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已知直线的极坐标方程为psin(θ+π/4)=√2／2求点A（2,7/4∏）到这条线的距离
题目详情
已知直线的极坐标方程为psin(θ+π/4)=√2／2求点A（2,7/4∏）到这条线的距离
▼优质解答
答案和解析
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
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
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