直线(2a平方-7a+3)x+(a平方-9)y+3a平方=0的倾斜角为π/4,则实数a为多少
问题描述:
直线(2a平方-7a+3)x+(a平方-9)y+3a平方=0的倾斜角为π/4,则实数a为多少
答
倾斜角为π/4
k=tanπ/4=1
则-(2a²-7a+3)/(a²-9)=1
且a²-9≠0
-(2a-1)(a-3)/(a+3)(a-3)=1
-(2a-1)/(a+3)=1
1-2a=a+3
a=-2/3