Math, asked by vibhu217, 1 year ago

बिंदु (a cos α, a sin α ) तथा (a cos β, a sinβ) को मिलाने वाली रेखा का समीकरण ज्ञात कीजिए ​

Answers

Answered by Swarnimkumar22
6

हल:-

सूत्र\boxed{\huge\bf \: y- y_{1} =  \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})}से

अभीष्ट रेखा का समीकरण निम्नलिखित है

\bf \: y-a \sin\alpha  =  \frac{a \sin\beta - a \sin \alpha   }{a \cos \beta  - a \cos\alpha    } (x - a \cos\alpha ) \\  \\ \implies \bf \: y - a \: sin \alpha  =  \frac{sin \beta  - sin \alpha }{cos \beta  - cos \alpha } (x - a \: cos \alpha ) \\  \\  \implies \bf \: y - a \: sin \alpha  =  \frac{2 \: cos \:  \frac{ \beta  +  \alpha }{2} . \: sin \:   \frac{ \beta -   \alpha }{2} }{ - 2 \: sin \:  \frac{ \beta  +  \alpha }{2}. \: sin \:  \frac{ \beta   - \alpha }{2}  } (x - a \: cos \alpha ) \\  \\  \implies \bf \:  - y \: sin \:  \frac{ \alpha   + \beta }{2}  + a \: sin \alpha . \: sin \frac{ \alpha  +  \beta }{2}  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   = \bf \: x \: cos \alpha \frac{ \alpha +   \beta }{2}   - a \: cos \alpha .cos \frac{ \alpha  +  \beta }{2}   \\  \\   \implies \bf \: x \: cos \frac{ \alpha  +  \beta }{2}  + y \: sin \frac{ \alpha   + \beta }{2}  \\  \bf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = a \{ \: cos \alpha . \: cos \frac{ \alpha  +  \beta }{2}  + sin \alpha .sin \frac{ \alpha  +  \beta }{2}  \} \\  \\  \implies \bf \: x \: cos \frac{ \alpha  +  \beta }{2}  + y \: sin \frac{ \alpha  +  \beta }{2}  = a \: cos \{ \: \:  \alpha  -  \frac{ \alpha  +  \beta }{2}  \} \\  \\  \implies \bf \: x \: cos \frac{ \alpha   + \beta }{2}  + y \: sin \frac{ \alpha   + \beta }{2}  = a \: cos \:  \frac{ \alpha  -  \beta }{2}


Swarnimkumar22: thanks
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