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NCERT Class XI Mathematics - Straight Lines - Solutions

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Question : 58 of 74
Marks: +1, -0
Find perpendicular distance from the origin of the line joining the points (cosθ, sinθ) and (cosϕ, sinϕ).
Solution:  
Let the points be A (cosθ, sinθ) and B(cosϕ, sinϕ)
Equation of the line AB is
y - sin θ = sin⁡ϕ−sin⁡θcos⁡ϕ−cos⁡ϕ\frac{\sin\phi-\sin\theta}{\cos\phi-\cos\phi} (x - cos θ)
⇒ y - sin θ =
2cos⁡(θ+ϕ2)sin⁡(ϕ−θ2)−2sin⁡(ϕ+θ2)sin⁡(ϕ−θ2)(x−cos⁡θ)\frac{2\cos\left(\frac{\theta+\phi}{2}\right)\sin\left(\frac{\phi-\theta}{2}\right)}{-2\sin\left(\frac{\phi+\theta}{2}\right)\sin\left(\frac{\phi-\theta}{2}\right)} (x - \cos\theta)
⇒ (y - sin θ) = −cos⁡(θ+ϕ2)sin⁡(θ+ϕ2)(x−cos⁡θ)-\frac{\cos\left(\frac{\theta+\phi}{2}\right)}{\sin\left(\frac{\theta+\phi}{2}\right)} (x-\cos\theta)
⇒ y sin (θ+ϕ2)\left(\frac{\theta+\phi}{2}\right) - sin θ sin (θ+ϕ2)\left(\frac{\theta+\phi}{2}\right) = - x cos (θ+ϕ2)\left(\frac{\theta+\phi}{2}\right) + cos θ . cos (θ+ϕ2)\left(\frac{\theta+\phi}{2}\right)
=
=xcos⁡(θ+ϕ2)= x\cos\left(\frac{\theta+\phi}{2}\right) + y sin (θ+ϕ2)\left(\frac{\theta+\phi}{2}\right) - [cos⁡θcos⁡(θ+ϕ2)+sin⁡θ⋅sin⁡(θ+ϕ2)]=0\left[ \cos\theta \cos\left(\frac{\theta+\phi}{2}\right) + \sin\theta \cdot \sin\left(\frac{\theta+\phi}{2}\right) \right] = 0
⇒ x cos (θ+ϕ2)\left(\frac{\theta+\phi}{2}\right) + y sin (θ+ϕ2)\left(\frac{\theta+\phi}{2}\right) - cos (θ−θ−ϕ2)\left( \theta - \frac{\theta-\phi}{2} \right) = 0
[By using cos A·cosB + sinA·sinB = cos (A – B)]
⇒ x cos (θ+ϕ2)\left(\frac{\theta+\phi}{2}\right) + y sin (θ+ϕ2)\left(\frac{\theta+\phi}{2}\right) - cos (θ−ϕ2)\left(\frac{\theta-\phi}{2}\right) = 0
Now distance of the above line from the origin
=∣0+0−cos⁡(θ−ϕ2)∣cos⁡2(θ+ϕ2)+sin⁡2(θ+ϕ2)= \frac{\left| 0+0-\cos\left(\frac{\theta-\phi}{2}\right) \right|}{\sqrt{ \cos^2\left(\frac{\theta+\phi}{2}\right) + \sin^2\left(\frac{\theta+\phi}{2}\right) }}
= ∣cos⁡(θ−ϕ2)∣\left| \cos\left(\frac{\theta-\phi}{2}\right) \right|.
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