Verified Solution • 2025–26 Session moderate

QJEEM22Ma6a4f: JEE Main 2022 Mathematics - Class 11 JEEM22Ma6a4f Solution: A line, with the slope greater than one, passes through the…

A line, with the slope greater than one, passes through the point A(4,3)A(4,3) and intersects the line x−y−2=0x-y-2=0 at the point B. If the length of the line segment ABA B is 293\frac{\sqrt{29}}{3} , then BB also lies on the line :

Multiple Choice Options

  • A 2x+y=92 x+y=9
  • B 3x−2y=73 x-2 y=7
  • C x+2y=6x+2 y=6
  • D 2x−3y=32 x-3 y=3

Step-by-Step Verified Solution

JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 54 English Explanation

Let inclination of required line is θ\theta ,

So the coordinates of point BB can be assumed as

(4−293cos⁡θ,3−293sin⁡θ)\left(4-\frac{\sqrt{29}}{3} \cos \theta, 3-\frac{\sqrt{29}}{3} \sin \theta\right)

Which satisfices x−y−2=0x-y-2=0

4−293cos⁡θ−3+293sin⁡θ−2=04-\frac{\sqrt{29}}{3} \cos \theta-3+\frac{\sqrt{29}}{3} \sin \theta-2=0

sin⁡θ−cos⁡θ=329\sin \theta-\cos \theta=\frac{3}{\sqrt{29}}

By squaring

sin⁡2θ=2029=2tan⁡θ1+tan⁡2θ\sin 2 \theta=\frac{20}{29}=\frac{2 \tan \theta}{1+\tan ^{2} \theta}

tan⁡θ=52\tan \theta=\frac{5}{2} only (because slope is greater than 1 )

sin⁡θ=529,cos⁡θ=229\sin \theta=\frac{5}{\sqrt{29}}, \cos \theta=\frac{2}{\sqrt{29}}

Point B:(103,43)B:\left(\frac{10}{3}, \frac{4}{3}\right)

Which also satisfies x+2y=6x+2 y=6

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