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Straight Lines and Pair of Straight Lines question

2024 · 29 Jan · Shift 1 · Q42
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  5. /2024 · 29 Jan · Shift 1 · Q42

Straight Lines and Pair of Straight Lines question

2024 · 29 Jan · Shift 1 · Q42

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
In a △ABC\triangle A B C△ABC, suppose y=xy=xy=x is the equation of the bisector of the angle BBB and the equation of the side ACA CAC is 2x−y=22 x-y=22x−y=2. If 2AB=BC2 A B=B C2AB=BC and the points AAA and BBB are respectively (4,6)(4,6)(4,6) and (α,β)(\alpha, \beta)(α,β), then α+2β\alpha+2 \betaα+2β is equal to
  1. A
    42
  2. B
    39
  3. C
    48
  4. D
    45
View written solutionFree

Correct answer: A

  1. Given data
  • A=(4,6)A=(4,6)A=(4,6)
  • B=(α,β)B=(\alpha,\beta)B=(α,β)
  • Side ACACAC lies on the line 2x−y=22x-y=22x−y=2
  • The angle bisector at BBB is y=xy=xy=x
  • Also, 2AB=BC2AB=BC2AB=BC

We need to find α+2β\alpha+2\betaα+2β.


  1. Find point CCC using the side ACACAC

Since AAA lies on side ACACAC, check that A=(4,6)A=(4,6)A=(4,6) satisfies 2x−y=22x-y=22x−y=2: 2(4)−6=8−6=22(4)-6=8-6=22(4)−6=8−6=2 So AAA is indeed on the line ACACAC.

Because y=xy=xy=x is the bisector of angle BBB, the two sides through BBB, namely BABABA and BCBCBC, must be symmetric about the line y=xy=xy=x.

So first reflect point A=(4,6)A=(4,6)A=(4,6) across the line y=xy=xy=x. Reflection across y=xy=xy=x swaps coordinates, hence the reflected point is A′=(6,4)A'=(6,4)A′=(6,4) Therefore, line BCBCBC is the line through BBB and A′A'A′, while line BABABA is the line through BBB and AAA.

Now point CCC lies on line BCBCBC and also on line AC:2x−y=2AC: 2x-y=2AC:2x−y=2.


  1. Use the ratio condition 2AB=BC2AB=BC2AB=BC

Since A,B,CA,B,CA,B,C are on the two rays forming angle BBB, and BABABA and BCBCBC are symmetric about y=xy=xy=x, we can use the reflection idea.

Let C′C'C′ be the reflection of CCC across y=xy=xy=x. Then C′C'C′ lies on line BABABA, and because reflection preserves distance from BBB, BC=BC′BC=BC'BC=BC′ Given BC=2ABBC=2ABBC=2AB So along the line BABABA, the point C′C'C′ must satisfy BC′=2ABBC'=2ABBC′=2AB Since AAA is on ray BABABA, this means C′C'C′ is the point on line BABABA such that BC′→=2BA→\overrightarrow{BC'}=2\overrightarrow{BA}BC′=2BA Hence C′=B+2(A−B)=2A−BC'=B+2(A-B)=2A-BC′=B+2(A−B)=2A−B Now reflect back across y=xy=xy=x to get CCC.

Let B=(α,β)B=(\alpha,\beta)B=(α,β). Then 2A−B=(8−α,12−β)2A-B=(8-\alpha,12-\beta)2A−B=(8−α,12−β) Reflecting across y=xy=xy=x gives C=(12−β, 8−α)C=(12-\beta,\,8-\alpha)C=(12−β,8−α)


  1. Use that CCC lies on AC:2x−y=2AC: 2x-y=2AC:2x−y=2

Substitute C=(12−β,8−α)C=(12-\beta,8-\alpha)C=(12−β,8−α) into 2x−y=22x-y=22x−y=2: 2(12−β)−(8−α)=22(12-\beta)-(8-\alpha)=22(12−β)−(8−α)=2 24−2β−8+α=224-2\beta-8+\alpha=224−2β−8+α=2 16+α−2β=216+\alpha-2\beta=216+α−2β=2 α−2β=−14(1)\alpha-2\beta=-14 \qquad (1)α−2β=−14(1)


  1. Use that BBB lies on the angle bisector condition

Since line BABABA reflects to line BCBCBC about y=xy=xy=x, the point BBB must lie on the bisector itself. Hence BBB lies on y=xy=xy=x, so α=β\alpha=\betaα=β Substitute into (1): α−2α=−14\alpha-2\alpha=-14α−2α=−14 −α=−14-\alpha=-14−α=−14 α=14\alpha=14α=14 Thus β=14\beta=14β=14

So B=(14,14)B=(14,14)B=(14,14)


  1. Compute the required value

α+2β=14+2(14)=42\alpha+2\beta=14+2(14)=42α+2β=14+2(14)=42


  1. Check with options

The value is 424242 So the correct option is A.


  1. Comparison with stored answer

Stored correct answer: A

Our derived answer also gives A.

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