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Question

# Assertion :Points P(−sin(β−α),−cosβ),Q(cos(β−α),sinβ) and R(cos(β−α+θ),sin(β−θ)), where β=π4+α2 are non-collinear. Reason: Three given points are non-collinear if they form a triangle of non-zero area.

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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C
Assertion is correct but Reason is incorrect
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D
Assertion is incorrect but the Reason is correct
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Solution

## The correct option is D Assertion is incorrect but the Reason is correctReason is wrong as three points are collinear when area of triangle formed is zero.Area of triangle PQR =12∣∣ ∣ ∣∣−sin(β−α)−cosβ1cos(β−α)sinβ1cos(β−α+θ)sin(β−θ)1∣∣ ∣ ∣∣=12∣∣ ∣ ∣ ∣ ∣ ∣∣−sin(π4+α2−α)−cos(π4+α2)1cos(π4+α2−α)sin(π4+α2)1cos(π4+α2−α+θ)sin(π4+α2−θ)1∣∣ ∣ ∣ ∣ ∣ ∣∣=12∣∣ ∣ ∣ ∣ ∣ ∣∣−sin(π4−α2)−cos(π4+α2)1cos(π4−α2)sin(π4+α2)1cos(π4−α2+θ)sin(π4+α2−θ)1∣∣ ∣ ∣ ∣ ∣ ∣∣Applying R2→R2−R1,R3→R3−R1=12∣∣ ∣ ∣ ∣ ∣ ∣∣−sin(π4−α2)−cos(π4+α2)1cos(π4−α2)+sin(π4−α2)sin(π4+α2)+cos(π4+α2)0cos(π4−α2+θ)+sin(π4−α2)sin(π4+α2−θ)+cos(π4+α2)0∣∣ ∣ ∣ ∣ ∣ ∣∣=12(cos(π4−α2)+sin(π4−α2))(cos(π4−α2+θ)+sin(π4−α2))×∣∣ ∣ ∣ ∣∣−sin(π4−α2)−cos(π4+α2)1110110∣∣ ∣ ∣ ∣∣=0Hence point P,Q and R are collinear Therefore assertion is wrong.

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