Assertion :Points P(−sin(β−α),−cosβ),Q(cos(β−α),sinβ) and R(cos(β−α+θ),sin(β−θ)), where β=4π+2α are non-collinear. Reason :Three given points are non-collinear if they form a triangle of non-zero area.

Question

Assertion :Points P(−sin(β−α),−cosβ),Q(cos(β−α),sinβ) and R(cos(β−α+θ),sin(β−θ)), where β=4π​+2α​ are non-collinear. Reason :Three given points are non-collinear if they form a triangle of non-zero area.

Filo · Accepted Answer

Reason is wrong as three points are collinear when area of triangle formed is zero.
Area of triangle

PQR

= \frac{1}{2} ∣ ∣ - sin (β - α) cos (β - α) cos (β - α + θ) - cos β sin β sin (β - θ) 111 ∣ ∣ = \frac{1}{2} ∣ ∣ - sin (\frac{π}{4} + \frac{α}{2} - α) cos (\frac{π}{4} + \frac{α}{2} - α) cos (\frac{π}{4} + \frac{α}{2} - α + θ) - cos (\frac{π}{4} + \frac{α}{2}) sin (\frac{π}{4} + \frac{α}{2}) sin (\frac{π}{4} + \frac{α}{2} - θ) 111 ∣ ∣ = \frac{1}{2} ∣ ∣ - sin (\frac{π}{4} - \frac{α}{2}) cos (\frac{π}{4} - \frac{α}{2}) cos (\frac{π}{4} - \frac{α}{2} + θ) - cos (\frac{π}{4} + \frac{α}{2}) sin (\frac{π}{4} + \frac{α}{2}) sin (\frac{π}{4} + \frac{α}{2} - θ) 111 ∣ ∣

Applying

R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}

= \frac{1}{2} ∣ ∣ - sin (\frac{π}{4} - \frac{α}{2}) cos (\frac{π}{4} - \frac{α}{2}) + sin (\frac{π}{4} - \frac{α}{2}) cos (\frac{π}{4} - \frac{α}{2} + θ) + sin (\frac{π}{4} - \frac{α}{2}) - cos (\frac{π}{4} + \frac{α}{2}) sin (\frac{π}{4} + \frac{α}{2}) + cos (\frac{π}{4} + \frac{α}{2}) sin (\frac{π}{4} + \frac{α}{2} - θ) + cos (\frac{π}{4} + \frac{α}{2}) 100 ∣ ∣

= \frac{1}{2} (cos (\frac{π}{4} - \frac{α}{2}) + sin (\frac{π}{4} - \frac{α}{2})) (cos (\frac{π}{4} - \frac{α}{2} + θ) + sin (\frac{π}{4} - \frac{α}{2})) \times

∣ ∣ - sin (\frac{π}{4} - \frac{α}{2}) 11 - cos (\frac{π}{4} + \frac{α}{2}) 11 100 ∣ ∣ = 0

Hence point

P, Q

and

R

are collinear
Therefore assertion is wrong.

Question Text	Assertion :Points $P (- sin (β - α), - cos β), Q (cos (β - α), sin β)$ and $R (cos (β - α + θ), sin (β - θ))$ , where $β = \frac{π}{4} + \frac{α}{2}$ are non-collinear. Reason :Three given points are non-collinear if they form a triangle of non-zero area.
Answer Type	Text solution:1
Upvotes	150

Assertion :Points $P (- sin (β - α), - cos β), Q (cos (β - α), sin β)$ and $R (cos (β - α + θ), sin (β - θ))$ , where $β = \frac{π}{4} + \frac{α}{2}$ are non-collinear.
Reason :Three given points are non-collinear if they form a triangle of non-zero area.

Text solutionVerified

Practice questions from similar books

Assertion :Points P(−sin(β−α),−cosβ),Q(cos(β−α),sinβ) and R(cos(β−α+θ),sin(β−θ)), where β=4π​+2α​ are non-collinear. Reason :Three given points are non-collinear if they form a triangle of non-zero area.

Text solutionVerified

Practice questions from similar books

Assertion :Points $P (- sin (β - α), - cos β), Q (cos (β - α), sin β)$ and $R (cos (β - α + θ), sin (β - θ))$ , where $β = \frac{π}{4} + \frac{α}{2}$ are non-collinear.
Reason :Three given points are non-collinear if they form a triangle of non-zero area.