AP EAMCET Engineering 18 Sep 2020 Shift 1 Paper

Let the angle of inclination of line θ, and as passed through point A(1,2), so equation of line is
‌

x−1

cos‌θ

=‌

y−2

sin‌θ

=±‌

√6

3

∴ General point on the line is
P(1±‌

√6

3

‌cos‌θ,2±‌

√6

3

‌sin‌θ)
Let the point P on the straight line x+y=4, so
3±‌

√6

3

(sin‌θ+cos‌θ)=4
⇒‌‌±‌

√6

3

(sin‌θ+cos‌θ)=1
On squaring both sides, we get
2(1+sin‌2‌θ)=3
⇒sin‌2‌θ=‌

1

2

⇒2θ=30∘ and 150∘⇒θ=15∘ and 75∘