Question

Consider three points \( P(\cos\alpha, \sin\beta) \), \( Q(\sin\alpha, \cos\beta) \) and \( R(0,0) \), where \( 0 < \alpha, \beta < \frac{\pi}{4} \). Then:

• A. \( P \) lies on the line segment \( RQ \).
• B. \( Q \) lies on the line segment \( PR \).
• C. \( R \) lies on the line segment \( PQ \).
• D. \( P, Q, R \) are non-collinear.

Hint:

For collinearity: \begin{itemize} \item Compare slopes. \item Trig coordinates often reduce to angle identities. \end{itemize}

Answer 1

The Correct Option is D

Concept: Three points are collinear if slopes are equal. Check slopes \( PR \) and \( QR \). Step 1: Slope of \( PR \). \[ m_{PR} = \frac{\sin\beta - 0}{\cos\alpha - 0} = \frac{\sin\beta}{\cos\alpha} \] Step 2: Slope of \( QR \). \[ m_{QR} = \frac{\cos\beta}{\sin\alpha} \] Step 3: Check equality. For collinearity: \[ \frac{\sin\beta}{\cos\alpha} = \frac{\cos\beta}{\sin\alpha} \] Cross multiply: \[ \sin\alpha \sin\beta = \cos\alpha \cos\beta \] \[ \Rightarrow \cos(\alpha + \beta) = 0 \] So: \[ \alpha + \beta = \frac{\pi}{2} \] But given: \[ 0 < \alpha, \beta < \frac{\pi}{4} \Rightarrow \alpha + \beta < \frac{\pi}{2} \] Hence slopes are not equal. Step 4: Conclusion. Points are non-collinear.