一條斜率既問題

a)
(b-0)/(0-a) = -1/3
3(b-0) = -1(0-a)
3b = a
所以 a = 3b

b) (^2 = 2次方; ^(1/2) = 開方)
AB = [(0-a)^2+(b-0)^2]^(1/2)
AB = (a^2+b^2)^(1/2)
AB = [(3b)^2+b^2)]^(1/2)
AB = (9b^2+b^2)^(1/2)
AB = (10b^2)^(1/2)
所以 AB = 10^(1/2)b

c)
2[10^(1/2)] = 10^(1/2)b
2 = b
所以 b = 2

y = mx + c (m = 斜率, c = y-intercept)
c = 2 (因為 B(0,2))
所以 y = (-1/3)x + 2
y - 2 = (-1/3)x
3y - 6 = -x
x + 3y - 6 = 0

x,y兩軸分別相交於 A(a.0) 和 B(0.b)兩點and 斜率為 -1/3 的直線
means 以兩點的坐標代入
Y=AX+B......(1)
0=Aa+B.......(2)
b=A(0)+B......(3)

part1
b=B
Aa=-b
A=-(1/3)
a(-1/3)=-b
a=3b

part2
AB^2=AO^2+BO^2
so AB=[(10)^(1/2)]b/3

part3
2[(10)^2]=[(10)^(1/2)]b/3
b=6

then AB: Y=(-1/3)X+6