मराठी

Find the Perpendicular Distance of the Line Joining the Points (Cos θ, Sin θ) and (Cos ϕ, Sin ϕ) from the Origin.

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प्रश्न

Find the perpendicular distance of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) from the origin.

थोडक्यात उत्तर
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उत्तर

The equation of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) is given below:

\[y - sin\theta = \frac{sin\phi - sin\theta}{cos\phi - cos\theta}\left( x - cos\theta \right)\]

\[ \Rightarrow \left( cos\phi - cos\theta \right)y - sin\theta\left( cos\phi - cos\theta \right) = \left( sin\phi - sin\theta \right)x - \left( sin\phi - sin\theta \right)cos\theta\]

\[ \Rightarrow \left( sin\phi - sin\theta \right)x - \left( cos\phi - cos\theta \right)y + sin \theta \ cos\phi - sin\phi \ cos\theta = 0\]

Let d be the perpendicular distance from the origin to the line \[\left( sin\phi - sin\theta \right)x - \left( cos\phi - cos\theta \right)y + sin\theta \ cos\phi - sin\phi \ cos\theta = 0\]

\[\therefore d = \left| \frac{sin\theta cos\phi - sin\phi cos\theta}{\sqrt{\left( sin\phi - sin\theta \right)^2 + \left( cos\phi - cos\theta \right)^2}} \right|\]

\[ \Rightarrow d = \left| \frac{\sin\left( \theta - \phi \right)}{\sqrt{\sin^2 \phi + \sin^2 \theta - 2sin\phi sin\theta + \cos^2 \phi + \cos^2 \theta - 2\cos\phi cos\theta}} \right|\]

\[ \Rightarrow d = \left| \frac{\sin\left( \theta - \phi \right)}{\sqrt{\sin^2 \phi + \cos^2 \phi + \sin^2 \theta + \cos^2 \theta - 2\cos\left( \theta - \phi \right)}} \right|\]

\[ \Rightarrow d = \frac{1}{\sqrt{2}}\left| \frac{\sin\left( \theta - \phi \right)}{\sqrt{1 - \cos\left( \theta - \phi \right)}} \right|\]

\[\Rightarrow d = \frac{1}{\sqrt{2}}\left| \frac{\sin\left( \theta - \phi \right)}{\sqrt{2 \sin^2 \left( \frac{\theta - \phi}{2} \right)}} \right| \]

\[ \Rightarrow d = \frac{1}{\sqrt{2} \times \sqrt{2}}\left| \frac{\sin\left( \theta - \phi \right)}{\sin\left( \frac{\theta - \phi}{2} \right)} \right| = \frac{1}{2}\left| \frac{2\sin\left( \frac{\theta - \phi}{2} \right)\cos\left( \frac{\theta - \phi}{2} \right)}{\sin\left( \frac{\theta - \phi}{2} \right)} \right|\]

\[ \Rightarrow d = \cos\left( \frac{\theta - \phi}{2} \right)\]

Hence, the required distance is \[\cos\left( \frac{\theta - \phi}{2} \right)\].

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पाठ 23: The straight lines - Exercise 23.15 [पृष्ठ १०७]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 23 The straight lines
Exercise 23.15 | Q 2 | पृष्ठ १०७

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