हिंदी

Find the Length of the Perpendicular from the Origin to the Straight Line Joining the Two Points Whose Coordinates Are (A Cos α, a Sin α) and (A Cos β, a Sin β).

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प्रश्न

Find the length of the perpendicular from the origin to the straight line joining the two points whose coordinates are (a cos α, a sin α) and (a cos β, a sin  β).

संक्षेप में उत्तर
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उत्तर

Equation of the line passing through (acosα, asinα) and (acosβ, asinβ) is

\[y - asin\alpha = \frac{asin\beta - asin\alpha}{acos\beta - acos\alpha}\left( x - acos\alpha \right)\]

\[ \Rightarrow y - asin\alpha = \frac{sin\beta - sin\alpha}{cos\beta - cos\alpha}\left( x - acos\alpha \right)\]

\[ \Rightarrow y - asin\alpha = \frac{2\cos\left( \frac{\beta + \alpha}{2} \right)\sin\left( \frac{\beta - \alpha}{2} \right)}{2\sin\left( \frac{\beta + \alpha}{2} \right)\sin\left( \frac{\alpha - \beta}{2} \right)}\left( x - acos\alpha \right)\]

\[ \Rightarrow y - asin\alpha = - \cot\left( \frac{\beta + \alpha}{2} \right)\left( x - acos\alpha \right)\]

\[ \Rightarrow y - asin\alpha = - \cot\left( \frac{\alpha + \beta}{2} \right)\left( x - acos\alpha \right)\]

\[\Rightarrow x\cot\left( \frac{\alpha + \beta}{2} \right) + y - asin\alpha - acos\alpha \cot\left( \frac{\alpha + \beta}{2} \right) = 0\]

The distance of the line from the origin is

\[d = \left| \frac{- asin\alpha - acos\alpha \cot\left( \frac{\alpha + \beta}{2} \right)}{\sqrt{\cot^2 \left( \frac{\alpha + \beta}{2} \right) + 1}} \right|\]

\[ \Rightarrow d = \left| \frac{asin\alpha + acos\alpha \cot\left( \frac{\alpha + \beta}{2} \right)}{\sqrt{{cosec}^2 \left( \frac{\alpha + \beta}{2} \right)}} \right| \left( \because {cosec}^2 \theta = 1 + \cot^2 \theta \right)\]

\[\Rightarrow d = a\left| \sin\left( \frac{\alpha + \beta}{2} \right)sin\alpha + cos\alpha \cos\left( \frac{\alpha + \beta}{2} \right) \right| \]

\[ \Rightarrow d = a\left| sin\alpha \sin\left( \frac{\alpha + \beta}{2} \right) + cos\alpha \cos\left( \frac{\alpha + \beta}{2} \right) \right| \]

\[ \Rightarrow d = a\left| \cos\left( \frac{\alpha + \beta}{2} - \alpha \right) \right| = a\cos\left( \frac{\beta - \alpha}{2} \right) = a\cos\left( \frac{\alpha - \beta}{2} \right)\]

Hence, the required distance is \[a\cos\left( \frac{\alpha - \beta}{2} \right)\]

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Equations of Line in Different Forms - Equation of Family of Lines Passing Through the Point of Intersection of Two Lines
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 23: The straight lines - Exercise 23.15 [पृष्ठ १०७]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 23 The straight lines
Exercise 23.15 | Q 3 | पृष्ठ १०७

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