JEE Mains 25-July-2022 Shift 1 Solved Paper

Let inclination of required line is θ,
So the coordinates of point B can be assumed as
(4−‌

√29

3

‌cos‌θ,3−‌

√29

3

‌sin‌θ)
Which satisfices x−y−2=0
4−‌

√29

3

‌cos‌θ−3+‌

√29

3

‌sin‌θ−2=0
sin‌θ−cos‌θ=‌

3

√29

By squaring
sin‌2‌θ=‌

20

29

=‌

2‌tan‌θ

1+tan2θ

tan‌θ=‌

5

2

only (because slope is greater than 1 )
sin‌θ=‌

5

√29

,cos‌θ=‌

2

√29

Point B:(‌

10

3

,‌

4

3

)
Which also satisfies x+2y=6