 + 2m)
1 + 2m = m - 2 ooro 1 + 2m = 2 - m
m = -3 ooro 3m = 1
m = -3 ooro m = 1/3

(b)
The line:
2x - 3y + 2 + k(x - y - 1) = 0
(2 + k)x - (3 + k)y + (2 - k) = 0
Slope m = (2 + k)/(3 + k)

m = 1/3 (positive):
(2 + k)/(3 + k) = 1/3
3(2 + k) = 3 + k
6 + 3k = 3 + k
2k = -3
k = -3/2

The line is:
[2 + (-3/2)]x - [3 + (-3/2)]y + [2 - (-3/2)] = 0
(1/2)x - (3/2)y + (7/2) = 0
x - 3y + 7 = 0
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