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Straight Lines and Pair of Straight Lines question

2021 · 27 Jul · Shift 2 · Q23
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  5. /2021 · 27 Jul · Shift 2 · Q23

Straight Lines and Pair of Straight Lines question

2021 · 27 Jul · Shift 2 · Q23

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
The point P (a, b) undergoes the following three transformations successively : (a) reflection about the line y = x. (b) translation through 2 units along the positive direction of x-axis. (c) rotation through angle π4{\pi \over 4}4π​ about the origin in the anti-clockwise direction. If the co-ordinates of the final position of the point P are (−12,72)\left( { - {1 \over {\sqrt 2 }},{7 \over {\sqrt 2 }}} \right)(−2​1​,2​7​), then the value of 2a + b is equal to :
  1. A
    13
  2. B
    9
  3. C
    5
  4. D
    7
View written solutionFree

Correct answer: B

  1. Initial point

    Let the original point be P(a,b).P(a,b).P(a,b).

  2. First transformation: reflection about the line y=xy=xy=x

    Reflection about y=xy=xy=x interchanges the coordinates.

    So, (a,b)→(b,a).(a,b) \to (b,a).(a,b)→(b,a).

  3. Second transformation: translation by 2 units along positive xxx-axis

    Adding 2 to the xxx-coordinate, (b,a)→(b+2,a).(b,a) \to (b+2,a).(b,a)→(b+2,a).

  4. Third transformation: rotation by π4\frac{\pi}{4}4π​ anticlockwise about the origin

    If a point (x,y)(x,y)(x,y) is rotated by angle θ\thetaθ, then the new coordinates are

    (xcos⁡θ−ysin⁡θ, xsin⁡θ+ycos⁡θ).(x\cos\theta-y\sin\theta,\, x\sin\theta+y\cos\theta).(xcosθ−ysinθ,xsinθ+ycosθ).

    Here θ=π4\theta=\frac{\pi}{4}θ=4π​, so cos⁡π4=sin⁡π4=12.\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{1}{\sqrt2}.cos4π​=sin4π​=2​1​.

    Thus, rotating (b+2,a)(b+2,a)(b+2,a) gives

    ((b+2)−a2, (b+2)+a2).\left(\frac{(b+2)-a}{\sqrt2},\, \frac{(b+2)+a}{\sqrt2}\right).(2​(b+2)−a​,2​(b+2)+a​).

  5. Match with the given final coordinates

    Given final point is (−12,72).\left(-\frac{1}{\sqrt2},\frac{7}{\sqrt2}\right).(−2​1​,2​7​).

    Therefore,

    b+2−a2=−12⇒b+2−a=−1⇒b−a=−3.\frac{b+2-a}{\sqrt2}=-\frac{1}{\sqrt2} \quad \Rightarrow \quad b+2-a=-1 \quad \Rightarrow \quad b-a=-3.2​b+2−a​=−2​1​⇒b+2−a=−1⇒b−a=−3.

    Also,

    b+2+a2=72⇒a+b+2=7⇒a+b=5.\frac{b+2+a}{\sqrt2}=\frac{7}{\sqrt2} \quad \Rightarrow \quad a+b+2=7 \quad \Rightarrow \quad a+b=5.2​b+2+a​=2​7​⇒a+b+2=7⇒a+b=5.

  6. Solve for aaa and bbb

    From b−a=−3b-a=-3b−a=−3 and a+b=5,a+b=5,a+b=5,

    adding the two equations: 2b=2⇒b=1.2b=2 \Rightarrow b=1.2b=2⇒b=1.

    Then a+b=5⇒a=4.a+b=5 \Rightarrow a=4.a+b=5⇒a=4.

  7. Find 2a+b2a+b2a+b

    2a+b=2(4)+1=9.2a+b=2(4)+1=9.2a+b=2(4)+1=9.

  8. Check options

    The correct option is 9\boxed{9}9​ which is Option B.

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