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,

(D) $P, Q, R$ are non-collinear

(A) $P$ lies on the line segment $RQ$

(B) $Q$ lies on the line segment $PR$

(C) $R$ lies on the line segment $QP$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Straight Lines

Consider three points

P = (- sin (β - α), - cos β)

Q = (cos (β - α), sin β)

and

R = (cos (β - α + θ), sin (β - θ))

, where

0 < α, β, θ < \frac{π}{4}

. Then,

Answer: (D)

Let

P \equiv (x_{1}, y_{1})

Q \equiv (x_{2}, y_{2})

, and

R \equiv (x_{2} cos θ + x_{1} sin θ, y_{2} cos θ + y_{1} sin θ)

. If we consider the point

T

that divides

P Q

in the ratio

sin θ : cos θ

, we get

T \equiv (\frac{x _{2} c o s θ + x _{1} s i n θ}{c o s θ + s i n θ}, \frac{y _{2} c o s θ + y _{1} s i n θ}{c o s θ + s i n θ})

. Since

P

Q

, and

T

are collinear but

R \neq = T

(as

R

has coordinates without the denominator

cos θ + sin θ

), the points

P

Q

R

are non-collinear.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option