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JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 22

From the point (−1,−1)(-1,-1), two rays are sent making angles of 45° 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, a,b,c,d∈Za,b,c,d\in\mathbf Z, then the value of ad+bcad+bc is ________.

Answer and worked solution

Official answer: 7

The rays. Lines at 45° to x+y=0x+y=0 through (−1,−1)(-1,-1) are y=−1y=-1 and x=−1x=-1. They meet the mirror at (3,−1)(3,-1) and (−1,1)(-1,1).

Image of the source in x+2y=1x+2y=1:
(−1,−1)−2(−4)5(1,2)=(35,115).(-1,-1)-\frac{2(-4)}5(1,2)=\left(\tfrac35,\tfrac{11}5\right).
Each reflected ray lies on the line from this image through the hit point.

The two reflected rays.

  • Through (3,−1)(3,-1): slope −43-\tfrac43, giving 4x+3y=94x+3y=9.
  • Through (−1,1)(-1,1): slope 34\tfrac34, giving −3x+4y=7-3x+4y=7.

Value. (a,b,c,d)=(4,3,−3,4)(a,b,c,d)=(4,3,-3,4), so ad+bc=16−9=7ad+bc=16-9=7.

Answer: 7

Topics:Equations of a lineAngle between lines
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