Straight LinesmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Angle Bisector Ratio BC²:AC² = 5:1 | JEE Main 2026

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

Question
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
A5:15:1correct
B1:51:5
C2:32:3
D3:23:2
Solution
Step 1: A=L1∩L2A=L_1\cap L_2: x=−3, y=x=−3⇒A(−3,−3)x=-3,\ y=x=-3\Rightarrow A(-3,-3). B=L1∩L3B=L_1\cap L_3: x=−3, y=−3x=9⇒B(−3,9)x=-3,\ y=-3x=9\Rightarrow B(-3,9). Step 2: The bisector from the origin OO meets L1L_1 at CC; by the property of the angle bisector at OO, BCAC=OBOA\dfrac{BC}{AC}=\dfrac{OB}{OA}. Step 3: OA2=(−3)2+(−3)2=18OA^2=(-3)^2+(-3)^2=18, OB2=(−3)2+92=90OB^2=(-3)^2+9^2=90, so
BC2AC2=(OBOA)2=9018=5.\frac{BC^2}{AC^2}=\left(\frac{OB}{OA}\right)^2=\frac{90}{18}=5.
 Step 4: Hence the ratio is 5:15:1. Correct answer: (1)
Solution working
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