点p（2,3）到直线ax+（a-1）y+3=0的距离是3,则a=

点p（2,3）到直线ax+（a-1）y+3=0的距离是3,则a=
点p（2,3）到直线ax+（a-1）y+3=0的距离是3,则a=

点p（2,3）到直线ax+（a-1）y+3=0的距离是3,则a=
距离=|2a+3(a-1)+3|/√[a²+(a-1)²]=3
两边平方
25a²=9(2a²-2a+1)
7a²+18a-9=0
(7a-3)(a+3)=0
a=3/7,a=-3

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