myxinyu
幼苗
共回答了22个问题采纳率:72.7% 举报
(1) y = -√3x/3 + 1
x = 0,y = 1,B(0,1)
y = 0,x = √3,A(√3,0)
tan∠OAB = OB/OA = 1/√3
∠OAB = 30˚,∠CAB = 60˚,AC与x轴垂直,AC = AB = 2,C(√3,2)
(2)S△CBA=S△PBA,底AB相同,只须其上的高相等,即C,P与AB的距离相等,CP与AB平行,斜率均为-√3/3
直线CP的解析式:y - 2 = (-√3/3)(x - √3)
y = 5 - √3x/3
1年前
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