Question

Find the equation of the straight lines passing through the following pair of point :

(a cos α, a sin α) and (a cos β, a sin β)

Answer 1

(a cos α, a sin α) and (a cos β, a sin β) 

\[\text { Here, } \left( x_1 , y_1 \right) \equiv \left( a\cos\alpha, a\sin\alpha \right) \]

\[\left( x_2 , y_2 \right) \equiv \left( a\cos\beta, a\sin\beta \right)\]

So, the equation of the line passing through the two points is

\[y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}\left( x - x_1 \right)\]

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

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

\[\Rightarrow y\left( \cos\beta - \cos\alpha \right) - x\left( \sin\beta - \sin\alpha \right) - a\sin\alpha\cos\beta + a\sin\alpha\cos\alpha + a\cos\alpha\sin\beta - a\cos\alpha\sin\alpha = 0\]

\[ \Rightarrow y\left( \cos\beta - \cos\alpha \right) - x\left( \sin\beta - \sin\alpha \right) = a\sin\alpha\cos\beta - a\cos\alpha\sin\beta\]

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

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

\[ \Rightarrow x\cos\left( \frac{\alpha + \beta}{2} \right) + y\sin\left( \frac{\alpha + \beta}{2} \right) = a\cos\left( \frac{\alpha - \beta}{2} \right) \left[ \text { dividing by } \sin\left( \frac{\alpha - \beta}{2} \right) \right]\]