JEE Maths›Straight Lines›Visual Solution
Visual SolutionPYQ 2024 · Jan Shift 1Tricky
Question
In a △ABC\triangle ABC, suppose y=xy = x is the equation of the bisector of angle B and the equation of the side AC is 2x−y=22x - y = 2. If 2AB=BC2AB = BC and the point A and B are respectively (4,6)(4, 6) and (α,β)(\alpha, \beta), then α+2β\alpha + 2\beta is equal to:
(A)4242
(B)3939
(C)4848
(D)4545
Solution Path
Reflect A(4,6)A(4,6) across y=xy=x to get A′(6,4)A'(6,4). Angle bisector theorem with 2AB=BC2AB=BC gives B=(14,14)B=(14,14). Answer: α+2β=42\alpha+2\beta = 42.
01Question Setup
1/4
In △ABC\triangle ABC, y=xy=x bisects ∠B\angle B, AC:2x−y=2AC: 2x-y=2, 2AB=BC2AB=BC, A=(4,6)A=(4,6). Find α+2β\alpha + 2\beta.
α+2β=  ?\alpha + 2\beta = \;?
02Reflection Property
2/4
Since BB lies on y=xy=x, reflect A(4,6)A(4,6) across y=xy=x to get A′(6,4)A'(6,4). Angle bisector theorem gives AD:DC=1:2AD:DC = 1:2.
A′=(6,4)A' = (6, 4)
03Solve for BKEY INSIGHT
3/4
Using reflection and section formula with AC:2x−y=2AC: 2x-y=2: α=14,β=14\alpha = 14, \beta = 14. So α+2β=42\alpha + 2\beta = 42.
14+28=4214 + 28 = 42
04Final Answer
4/4
α+2β=42\alpha + 2\beta = 42.
42\boxed{42}
Concepts from this question1 concepts unlocked
★

Angle Between Two Lines

EASY

Given slopes m₁ and m₂, the tangent of the acute angle between the lines is |m₁-m₂|/(1+m₁m₂).

tan⁡θ=∣m1−m2∣1+m1m2\tan\theta = \frac{|m_1 - m_2|}{1 + m_1 m_2}

Appears directly in 2-3 JEE questions every year. Quick formula application once you have the slopes.

Angle problemsBisector problemsTrisection
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