Answers by visitors named: Anish Dhakal

here, (bx+ay)/ab=1 or, (bx+ay0=ab Dividing both sides by +-√(x-coefficient)^2+(y-coefficient)^2 =+-√(b)^2+(a)^2 =+-√(a^2)+(b^2) ∴ (bx)/{+-√(a^2)+(b^2)} +(ay)/{+-√(a^2)+(b^2)} =(ab)/{+-√(a^2)+(b^2)} Comparing with x-cos alpha + y-sin alpha = perpendicular perperndicular=(ab)/{+-√(a^2)+(b^2)} Squaring on both sides, P^2=[(ab)/{+-√(a^2)+(b^2)}]^2 or, P^2=(a^2×b^2)/{(a^2) + (b^2)} or, {(a^2) + (b^2)}/(a^2×b^2)=1/p^2 or, (a^2)/(a^2×b^2)+(b^2)/(a^2×b^2)=1/p^2 or, 1/p^2 = 1/(b^2) + 1/(a^2) HENCE, 1/p^2 = 1/(b^2) + 1/(a^2) PROVED#

here, (bx+ay)/ab=1 or, (bx+ay)=ab Dividing both sides by +-√(x-coefficient)^2+(y-coefficient)^2 =+-√(b)^2+(a)^2 =+-√(a^2)+(b^2) ∴ (bx)/{+-√(a^2)+(b^2)} +(ay)/{+-√(a^2)+(b^2)} =(ab)/{+-√(a^2)+(b^2)} Comparing with x-cos alpha + y-sin alpha = perpendicular perperndicular=(ab)/{+-√(a^2)+(b^2)} Squaring on both sides, P^2=[(ab)/{+-√(a^2)+(b^2)}]^2 or, P^2=(a^2×b^2)/{(a^2) + (b^2)} or, {(a^2) + (b^2)}/(a^2×b^2)=1/p^2 or, (a^2)/(a^2×b^2)+(b^2)/(a^2×b^2)=1/p^2 or, 1/p^2 = 1/(b^2) + 1/(a^2) HENCE, 1/p^2 = 1/(b^2) + 1/(a^2) PROVED#