已知直线的极坐标方程为psin(θ+π⼀4)=√2/2求点A(2,7⼀4∏)到这条线的距离

2024-12-02 20:35:06
推荐回答(2个)
回答1:

解:利用公式:pcosm=x,psinm=y
将极坐标化为直角坐标,得:
直线方程为:x+y-1=0
点A(-√2,-√2)
利用点到直线距离公式,得:d=|-√2-√2-1|/(√2)=2+√2/2

回答2:

psin(θ+π/4)=√2/2
p(sinθcosπ/4+cosθsinπ/4)=√2/2
√2/2p(sinθ+cosθ)=√2/2
p(sinθ+cosθ)=1
psinθ+pcosθ=1
x+y=1
x+y-1=0
d=∣2+7π/4-4∣/√(1^2+1^2)
=∣7π/4-2∣/√2
=(7π/4-2)/√2
=(7π√2/4-2√2)/2