BC直线的斜率k = (-3-3)/(-1-2) = 2
BC直线方程：y - 2 = 2(x-3)
由AD垂直BC,得AD直线斜率k' = -1/2
AD直线方程：y+1 = -1/2*(x-2)
AD直线与BC直线联立,解得x = 8/5,y = -4/5
D(8/5,-4/5)
三角形ABC A(2,-1) B(3,2) C(-3,-1) BC上的高为AD 求点D坐标
三角形ABC A(2,-1) B(3,2) C(-3,-1) BC上的高为AD 求点D坐标
数学人气：546 ℃时间：2020-05-22 11:34:59
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