∴ 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=