JEE Main 18 March 2021 Shift 1 Solved Paper

Let m= Slope of required line ∴ Equation of required line
y−3=m(x−1)...(i)
Given, equation of line is
3√2x−y−1=0...(ii)
Since, angle θ between Eqs. (i) and (ii) is tan−1(√2)
i.e. tan‌θ=√2
⇒‌‌|‌

m−3√2

1+3√2m

|=√2
(∵ Slope of Eq. (i) =m and slope of Eq. (ii) =3√2)
Squaring on both sides,
m2−6√2m+18=2(1+18m2+6√2m)
⇒35m2+18√2m−16=0
∴‌‌m=‌

−18√2±√648+2240

70

=‌

−18√2±38√2

70

⇒‌‌m=‌

2√2

7

,−‌

4

5

√2
For m=‌

2√2

7

, equation of required line will be
y−3=‌

2√2

7

(x−1)
⇒2√2x−7y+21−2√2=0
(options are not matching so, neglect this)
For m=‌

−4√2

5

, equation of required line will be
y−3=‌

−4√2

5

(x−1)
⇒‌5y−15=−4√2x+4√2
⇒‌4√2x+5y−15−4√2=0
⇒‌4√2x+5y−(15+4√2)=0