If points a (a, 0), B (0, b), C (1, - 1), (a > 0, B < 0) are collinear, then the minimum value of A-B is equal to

If points a (a, 0), B (0, b), C (1, - 1), (a > 0, B < 0) are collinear, then the minimum value of A-B is equal to

A. B, C straight line has slope
Y=kX+b  
0=ak+b    k=-b/a
In addition, y = - B / A * x + B
A little more (1, - 1)

If AB > 0 and a (a, 0), B (0, b), C (- 2, - 2) are collinear, the minimum value of AB is obtained

∵ a, B and C are collinear, ∵ KAB = KAC, that is, B − 00 − a = − 2 − 0 − 2 − a, ∵ 1A + 1b = - 12, ∵ 12 = | 1A + 1b | = | 1a | + | 1b | ≥ 2Ab (take the equal sign when a = b), ∵ ab ≥ 4, ab ≥ 16ab minimum value is: 16

If three points a (1,9) B (a, 0) C (0, b) are known to be collinear, then the minimum value of a + B is

A (1,9) B (a, 0) C (0, b) three points are collinear, ∧ - 9 / (A-1) = - (B-9), ∧ (A-1) (B-9) = 9, let a + B = x, then B = x-a, ∧ x = a + 9 + 9 / (A-1) = A-1 + 9 / (A-1) + 10, when a → 1 -, X → - ∞, a + B has no minimum value. If a > 1, then x > = 6 + 10 = 16, when a = 4, take the equal sign, the minimum value of a + B is 16

Given that a (1,2), B (2, - 4), C (x, - 3) are collinear, then the minimum value of X is?

Let the line be y = KX + C
Over a (1,2), B (2, - 4) C (x, - 3)
2=k+c
-4=2k+c
k=-6
c=8
y=-6x+8
-3=-6x+8
x=11/6