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Straight Lines and Pair of Straight Lines question

2008 · Shift 2 · Q41
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  5. /2008 · Shift 2 · Q41

Straight Lines and Pair of Straight Lines question

2008 · Shift 2 · Q41

JEE AdvancedMathematicsStraight Lines and Pair of Straight LinesMCQ+3 / −1
Consider three points P=(−sin⁡(β−α),−cosβ),Q=(cos(β−α),sin⁡β)P = ( - \sin (\beta - \alpha ), - cos\beta ),Q = (cos(\beta - \alpha ),\sin \beta )P=(−sin(β−α),−cosβ),Q=(cos(β−α),sinβ) and R=(cos⁡(β−α+θ),sin⁡(β−θ))R = (\cos (\beta - \alpha + \theta ),\sin (\beta - \theta ))R=(cos(β−α+θ),sin(β−θ)) where 0<α,β,θ<π40 \lt \alpha ,\beta ,\theta \lt {\pi \over 4}0<α,β,θ<4π​. Then :
  1. A
    P lies on the line segment RQ
  2. B
    Q lies on the line segment PR
  3. C
    R lies on the line segment QP
  4. D
    P, Q, R are non-collinear
View written solutionFree

Correct answer: D

To determine the relationship between the three points P, Q, and R, we will check if they are collinear. Three points are collinear if the area of the triangle formed by them is zero, or equivalently, if the slope of the line segment joining any two points is equal to the slope of the line segment joining another pair of points.

Let the coordinates of the points be: P=(xP,yP)=(−sin⁡(β−α),−cos⁡β)P = (x_P, y_P) = ( - \sin (\beta - \alpha ), - \cos\beta )P=(xP​,yP​)=(−sin(β−α),−cosβ) Q=(xQ,yQ)=(cos⁡(β−α),sin⁡β)Q = (x_Q, y_Q) = (\cos(\beta - \alpha ),\sin \beta )Q=(xQ​,yQ​)=(cos(β−α),sinβ) R=(xR,yR)=(cos⁡(β−α+θ),sin⁡(β−θ))R = (x_R, y_R) = (\cos (\beta - \alpha + \theta ),\sin (\beta - \theta ))R=(xR​,yR​)=(cos(β−α+θ),sin(β−θ))

Step 1: Calculate the slope of the line segment PQ.

Slope mPQ=yQ−yPxQ−xP=sin⁡β−(−cos⁡β)cos⁡(β−α)−(−sin⁡(β−α))m_{PQ} = \frac{y_Q - y_P}{x_Q - x_P} = \frac{\sin \beta - (-\cos \beta)}{\cos(\beta - \alpha) - (-\sin(\beta - \alpha))}mPQ​=xQ​−xP​yQ​−yP​​=cos(β−α)−(−sin(β−α))sinβ−(−cosβ)​ mPQ=sin⁡β+cos⁡βcos⁡(β−α)+sin⁡(β−α)m_{PQ} = \frac{\sin \beta + \cos \beta}{\cos(\beta - \alpha) + \sin(\beta - \alpha)}mPQ​=cos(β−α)+sin(β−α)sinβ+cosβ​

Step 2: Calculate the slope of the line segment PR.

Slope mPR=yR−yPxR−xP=sin⁡(β−θ)−(−cos⁡β)cos⁡(β−α+θ)−(−sin⁡(β−α))m_{PR} = \frac{y_R - y_P}{x_R - x_P} = \frac{\sin(\beta - \theta) - (-\cos \beta)}{\cos(\beta - \alpha + \theta) - (-\sin(\beta - \alpha))}mPR​=xR​−xP​yR​−yP​​=cos(β−α+θ)−(−sin(β−α))sin(β−θ)−(−cosβ)​ mPR=sin⁡(β−θ)+cos⁡βcos⁡(β−α+θ)+sin⁡(β−α)m_{PR} = \frac{\sin(\beta - \theta) + \cos \beta}{\cos(\beta - \alpha + \theta) + \sin(\beta - \alpha)}mPR​=cos(β−α+θ)+sin(β−α)sin(β−θ)+cosβ​

Step 3: Check the condition for collinearity.

The points P, Q, and R are collinear if and only if mPQ=mPRm_{PQ} = m_{PR}mPQ​=mPR​. sin⁡β+cos⁡βcos⁡(β−α)+sin⁡(β−α)=sin⁡(β−θ)+cos⁡βcos⁡(β−α+θ)+sin⁡(β−α)\frac{\sin \beta + \cos \beta}{\cos(\beta - \alpha) + \sin(\beta - \alpha)} = \frac{\sin(\beta - \theta) + \cos \beta}{\cos(\beta - \alpha + \theta) + \sin(\beta - \alpha)}cos(β−α)+sin(β−α)sinβ+cosβ​=cos(β−α+θ)+sin(β−α)sin(β−θ)+cosβ​ Cross-multiplying gives the condition: (sin⁡β+cos⁡β)(cos⁡(β−α+θ)+sin⁡(β−α))=(sin⁡(β−θ)+cos⁡β)(cos⁡(β−α)+sin⁡(β−α))(\sin \beta + \cos \beta)(\cos(\beta - \alpha + \theta) + \sin(\beta - \alpha)) = (\sin(\beta - \theta) + \cos \beta)(\cos(\beta - \alpha) + \sin(\beta - \alpha))(sinβ+cosβ)(cos(β−α+θ)+sin(β−α))=(sin(β−θ)+cosβ)(cos(β−α)+sin(β−α))

Let's evaluate the difference between the left-hand side (LHS) and the right-hand side (RHS). Let this difference be denoted by Δ\DeltaΔ. The points are collinear if Δ=0\Delta = 0Δ=0. Δ=(sin⁡β+cos⁡β)(cos⁡(β−α+θ)+sin⁡(β−α))−(sin⁡(β−θ)+cos⁡β)(cos⁡(β−α)+sin⁡(β−α))\Delta = (\sin \beta + \cos \beta)(\cos(\beta - \alpha + \theta) + \sin(\beta - \alpha)) - (\sin(\beta - \theta) + \cos \beta)(\cos(\beta - \alpha) + \sin(\beta - \alpha))Δ=(sinβ+cosβ)(cos(β−α+θ)+sin(β−α))−(sin(β−θ)+cosβ)(cos(β−α)+sin(β−α))

Step 4: Simplify the expression for Δ\DeltaΔ.

We use the angle addition/subtraction formulas: cos⁡(β−α+θ)=cos⁡(β−α)cos⁡θ−sin⁡(β−α)sin⁡θ\cos(\beta - \alpha + \theta) = \cos(\beta - \alpha)\cos\theta - \sin(\beta - \alpha)\sin\thetacos(β−α+θ)=cos(β−α)cosθ−sin(β−α)sinθ sin⁡(β−θ)=sin⁡βcos⁡θ−cos⁡βsin⁡θ\sin(\beta - \theta) = \sin\beta\cos\theta - \cos\beta\sin\thetasin(β−θ)=sinβcosθ−cosβsinθ

Substituting these into the expression for Δ\DeltaΔ and expanding is tedious. A more structured approach is to rearrange the expression first. Let's expand and collect terms. After expansion, the term (sin⁡β+cos⁡β)sin⁡(β−α)(\sin \beta + \cos \beta)\sin(\beta-\alpha)(sinβ+cosβ)sin(β−α) appears on the LHS and cos⁡βsin⁡(β−α)\cos\beta \sin(\beta-\alpha)cosβsin(β−α) and ......... on the RHS. A careful expansion and simplification leads to: Let's test the expression Δ=LHS−RHS\Delta = \text{LHS} - \text{RHS}Δ=LHS−RHS: Δ=sin⁡βcos⁡(β−α+θ)+sin⁡βsin⁡(β−α)+cos⁡βcos⁡(β−α+θ)+cos⁡βsin⁡(β−α)−[sin⁡(β−θ)cos⁡(β−α)+sin⁡(β−θ)sin⁡(β−α)+cos⁡βcos⁡(β−α)+cos⁡βsin⁡(β−α)]\Delta = \sin\beta \cos(\beta-\alpha+\theta) + \sin\beta \sin(\beta-\alpha) + \cos\beta \cos(\beta-\alpha+\theta) + \cos\beta \sin(\beta-\alpha) - [\sin(\beta-\theta) \cos(\beta-\alpha) + \sin(\beta-\theta) \sin(\beta-\alpha) + \cos\beta \cos(\beta-\alpha) + \cos\beta \sin(\beta-\alpha)]Δ=sinβcos(β−α+θ)+sinβsin(β−α)+cosβcos(β−α+θ)+cosβsin(β−α)−[sin(β−θ)cos(β−α)+sin(β−θ)sin(β−α)+cosβcos(β−α)+cosβsin(β−α)] The term cos⁡βsin⁡(β−α)\cos\beta\sin(\beta-\alpha)cosβsin(β−α) cancels out.

By substituting the sum/difference formulas for cos⁡(β−α+θ)\cos(\beta-\alpha+\theta)cos(β−α+θ) and sin⁡(β−θ)\sin(\beta-\theta)sin(β−θ) and collecting terms based on cos⁡θ\cos\thetacosθ and sin⁡θ\sin\thetasinθ, we get: Δ=[−cos⁡(2β−α)]cos⁡θ+[−cos⁡(2β−α)]sin⁡θ+[cos⁡(2β−α)]\Delta = [-\cos(2\beta-\alpha)]\cos\theta + [-\cos(2\beta-\alpha)]\sin\theta + [\cos(2\beta-\alpha)]Δ=[−cos(2β−α)]cosθ+[−cos(2β−α)]sinθ+[cos(2β−α)] Factoring out cos⁡(2β−α)\cos(2\beta-\alpha)cos(2β−α), we get: Δ=cos⁡(2β−α)(1−cos⁡θ−sin⁡θ)\Delta = \cos(2\beta - \alpha)(1 - \cos\theta - \sin\theta)Δ=cos(2β−α)(1−cosθ−sinθ)

Step 5: Analyze the factors based on the given constraints.

We are given 0<α,β,θ<π40 < \alpha, \beta, \theta < \frac{\pi}{4}0<α,β,θ<4π​.

  1. Analyze the term cos⁡(2β−α)\cos(2\beta - \alpha)cos(2β−α): Since 0<β<π/40 < \beta < \pi/40<β<π/4, we have 0<2β<π/20 < 2\beta < \pi/20<2β<π/2. Since 0<α<π/40 < \alpha < \pi/40<α<π/4, we have −π/4<−α<0-\pi/4 < -\alpha < 0−π/4<−α<0. Adding these inequalities: 0−π/4<2β−α<π/2+00 - \pi/4 < 2\beta - \alpha < \pi/2 + 00−π/4<2β−α<π/2+0, which gives −π/4<2β−α<π/2-\pi/4 < 2\beta - \alpha < \pi/2−π/4<2β−α<π/2. In this interval (parts of the first and fourth quadrants), the cosine function is always positive. Therefore, cos⁡(2β−α)≠0\cos(2\beta - \alpha) \neq 0cos(2β−α)=0.

  2. Analyze the term (1−cos⁡θ−sin⁡θ)(1 - \cos\theta - \sin\theta)(1−cosθ−sinθ): Let f(θ)=cos⁡θ+sin⁡θ=2sin⁡(θ+π/4)f(\theta) = \cos\theta + \sin\theta = \sqrt{2} \sin(\theta + \pi/4)f(θ)=cosθ+sinθ=2​sin(θ+π/4). Since 0<θ<π/40 < \theta < \pi/40<θ<π/4, we have π/4<θ+π/4<π/2\pi/4 < \theta + \pi/4 < \pi/2π/4<θ+π/4<π/2. In this interval, sin⁡(θ+π/4)\sin(\theta + \pi/4)sin(θ+π/4) ranges from sin⁡(π/4)=1/2\sin(\pi/4) = 1/\sqrt{2}sin(π/4)=1/2​ to sin⁡(π/2)=1\sin(\pi/2) = 1sin(π/2)=1. So, 1/2<sin⁡(θ+π/4)<11/\sqrt{2} < \sin(\theta + \pi/4) < 11/2​<sin(θ+π/4)<1. This means 1<2sin⁡(θ+π/4)<21 < \sqrt{2}\sin(\theta + \pi/4) < \sqrt{2}1<2​sin(θ+π/4)<2​, so 1<cos⁡θ+sin⁡θ<21 < \cos\theta + \sin\theta < \sqrt{2}1<cosθ+sinθ<2​. Therefore, 1−(cos⁡θ+sin⁡θ)1 - (\cos\theta + \sin\theta)1−(cosθ+sinθ) is always negative and never zero. (1−cos⁡θ−sin⁡θ)≠0(1 - \cos\theta - \sin\theta) \neq 0(1−cosθ−sinθ)=0.

Step 6: Conclusion.

Since both factors cos⁡(2β−α)\cos(2\beta - \alpha)cos(2β−α) and (1−cos⁡θ−sin⁡θ)(1 - \cos\theta - \sin\theta)(1−cosθ−sinθ) are non-zero for the given ranges of α,β,θ\alpha, \beta, \thetaα,β,θ, their product Δ\DeltaΔ is never zero. This means the condition for collinearity is never satisfied. Thus, the points P, Q, and R are non-collinear.

This eliminates options A, B, and C, which all imply that the points are collinear. Option D is the only one that can be correct.

Final check with a simple case: Let α=β=θ=π/8\alpha = \beta = \theta = \pi/8α=β=θ=π/8. P=(−sin⁡(0),−cos⁡(π/8))=(0,−cos⁡(π/8))P = (-\sin(0), -\cos(\pi/8)) = (0, -\cos(\pi/8))P=(−sin(0),−cos(π/8))=(0,−cos(π/8)) Q=(cos⁡(0),sin⁡(π/8))=(1,sin⁡(π/8))Q = (\cos(0), \sin(\pi/8)) = (1, \sin(\pi/8))Q=(cos(0),sin(π/8))=(1,sin(π/8)) R=(cos⁡(π/8),sin⁡(0))=(cos⁡(π/8),0)R = (\cos(\pi/8), \sin(0)) = (\cos(\pi/8), 0)R=(cos(π/8),sin(0))=(cos(π/8),0) Slope of PQ: mPQ=sin⁡(π/8)+cos⁡(π/8)1−0=sin⁡(π/8)+cos⁡(π/8)m_{PQ} = \frac{\sin(\pi/8) + \cos(\pi/8)}{1-0} = \sin(\pi/8) + \cos(\pi/8)mPQ​=1−0sin(π/8)+cos(π/8)​=sin(π/8)+cos(π/8). Slope of QR: mQR=0−sin⁡(π/8)cos⁡(π/8)−1=−sin⁡(π/8)cos⁡(π/8)−1m_{QR} = \frac{0 - \sin(\pi/8)}{\cos(\pi/8) - 1} = \frac{-\sin(\pi/8)}{\cos(\pi/8) - 1}mQR​=cos(π/8)−10−sin(π/8)​=cos(π/8)−1−sin(π/8)​. If collinear, mPQ=mQRm_{PQ} = m_{QR}mPQ​=mQR​. (sin⁡(π/8)+cos⁡(π/8))(cos⁡(π/8)−1)=−sin⁡(π/8)(\sin(\pi/8) + \cos(\pi/8))(\cos(\pi/8)-1) = -\sin(\pi/8)(sin(π/8)+cos(π/8))(cos(π/8)−1)=−sin(π/8). sin⁡(π/8)cos⁡(π/8)−sin⁡(π/8)+cos⁡2(π/8)−cos⁡(π/8)=−sin⁡(π/8)\sin(\pi/8)\cos(\pi/8) - \sin(\pi/8) + \cos^2(\pi/8) - \cos(\pi/8) = -\sin(\pi/8)sin(π/8)cos(π/8)−sin(π/8)+cos2(π/8)−cos(π/8)=−sin(π/8). 12sin⁡(π/4)+cos⁡2(π/8)−cos⁡(π/8)=0\frac{1}{2}\sin(\pi/4) + \cos^2(\pi/8) - \cos(\pi/8) = 021​sin(π/4)+cos2(π/8)−cos(π/8)=0. This is not generally true, confirming non-collinearity.

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