Question

已知直线l经过点P(2,-1),它的倾斜角为α,直线l'经过点A(-2,3),B(4,-5),它的倾斜角为α',且α=1

Answer 1

K(l')=tanα'=(3+5)/(-2-4)=-4/3

∴ 方程是 y-3=(-4/3)(x+2)

即 4x+3y-1=0

∵ α=(1/2)α'

∴ tanα'=2tanα/[1-(tanα)^2]

∴ (-4/3)[1-(tanα)^2]=2tanα

∴ -2[1-(tanα)^2]=3tanα

∴ 2(tanα)^2-3tanα-2=0

即 (tanα-2)*(2tanα+1)=0

∴ tanα=2或tanα=-1/2

∵ α是锐角,

∴ tanα=2

即 k(l)=2

∴ 方程是 y+1=2(x-2)

即 2x-y-5=0