If the points (−2,0),(−1,31) and (cosθ,sinθ) are collinear, then the number of values of θ ∈[0,2π] :

Given points (−2,0),(−1,31) and (cosθ,sinθ)For collinearity,∣∣x1x2x3 y1 y2y3 111∣∣=0∣∣−2−1cosθ 03 1 sinθ 111∣∣=0sinθ −3 1cosθ =3 2⇒ 23 sinθ −21cosθ =1⇒sin(θ−6π )=sin2π ⇒ θ−6π =nπ+(−1)n2π (n∈Z)⇒θ=32π (when θ∈[0,2π],n=1)

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If the points (-2,0),(-1,{1}{{3}}) and (, ) are collinear, the

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If the points $(- 2, 0), (- 1, \frac{1}{3})$ and $(cos θ, sin θ)$ are collinear, then the number of values of $θ \in [0, 2 π]$ :

0

1

2

infinite

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Charlotte Discussed

If the points

(- 2, 0), (- 1, \frac{1}{3})

and

(cos θ, sin θ)

are collinear, then the number of values of

θ \in [0, 2 π]

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

Given points

(- 2, 0), (- 1, \frac{1}{3})

and

(cos θ, sin θ)

For collinearity,

∣ ∣ x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111 ∣ ∣ = 0

∣ ∣ - 2 - 1 cos θ 0 \frac{1}{3} sin θ 111 ∣ ∣ = 0

sin θ - \frac{1}{3} cos θ = \frac{2}{3}

\Rightarrow \frac{3}{2} sin θ - \frac{1}{2} cos θ = 1

\Rightarrow sin (θ - \frac{π}{6}) = sin \frac{π}{2}

\Rightarrow θ - \frac{π}{6} = nπ + (- 1)^{n} \frac{π}{2}

(

n \in Z

)

\Rightarrow θ = \frac{2 π}{3}

(when

θ \in [0, 2 π], n = 1

)

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Question Text	If the points $(- 2, 0), (- 1, \frac{1}{3})$ and $(cos θ, sin θ)$ are collinear, then the number of values of $θ \in [0, 2 π]$ :
Answer Type	Text solution:1
Upvotes	150

If the points (−2,0),(−1,3​1​) and (cosθ,sinθ) are collinear, then the number of values of θ ∈[0,2π] :

Text solutionVerified

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If the points $(- 2, 0), (- 1, \frac{1}{3})$ and $(cos θ, sin θ)$ are collinear, then the number of values of $θ \in [0, 2 π]$ :