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Questions and answers by
Anish Dhakal
Answers (2)
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 =
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 =
