Straight LinesmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Straight Lines: Point Rays Sent Making Angles Line These Rays (JEE Main 2026)

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

Question
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
Solution
Answer: 7
Step 1: Determine the two incident rays The line x+y=0x+y=0 has slope −1-1. Applying the angle formula tan⁡45∘=∣m+11−m∣=1\tan45^\circ=\left|\dfrac{m+1}{1-m}\right|=1: m+1=±(1−m)m+1=\pm(1-m), giving slopes m=0m=0 (horizontal) and the vertical direction. The two rays through (−1,−1)(-1,-1) are y=−1y=-1 and x=−1x=-1 Step 2: Reflect the ray y=−1y=-1 The image of (−1,−1)(-1,-1) in the mirror x+2y−1=0x+2y-1=0 (using the reflection formula with a2+b2=5a^2+b^2=5):
A′=(−1+85, −1+165)=(35,115)A' = \left(-1+\frac{8}{5},\ -1+\frac{16}{5}\right) = \left(\frac{3}{5},\frac{11}{5}\right)
 The ray y=−1y=-1 meets the mirror at (3,−1)(3,-1). The reflected ray passes through (3,−1)(3,-1) with slope −1−11/53−3/5=−43\dfrac{-1-11/5}{3-3/5} = -\dfrac{4}{3}:
y+1=−43(x−3)  ⟹  4x+3y=9y+1=-\frac{4}{3}(x-3) \implies 4x+3y=9
 Step 3: Reflect the ray x=−1x=-1 The ray x=−1x=-1 meets the mirror at (−1,1)(-1,1). Using the same image A′=(35,115)A'=\left(\frac{3}{5},\frac{11}{5}\right), the reflected ray through (−1,1)(-1,1) has slope 1−11/5−1−3/5=34\dfrac{1-11/5}{-1-3/5} = \dfrac{3}{4}:
y−1=34(x+1)  ⟹  −3x+4y=7y-1=\frac{3}{4}(x+1) \implies -3x+4y=7
 Step 4: Compute ad+bcad+bc From 4x+3y=94x+3y=9: a=4a=4, b=3b=3. From −3x+4y=7-3x+4y=7: c=−3c=-3, d=4d=4.
ad+bc=(4)(4)+(3)(−3)=16−9=7ad+bc = (4)(4)+(3)(-3) = 16-9 = 7
 Answer: 7
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