If (x 2 + y 2 …

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Sia ? 6 years, 6 months ago
Given, {tex}\left( x ^ { 2 } + y ^ { 2 } \right) \left( a ^ { 2 } + b ^ { 2 } \right) = ( a x + b y ) ^ { 2 }{/tex}
or,{tex}x ^ { 2 } a ^ { 2 } + x ^ { 2 } b ^ { 2 } + y ^ { 2 } a ^ { 2 } + y ^ { 2 } b ^ { 2 } = a ^ { 2 } x ^ { 2 } + b ^ { 2 } y ^ { 2 } + 2 a b x y{/tex}
or, {tex}x ^ { 2 } b ^ { 2 } + y ^ { 2 } a ^ { 2 } - 2 a b x y = 0{/tex}
or,{tex}( x b - y a ) ^ { 2 } = 0{/tex} {tex}\left[ \because ( a - b ) ^ { 2 } = a ^ { 2 } + b ^ { 2 } - 2 a b \right]{/tex}
or, xb = ya
{tex}\therefore \quad \frac { x } { a } = \frac { y } { b }{/tex} Hence Proved.
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