Given, L≡x‌cos‌α+ysin‌α−p=0 Lis perpendicular to the line x+y+1=0 pis positive,α∈(270∘,360∘)and Distance of point(√2,√2)to the line L is 5 units. Slope ofL=‌
−cos‌α
sin‌α
=−cot‌α Slope of linex+y+1=0is -1 ∴‌‌(−cot‌α)(−1)=−1⇒cot‌α=−1 Now,L≡x‌cos‌315∘+ysin‌315∘−p=0 ‌⇒‌‌x(‌
1
√2
)+y(−‌
1
√2
)−p=0 ‌⇒‌‌x−y−p√2=0 According to the question, ‌
