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 4545^\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,dZa,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 tan45=m+11m=1\tan45^\circ=\left|\dfrac{m+1}{1-m}\right|=1: m+1=±(1m)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+2y1=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 111/533/5=43\dfrac{-1-11/5}{3-3/5} = -\dfrac{4}{3}:
y+1=43(x3)    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 111/513/5=34\dfrac{1-11/5}{-1-3/5} = \dfrac{3}{4}:
y1=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)=169=7ad+bc = (4)(4)+(3)(-3) = 16-9 = 7
Answer: 7
Still stuck on this question?Ask your doubt on WhatsApp

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.