wx0516
幼苗
共回答了25个问题采纳率:92% 举报
y = 0.5x + 2
x= 0, y = 2, B(0, 2)
y = 0, x = -4, A(-4, 0)
用推导挺复杂,但用观察法容易猜到二者均为直角三角形.
直线l斜率为k = (2 - 0)/(0 + 4) = 1/2
直线m斜率为k' = -1/k = -2
直线m: y = -2(x + 3)
与直线l联立得交点D(-16/5, 2/5), 直线m与y轴交于C(0, -6)
CB = 8, CD = √[(-16/5 - 0)² + (2/5 + 6)²] = 16√5/5,DB = 8√5/5
AM = 1, AD =√[(-16/5 + 4)² + (2/5 -0)²] = 2√5/5,DP = √5/5
CB:AM = CD :AD = DB :DP = 8 :1
三角形相似
后来想还有一种可能,即直线与y轴交点C(0, c)在x轴上方,此时二三角形有一对顶角,要相似只须角BAO = 角OCB
tanBAO = 直线l斜率 = 1/2
tanOCB = 1/2 = MO/OC = 3/c
c = 6
C(0, 6)
x/(-3) + y/6 = 1
y = 2x + 6
1年前
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