F.4 A.maths Coordinates 2

1.
The required line:
ax - by - 1 + k(bx - ay - 1) = 0
(a + kb)x - (b + ka)y - (1 + k) = 0

The required line passes through origin (0, 0):
(a + kb)(0) - (b + ka)(0) - (1 + k) = 0
k = -1

Hence, the required line is:
(a - b)x + (a - b)y = 0
x + y = 0


2,
(a)
2x + 4y - 3 + k(x - y + 5) = 0
(2 + k)x + (4 - k)y + (5k - 3) = 0

(b)
(i)
If the required line // x - axis, then (2 + k) = 0
k = -2

Hence, the required line is:
[4 - (-2)]y + [5(-2) - 3] = 0
6y - 13 = 0

(ii)
Slope of the line (x = 2y) = 2

Slope of the required line:
-(2 + k)/(4 - k) = -1/2
k = 0

Hence, the required line is:
(2 + 0)x + (4 - 0)y + (0 - 3) = 0
2x + 4y - 3 = 0


3.
(a)
(2x + y - 8) + k(x + 2y) = 0
(2 + k)x + (1 + 2k)y - 8 = 0

(b)
(i)
The line passes through (2, 0):
(2 + k)(2) + (1 + 2k)(0) - 8 = 0
k = 2

The required line is:
[2 + (2)]x + [1 + 2(2)]y - 8 = 0
4x + 5y - 8 = 0

(ii)
The required line passes through (a, 0) and (0, 2a).

Slope of the required line:
-(2 + k)/(1 + 2k) = (0 - 2a)/(a - 0)
-(2 + k)/(1 + 2k) = -2
k = 0

The required line is:
[2 + (0)]x + [1 + 2(0)]y - 8 = 0
2x + y - 8 = 0


4.
(a)
The family of st. lines pass through the intersection of the two given lines below:
2x - y + 8 = 0 ...... (1)
x - 4y + 7 = 0 ...... (2)

(2) x 2:
2x - 8y + 14 = 0 ...... (3)

(1)-(3):
7y = 6
y = 6/7

(2):
x - 4(6/7) + 7 = 0
x = -25/7

P = (-25/7, 6/7)

(b)
L: (2x - y + 8) + k(x - 4y + 7) = 0
L: (2 + k)x - (1 + 4k)y + (8 + 7k) = 0

Slope of the line (x + 3y - 6 = 0) = -1/3

Slope of L:
(2 + k)/(1 + 4k) = -1/3
k = -1

L : [2 + (-1)]x - [1 + 4(-1)]y + [8 + 7(-1)] = 0
L : x + 3y + 1 = 0

(b)
Let m be the slope of the required line:
Slope of the line (y = 2x + 3) = 2

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

Slope of L:
(2 + k)/(1 + 4k) = -3
k = -5/13

L: [2 + (-5/13)]x - [1 + 4(-5/13)]y + [8 + 7(-5/13)] = 0
L: 21x + 7y + 69 = 0

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

Slope of L:
(2 + k)/(1 + 4k) = 1/3
k = 5

L: [2 + (5)]x - [1 + 4(5)]y + [8 + 7(5)] = 0
L: 7x - 21y + 43 = 0
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