急 ! 急! F5 Locus 2 題 9q5?

9.
P maintains an equal distance from the vertical lines x = 1 and x = 7 :
|x - 1| = |x - 7|
x - 1 = -(x - 7) or x - 1 = x - 7
x - 1 = -x + 7 or -1 = -7 (rejected)
2x = 8
x = 4

The equation of the locus of P :
x = 4


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21.
(a)
PA = PB
√[(x - 2)² + (y - 5)²] = √[(x + 4)² + (y - 2)²]
(x - 2)² + (y - 5)² = (x + 4)² + (y - 2)²
x² - 4x + 4 + y² - 10y + 25 = x² + 8x + 16 + y² - 4y + 4
-4x - 10y + 29 = 8x - 4y + 20
12x + 6y - 9 = 0
4x + 2y - 3 = 0

The equation of the locus of P :
4x + 2y - 3 = 0

(b)(i)
The slope of the locus of P
= -4/2
= -2

(b)(ii)
4x + 2y - 3 = 0

When x = 1 and y = -1/2 :
L.S.
= 4(1) + 2(-1/2) - 3
= 0
= R.S.

Hence, (1, -1/2) is a point on thelocus of P.

(c)
Slope of L = -1/(-2) = 1/2

Equation of L :
[y + (1/2)] = (1/2)(x - 1)
2y + 1 = x - 1
x - 2y - 2 = 0

2013-03-01 13:07:57 補充：
From the figure, P(x, y) lies between the two vertical lines.
Hence, 1 < x < 7
x - 1 > 0 and x - 7 < 0
Hence x - 1 = -(x - 7)

2013-03-01 13:09:01 補充：
From the figure, P(x, y) lies between the two vertical lines.
Hence, 1 < x < 7
x - 1 > 0 and x - 7 < 0
Hence x - 1 = -(x - 7)