连 OB ,则由题意知:
OB = 1,OB ⊥ AB .
又∵ OA = 2 = 2 OB
(∴根据勾股定理,AB = 根号3)
∴∠OAB = 30°
∴直线 AB 斜率为(根号3/3)
根据点斜式,AB(也就是 AC)函数解析式为: y = (x-2)*(根号3/3)
连 OB ,则由题意知:
OB = 1,OB ⊥ AB .
又∵ OA = 2 = 2 OB
(∴根据勾股定理,AB = 根号3)
∴∠OAB = 30°
∴直线 AB 斜率为(根号3/3)
根据点斜式,AB(也就是 AC)函数解析式为: y = (x-2)*(根号3/3)