數學填充題

1. a - b = 2 => a = 2 + b
P = (2+b, b)至A(6,5).B(7,-2)兩點的距離相等
(2 + b - 6)2 + (b - 5)2 = (2 + b - 7)2 + (b + 2)2
(b - 4)2 + (b - 5)2 = (b - 5)2 + (b + 2)2
(b - 4)2 = (b + 2)2
(b - 4)2 - (b + 2)2 = 0
(b - 4 + b + 2)(b - 4 - b - 2) = 0
(2b - 2)(-6) = 0
b = 1 => a = 3
ab = 3
2. 設 P = (a,b)
a - 2b + 4 = 0 => a = 2b - 4
到A(2,0)和B(-6,0)的距離相等
=> (2b - 4 - 2)2 + (b - 0)2 = (2b - 4 + 6)2 + (b - 0)2
=> (2b - 6)2 = (2b + 2)2
=> (2b - 6)2 - (2b + 2)2 = 0
=> (2b - 6 + 2b + 2)(2b - 6 - 2b - 2) = 0
=> (4b - 4)(-8) = 0
=> b = 1 => a = -2
P = (-2, 1)
3. y = x2 + 1 ... (1)
x - y + 3 = 0 ... (2)
(2) => y = x + 3
代入(1) => x + 3 = x2 + 1
x2 - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2 (=> y = 5) 或x = -1 (=> y = 2)
P = (2,5), Q = (-1,2)
直線PQ為 √[(2 + 1)2 + (5 - 2)2 ] = 3√2
4. y = 2x 和 y = 2 => x = 1
P = (1,2)
y = 2x 和 y = 8 => x = 4
Q = (4, 8)
PQ = √[(8 - 2)2 + (4 - 1)2 ] = √(36 + 9) = 3√5
5. y = x2 - 5 ... (1)
y = x - 3 ... (2)
代(2)入(1) => x - 3 = x2 - 5
=> x2 - x - 2 = 0
=> (x - 2)(x + 1) = 0
=> x = 2 (=> y = -1) 或x = -1 (=> y = -4)
P = (2, -1), Q = (-1, -4)
PQ = √[(2 + 1)2 + (-1 + 4)2 ] = 3√2

1.
P至A(6,5).B(7,-2)兩點的距離相等，則P在線段AB的中垂線x-7y=-4上，
x-7y=-4再與L:x-y=2解交點，得P(3,1)。