Straight LinesmediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Straight Lines: Let Points Half Lines Distance Their Point Intersection (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let A,BA,B be points on the two half-lines x−3 ∣y∣=αx-\sqrt3\,|y|=\alpha, α>0\alpha>0, at a distance α\alpha from their point of intersection PP. The line segment ABAB meets the angle bisector of the given half-lines at the point QQ. If PQ=92PQ=\dfrac92 and RR is the radius of the circumcircle of △PAB\triangle PAB, then α2R\dfrac{\alpha^2}{R} is equal to
Solution
Answer: 9 (± 0.01)
Step 1: x−3∣y∣=αx-\sqrt3|y|=\alpha splits into x−3y=αx-\sqrt3y=\alpha (30∘30^\circ) and x+3y=αx+\sqrt3y=\alpha (−30∘-30^\circ), meeting at P(α,0)P(\alpha,0). Apex angle =60∘=60^\circ; PA=PB=αPA=PB=\alpha ⇒△PAB\Rightarrow \triangle PAB equilateral with side α\alpha. Step 2: QQ = midpoint of ABAB; PQPQ = median =αcos⁡30∘=32α=\alpha\cos30^\circ=\dfrac{\sqrt3}{2}\alpha. PQ=92PQ=\dfrac92:
32α=92⇒α=93=33.\frac{\sqrt3}{2}\alpha=\frac92\Rightarrow \alpha=\frac{9}{\sqrt3}=3\sqrt3.
 Step 3:
R=α3=333=3.R=\frac{\alpha}{\sqrt3}=\frac{3\sqrt3}{\sqrt3}=3.
 Step 4:
α2R=(33)23=273=9.\frac{\alpha^2}{R}=\frac{(3\sqrt3)^2}{3}=\frac{27}{3}=9.
 Correct answer: 9
Solution working
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Straight Lines · medium
From the point (−1,−1)(-1,-1), two rays are sent making angles of 45∘45^\circ with the line x+y=0x+y=0. These rays get reflected from the mirror x+2y=1x+2y=1. If the equations of the reflected rays are ax+by=9ax+by=9 and cx+dy=7cx+dy=7 with a,b,c,d∈Za,b,c,d\in\mathbb{Z}, then ad+bcad+bc is
Straight Lines · medium
In an equilateral triangle PQRPQR, let the vertex PP be at (3,5)(3,5) and the side QRQR be along the line x+y=4x+y=4. If the orthocentre of the triangle is (α,β)(\alpha,\beta), then 9(α+β)9(\alpha+\beta) is equal to
Straight Lines · medium
Let the vertex AA of a triangle ABCABC be (1,2)(1,2), and the mid-point of the side ABAB be (5,−1)(5,-1). If the centroid of the triangle is (3,4)(3,4) and its circumcenter is (α,β)(\alpha,\beta), then 21(α+β)21(\alpha+\beta) is equal to
Straight Lines · medium
Let the line L1:x+3=0L_1:x+3=0 intersect the lines L2:x−y=0L_2:x-y=0 and L3:3x+y=0L_3:3x+y=0 at the points AA and BB, respectively. Let the bisector of the obtuse angle between the lines L2L_2 and L3L_3 intersect the line L1L_1 at the point CC. Then BC2AC2\dfrac{BC^2}{AC^2} is equal to

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath