设：直线方程为 x/a + y/b =1∵|a|=|b|∴1/|a| + 2/|a|=1|a|=3a=±3 ,则b=±3直线有4条:x/3 + y/3 =1 ,即：x+y-3=0x/3 + y/-3=1 ,即：x-y-3=0x/-3 + y/3=1 ,即：x-y+3=0x/-3 + y/-3=1 ,即：x+y+3=0...