F4 Math - Equations

1.
Let AP = y

AP² + PB²
= y² + (8 - y²)
= y² + (64 - 16y + y²)
= 2y² - 16y + 64
= 2(y² - 8y) + 64
= 2(y² - 8y + 16) - 32 + 64
= 2(y - 4)² + 32

For any real number y, (y - 4)² ≥ 0
Hence, AP² + PB² is a minimum when (y - 4)² = 0
i.e. Distance of P from A = y = 4


2)
Let y be the length of one of the pieces of strength.

Total area of the two squares
= y² + (l - y)²
= y² + l ² - 2 l y + y²
= 2y² - 2l y + l ²
= 2(y² - l y) + l ²
= 2(y² - l y + l /4) - 2l ²/4+ l ²
= 2(y - l /2)² + l ²/2

For any real number y, (y - l /2)² ≥ 0
Hence, when y = l /2,
the minimum total area of the two squares = l ²/2


3)
cosθ = 1/2
sinθ = √(1 - cos²θ) = √[1 - (1/2)²] = (√3)/2
Slope of the line = tanθ = sinθ/cosθ = √3

x-intercept = 2
The line passes through (2, 0)

Equation of the line:
(y - 0) = (√3)(x - 2)
(√3)x - y - 2√3 = 0


4)
Let A = (a, 0) and B = (0, b)

Slope of the line l = 1
(0 - b)/(a - 0) = 1
a = -b …… (1)

Area of the ΔOAB = 18
(1/2)|ab| = 18
ab = -36 …… (2)
or ab = 36 (rejected for a = -b)

Subst. (1) into (2)
(-b)b = -36
b = 6 or b = -6

When b = 6, B(0, 6)
Hence, equation of the line l :
y - 6 = 1(x - 0)
x - y + 6 = 0

When b = -6, B(0, -6)
Hence, equation of the line l :
y + 6 = 1(x - 0)
x - y - 6 = 0

2010-08-23 01:13:18 補充：
In Q.1, the unit is "cm".

2010-08-23 01:22:42 補充：
Amendment of Q.2
Total area of the two squares
= y² + [(l - 4y)/4]²
= 2y² - l y/2 + l ²/16
= 2(y² - l y/4 + l ²/64) - 2l ²/64+ l ²/16
= 2(y - l /8)² + l ²/32

For any real number y, (y - l /8)² ≥ 0
Hence, when y = l /8,
the minimum total area of the two squares = l ²/32

2010-08-23 16:22:41 補充：
Q.4
x - y + 6 = 0 可寫成 y = x + 6
x - y - 6 = 0 可寫成 y = x - 6