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Straight Lines and Pair of Straight Lines question

2003 · Shift 0 · Q109
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  5. /2003 · Shift 0 · Q109

Straight Lines and Pair of Straight Lines question

2003 · Shift 0 · Q109

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
A square of side a lies above the xxx-axis and has one vertex at the origin. The side passing through the origin makes an angle α(0<α<π4)\alpha \left( {0 \lt \alpha \lt {\pi \over 4}} \right)α(0<α<4π​) with the positive direction of x-axis. The equation of its diagonal not passing through the origin is :
  1. A
    y(cos⁡α+sin⁡α)+x(cos⁡α−sin⁡α)=ay\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = ay(cosα+sinα)+x(cosα−sinα)=a
  2. B
    y(cos⁡α−sin⁡α)−x(sin⁡α−cos⁡α)=ay\left( {\cos \alpha - \sin \alpha } \right) - x\left( {\sin \alpha - \cos \alpha } \right) = ay(cosα−sinα)−x(sinα−cosα)=a
  3. C
    y(cos⁡α+sin⁡α)+x(sin⁡α−cos⁡α)=ay\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha - \cos \alpha } \right) = ay(cosα+sinα)+x(sinα−cosα)=a
  4. D
    y(cos⁡α+sin⁡α)+x(sin⁡α+cos⁡α)=ay\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha + \cos \alpha } \right) = ay(cosα+sinα)+x(sinα+cosα)=a
View written solutionFree

Correct answer: A

  1. Set up the square using vectors

Let the vertex at the origin be O=(0,0)O=(0,0)O=(0,0).

One side through the origin makes angle α\alphaα with the positive xxx-axis, so the side vector of length aaa is

u⃗=a(cos⁡α,sin⁡α).\vec{u}=a(\cos\alpha,\sin\alpha).u=a(cosα,sinα).

Since the square lies above the xxx-axis and 0<α<π40<\alpha<\frac{\pi}{4}0<α<4π​, the adjacent side must be perpendicular to this and directed into the upper half-plane. Rotating u⃗\vec{u}u by 90∘90^\circ90∘ anticlockwise gives

v⃗=a(−sin⁡α,cos⁡α).\vec{v}=a(-\sin\alpha,\cos\alpha).v=a(−sinα,cosα).

So the four vertices are:

O=(0,0),O=(0,0),O=(0,0), A=(acos⁡α,asin⁡α),A=(a\cos\alpha,a\sin\alpha),A=(acosα,asinα), B=(−asin⁡α,acos⁡α),B=(-a\sin\alpha,a\cos\alpha),B=(−asinα,acosα), C=A+B=(a(cos⁡α−sin⁡α),a(sin⁡α+cos⁡α)).C=A+B=(a(\cos\alpha-\sin\alpha),a(\sin\alpha+\cos\alpha)).C=A+B=(a(cosα−sinα),a(sinα+cosα)).
  1. Identify the diagonal not passing through the origin

The two diagonals are:

  • OCOCOC, which passes through the origin,
  • ABABAB, which does not pass through the origin.

So we need the equation of the line through AAA and BBB.

  1. Use midpoint/parallel property of diagonals

In a square, diagonal ABABAB is perpendicular to diagonal OCOCOC.

Slope direction of OCOCOC is along

(cos⁡α−sin⁡α,sin⁡α+cos⁡α).(\cos\alpha-\sin\alpha,\sin\alpha+\cos\alpha).(cosα−sinα,sinα+cosα).

Hence a normal vector to line ABABAB can be taken as

(cos⁡α−sin⁡α,sin⁡α+cos⁡α).(\cos\alpha-\sin\alpha,\sin\alpha+\cos\alpha).(cosα−sinα,sinα+cosα).

Therefore equation of ABABAB is of the form

(cos⁡α−sin⁡α)x+(sin⁡α+cos⁡α)y=k.(\cos\alpha-\sin\alpha)x+(\sin\alpha+\cos\alpha)y=k.(cosα−sinα)x+(sinα+cosα)y=k.
  1. Find the constant using point AAA

Substitute A=(acos⁡α,asin⁡α)A=(a\cos\alpha,a\sin\alpha)A=(acosα,asinα):

(cos⁡α−sin⁡α)(acos⁡α)+(sin⁡α+cos⁡α)(asin⁡α)=k.(\cos\alpha-\sin\alpha)(a\cos\alpha)+(\sin\alpha+\cos\alpha)(a\sin\alpha)=k.(cosα−sinα)(acosα)+(sinα+cosα)(asinα)=k.

Simplify:

k=a(cos⁡2α−sin⁡αcos⁡α+sin⁡2α+sin⁡αcos⁡α)=a.k=a\left(\cos^2\alpha-\sin\alpha\cos\alpha+\sin^2\alpha+\sin\alpha\cos\alpha\right)=a.k=a(cos2α−sinαcosα+sin2α+sinαcosα)=a.

Thus the equation is

(cos⁡α−sin⁡α)x+(cos⁡α+sin⁡α)y=a.(\cos\alpha-\sin\alpha)x+(\cos\alpha+\sin\alpha)y=a.(cosα−sinα)x+(cosα+sinα)y=a.

Rewriting,

y(cos⁡α+sin⁡α)+x(cos⁡α−sin⁡α)=a.y(\cos\alpha+\sin\alpha)+x(\cos\alpha-\sin\alpha)=a.y(cosα+sinα)+x(cosα−sinα)=a.
  1. Match with the options

This is exactly Option A.

  1. Check other options briefly
  • A: Matches derived equation. Correct.
  • B: Gives coefficients of xxx and yyy both as (cos⁡α−sin⁡α)(\cos\alpha-\sin\alpha)(cosα−sinα) after simplification, not our line.
  • C: Coefficient of xxx is (sin⁡α−cos⁡α)=−(cos⁡α−sin⁡α)(\sin\alpha-\cos\alpha)=-(\cos\alpha-\sin\alpha)(sinα−cosα)=−(cosα−sinα), so different line.
  • D: Coefficients of both xxx and yyy are positive and equal to (sin⁡α+cos⁡α)(\sin\alpha+\cos\alpha)(sinα+cosα), not correct.

Hence the correct answer is A.

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