Question

初二数学，急急急

Answer 1

(1)
A(1, 4), m = 4
ab = 4 (i)
△ABD的面积 = (1/2)*BD*A到BD的距离 = (1/2)*(B的横坐标)(A的纵坐标 - B的纵坐标)
= (1/2)a(4 - b) = 2a - ab/2 = 2a - 4/2 = 2a - 2 = 4
a = 3
B(3, 4/3)

(2)
C(1, 0), D(0, 4/3), CD的斜率为(4/3 - 0)/(0 - 1) = -4/3
AB的斜率(4 - 4/3)/(1 - 3) = -4/3
二者相等，相互平行

(3)
AD² = (1 - 0)² + (4 - b)² = b² - 8b + 17
BC² = (a - 1)² + (b - 0)² = a² - 2a + 1 + b²
b² - 8b + 17 = a² - 2a + 1 + b²
消去b², 并利用b = 4/a得 a³ - 2a² - 16a + 32 = 0
(a - 2)(a - 4)(a + 4) = 0
a = 2, B(2, 2), AB: (y - 2)/(4 - 2) = (x - 2)/(1 - 2), y = 6 - 2x
a = 4, B(4, 1), AB: (y - 1)/(4 - 1) = (x - 4)/(1 - 4), y = 5 - x
舍去a = -4 < 0