Consider three points $P=(-\sin (\beta-\alpha),-\cos \beta), Q=(\cos (\beta-\alpha), \sin \beta)$ and…
Consider three points $P=(-\sin (\beta-\alpha),-\cos \beta), Q=(\cos (\beta-\alpha), \sin \beta)$ and $R=(\cos (\beta-\alpha+\theta), \sin (\beta-\theta)$, where $0 < \alpha, \beta, \theta < \frac{\pi}{4}$. Then,
- $P$ lies on the line segment $R Q$
- $Q$ lies on the line segment $P R$
- $R$ lies on the line segment $Q P$
- $P, Q, R$ are non- collinear
Solution
For collinear points
$
\Delta=\left|\begin{array}{ccc}
-\sin (\beta-\alpha) & -\cos \beta & 1 \\
\cos (\beta-\alpha) & \sin \beta & 1 \\
\cos (\beta-\alpha+\theta) & \sin (\beta-\theta) & 1
\end{array}\right|
$
Clearly, $\Delta \neq 0$ for any value of $\alpha, \beta, \theta$, hence points are non-collinear
Asked in: JEE Advanced 2008 (Paper 2)
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