Transform the equation x/a + y/b = 1 into the normal form when a>0, b>0. If the perpendicular distance of the straight line from the origin is P, deduce that 1/p= 1/a+ 1/b2​.

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xa+yb=1....i, a>0, b>01a2+1b2=a2+b2a2b2=a2+b2abDivide i through out by  a2+b2abxa.a2+b2ab+yb.a2+b2ab=1a2+b2abxba2+b2+yaa2+b2=aba2+b2....iSlope of i is given bytanα=-Ceofficient of xCoefficient of y=-1a1b=-ba which is negative, hence α is obtuse.tanα=-ba=Perendicular BaseHypotenuse=Base2+Perpendicular2=a2+b2For obtuse angles sin is positive and cos in negative.sinα=ba2+b2cosα=-ba2+b2Hence i becomesx-cosα+ysinα=aba2+b2We know that sinA=sinπ-A and cosA=-cosπ-Axcosπ-α+ysinπ-α=aba2+b2which is the required normal form where tanα is slope of given line.Hence right hand side must give  perpendicular distance of the straight line from the origin.p=aba2+b2p2=a2b2a2+b21p2=a2+b2a2b21p2=a2a2b2+b2a2b21p2=1a2+1b2

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