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Miscellaneous Exercise · Q4

Q.Find perpendicular distance from the origin to the line joining the points (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) and (cos⁡ϕ,sin⁡ϕ)(\cos\phi, \sin\phi).

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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The perpendicular distance from the origin to the line through (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) and (cos⁡ϕ,sin⁡ϕ)(\cos\phi, \sin\phi) is ∣cos⁡θ−ϕ2∣\left|\cos\frac{\theta-\phi}{2}\right|.

The key idea is that both given points lie on the unit circle x2+y2=1x^2 + y^2 = 1, since cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1 and similarly for ϕ\phi. So the line joining them is a chord of the unit circle. The perpendicular distance from the centre (the origin) to a chord is a standard geometric quantity — it’s the distance from the centre to the chord’s midpoint, which relates directly to the angle subtended at the centre.

Let’s work through it step by step.

  1. Find the equation of the line through the two points. The two points are A(cos⁡θ,sin⁡θ)A(\cos\theta, \sin\theta) and B(cos⁡ϕ,sin⁡ϕ)B(\cos\phi, \sin\phi). The slope of ABAB is

m=sin⁡ϕ−sin⁡θcos⁡ϕ−cos⁡θ.m = \frac{\sin\phi - \sin\theta}{\cos\phi - \cos\theta}.

Using the sum-to-product identities:

sin⁡ϕ−sin⁡θ=2cos⁡θ+ϕ2sin⁡ϕ−θ2,\sin\phi - \sin\theta = 2\cos\frac{\theta+\phi}{2}\sin\frac{\phi-\theta}{2},

cos⁡ϕ−cos⁡θ=−2sin⁡θ+ϕ2sin⁡ϕ−θ2.\cos\phi - \cos\theta = -2\sin\frac{\theta+\phi}{2}\sin\frac{\phi-\theta}{2}.

So

m=2cos⁡θ+ϕ2sin⁡ϕ−θ2−2sin⁡θ+ϕ2sin⁡ϕ−θ2=−cot⁡θ+ϕ2.m = \frac{2\cos\frac{\theta+\phi}{2}\sin\frac{\phi-\theta}{2}}{-2\sin\frac{\theta+\phi}{2}\sin\frac{\phi-\theta}{2}} = -\cot\frac{\theta+\phi}{2}.

The line equation in point-slope form using AA:

y−sin⁡θ=−cot⁡θ+ϕ2 (x−cos⁡θ).y - \sin\theta = -\cot\frac{\theta+\phi}{2}\,(x - \cos\theta).

  1. Convert to standard form ax+by+c=0ax + by + c = 0. Multiply through by sin⁡θ+ϕ2\sin\frac{\theta+\phi}{2} to avoid fractions:

(y−sin⁡θ)sin⁡θ+ϕ2=−cos⁡θ+ϕ2 (x−cos⁡θ).(y - \sin\theta)\sin\frac{\theta+\phi}{2} = -\cos\frac{\theta+\phi}{2}\,(x - \cos\theta).

Expand:

ysin⁡θ+ϕ2−sin⁡θsin⁡θ+ϕ2=−xcos⁡θ+ϕ2+cos⁡θcos⁡θ+ϕ2.y\sin\frac{\theta+\phi}{2} - \sin\theta\sin\frac{\theta+\phi}{2} = -x\cos\frac{\theta+\phi}{2} + \cos\theta\cos\frac{\theta+\phi}{2}.

Bring all terms to one side:

xcos⁡θ+ϕ2+ysin⁡θ+ϕ2−(sin⁡θsin⁡θ+ϕ2+cos⁡θcos⁡θ+ϕ2)=0.x\cos\frac{\theta+\phi}{2} + y\sin\frac{\theta+\phi}{2} - \left(\sin\theta\sin\frac{\theta+\phi}{2} + \cos\theta\cos\frac{\theta+\phi}{2}\right) = 0.

The bracket simplifies using the cosine difference identity:

cos⁡θcos⁡θ+ϕ2+sin⁡θsin⁡θ+ϕ2=cos⁡(θ−θ+ϕ2)=cos⁡θ−ϕ2.\cos\theta\cos\frac{\theta+\phi}{2} + \sin\theta\sin\frac{\theta+\phi}{2} = \cos\left(\theta - \frac{\theta+\phi}{2}\right) = \cos\frac{\theta-\phi}{2}.

So the line is:

xcos⁡θ+ϕ2+ysin⁡θ+ϕ2−cos⁡θ−ϕ2=0.x\cos\frac{\theta+\phi}{2} + y\sin\frac{\theta+\phi}{2} - \cos\frac{\theta-\phi}{2} = 0.

  1. Apply the distance formula from the origin (0,0)(0,0) to a line ax+by+c=0ax+by+c=0. The perpendicular distance is d=∣a⋅0+b⋅0+c∣a2+b2=∣c∣a2+b2.d = \frac{|a\cdot0 + b\cdot0 + c|}{\sqrt{a^2 + b^2}} = \frac{|c|}{\sqrt{a^2 + b^2}}. …
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