F4 Equation of Staright Lines

19.
Let b be the y-intercept. (b > 0)
Then, the x-intercept = 2b

The st. line passes through the point (2b, 0), (1, 2) and (0, b).

Slope of the st. line.
(2- 0)/(1 - 2b) = (2 - b)/(1 - 0)
2/(1 - 2b) = (2 - b)
(1 - 2b)(2 - b) = 2
2b² - 5b + 2 = 2
2b² - 5b = 0
b(2b - 5) = 0
b = 0 (rejected for b > 0) or b = 5/2

The st. line passes through the point (1, 2) and (0, 5/2).
Hence, equation of the st. line :
y - 2 = {[2 - (5/2)]/(1 - 0)} (x - 1)
y - 2 = (-1/2)(x - 1)
2(y - 2) = -(x - 1)
2y - 4 = -x + 1
x + 2y - 5 = 0


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20.
Let A = (a, 0) and (0, b)

AB = 10
√[(a - 0)² + (0 - b)²] = 10
√(a² + b²) = 10
a² + b² = 100

There are infinite solutions.

One of the solutions is that : a = 6 and b = 8
The st. line passes through A(6, 0) and B(0, 8)
y - 0 = [(0 - 8)/(6 - 0)] (x - 6)
y = (-4/3)(x - 6)
3y = -4x + 24
4x + 3y - 24 = 0

One more of the solutions is that : a = 8 and b = 6
The st. line passes through A(8, 0) and B(0, 6)
y - 0 = [(0 - 6)/(8 - 0)] (x - 8)
y = (-3/4)(x - 8)
4y = -3x + 24
3x + 4y - 24 = 0


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22.
(a)
x + 2y - 6 + k(2x - y + 1) = 0
x + 2y - 6 + 2kx - ky + k = 0
(1 + 2k)x + (2 - k)y + (k - 6) = 0

The equation is in the form of a equation of straight line ax + by + c = 0
Hence, the equation must be astraight line.

(b)
(i)
y-intercept = -3
Put x = 0 and y = -3 into the equation :
(1 + 2k)(0) + (2 - k)(-3) + (k - 6) = 0
-6 + 3k + k - 6 = 0
4k = 12
k = 3

(ii)
Put x = 1 and y = 1 into the equation :
(1 + 2k)(q) + (2 - k)(1) + (k - 6) = 0
1 + 2k + 2 - k + k - 6 = 0
2k = 3
k = 3/2